From 5dce9076cd26555c5f3713a1b5db53e7571dfae6 Mon Sep 17 00:00:00 2001 From: algorithmzuo Date: Wed, 27 Jul 2022 12:47:27 +0800 Subject: [PATCH] first commit --- .../class01/Code01_SelectionSort.java | 115 ++++ .../class01/Code02_BubbleSort.java | 110 ++++ .../class01/Code03_InsertionSort.java | 116 ++++ 体系学习班/class01/Code04_BSExist.java | 65 +++ .../class01/Code05_BSNearLeft.java | 75 +++ .../class01/Code05_BSNearRight.java | 75 +++ 体系学习班/class01/Code06_BSAwesome.java | 76 +++ 体系学习班/class02/Code01_Swap.java | 71 +++ .../class02/Code02_EvenTimesOddTimes.java | 83 +++ 体系学习班/class02/Code03_KM.java | 159 ++++++ .../class03/Code01_ReverseList.java | 221 ++++++++ .../class03/Code02_DeleteGivenValue.java | 38 ++ ...Code03_DoubleEndsQueueToStackAndQueue.java | 183 +++++++ 体系学习班/class03/Code04_RingArray.java | 50 ++ .../class03/Code05_GetMinStack.java | 105 ++++ .../Code06_TwoStacksImplementQueue.java | 60 ++ .../Code07_TwoQueueImplementStack.java | 90 +++ 体系学习班/class03/Code08_GetMax.java | 24 + .../class03/HashMapAndSortedMap.java | 128 +++++ 体系学习班/class04/Code01_MergeSort.java | 147 +++++ 体系学习班/class04/Code02_SmallSum.java | 137 +++++ .../class04/Code03_ReversePair.java | 130 +++++ .../class04/Code04_BiggerThanRightTwice.java | 135 +++++ .../class05/Code01_CountOfRangeSum.java | 63 +++ .../class05/Code02_PartitionAndQuickSort.java | 197 +++++++ ...de03_QuickSortRecursiveAndUnrecursive.java | 195 +++++++ .../Code04_DoubleLinkedListQuickSort.java | 300 ++++++++++ .../class06/Code01_Comparator.java | 168 ++++++ 体系学习班/class06/Code02_Heap.java | 192 +++++++ 体系学习班/class06/Code03_HeapSort.java | 158 ++++++ .../Code04_SortArrayDistanceLessK.java | 127 +++++ 体系学习班/class07/Code01_CoverMax.java | 155 ++++++ .../class07/Code02_EveryStepShowBoss.java | 303 +++++++++++ 体系学习班/class07/HeapGreater.java | 112 ++++ 体系学习班/class07/Inner.java | 9 + 体系学习班/class08/Code01_TrieTree.java | 299 ++++++++++ 体系学习班/class08/Code02_TrieTree.java | 306 +++++++++++ 体系学习班/class08/Code03_CountSort.java | 111 ++++ 体系学习班/class08/Code04_RadixSort.java | 148 +++++ .../class09/Code01_LinkedListMid.java | 162 ++++++ .../class09/Code02_IsPalindromeList.java | 204 +++++++ .../class09/Code03_SmallerEqualBigger.java | 138 +++++ .../class09/Code04_CopyListWithRandom.java | 79 +++ .../Code01_FindFirstIntersectNode.java | 170 ++++++ .../class10/Code02_RecursiveTraversalBT.java | 72 +++ .../Code03_UnRecursiveTraversalBT.java | 118 ++++ .../class11/Code01_LevelTraversalBT.java | 49 ++ .../Code02_SerializeAndReconstructTree.java | 264 +++++++++ .../Code03_EncodeNaryTreeToBinaryTree.java | 86 +++ .../class11/Code04_PrintBinaryTree.java | 74 +++ .../class11/Code05_TreeMaxWidth.java | 114 ++++ .../class11/Code06_SuccessorNode.java | 86 +++ .../class11/Code07_PaperFolding.java | 28 + 体系学习班/class12/Code01_IsCBT.java | 149 +++++ 体系学习班/class12/Code02_IsBST.java | 125 +++++ .../class12/Code03_IsBalanced.java | 102 ++++ 体系学习班/class12/Code04_IsFull.java | 109 ++++ .../class12/Code05_MaxSubBSTSize.java | 157 ++++++ .../class12/Code06_MaxDistance.java | 180 ++++++ 体系学习班/class13/Code01_IsCBT.java | 120 ++++ .../class13/Code02_MaxSubBSTHead.java | 136 +++++ .../class13/Code03_lowestAncestor.java | 141 +++++ 体系学习班/class13/Code04_MaxHappy.java | 116 ++++ .../class13/Code05_LowestLexicography.java | 116 ++++ 体系学习班/class14/Code01_Light.java | 105 ++++ .../class14/Code02_LessMoneySplitGold.java | 79 +++ .../class14/Code03_BestArrange.java | 111 ++++ 体系学习班/class14/Code04_IPO.java | 58 ++ 体系学习班/class14/Code05_UnionFind.java | 70 +++ .../class15/Code01_FriendCircles.java | 78 +++ .../class15/Code02_NumberOfIslands.java | 317 +++++++++++ .../class15/Code03_NumberOfIslandsII.java | 165 ++++++ 体系学习班/class16/Code01_BFS.java | 30 + 体系学习班/class16/Code02_DFS.java | 34 ++ .../class16/Code03_TopologicalOrderBFS.java | 53 ++ .../class16/Code03_TopologicalOrderDFS1.java | 70 +++ .../class16/Code03_TopologicalOrderDFS2.java | 76 +++ .../class16/Code03_TopologySort.java | 36 ++ 体系学习班/class16/Code04_Kruskal.java | 101 ++++ 体系学习班/class16/Code05_Prim.java | 94 ++++ 体系学习班/class16/Code06_Dijkstra.java | 160 ++++++ .../class16/Code06_NetworkDelayTime.java | 151 ++++++ 体系学习班/class16/Edge.java | 14 + 体系学习班/class16/Graph.java | 14 + 体系学习班/class16/GraphGenerator.java | 37 ++ 体系学习班/class16/Node.java | 20 + 体系学习班/class17/Code01_Dijkstra.java | 162 ++++++ 体系学习班/class17/Code02_Hanoi.java | 139 +++++ .../class17/Code03_PrintAllSubsquences.java | 74 +++ .../class17/Code04_PrintAllPermutations.java | 110 ++++ .../Code05_ReverseStackUsingRecursive.java | 44 ++ 体系学习班/class17/Edge.java | 14 + 体系学习班/class17/Graph.java | 14 + 体系学习班/class17/Node.java | 20 + 体系学习班/class18/Code01_RobotWalk.java | 94 ++++ .../class18/Code02_CardsInLine.java | 116 ++++ 体系学习班/class19/Code01_Knapsack.java | 63 +++ .../class19/Code02_ConvertToLetterString.java | 123 +++++ .../class19/Code03_StickersToSpellWord.java | 151 ++++++ .../Code04_LongestCommonSubsequence.java | 124 +++++ .../class20/Code01_PalindromeSubsequence.java | 121 +++++ 体系学习班/class20/Code02_HorseJump.java | 109 ++++ 体系学习班/class20/Code03_Coffee.java | 204 +++++++ .../class21/Code01_MinPathSum.java | 79 +++ .../Code02_CoinsWayEveryPaperDifferent.java | 77 +++ .../class21/Code03_CoinsWayNoLimit.java | 108 ++++ .../Code04_CoinsWaySameValueSamePapper.java | 148 +++++ 体系学习班/class21/Code05_BobDie.java | 58 ++ .../class22/Code01_KillMonster.java | 100 ++++ .../class22/Code02_MinCoinsNoLimit.java | 121 +++++ .../class22/Code03_SplitNumber.java | 85 +++ .../class23/Code01_SplitSumClosed.java | 94 ++++ .../Code02_SplitSumClosedSizeHalf.java | 243 +++++++++ 体系学习班/class23/Code03_NQueens.java | 89 +++ .../class24/Code01_SlidingWindowMaxArray.java | 99 ++++ .../class24/Code02_AllLessNumSubArray.java | 109 ++++ .../class24/Code03_GasStation.java | 53 ++ .../class24/Code04_MinCoinsOnePaper.java | 250 +++++++++ .../class25/Code01_MonotonousStack.java | 165 ++++++ .../Code01_MonotonousStackForNowcoder.java | 71 +++ .../class25/Code02_AllTimesMinToMax.java | 98 ++++ .../Code03_LargestRectangleInHistogram.java | 58 ++ .../class25/Code04_MaximalRectangle.java | 48 ++ .../Code05_CountSubmatricesWithAllOnes.java | 81 +++ .../class26/Code01_SumOfSubarrayMinimums.java | 156 ++++++ .../class26/Code02_FibonacciProblem.java | 189 +++++++ .../Code03_ZeroLeftOneStringNumber.java | 113 ++++ 体系学习班/class27/Code01_KMP.java | 75 +++ 体系学习班/class27/Code02_TreeEqual.java | 172 ++++++ .../class27/Code03_IsRotation.java | 64 +++ 体系学习班/class28/Code01_Manacher.java | 90 +++ .../class28/Code02_AddShortestEnd.java | 54 ++ .../class29/Code01_FindMinKth.java | 175 ++++++ 体系学习班/class29/Code02_MaxTopK.java | 223 ++++++++ .../class29/Code03_ReservoirSampling.java | 94 ++++ .../class30/Code01_MorrisTraversal.java | 230 ++++++++ 体系学习班/class30/Code02_MinDepth.java | 88 +++ .../class31/Code01_SegmentTree.java | 245 +++++++++ .../class31/Code02_FallingSquares.java | 109 ++++ 体系学习班/class32/Code01_IndexTree.java | 83 +++ .../class32/Code02_IndexTree2D.java | 57 ++ 体系学习班/class32/Code03_AC1.java | 105 ++++ 体系学习班/class32/Code04_AC2.java | 125 +++++ 体系学习班/class33/Hash.java | 49 ++ 体系学习班/class34/ReadMe.java | 2 + .../class35/Code01_AVLTreeMap.java | 257 +++++++++ .../class36/Code01_SizeBalancedTreeMap.java | 360 ++++++++++++ .../class36/Code02_SkipListMap.java | 248 +++++++++ .../class37/Code01_CountofRangeSum.java | 223 ++++++++ .../class37/Code02_SlidingWindowMedian.java | 228 ++++++++ .../Code03_AddRemoveGetIndexGreat.java | 259 +++++++++ .../Code04_QueueReconstructionByHeight.java | 265 +++++++++ 体系学习班/class37/Compare.java | 408 ++++++++++++++ .../class38/Code01_AppleMinBags.java | 41 ++ 体系学习班/class38/Code02_EatGrass.java | 57 ++ 体系学习班/class38/Code03_MSumToN.java | 44 ++ .../class38/Code04_MoneyProblem.java | 169 ++++++ .../class39/Code01_SubsquenceMaxModM.java | 141 +++++ .../class39/Code02_SnacksWays.java | 79 +++ .../class39/Code02_SnacksWaysMain1.java | 126 +++++ .../class39/Code02_SnacksWaysMain2.java | 135 +++++ 体系学习班/class39/Code03_10Ways.java | 92 ++++ .../class39/Code04_DifferentBTNum.java | 66 +++ 体系学习班/class39/IsSum.java | 180 ++++++ ...ngestSumSubArrayLengthInPositiveArray.java | 91 ++++ .../Code02_LongestSumSubArrayLength.java | 92 ++++ .../Code03_LongestLessSumSubArrayLength.java | 103 ++++ ...de04_AvgLessEqualValueLongestSubarray.java | 152 ++++++ .../Code05_PrintMatrixSpiralOrder.java | 53 ++ .../class40/Code06_RotateMatrix.java | 44 ++ .../class40/Code07_ZigZagPrintMatrix.java | 42 ++ 体系学习班/class40/Code08_PrintStar.java | 46 ++ .../class41/Code01_BestSplitForAll.java | 72 +++ .../Code02_BestSplitForEveryPosition.java | 130 +++++ .../class41/Code03_StoneMerge.java | 117 ++++ .../class41/Code04_SplitArrayLargestSum.java | 193 +++++++ .../class42/Code01_PostOfficeProblem.java | 115 ++++ .../Code02_ThrowChessPiecesProblem.java | 166 ++++++ 体系学习班/class43/Code01_CanIWin.java | 116 ++++ 体系学习班/class43/Code02_TSP.java | 295 ++++++++++ .../class43/Code03_PavingTile.java | 215 ++++++++ ...1_LastSubstringInLexicographicalOrder.java | 136 +++++ 体系学习班/class44/DC3.java | 168 ++++++ 体系学习班/class44/DC3_Algorithm.pdf | Bin 0 -> 192716 bytes ...e01_InsertS2MakeMostAlphabeticalOrder.java | 261 +++++++++ .../class45/Code02_CreateMaximumNumber.java | 250 +++++++++ ...LongestCommonSubstringConquerByHeight.java | 286 ++++++++++ .../Code04_LongestRepeatingSubstring.java | 196 +++++++ .../class46/Code01_BurstBalloons.java | 107 ++++ .../class46/Code02_RemoveBoxes.java | 81 +++ .../Code03_DeleteAdjacentSameCharacter.java | 176 ++++++ .../class46/Code04_MaxSumLengthNoMore.java | 100 ++++ .../class46/Code05_HuffmanTree.java | 214 ++++++++ .../class47/Code01_StrangePrinter.java | 78 +++ .../class47/Code02_RestoreWays.java | 246 +++++++++ .../class47/Code03_DinicAlgorithm.java | 139 +++++ .../class01/Code01_CordCoverMaxPoint.java | 87 +++ .../class01/Code02_CountFiles.java | 40 ++ .../class01/Code03_Near2Power.java | 22 + .../class01/Code04_MinSwapStep.java | 74 +++ .../class01/Code05_LongestIncreasingPath.java | 54 ++ 大厂刷题班/class01/Code06_AOE.java | 304 +++++++++++ 大厂刷题班/class01/Code07_TargetSum.java | 127 +++++ .../class02/Code01_ChooseWork.java | 48 ++ 大厂刷题班/class02/Code02_Cola.java | 151 ++++++ .../Code03_ReceiveAndPrintOrderLine.java | 89 +++ 大厂刷题班/class02/Code04_Drive.java | 116 ++++ 大厂刷题班/class02/Code05_SetAll.java | 48 ++ .../class02/Code06_MinLengthForSort.java | 30 + ...stSubstringWithoutRepeatingCharacters.java | 27 + .../class03/Code02_HowManyTypes.java | 82 +++ .../Code03_Largest1BorderedSquare.java | 59 ++ .../class03/Code04_MaxPairNumber.java | 126 +++++ .../class03/Code05_BoatsToSavePeople.java | 74 +++ .../class03/Code06_ClosestSubsequenceSum.java | 46 ++ .../class03/Code07_FreedomTrail.java | 62 +++ .../class03/Code08_DistanceKNodes.java | 105 ++++ .../class04/Code01_QueryHobby.java | 115 ++++ .../class04/Code02_SubArrayMaxSum.java | 35 ++ .../class04/Code03_SubMatrixMaxSum.java | 75 +++ .../Code04_SubArrayMaxSumFollowUp.java | 59 ++ .../class04/Code05_CandyProblem.java | 145 +++++ 大厂刷题班/class04/Code06_MakeNo.java | 59 ++ .../class04/Code07_InterleavingString.java | 53 ++ .../class04/Code08_TheSkylineProblem.java | 73 +++ ...BinarySearchTreeFromPreorderTraversal.java | 115 ++++ .../Code02_LeftRightSameTreeNumber.java | 95 ++++ 大厂刷题班/class05/Code03_EditCost.java | 89 +++ .../class05/Code04_DeleteMinCost.java | 258 +++++++++ 大厂刷题班/class05/Hash.java | 49 ++ 大厂刷题班/class06/Code01_MaxXOR.java | 130 +++++ ...ode02_MaximumXorOfTwoNumbersInAnArray.java | 50 ++ ...de03_MaximumXorWithAnElementFromArray.java | 67 +++ .../class06/Code04_MostXorZero.java | 122 +++++ 大厂刷题班/class06/Code05_Nim.java | 18 + .../class07/Code01_MaxAndValue.java | 102 ++++ .../class07/Code02_MinCameraCover.java | 113 ++++ 大厂刷题班/class07/Code03_MaxGap.java | 46 ++ .../class07/Code04_Power2Diffs.java | 97 ++++ .../class07/Code05_WorldBreak.java | 237 ++++++++ .../class07/Code06_SplitStringMaxValue.java | 137 +++++ 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135 +++++ .../class13/Code01_NCardsABWin.java | 169 ++++++ .../class13/Code02_SuperWashingMachines.java | 34 ++ .../class13/Code03_ScrambleString.java | 186 +++++++ .../class13/Code04_BricksFallingWhenHit.java | 149 +++++ .../class14/Code01_Parentheses.java | 97 ++++ .../Code02_MaxSubArraySumLessOrEqualK.java | 28 + .../Code03_BiggestBSTTopologyInTree.java | 65 +++ .../Code04_CompleteTreeNodeNumber.java | 45 ++ .../Code05_RecoverBinarySearchTree.java | 511 ++++++++++++++++++ .../class14/Code06_MissingNumber.java | 29 + .../Code01_BestTimeToBuyAndSellStock.java | 21 + .../Code02_BestTimeToBuyAndSellStockII.java | 17 + .../Code03_BestTimeToBuyAndSellStockIII.java | 23 + .../Code04_BestTimeToBuyAndSellStockIV.java | 65 +++ ...BestTimeToBuyAndSellStockWithCooldown.java | 101 ++++ ...meToBuyAndSellStockWithTransactionFee.java | 27 + 大厂刷题班/class16/Code01_IsSum.java | 206 +++++++ .../class16/Code02_SmallestUnFormedSum.java | 123 +++++ .../class16/Code03_MinPatches.java | 82 +++ .../class16/Code04_MergeRecord.java | 220 ++++++++ .../class16/Code05_JosephusProblem.java | 120 ++++ .../class17/Code01_FindNumInSortedMatrix.java | 34 ++ ...de02_KthSmallestElementInSortedMatrix.java | 110 ++++ .../class17/Code03_PalindromePairs.java | 104 ++++ .../class17/Code04_DistinctSubseq.java | 101 ++++ .../class17/Code05_DistinctSubseqValue.java | 50 ++ .../class18/Code01_HanoiProblem.java | 97 ++++ .../class18/Code02_ShortestBridge.java | 96 ++++ .../class18/Code03_CherryPickup.java | 70 +++ .../class18/Code04_TopKSumCrossTwoArrays.java | 93 ++++ 大厂刷题班/class19/Code01_LRUCache.java | 142 +++++ 大厂刷题班/class19/Code02_LFUCache.java | 204 +++++++ 大厂刷题班/class19/Code03_OneNumber.java | 83 +++ ...allestRangeCoveringElementsfromKLists.java | 58 ++ .../class19/Code05_CardsProblem.java | 166 ++++++ .../Code01_PreAndInArrayToPosArray.java | 233 ++++++++ ...02_LargestComponentSizebyCommonFactor.java | 109 ++++ .../class20/Code03_ShuffleProblem.java | 154 ++++++ .../class20/Code04_PalindromeWays.java | 91 ++++ .../class21/TreeChainPartition.java | 416 ++++++++++++++ ..._MaximumSumof3NonOverlappingSubarrays.java | 102 ++++ .../class22/Code02_TrappingRainWater.java | 28 + .../class22/Code03_TrappingRainWaterII.java | 76 +++ .../class22/Code04_VisibleMountains.java | 194 +++++++ .../class22/Code05_TallestBillboard.java | 38 ++ ...Code01_LCATarjanAndTreeChainPartition.java | 352 ++++++++++++ .../Code02_MaxABSBetweenLeftAndRight.java | 63 +++ .../Code03_LongestIntegratedLength.java | 106 ++++ .../class23/Code04_FindKMajority.java | 113 ++++ .../Code05_MinimumCostToMergeStones.java | 159 ++++++ .../class24/Code01_Split4Parts.java | 89 +++ .../class24/Code02_KthMinPair.java | 180 ++++++ .../class24/Code03_NotContains4.java | 144 +++++ 大厂刷题班/class24/Code04_Painting.java | 75 +++ .../class24/Code05_MinWindowLength.java | 80 +++ ...Code06_RemoveDuplicateLettersLessLexi.java | 74 +++ 大厂刷题班/class25/Code01_IPToCIDR.java | 90 +++ 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..._0026_RemoveDuplicatesFromSortedArray.java | 21 + ...AndLastPositionOfElementInSortedArray.java | 33 ++ .../class28/Problem_0036_ValidSudoku.java | 26 + .../class28/Problem_0037_SudokuSolver.java | 61 +++ .../class28/Problem_0038_CountAndSay.java | 33 ++ .../class28/Problem_0049_GroupAnagrams.java | 52 ++ 大厂刷题班/class28/说明 | 33 ++ ...oblem_0033_SearchInRotatedSortedArray.java | 67 +++ .../class29/Problem_0050_PowXN.java | 39 ++ .../class29/Problem_0056_MergeIntervals.java | 30 + .../class29/Problem_0062_UniquePaths.java | 30 + .../class29/Problem_0066_PlusOne.java | 19 + .../class29/Problem_0069_SqrtX.java | 30 + .../class29/Problem_0073_SetMatrixZeroes.java | 79 +++ 大厂刷题班/class29/说明 | 20 + .../class30/Problem_0079_WordSearch.java | 41 ++ .../Problem_0088_MergeSortedArray.java | 22 + .../class30/Problem_0091_DecodeWays.java | 87 +++ ...Problem_0098_ValidateBinarySearchTree.java | 42 ++ .../class30/Problem_0101_SymmetricTree.java | 30 + ...3_BinaryTreeZigzagLevelOrderTraversal.java | 52 ++ ..._ConvertSortedArrayToBinarySearchTree.java | 33 ++ ...PopulatingNextRightPointersInEachNode.java | 77 +++ .../class30/Problem_0118_PascalTriangle.java | 23 + .../Problem_0119_PascalTriangleII.java | 19 + ...Problem_0124_BinaryTreeMaximumPathSum.java | 206 +++++++ .../class30/Problem_0639_DecodeWaysII.java | 161 ++++++ 大厂刷题班/class30/说明 | 26 + .../class31/Problem_0125_ValidPalindrome.java | 52 ++ .../class31/Problem_0127_WordLadder.java | 124 +++++ .../Problem_0130_SurroundedRegions.java | 115 ++++ .../class31/Problem_0139_WordBreak.java | 97 ++++ .../class31/Problem_0140_WordBreakII.java | 104 ++++ .../class31/Problem_0148_SortList.java | 227 ++++++++ ...em_0150_EvaluateReversePolishNotation.java | 40 ++ 大厂刷题班/class31/说明 | 21 + .../Problem_0152_MaximumProductSubarray.java | 45 ++ .../class32/Problem_0163_MissingRanges.java | 35 ++ ...oblem_0166_FractionToRecurringDecimal.java | 46 ++ .../Problem_0171_ExcelSheetColumnNumber.java | 15 + .../Problem_0172_FactorialTrailingZeroes.java | 14 + .../class32/Problem_0189_RotateArray.java | 60 ++ .../class32/Problem_0190_ReverseBits.java | 48 ++ .../class32/Problem_0191_NumberOf1Bits.java | 26 + .../class32/Problem_0202_HappyNumber.java | 59 ++ .../class32/Problem_0204_CountPrimes.java | 30 + 大厂刷题班/class32/SequenceM.java | 162 ++++++ 大厂刷题班/class32/说明 | 25 + .../class33/Problem_0207_CourseSchedule.java | 113 ++++ .../Problem_0210_CourseScheduleII.java | 71 +++ .../class33/Problem_0213_HouseRobberII.java | 50 ++ .../Problem_0237_DeleteNodeInLinkedList.java | 15 + ...Problem_0238_ProductOfArrayExceptSelf.java | 24 + .../class33/Problem_0242_ValidAnagram.java | 23 + .../class33/Problem_0251_Flatten2DVector.java | 53 ++ .../class33/Problem_0269_AlienDictionary.java | 65 +++ .../Problem_0277_FindTheCelebrity.java | 38 ++ .../class33/Problem_0279_PerfectSquares.java | 70 +++ .../class33/Problem_0283_MoveZeroes.java | 20 + 大厂刷题班/class33/说明 | 32 ++ .../Problem_0287_FindTheDuplicateNumber.java | 23 + .../class34/Problem_0289_GameOfLife.java | 43 ++ ...Problem_0295_FindMedianFromDataStream.java | 46 ++ ...m_0315_CountOfSmallerNumbersAfterSelf.java | 69 +++ .../class34/Problem_0324_WiggleSortII.java | 186 +++++++ .../class34/Problem_0326_PowerOfThree.java | 14 + .../Problem_0328_OddEvenLinkedList.java | 43 ++ ...ubstringWithAtMostKDistinctCharacters.java | 29 + ...roblem_0341_FlattenNestedListIterator.java | 110 ++++ .../class34/Problem_0348_DesignTicTacToe.java | 47 ++ .../Problem_0380_InsertDeleteGetRandom.java | 52 ++ .../class34/Problem_0384_ShuffleAnArray.java | 34 ++ 大厂刷题班/class34/说明 | 30 + 大厂刷题班/class35/Code01_StringKth.java | 96 ++++ .../class35/Code02_MagicStone.java | 49 ++ .../class35/Code03_WatchMovieMaxTime.java | 136 +++++ 大厂刷题班/class35/Code04_WalkToEnd.java | 53 ++ .../class35/Code05_CircleCandy.java | 58 ++ .../Problem_0347_TopKFrequentElements.java | 54 ++ ...stringWithAtLeastKRepeatingCharacters.java | 110 ++++ .../class35/Problem_0412_FizzBuzz.java | 24 + .../class35/Problem_0454_4SumII.java | 32 ++ ..._NumberOfLongestIncreasingSubsequence.java | 90 +++ .../Problem_0687_LongestUnivaluePath.java | 62 +++ 大厂刷题班/class35/说明 | 21 + .../class36/Code01_ReverseInvertString.java | 108 ++++ .../class36/Code02_Ratio01Split.java | 68 +++ .../class36/Code03_MatchCount.java | 66 +++ .../Code04_ComputeExpressionValue.java | 58 ++ .../class36/Code05_Query3Problems.java | 132 +++++ .../class36/Code06_NodeWeight.java | 48 ++ .../class36/Code07_PickAddMax.java | 80 +++ .../class36/Code08_MinBoatEvenNumbers.java | 98 ++++ .../class36/Code09_MaxKLenSequence.java | 87 +++ .../class36/Code10_StoneGameIV.java | 63 +++ 大厂刷题班/class36/Code11_BusRoutes.java | 59 ++ 大厂刷题班/class36/说明 | 12 + .../class37/Code01_ArrangeProject.java | 50 ++ .../class37/Code02_GameForEveryStepWin.java | 111 ++++ ...em_0114_FlattenBinaryTreeToLinkedList.java | 93 ++++ .../class37/Problem_0221_MaximalSquare.java | 40 ++ .../Problem_0226_InvertBinaryTree.java | 21 + .../class37/Problem_0337_HouseRobberIII.java | 37 ++ .../class37/Problem_0394_DecodeString.java | 50 ++ ...blem_0406_QueueReconstructionByHeight.java | 264 +++++++++ .../class37/Problem_0437_PathSumIII.java | 44 ++ 大厂刷题班/class37/说明 | 30 + .../class38/Code01_FillGapMinStep.java | 103 ++++ 大厂刷题班/class38/Code02_GreatWall.java | 183 +++++++ ...Problem_0438_FindAllAnagramsInAString.java | 58 ++ ...48_FindAllNumbersDisappearedInAnArray.java | 42 ++ .../Problem_0617_MergeTwoBinaryTrees.java | 31 ++ .../class38/Problem_0621_TaskScheduler.java | 33 ++ .../Problem_0647_PalindromicSubstrings.java | 49 ++ .../Problem_0739_DailyTemperatures.java | 34 ++ .../class38/Problem_0763_PartitionLabels.java | 28 + 大厂刷题班/class38/说明 | 25 + .../class39/Code01_01AddValue.java | 46 ++ .../class39/Code02_ValidSequence.java | 113 ++++ .../Code03_SequenceKDifferentKinds.java | 63 +++ .../class39/Code04_JumpGameOnMatrix.java | 239 ++++++++ .../class39/Code05_0123Disappear.java | 82 +++ 大厂刷题班/class40/Code01_SplitTo01.java | 195 +++++++ 大厂刷题班/class40/Code02_Mod3Max.java | 247 +++++++++ .../class40/Code03_MaxMeetingScore.java | 109 ++++ .../class40/Code04_LetASorted.java | 54 ++ 大厂刷题班/class40/Code05_AllSame.java | 135 +++++ .../class41/Code01_MinSwapTimes.java | 88 +++ .../class41/Code02_PoemProblem.java | 381 +++++++++++++ .../class41/Code03_MagicGoToAim.java | 71 +++ .../class41/Problem_0031_NextPermutation.java | 44 ++ .../class42/Problem_0265_PaintHouseII.java | 56 ++ ...m_0272_ClosestBinarySearchTreeValueII.java | 109 ++++ .../Problem_0273_IntegerToEnglishWords.java | 78 +++ .../Problem_0296_BestMeetingPoint.java | 48 ++ .../class42/Problem_0335_SelfCrossing.java | 28 + .../class43/Code01_SumNoPositiveMinCost.java | 197 +++++++ .../class43/Code02_MinCostToYeahArray.java | 304 +++++++++++ ...Problem_0248_StrobogrammaticNumberIII.java | 216 ++++++++ ...0317_ShortestDistanceFromAllBuildings.java | 237 ++++++++ ..._0992_SubarraysWithKDifferentIntegers.java | 71 +++ .../class45/Code01_SplitBuildingBlock.java | 87 +++ .../class45/Problem_0291_WordPatternII.java | 62 +++ .../class45/Problem_0403_FrogJump.java | 40 ++ ...yIntoTwoArraysToMinimizeSumDifference.java | 58 ++ ...m_0363_MaxSumOfRectangleNoLargerThanK.java | 77 +++ .../Problem_0391_PerfectRectangle.java | 59 ++ ...em_0411_MinimumUniqueWordAbbreviation.java | 174 ++++++ .../class46/Problem_0425_WordSquares.java | 87 +++ .../class47/Code01_DynamicSegmentTree.java | 135 +++++ .../class47/Code02_DynamicSegmentTree.java | 198 +++++++ ...m_0315_CountOfSmallerNumbersAfterSelf.java | 94 ++++ ...em_0358_RearrangeStringKDistanceApart.java | 64 +++ ..._0428_SerializeAndDeserializeNaryTree.java | 103 ++++ .../Problem_0465_OptimalAccountBalancing.java | 149 +++++ .../class47/Problem_0475_Heaters.java | 113 ++++ .../class48/Code01_MinKthPairMinusABS.java | 122 +++++ .../Problem_0472_ConcatenatedWords.java | 124 +++++ .../Problem_0483_SmallestGoodBase.java | 33 ++ .../class48/Problem_0499_TheMazeIII.java | 87 +++ .../Problem_0377_CombinationSumIV.java | 64 +++ ...440_KthSmallestInLexicographicalOrder.java | 88 +++ ...em_0446_ArithmeticSlicesIISubsequence.java | 30 + .../Problem_0489_RobotRoomCleaner.java | 57 ++ .../Problem_0527_WordAbbreviation.java | 48 ++ .../Problem_0548_SplitArrayEithEqualSum.java | 45 ++ ...Problem_0564_FindTheClosestPalindrome.java | 72 +++ .../Problem_0568_MaximumVacationDays.java | 50 ++ .../class50/Problem_0587_ErectTheFence.java | 55 ++ ...Problem_0588_DesignInMemoryFileSystem.java | 107 ++++ ...egativeIntegersWithoutConsecutiveOnes.java | 80 +++ 大厂刷题班/class51/LCP_0003_Robot.java | 144 +++++ .../Problem_0630_CourseScheduleIII.java | 33 ++ ...m_0642_DesignSearchAutocompleteSystem.java | 134 +++++ .../Problem_0875_KokoEatingBananas.java | 34 ++ .../class51/Problem_1035_UncrossedLines.java | 86 +++ .../class52/Problem_0656_CoinPath.java | 67 +++ .../class52/Problem_0683_KEmptySlots.java | 75 +++ .../Problem_1488_AvoidFloodInTheCity.java | 81 +++ .../Code01_RetainTree.java | 89 +++ .../Code02_GuessNumberHigherOrLowerII.java | 107 ++++ .../Code03_StartToEndBinaryOneTarget.java | 337 ++++++++++++ .../Code01_XtoYMinDistance.java | 171 ++++++ .../Code02_4KeysKeyboard.java | 38 ++ .../Code03_RedundantConnectionII.java | 102 ++++ .../Code01_FindAllPeopleWithSecret.java | 102 ++++ .../Code02_AwayFromBlackHole.java | 122 +++++ .../class_2021_12_2_week/Code03_MagicSum.java | 301 +++++++++++ ...4_LowestCommonAncestorOfABinaryTreeIV.java | 57 ++ .../class_2021_12_2_week/Code05_Colors.java | 325 +++++++++++ .../Code01_RightMoveInBinaryTree.java | 74 +++ .../Code02_BinaryNegate.java | 25 + .../Code03_OneCountsInKSystem.java | 58 ++ .../Code04_CutOffTreesForGolfEvent.java | 79 +++ .../Code05_MinContinuousFragment.java | 162 ++++++ .../Code01_FiveNodesListNumbers.java | 58 ++ .../Code02_MergeArea.java | 143 +++++ .../Code03_HowManyObtuseAngles.java | 42 ++ .../Code04_MaximumNumberOfVisiblePoints.java | 43 ++ .../Code05_SplitApples.java | 123 +++++ .../Code01_LoudAndRich.java | 68 +++ .../Code02_DoAllJobs.java | 129 +++++ .../Code03_WaysToBuildWall.java | 64 +++ .../Code01_ABDisappear.java | 131 +++++ .../Code02_CatAndMouse.java | 233 ++++++++ ...romPerformingMultiplicationOperations.java | 58 ++ ...eFromLeftUpToRightDownWalk4Directions.java | 119 ++++ .../Problem_0913_CatAndMouse.java | 93 ++++ .../Code01_StringCounts.java | 13 + .../class_2022_01_2_week/Code02_BrickAll.java | 49 ++ ...ringNumberConvertBinaryAndHexadecimal.java | 119 ++++ ...umOperationsToMakeTheArrayKIncreasing.java | 49 ++ .../Code05_MagicTowSubarrayMakeMaxSum.java | 129 +++++ .../class_2022_01_2_week/Code06_QuietSum.java | 290 ++++++++++ .../Code01_AStarAlgorithm.java | 195 +++++++ .../Code02_EscapeALargeMaze.java | 74 +++ ...ode03_ShortestSubarrayWithSumAtLeastK.java | 74 +++ .../Code01_BuyThingsAboutCollocation.java | 172 ++++++ .../Code02_SplitToMArraysMinScore.java | 178 ++++++ .../Code03_RandomPickWithBlacklist.java | 52 ++ .../Code04_BattleshipsInABoard.java | 21 + .../class_2022_02_2_week/Code01_24Game.java | 97 ++++ .../Code02_DesignBitset.java | 95 ++++ .../Code03_FindKthSmallestPairDistance.java | 64 +++ .../Code04_ReachingPoints.java | 38 ++ .../Code05_RecoverTheOriginalArray.java | 48 ++ .../Code01_CheapestFlightsWithinKStops.java | 75 +++ ...de02_MinimumNumberOfDaysToEatNOranges.java | 35 ++ .../Code03_RobotBoundedInCircle.java | 40 ++ .../Code04_MaxTeamNumber.java | 94 ++++ .../Code05_StoneGameIX.java | 16 + .../Code01_SplitSameNumberWays.java | 36 ++ .../Code02_NearBiggerNoSameNeighbour.java | 69 +++ .../Code03_PartitionArrayForMaximumSum.java | 53 ++ .../Code04_NumberOfDescendingTriples.java | 134 +++++ .../Code05_GroupsOfStrings.java | 195 +++++++ .../Code01_StronglyConnectedComponents.java | 105 ++++ .../Code02_NetworkOfSchools.java | 148 +++++ .../Code03_PopularCows.java | 150 +++++ .../Code04_IgniteMinBombs.java | 328 +++++++++++ .../Code01_MeetingCheck.java | 202 +++++++ .../Code02_StringCheck.java | 65 +++ .../class_2022_03_2_week/Code03_AiFill.java | 112 ++++ .../Code04_SameTeams.java | 62 +++ .../Code05_NumberOfDivisibleByM.java | 69 +++ .../Code06_JobMinDays.java | 100 ++++ .../Code07_MinWaitingTime.java | 95 ++++ ...ode08_TimeNSpace1LowestCommonAncestor.java | 111 ++++ .../Code01_LongestUncontinuousSet.java | 100 ++++ .../class_2022_03_3_week/Code02_CutDouFu.java | 54 ++ .../Code03_MaxSumOnReverseArray.java | 105 ++++ .../Code04_ArrangeAddGetMax.java | 150 +++++ .../class_2022_03_3_week/Code05_EatFish.java | 106 ++++ .../Code06_FinancialProduct.java | 51 ++ .../Code07_CoopDevelop.java | 51 ++ .../Code01_ArrangeJob.java | 66 +++ .../Code02_BuyGoodsHaveDiscount.java | 89 +++ .../Code03_MinTowNumberSumABS.java | 159 ++++++ .../Code04_JumpToTargets.java | 45 ++ .../Code05_HowManyWaysFromBottomToTop.java | 81 +++ .../Code06_LongestContinuousTrees.java | 21 + .../Code07_IrregularSudoku.java | 127 +++++ .../class_2022_03_4_week/Code08_EggXtoY.java | 82 +++ .../Code01_KMAlgorithm.java | 166 ++++++ .../Code02_ToAllSpace.java | 213 ++++++++ .../Code03_MaximumAndSumOfArray.java | 90 +++ .../Code04_KillAllSameTime.java | 18 + .../Code01_FourNumbersMinusOne.java | 58 ++ .../Code02_MaxOrSmallestSubarray.java | 116 ++++ .../Code03_ArrangeMeetingPosCancelPre.java | 219 ++++++++ .../Code04_MaxScoreMoveInBoard.java | 88 +++ ...de05_PickKnumbersNearTowNumberMaxDiff.java | 57 ++ .../Code06_TopMinSubsquenceSum.java | 114 ++++ .../Code07_TopMaxSubsquenceSum.java | 124 +++++ .../Code01_SumOfValuesAboutPrimes.java | 114 ++++ ...de02_MinDistanceFromLeftUpToRightDown.java | 125 +++++ .../Code03_MaxSumDividedBy7.java | 82 +++ .../Code04_AllJobFinishTime.java | 48 ++ ...Code05_TowLongestSubarraySame01Number.java | 95 ++++ .../Code06_PerfectPairNumber.java | 53 ++ .../Code07_MaxMoneyMostMin.java | 236 ++++++++ .../Code01_MaxOneNumbers.java | 79 +++ .../class_2022_04_3_week/Code02_RMQ.java | 118 ++++ .../Code03_ValidSortedArrayWays.java | 125 +++++ .../Code04_SumEvenSubNumber.java | 84 +++ .../Code05_ModKSubstringNumbers.java | 83 +++ .../Code01_JumMinSameValue.java | 79 +++ .../Code02_WhoWin21Balls.java | 160 ++++++ .../Code03_FindDuplicateOnlyOne.java | 114 ++++ .../Code04_SumOfQuadraticSum.java | 113 ++++ ...eStringNoLessKLenNoOverlapingMaxParts.java | 131 +++++ .../Code01_TwoObjectMaxValue.java | 155 ++++++ .../Code02_ModifyOneNumberModXWays.java | 84 +++ .../Code03_SortedSubsequenceMaxSum.java | 162 ++++++ .../Code04_OneEdgeMagicMinPathSum.java | 173 ++++++ .../Code05_RedAndWhiteSquares.java | 211 ++++++++ .../Code01_MaxNumberUnderLimit.java | 188 +++++++ .../Code02_RemoveNumbersNotIncreasingAll.java | 162 ++++++ .../Code03_NumberOfCannon.java | 97 ++++ .../Code04_MinJumpUsePre.java | 154 ++++++ .../Code01_SomeDPFromVT.java | 246 +++++++++ .../Code02_MinSetForEveryRange.java | 48 ++ ...easingSubarrayCanDeleteContinuousPart.java | 221 ++++++++ .../Code04_ABCSameNumber.java | 169 ++++++ .../Code01_WhereWillTheBallFall.java | 46 ++ ...02_UniqueSubstringsInWraparoundString.java | 34 ++ .../Code03_NumberOfAtoms.java | 104 ++++ .../Code04_SubstringWithLargestVariance.java | 98 ++++ ..._MostStonesRemovedWithSameRowOrColumn.java | 87 +++ .../class_2022_06_2_week/Code02_Solution.HEIC | Bin 0 -> 2358360 bytes .../Code02_SumOfTotalStrengthOfWizards.java | 53 ++ ...e03_NumberOfDifferentSubsequencesGCDs.java | 60 ++ .../Code04_ConsecutiveNumbersSum.java | 101 ++++ .../Code01_MaxChunksToMakeSortedII.java | 42 ++ .../Code02_SellingPiecesOfWood.java | 151 ++++++ .../Code03_RangeModule1.java | 196 +++++++ .../Code03_RangeModule2.java | 112 ++++ .../Code04_StarNumber.java | 87 +++ .../Code01_MinimumWindowSubsequence.java | 267 +++++++++ .../Code02_StackNotSplit.java | 177 ++++++ .../Code03_MaxAnimalNumber.java | 117 ++++ ...ode04_MinimizeMaxDistanceToGasStation.java | 45 ++ .../Code01_WindPrevent.java | 149 +++++ ...de02_MinimumScoreAfterRemovalsOnATree.java | 136 +++++ .../Code03_NumberOfPeopleAwareOfASecret.java | 52 ++ .../Code01_DistinctSubseqValue.java | 68 +++ .../Code02_WaysSubsqenceXToY.java | 207 +++++++ .../Code03_SwimInRisingWater.java | 155 ++++++ .../Code04_EmployeeFreeTime.java | 57 ++ .../Code05_LineSweepAlgorithm1.java | 129 +++++ .../Code05_LineSweepAlgorithm2.java | 98 ++++ .../Code01_SetIntersectionSizeAtLeastTwo.java | 57 ++ .../Code02_ValidParenthesisString.java | 119 ++++ .../Code03_TopKFrequentElements.java | 68 +++ .../Code04_SpecialBinaryString.java | 59 ++ .../Code01_WaysWiggle.java | 146 +++++ .../Code02_SidingPuzzle1.java | 103 ++++ .../Code02_SidingPuzzle2.java | 120 ++++ .../Code03_TheNumberOfGoodSubsets.java | 49 ++ .../Code04_MatchsticksToSquare.java | 40 ++ .../class01/Code01_PrintBinary.java | 65 +++ .../class01/Code02_SumOfFactorial.java | 37 ++ 算法新手班/class01/Code03_Sort.java | 79 +++ .../class01/Code04_SelectionSort.java | 114 ++++ .../class01/Code05_BubbleSort.java | 107 ++++ .../class01/Code06_InsertionSort.java | 111 ++++ 算法新手班/class02/Code01_PreSum.java | 42 ++ .../class02/Code02_RandToRand.java | 231 ++++++++ 算法新手班/class02/Code03_Comp.java | 90 +++ .../Code03_EqualProbabilityRandom.java | 67 +++ 算法新手班/class03/Code01_BSExist.java | 64 +++ .../class03/Code02_BSNearLeft.java | 78 +++ .../class03/Code03_BSNearRight.java | 75 +++ 算法新手班/class03/Code04_BSAwesome.java | 90 +++ .../class03/Code05_HashMapTreeMap.java | 93 ++++ .../class04/Code01_ReverseList.java | 223 ++++++++ .../Code02_LinkedListToQueueAndStack.java | 222 ++++++++ .../Code03_DoubleLinkedListToDeque.java | 189 +++++++ .../class04/Code04_ReverseNodesInKGroup.java | 59 ++ .../class04/Code05_AddTwoNumbers.java | 62 +++ .../Code06_MergeTwoSortedLinkedList.java | 34 ++ 算法新手班/class05/Code01_BitMap1.java | 61 +++ 算法新手班/class05/Code02_BitMap2.java | 60 ++ .../class05/Code03_BitAddMinusMultiDiv.java | 70 +++ .../class06/Code01_MergeKSortedLists.java | 52 ++ 算法新手班/class06/Code02_SameTree.java | 23 + .../class06/Code03_SymmetricTree.java | 26 + .../Code04_MaximumDepthOfBinaryTree.java | 20 + ...ryTreeFromPreorderAndInorderTraversal.java | 72 +++ 算法新手班/class06/ShowComparator.java | 106 ++++ 算法新手班/class06/ShowComparator2.java | 78 +++ .../class06/TraversalBinaryTree.java | 72 +++ ...ode01_BinaryTreeLevelOrderTraversalII.java | 45 ++ .../class07/Code02_BalancedBinaryTree.java | 42 ++ 算法新手班/class07/Code03_PathSum.java | 60 ++ 算法新手班/class07/Code04_PathSumII.java | 58 ++ .../class07/Code05_IsBinarySearchTree.java | 85 +++ 算法新手班/class08/Code01_GetMax.java | 43 ++ 算法新手班/class08/Code02_MergeSort.java | 183 +++++++ .../class08/Code03_PartitionAndQuickSort.java | 231 ++++++++ 718 files changed, 73447 insertions(+) create mode 100644 体系学习班/class01/Code01_SelectionSort.java create mode 100644 体系学习班/class01/Code02_BubbleSort.java create mode 100644 体系学习班/class01/Code03_InsertionSort.java create mode 100644 体系学习班/class01/Code04_BSExist.java create mode 100644 体系学习班/class01/Code05_BSNearLeft.java create mode 100644 体系学习班/class01/Code05_BSNearRight.java create mode 100644 体系学习班/class01/Code06_BSAwesome.java create mode 100644 体系学习班/class02/Code01_Swap.java create mode 100644 体系学习班/class02/Code02_EvenTimesOddTimes.java create mode 100644 体系学习班/class02/Code03_KM.java create mode 100644 体系学习班/class03/Code01_ReverseList.java create mode 100644 体系学习班/class03/Code02_DeleteGivenValue.java create mode 100644 体系学习班/class03/Code03_DoubleEndsQueueToStackAndQueue.java create mode 100644 体系学习班/class03/Code04_RingArray.java create mode 100644 体系学习班/class03/Code05_GetMinStack.java create mode 100644 体系学习班/class03/Code06_TwoStacksImplementQueue.java create mode 100644 体系学习班/class03/Code07_TwoQueueImplementStack.java create mode 100644 体系学习班/class03/Code08_GetMax.java create mode 100644 体系学习班/class03/HashMapAndSortedMap.java create mode 100644 体系学习班/class04/Code01_MergeSort.java create mode 100644 体系学习班/class04/Code02_SmallSum.java create mode 100644 体系学习班/class04/Code03_ReversePair.java create mode 100644 体系学习班/class04/Code04_BiggerThanRightTwice.java create mode 100644 体系学习班/class05/Code01_CountOfRangeSum.java create mode 100644 体系学习班/class05/Code02_PartitionAndQuickSort.java create mode 100644 体系学习班/class05/Code03_QuickSortRecursiveAndUnrecursive.java create mode 100644 体系学习班/class05/Code04_DoubleLinkedListQuickSort.java create mode 100644 体系学习班/class06/Code01_Comparator.java create mode 100644 体系学习班/class06/Code02_Heap.java create mode 100644 体系学习班/class06/Code03_HeapSort.java create mode 100644 体系学习班/class06/Code04_SortArrayDistanceLessK.java create mode 100644 体系学习班/class07/Code01_CoverMax.java create mode 100644 体系学习班/class07/Code02_EveryStepShowBoss.java create mode 100644 体系学习班/class07/HeapGreater.java create mode 100644 体系学习班/class07/Inner.java create mode 100644 体系学习班/class08/Code01_TrieTree.java create mode 100644 体系学习班/class08/Code02_TrieTree.java create mode 100644 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/dev/null +++ b/体系学习班/class01/Code01_SelectionSort.java @@ -0,0 +1,115 @@ +package class01; + +import java.util.Arrays; + +public class Code01_SelectionSort { + + public static void selectionSort(int[] arr) { + if (arr == null || arr.length < 2) { + return; + } + // 0 ~ N-1 找到最小值，在哪，放到0位置上 + // 1 ~ n-1 找到最小值，在哪，放到1 位置上 + // 2 ~ n-1 找到最小值，在哪，放到2 位置上 + for (int i = 0; i < arr.length - 1; i++) { + int minIndex = i; + for (int j = i + 1; j < arr.length; j++) { // i ~ N-1 上找最小值的下标 + minIndex = arr[j] < arr[minIndex] ? j : minIndex; + } + swap(arr, i, minIndex); + } + } + + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + + // for test + public static void comparator(int[] arr) { + Arrays.sort(arr); + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + // Math.random() [0,1) + // Math.random() * N [0,N) + // (int)(Math.random() * N) [0, N-1] + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + // [-? , +?] + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + return arr; + } + + // for test + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 100; + int maxValue = 100; + boolean succeed = true; + for (int i = 0; i < testTime; i++) { + int[] arr1 = generateRandomArray(maxSize, maxValue); + int[] arr2 = copyArray(arr1); + selectionSort(arr1); + comparator(arr2); + if (!isEqual(arr1, arr2)) { + succeed = false; + printArray(arr1); + printArray(arr2); + break; + } + } + System.out.println(succeed ? "Nice!" : "Fucking fucked!"); + + int[] arr = generateRandomArray(maxSize, maxValue); + printArray(arr); + selectionSort(arr); + printArray(arr); + } + +} diff --git a/体系学习班/class01/Code02_BubbleSort.java b/体系学习班/class01/Code02_BubbleSort.java new file mode 100644 index 0000000..14b6e47 --- /dev/null +++ b/体系学习班/class01/Code02_BubbleSort.java @@ -0,0 +1,110 @@ +package class01; + +import java.util.Arrays; + +public class Code02_BubbleSort { + + public static void bubbleSort(int[] arr) { + if (arr == null || arr.length < 2) { + return; + } + // 0 ~ N-1 + // 0 ~ N-2 + // 0 ~ N-3 + for (int e = arr.length - 1; e > 0; e--) { // 0 ~ e + for (int i = 0; i < e; i++) { + if (arr[i] > arr[i + 1]) { + swap(arr, i, i + 1); + } + } + } + } + + // 交换arr的i和j位置上的值 + public static void swap(int[] arr, int i, int j) { + arr[i] = arr[i] ^ arr[j]; + arr[j] = arr[i] ^ arr[j]; + arr[i] = arr[i] ^ arr[j]; + } + + // for test + public static void comparator(int[] arr) { + Arrays.sort(arr); + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + return arr; + } + + // for test + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 100; + int maxValue = 100; + boolean succeed = true; + for (int i = 0; i < testTime; i++) { + int[] arr1 = generateRandomArray(maxSize, maxValue); + int[] arr2 = copyArray(arr1); + bubbleSort(arr1); + comparator(arr2); + if (!isEqual(arr1, arr2)) { + succeed = false; + break; + } + } + System.out.println(succeed ? "Nice!" : "Fucking fucked!"); + + int[] arr = generateRandomArray(maxSize, maxValue); + printArray(arr); + bubbleSort(arr); + printArray(arr); + } + +} diff --git a/体系学习班/class01/Code03_InsertionSort.java b/体系学习班/class01/Code03_InsertionSort.java new file mode 100644 index 0000000..7283369 --- /dev/null +++ b/体系学习班/class01/Code03_InsertionSort.java @@ -0,0 +1,116 @@ +package class01; + +import java.util.Arrays; + +public class Code03_InsertionSort { + + public static void insertionSort(int[] arr) { + if (arr == null || arr.length < 2) { + return; + } + // 不只1个数 + for (int i = 1; i < arr.length; i++) { // 0 ~ i 做到有序 + for (int j = i - 1; j >= 0 && arr[j] > arr[j + 1]; j--) { + swap(arr, j, j + 1); + } + } + } + + // i和j是一个位置的话，会出错 + public static void swap(int[] arr, int i, int j) { + arr[i] = arr[i] ^ arr[j]; + arr[j] = arr[i] ^ arr[j]; + arr[i] = arr[i] ^ arr[j]; + } + + // for test + public static void comparator(int[] arr) { + Arrays.sort(arr); + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + // Math.random() -> [0,1) 所有的小数，等概率返回一个 + // Math.random() * N -> [0,N) 所有小数，等概率返回一个 + // (int)(Math.random() * N) -> [0,N-1] 所有的整数，等概率返回一个 + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; // 长度随机 + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + return arr; + } + + // for test + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 100; // 随机数组的长度0～100 + int maxValue = 100;// 值：-100～100 + boolean succeed = true; + for (int i = 0; i < testTime; i++) { + int[] arr = generateRandomArray(maxSize, maxValue); + int[] arr1 = copyArray(arr); + int[] arr2 = copyArray(arr); + insertionSort(arr1); + comparator(arr2); + if (!isEqual(arr1, arr2)) { + // 打印arr1 + // 打印arr2 + succeed = false; + for (int j = 0; j < arr.length; j++) { + System.out.print(arr[j] + " "); + } + System.out.println(); + break; + } + } + System.out.println(succeed ? "Nice!" : "Fucking fucked!"); + + int[] arr = generateRandomArray(maxSize, maxValue); + printArray(arr); + insertionSort(arr); + printArray(arr); + } + +} diff --git a/体系学习班/class01/Code04_BSExist.java b/体系学习班/class01/Code04_BSExist.java new file mode 100644 index 0000000..d89be00 --- /dev/null +++ b/体系学习班/class01/Code04_BSExist.java @@ -0,0 +1,65 @@ +package class01; + +import java.util.Arrays; + +public class Code04_BSExist { + + public static boolean exist(int[] sortedArr, int num) { + if (sortedArr == null || sortedArr.length == 0) { + return false; + } + int L = 0; + int R = sortedArr.length - 1; + int mid = 0; + // L..R + while (L < R) { // L..R 至少两个数的时候 + mid = L + ((R - L) >> 1); + if (sortedArr[mid] == num) { + return true; + } else if (sortedArr[mid] > num) { + R = mid - 1; + } else { + L = mid + 1; + } + } + return sortedArr[L] == num; + } + + // for test + public static boolean test(int[] sortedArr, int num) { + for(int cur : sortedArr) { + if(cur == num) { + return true; + } + } + return false; + } + + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + return arr; + } + + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 10; + int maxValue = 100; + boolean succeed = true; + for (int i = 0; i < testTime; i++) { + int[] arr = generateRandomArray(maxSize, maxValue); + Arrays.sort(arr); + int value = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + if (test(arr, value) != exist(arr, value)) { + succeed = false; + break; + } + } + System.out.println(succeed ? "Nice!" : "Fucking fucked!"); + } + +} diff --git a/体系学习班/class01/Code05_BSNearLeft.java b/体系学习班/class01/Code05_BSNearLeft.java new file mode 100644 index 0000000..eaf5804 --- /dev/null +++ b/体系学习班/class01/Code05_BSNearLeft.java @@ -0,0 +1,75 @@ +package class01; + +import java.util.Arrays; + +public class Code05_BSNearLeft { + + // 在arr上，找满足>=value的最左位置 + public static int nearestIndex(int[] arr, int value) { + int L = 0; + int R = arr.length - 1; + int index = -1; // 记录最左的对号 + while (L <= R) { // 至少一个数的时候 + int mid = L + ((R - L) >> 1); + if (arr[mid] >= value) { + index = mid; + R = mid - 1; + } else { + L = mid + 1; + } + } + return index; + } + + // for test + public static int test(int[] arr, int value) { + for (int i = 0; i < arr.length; i++) { + if (arr[i] >= value) { + return i; + } + } + return -1; + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + return arr; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 10; + int maxValue = 100; + boolean succeed = true; + for (int i = 0; i < testTime; i++) { + int[] arr = generateRandomArray(maxSize, maxValue); + Arrays.sort(arr); + int value = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + if (test(arr, value) != nearestIndex(arr, value)) { + printArray(arr); + System.out.println(value); + System.out.println(test(arr, value)); + System.out.println(nearestIndex(arr, value)); + succeed = false; + break; + } + } + System.out.println(succeed ? "Nice!" : "Fucking fucked!"); + } + +} diff --git a/体系学习班/class01/Code05_BSNearRight.java b/体系学习班/class01/Code05_BSNearRight.java new file mode 100644 index 0000000..f3b6f25 --- /dev/null +++ b/体系学习班/class01/Code05_BSNearRight.java @@ -0,0 +1,75 @@ +package class01; + +import java.util.Arrays; + +public class Code05_BSNearRight { + + // 在arr上，找满足<=value的最右位置 + public static int nearestIndex(int[] arr, int value) { + int L = 0; + int R = arr.length - 1; + int index = -1; // 记录最右的对号 + while (L <= R) { + int mid = L + ((R - L) >> 1); + if (arr[mid] <= value) { + index = mid; + L = mid + 1; + } else { + R = mid - 1; + } + } + return index; + } + + // for test + public static int test(int[] arr, int value) { + for (int i = arr.length - 1; i >= 0; i--) { + if (arr[i] <= value) { + return i; + } + } + return -1; + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + return arr; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 10; + int maxValue = 100; + boolean succeed = true; + for (int i = 0; i < testTime; i++) { + int[] arr = generateRandomArray(maxSize, maxValue); + Arrays.sort(arr); + int value = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + if (test(arr, value) != nearestIndex(arr, value)) { + printArray(arr); + System.out.println(value); + System.out.println(test(arr, value)); + System.out.println(nearestIndex(arr, value)); + succeed = false; + break; + } + } + System.out.println(succeed ? "Nice!" : "Fucking fucked!"); + } + +} diff --git a/体系学习班/class01/Code06_BSAwesome.java b/体系学习班/class01/Code06_BSAwesome.java new file mode 100644 index 0000000..c167629 --- /dev/null +++ b/体系学习班/class01/Code06_BSAwesome.java @@ -0,0 +1,76 @@ +package class01; + +public class Code06_BSAwesome { + + // 课上的代码 + public static int getLessIndex(int[] arr) { + if (arr == null || arr.length == 0) { + return -1; + } + if (arr.length == 1 || arr[0] < arr[1]) { + return 0; + } + if (arr[arr.length - 1] < arr[arr.length - 2]) { + return arr.length - 1; + } + int left = 1; + int right = arr.length - 2; + int mid = 0; + while (left < right) { + mid = (left + right) / 2; + if (arr[mid] > arr[mid - 1]) { + right = mid - 1; + } else if (arr[mid] > arr[mid + 1]) { + left = mid + 1; + } else { + return mid; + } + } + return left; + } + + // 验证得到的结果，是不是局部最小 + public static boolean isRight(int[] arr, int index) { + if (arr.length <= 1) { + return true; + } + if (index == 0) { + return arr[index] < arr[index + 1]; + } + if (index == arr.length - 1) { + return arr[index] < arr[index - 1]; + } + return arr[index] < arr[index - 1] && arr[index] < arr[index + 1]; + } + + // 为了测试 + // 生成相邻不相等的数组 + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) (Math.random() * maxSize) + 1]; + arr[0] = (int) (Math.random() * maxValue) - (int) (Math.random() * maxValue); + for (int i = 1; i < arr.length; i++) { + do { + arr[i] = (int) (Math.random() * maxValue) - (int) (Math.random() * maxValue); + } while (arr[i] == arr[i - 1]); + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 30; + int maxValue = 100; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int[] arr = generateRandomArray(maxSize, maxValue); + int ans = getLessIndex(arr); + if (!isRight(arr, ans)) { + System.out.println("出错了！"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class02/Code01_Swap.java b/体系学习班/class02/Code01_Swap.java new file mode 100644 index 0000000..c625614 --- /dev/null +++ b/体系学习班/class02/Code01_Swap.java @@ -0,0 +1,71 @@ +package class02; + +public class Code01_Swap { + + public static void main(String[] args) { + + + + + + + int a = 16; + int b = 603; + + System.out.println(a); + System.out.println(b); + + + a = a ^ b; + b = a ^ b; + a = a ^ b; + + + System.out.println(a); + System.out.println(b); + + + + + int[] arr = {3,1,100}; + + int i = 0; + int j = 0; + + arr[i] = arr[i] ^ arr[j]; + arr[j] = arr[i] ^ arr[j]; + arr[i] = arr[i] ^ arr[j]; + + System.out.println(arr[i] + " , " + arr[j]); + + + + + + + + + + System.out.println(arr[0]); + System.out.println(arr[2]); + + swap(arr, 0, 0); + + System.out.println(arr[0]); + System.out.println(arr[2]); + + + + } + + + public static void swap (int[] arr, int i, int j) { + // arr[0] = arr[0] ^ arr[0]; + arr[i] = arr[i] ^ arr[j]; + arr[j] = arr[i] ^ arr[j]; + arr[i] = arr[i] ^ arr[j]; + } + + + +} diff --git a/体系学习班/class02/Code02_EvenTimesOddTimes.java b/体系学习班/class02/Code02_EvenTimesOddTimes.java new file mode 100644 index 0000000..13ac877 --- /dev/null +++ b/体系学习班/class02/Code02_EvenTimesOddTimes.java @@ -0,0 +1,83 @@ +package class02; + +public class Code02_EvenTimesOddTimes { + + // arr中，只有一种数，出现奇数次 + public static void printOddTimesNum1(int[] arr) { + int eor = 0; + for (int i = 0; i < arr.length; i++) { + eor ^= arr[i]; + } + System.out.println(eor); + } + + // arr中，有两种数，出现奇数次 + public static void printOddTimesNum2(int[] arr) { + int eor = 0; + for (int i = 0; i < arr.length; i++) { + eor ^= arr[i]; + } + // a 和 b是两种数 + // eor != 0 + // eor最右侧的1，提取出来 + // eor : 00110010110111000 + // rightOne :00000000000001000 + int rightOne = eor & (-eor); // 提取出最右的1 + + + int onlyOne = 0; // eor' + for (int i = 0 ; i < arr.length;i++) { + // arr[1] = 111100011110000 + // rightOne= 000000000010000 + if ((arr[i] & rightOne) != 0) { + onlyOne ^= arr[i]; + } + } + System.out.println(onlyOne + " " + (eor ^ onlyOne)); + } + + + public static int bit1counts(int N) { + int count = 0; + + // 011011010000 + // 000000010000 1 + + // 011011000000 + // + + + + while(N != 0) { + int rightOne = N & ((~N) + 1); + count++; + N ^= rightOne; + // N -= rightOne + } + + + return count; + + } + + + public static void main(String[] args) { + int a = 5; + int b = 7; + + a = a ^ b; + b = a ^ b; + a = a ^ b; + + System.out.println(a); + System.out.println(b); + + int[] arr1 = { 3, 3, 2, 3, 1, 1, 1, 3, 1, 1, 1 }; + printOddTimesNum1(arr1); + + int[] arr2 = { 4, 3, 4, 2, 2, 2, 4, 1, 1, 1, 3, 3, 1, 1, 1, 4, 2, 2 }; + printOddTimesNum2(arr2); + + } + +} diff --git a/体系学习班/class02/Code03_KM.java b/体系学习班/class02/Code03_KM.java new file mode 100644 index 0000000..a530445 --- /dev/null +++ b/体系学习班/class02/Code03_KM.java @@ -0,0 +1,159 @@ +package class02; + +import java.util.HashMap; +import java.util.HashSet; + +// 输入一定能够保证，数组中所有的数都出现了M次，只有一种数出现了K次 +// 1 <= K < M +// 返回这种数 +public class Code03_KM { + + public static int test(int[] arr, int k, int m) { + HashMap map = new HashMap<>(); + for (int num : arr) { + if (map.containsKey(num)) { + map.put(num, map.get(num) + 1); + } else { + map.put(num, 1); + } + } + int ans = 0; + for (int num : map.keySet()) { + if (map.get(num) == k) { + ans = num; + break; + } + } + return ans; + } + + public static HashMap map = new HashMap<>(); + + // 请保证arr中，只有一种数出现了K次，其他数都出现了M次 + public static int onlyKTimes(int[] arr, int k, int m) { + if (map.size() == 0) { + mapCreater(map); + } + int[] t = new int[32]; + // t[0] 0位置的1出现了几个 + // t[i] i位置的1出现了几个 + for (int num : arr) { + while (num != 0) { + int rightOne = num & (-num); + t[map.get(rightOne)]++; + num ^= rightOne; + } + } + int ans = 0; + // 如果这个出现了K次的数，就是0 + // 那么下面代码中的 : ans |= (1 << i); + // 就不会发生 + // 那么ans就会一直维持0，最后返回0，也是对的！ + for (int i = 0; i < 32; i++) { + if (t[i] % m != 0) { + ans |= (1 << i); + } + } + return ans; + } + + public static void mapCreater(HashMap map) { + int value = 1; + for (int i = 0; i < 32; i++) { + map.put(value, i); + value <<= 1; + } + } + + // 更简洁的写法 + public static int km(int[] arr, int k, int m) { + int[] help = new int[32]; + for (int num : arr) { + for (int i = 0; i < 32; i++) { + help[i] += (num >> i) & 1; + } + } + int ans = 0; + for (int i = 0; i < 32; i++) { + help[i] %= m; + if (help[i] != 0) { + ans |= 1 << i; + } + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int maxKinds, int range, int k, int m) { + int ktimeNum = randomNumber(range); + // 真命天子出现的次数 + int times = k; + // 2 + int numKinds = (int) (Math.random() * maxKinds) + 2; + // k * 1 + (numKinds - 1) * m + int[] arr = new int[times + (numKinds - 1) * m]; + int index = 0; + for (; index < times; index++) { + arr[index] = ktimeNum; + } + numKinds--; + HashSet set = new HashSet<>(); + set.add(ktimeNum); + while (numKinds != 0) { + int curNum = 0; + do { + curNum = randomNumber(range); + } while (set.contains(curNum)); + set.add(curNum); + numKinds--; + for (int i = 0; i < m; i++) { + arr[index++] = curNum; + } + } + // arr 填好了 + for (int i = 0; i < arr.length; i++) { + // i 位置的数，我想随机和j位置的数做交换 + int j = (int) (Math.random() * arr.length);// 0 ~ N-1 + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + return arr; + } + + // 为了测试 + // [-range, +range] + public static int randomNumber(int range) { + return (int) (Math.random() * (range + 1)) - (int) (Math.random() * (range + 1)); + } + + // 为了测试 + public static void main(String[] args) { + int kinds = 5; + int range = 30; + int testTime = 100000; + int max = 9; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int a = (int) (Math.random() * max) + 1; // a 1 ~ 9 + int b = (int) (Math.random() * max) + 1; // b 1 ~ 9 + int k = Math.min(a, b); + int m = Math.max(a, b); + // k < m + if (k == m) { + m++; + } + int[] arr = randomArray(kinds, range, k, m); + int ans1 = test(arr, k, m); + int ans2 = onlyKTimes(arr, k, m); + int ans3 = km(arr, k, m); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println(ans1); + System.out.println(ans3); + System.out.println("出错了！"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class03/Code01_ReverseList.java b/体系学习班/class03/Code01_ReverseList.java new file mode 100644 index 0000000..4b0318a --- /dev/null +++ b/体系学习班/class03/Code01_ReverseList.java @@ -0,0 +1,221 @@ +package class03; + +import java.util.ArrayList; +import java.util.List; + +public class Code01_ReverseList { + + public static class Node { + public int value; + public Node next; + + public Node(int data) { + value = data; + } + } + + public static class DoubleNode { + public int value; + public DoubleNode last; + public DoubleNode next; + + public DoubleNode(int data) { + value = data; + } + } + + // head + // a -> b -> c -> null + // c -> b -> a -> null + public static Node reverseLinkedList(Node head) { + Node pre = null; + Node next = null; + while (head != null) { + next = head.next; + head.next = pre; + pre = head; + head = next; + } + return pre; + } + + public static DoubleNode reverseDoubleList(DoubleNode head) { + DoubleNode pre = null; + DoubleNode next = null; + while (head != null) { + next = head.next; + head.next = pre; + head.last = next; + pre = head; + head = next; + } + return pre; + } + + public static Node testReverseLinkedList(Node head) { + if (head == null) { + return null; + } + ArrayList list = new ArrayList<>(); + while (head != null) { + list.add(head); + head = head.next; + } + list.get(0).next = null; + int N = list.size(); + for (int i = 1; i < N; i++) { + list.get(i).next = list.get(i - 1); + } + return list.get(N - 1); + } + + public static DoubleNode testReverseDoubleList(DoubleNode head) { + if (head == null) { + return null; + } + ArrayList list = new ArrayList<>(); + while (head != null) { + list.add(head); + head = head.next; + } + list.get(0).next = null; + DoubleNode pre = list.get(0); + int N = list.size(); + for (int i = 1; i < N; i++) { + DoubleNode cur = list.get(i); + cur.last = null; + cur.next = pre; + pre.last = cur; + pre = cur; + } + return list.get(N - 1); + } + + // for test + public static Node generateRandomLinkedList(int len, int value) { + int size = (int) (Math.random() * (len + 1)); + if (size == 0) { + return null; + } + size--; + Node head = new Node((int) (Math.random() * (value + 1))); + Node pre = head; + while (size != 0) { + Node cur = new Node((int) (Math.random() * (value + 1))); + pre.next = cur; + pre = cur; + size--; + } + return head; + } + + // for test + public static DoubleNode generateRandomDoubleList(int len, int value) { + int size = (int) (Math.random() * (len + 1)); + if (size == 0) { + return null; + } + size--; + DoubleNode head = new DoubleNode((int) (Math.random() * (value + 1))); + DoubleNode pre = head; + while (size != 0) { + DoubleNode cur = new DoubleNode((int) (Math.random() * (value + 1))); + pre.next = cur; + cur.last = pre; + pre = cur; + size--; + } + return head; + } + + // for test + public static List getLinkedListOriginOrder(Node head) { + List ans = new ArrayList<>(); + while (head != null) { + ans.add(head.value); + head = head.next; + } + return ans; + } + + // for test + public static boolean checkLinkedListReverse(List origin, Node head) { + for (int i = origin.size() - 1; i >= 0; i--) { + if (!origin.get(i).equals(head.value)) { + return false; + } + head = head.next; + } + return true; + } + + // for test + public static List getDoubleListOriginOrder(DoubleNode head) { + List ans = new ArrayList<>(); + while (head != null) { + ans.add(head.value); + head = head.next; + } + return ans; + } + + // for test + public static boolean checkDoubleListReverse(List origin, DoubleNode head) { + DoubleNode end = null; + for (int i = origin.size() - 1; i >= 0; i--) { + if (!origin.get(i).equals(head.value)) { + return false; + } + end = head; + head = head.next; + } + for (int i = 0; i < origin.size(); i++) { + if (!origin.get(i).equals(end.value)) { + return false; + } + end = end.last; + } + return true; + } + + // for test + public static void main(String[] args) { + int len = 50; + int value = 100; + int testTime = 100000; + System.out.println("test begin!"); + for (int i = 0; i < testTime; i++) { + Node node1 = generateRandomLinkedList(len, value); + List list1 = getLinkedListOriginOrder(node1); + node1 = reverseLinkedList(node1); + if (!checkLinkedListReverse(list1, node1)) { + System.out.println("Oops1!"); + } + + Node node2 = generateRandomLinkedList(len, value); + List list2 = getLinkedListOriginOrder(node2); + node2 = testReverseLinkedList(node2); + if (!checkLinkedListReverse(list2, node2)) { + System.out.println("Oops2!"); + } + + DoubleNode node3 = generateRandomDoubleList(len, value); + List list3 = getDoubleListOriginOrder(node3); + node3 = reverseDoubleList(node3); + if (!checkDoubleListReverse(list3, node3)) { + System.out.println("Oops3!"); + } + + DoubleNode node4 = generateRandomDoubleList(len, value); + List list4 = getDoubleListOriginOrder(node4); + node4 = reverseDoubleList(node4); + if (!checkDoubleListReverse(list4, node4)) { + System.out.println("Oops4!"); + } + + } + System.out.println("test finish!"); + + } + +} \ No newline at end of file diff --git a/体系学习班/class03/Code02_DeleteGivenValue.java b/体系学习班/class03/Code02_DeleteGivenValue.java new file mode 100644 index 0000000..647047a --- /dev/null +++ b/体系学习班/class03/Code02_DeleteGivenValue.java @@ -0,0 +1,38 @@ +package class03; + +public class Code02_DeleteGivenValue { + + public static class Node { + public int value; + public Node next; + + public Node(int data) { + this.value = data; + } + } + + // head = removeValue(head, 2); + public static Node removeValue(Node head, int num) { + // head来到第一个不需要删的位置 + while (head != null) { + if (head.value != num) { + break; + } + head = head.next; + } + // 1 ) head == null + // 2 ) head != null + Node pre = head; + Node cur = head; + while (cur != null) { + if (cur.value == num) { + pre.next = cur.next; + } else { + pre = cur; + } + cur = cur.next; + } + return head; + } + +} diff --git a/体系学习班/class03/Code03_DoubleEndsQueueToStackAndQueue.java b/体系学习班/class03/Code03_DoubleEndsQueueToStackAndQueue.java new file mode 100644 index 0000000..e9b30a9 --- /dev/null +++ b/体系学习班/class03/Code03_DoubleEndsQueueToStackAndQueue.java @@ -0,0 +1,183 @@ +package class03; + +import java.util.LinkedList; +import java.util.Queue; +import java.util.Stack; + +public class Code03_DoubleEndsQueueToStackAndQueue { + + public static class Node { + public T value; + public Node last; + public Node next; + + public Node(T data) { + value = data; + } + } + + public static class DoubleEndsQueue { + public Node head; + public Node tail; + + public void addFromHead(T value) { + Node cur = new Node(value); + if (head == null) { + head = cur; + tail = cur; + } else { + cur.next = head; + head.last = cur; + head = cur; + } + } + + public void addFromBottom(T value) { + Node cur = new Node(value); + if (head == null) { + head = cur; + tail = cur; + } else { + cur.last = tail; + tail.next = cur; + tail = cur; + } + } + + public T popFromHead() { + if (head == null) { + return null; + } + Node cur = head; + if (head == tail) { + head = null; + tail = null; + } else { + head = head.next; + cur.next = null; + head.last = null; + } + return cur.value; + } + + public T popFromBottom() { + if (head == null) { + return null; + } + Node cur = tail; + if (head == tail) { + head = null; + tail = null; + } else { + tail = tail.last; + tail.next = null; + cur.last = null; + } + return cur.value; + } + + public boolean isEmpty() { + return head == null; + } + + } + + public static class MyStack { + private DoubleEndsQueue queue; + + public MyStack() { + queue = new DoubleEndsQueue(); + } + + public void push(T value) { + queue.addFromHead(value); + } + + public T pop() { + return queue.popFromHead(); + } + + public boolean isEmpty() { + return queue.isEmpty(); + } + + } + + public static class MyQueue { + private DoubleEndsQueue queue; + + public MyQueue() { + queue = new DoubleEndsQueue(); + } + + public void push(T value) { + queue.addFromHead(value); + } + + public T poll() { + return queue.popFromBottom(); + } + + public boolean isEmpty() { + return queue.isEmpty(); + } + + } + + public static boolean isEqual(Integer o1, Integer o2) { + if (o1 == null && o2 != null) { + return false; + } + if (o1 != null && o2 == null) { + return false; + } + if (o1 == null && o2 == null) { + return true; + } + return o1.equals(o2); + } + + public static void main(String[] args) { + int oneTestDataNum = 100; + int value = 10000; + int testTimes = 100000; + for (int i = 0; i < testTimes; i++) { + MyStack myStack = new MyStack<>(); + MyQueue myQueue = new MyQueue<>(); + Stack stack = new Stack<>(); + Queue queue = new LinkedList<>(); + for (int j = 0; j < oneTestDataNum; j++) { + int nums = (int) (Math.random() * value); + if (stack.isEmpty()) { + myStack.push(nums); + stack.push(nums); + } else { + if (Math.random() < 0.5) { + myStack.push(nums); + stack.push(nums); + } else { + if (!isEqual(myStack.pop(), stack.pop())) { + System.out.println("oops!"); + } + } + } + int numq = (int) (Math.random() * value); + if (queue.isEmpty()) { + myQueue.push(numq); + queue.offer(numq); + } else { + if (Math.random() < 0.5) { + myQueue.push(numq); + queue.offer(numq); + } else { + if (!isEqual(myQueue.poll(), queue.poll())) { + System.out.println("oops!"); + } + } + } + } + } + System.out.println("finish!"); + } + +} diff --git a/体系学习班/class03/Code04_RingArray.java b/体系学习班/class03/Code04_RingArray.java new file mode 100644 index 0000000..cad73b4 --- /dev/null +++ b/体系学习班/class03/Code04_RingArray.java @@ -0,0 +1,50 @@ +package class03; + +public class Code04_RingArray { + + public static class MyQueue { + private int[] arr; + private int pushi;// end + private int polli;// begin + private int size; + private final int limit; + + public MyQueue(int limit) { + arr = new int[limit]; + pushi = 0; + polli = 0; + size = 0; + this.limit = limit; + } + + public void push(int value) { + if (size == limit) { + throw new RuntimeException("队列满了，不能再加了"); + } + size++; + arr[pushi] = value; + pushi = nextIndex(pushi); + } + + public int pop() { + if (size == 0) { + throw new RuntimeException("队列空了，不能再拿了"); + } + size--; + int ans = arr[polli]; + polli = nextIndex(polli); + return ans; + } + + public boolean isEmpty() { + return size == 0; + } + + // 如果现在的下标是i，返回下一个位置 + private int nextIndex(int i) { + return i < limit - 1 ? i + 1 : 0; + } + + } + +} diff --git a/体系学习班/class03/Code05_GetMinStack.java b/体系学习班/class03/Code05_GetMinStack.java new file mode 100644 index 0000000..91b20ae --- /dev/null +++ b/体系学习班/class03/Code05_GetMinStack.java @@ -0,0 +1,105 @@ +package class03; + +import java.util.Stack; + +public class Code05_GetMinStack { + + public static class MyStack1 { + private Stack stackData; + private Stack stackMin; + + public MyStack1() { + this.stackData = new Stack(); + this.stackMin = new Stack(); + } + + public void push(int newNum) { + if (this.stackMin.isEmpty()) { + this.stackMin.push(newNum); + } else if (newNum <= this.getmin()) { + this.stackMin.push(newNum); + } + this.stackData.push(newNum); + } + + public int pop() { + if (this.stackData.isEmpty()) { + throw new RuntimeException("Your stack is empty."); + } + int value = this.stackData.pop(); + if (value == this.getmin()) { + this.stackMin.pop(); + } + return value; + } + + public int getmin() { + if (this.stackMin.isEmpty()) { + throw new RuntimeException("Your stack is empty."); + } + return this.stackMin.peek(); + } + } + + public static class MyStack2 { + private Stack stackData; + private Stack stackMin; + + public MyStack2() { + this.stackData = new Stack(); + this.stackMin = new Stack(); + } + + public void push(int newNum) { + if (this.stackMin.isEmpty()) { + this.stackMin.push(newNum); + } else if (newNum < this.getmin()) { + this.stackMin.push(newNum); + } else { + int newMin = this.stackMin.peek(); + this.stackMin.push(newMin); + } + this.stackData.push(newNum); + } + + public int pop() { + if (this.stackData.isEmpty()) { + throw new RuntimeException("Your stack is empty."); + } + this.stackMin.pop(); + return this.stackData.pop(); + } + + public int getmin() { + if (this.stackMin.isEmpty()) { + throw new RuntimeException("Your stack is empty."); + } + return this.stackMin.peek(); + } + } + + public static void main(String[] args) { + MyStack1 stack1 = new MyStack1(); + stack1.push(3); + System.out.println(stack1.getmin()); + stack1.push(4); + System.out.println(stack1.getmin()); + stack1.push(1); + System.out.println(stack1.getmin()); + System.out.println(stack1.pop()); + System.out.println(stack1.getmin()); + + System.out.println("============="); + + MyStack1 stack2 = new MyStack1(); + stack2.push(3); + System.out.println(stack2.getmin()); + stack2.push(4); + System.out.println(stack2.getmin()); + stack2.push(1); + System.out.println(stack2.getmin()); + System.out.println(stack2.pop()); + System.out.println(stack2.getmin()); + } + +} diff --git a/体系学习班/class03/Code06_TwoStacksImplementQueue.java b/体系学习班/class03/Code06_TwoStacksImplementQueue.java new file mode 100644 index 0000000..483d7a7 --- /dev/null +++ b/体系学习班/class03/Code06_TwoStacksImplementQueue.java @@ -0,0 +1,60 @@ +package class03; + +import java.util.Stack; + +public class Code06_TwoStacksImplementQueue { + + public static class TwoStacksQueue { + public Stack stackPush; + public Stack stackPop; + + public TwoStacksQueue() { + stackPush = new Stack(); + stackPop = new Stack(); + } + + // push栈向pop栈倒入数据 + private void pushToPop() { + if (stackPop.empty()) { + while (!stackPush.empty()) { + stackPop.push(stackPush.pop()); + } + } + } + + public void add(int pushInt) { + stackPush.push(pushInt); + pushToPop(); + } + + public int poll() { + if (stackPop.empty() && stackPush.empty()) { + throw new RuntimeException("Queue is empty!"); + } + pushToPop(); + return stackPop.pop(); + } + + public int peek() { + if (stackPop.empty() && stackPush.empty()) { + throw new RuntimeException("Queue is empty!"); + } + pushToPop(); + return stackPop.peek(); + } + } + + public static void main(String[] args) { + TwoStacksQueue test = new TwoStacksQueue(); + test.add(1); + test.add(2); + test.add(3); + System.out.println(test.peek()); + System.out.println(test.poll()); + System.out.println(test.peek()); + System.out.println(test.poll()); + System.out.println(test.peek()); + System.out.println(test.poll()); + } + +} diff --git a/体系学习班/class03/Code07_TwoQueueImplementStack.java b/体系学习班/class03/Code07_TwoQueueImplementStack.java new file mode 100644 index 0000000..a2a3c54 --- /dev/null +++ b/体系学习班/class03/Code07_TwoQueueImplementStack.java @@ -0,0 +1,90 @@ +package class03; + +import java.util.LinkedList; +import java.util.Queue; +import java.util.Stack; + +public class Code07_TwoQueueImplementStack { + + public static class TwoQueueStack { + public Queue queue; + public Queue help; + + public TwoQueueStack() { + queue = new LinkedList<>(); + help = new LinkedList<>(); + } + + public void push(T value) { + queue.offer(value); + } + + public T poll() { + while (queue.size() > 1) { + help.offer(queue.poll()); + } + T ans = queue.poll(); + Queue tmp = queue; + queue = help; + help = tmp; + return ans; + } + + public T peek() { + while (queue.size() > 1) { + help.offer(queue.poll()); + } + T ans = queue.poll(); + help.offer(ans); + Queue tmp = queue; + queue = help; + help = tmp; + return ans; + } + + public boolean isEmpty() { + return queue.isEmpty(); + } + + } + + public static void main(String[] args) { + System.out.println("test begin"); + TwoQueueStack myStack = new TwoQueueStack<>(); + Stack test = new Stack<>(); + int testTime = 1000000; + int max = 1000000; + for (int i = 0; i < testTime; i++) { + if (myStack.isEmpty()) { + if (!test.isEmpty()) { + System.out.println("Oops"); + } + int num = (int) (Math.random() * max); + myStack.push(num); + test.push(num); + } else { + if (Math.random() < 0.25) { + int num = (int) (Math.random() * max); + myStack.push(num); + test.push(num); + } else if (Math.random() < 0.5) { + if (!myStack.peek().equals(test.peek())) { + System.out.println("Oops"); + } + } else if (Math.random() < 0.75) { + if (!myStack.poll().equals(test.pop())) { + System.out.println("Oops"); + } + } else { + if (myStack.isEmpty() != test.isEmpty()) { + System.out.println("Oops"); + } + } + } + } + + System.out.println("test finish!"); + + } + +} diff --git a/体系学习班/class03/Code08_GetMax.java b/体系学习班/class03/Code08_GetMax.java new file mode 100644 index 0000000..81661d8 --- /dev/null +++ b/体系学习班/class03/Code08_GetMax.java @@ -0,0 +1,24 @@ +package class03; + +public class Code08_GetMax { + + // 求arr中的最大值 + public static int getMax(int[] arr) { + return process(arr, 0, arr.length - 1); + } + + // arr[L..R]范围上求最大值 L ... R N + public static int process(int[] arr, int L, int R) { + // arr[L..R]范围上只有一个数，直接返回，base case + if (L == R) { + return arr[L]; + } + // L...R 不只一个数 + // mid = (L + R) / 2 + int mid = L + ((R - L) >> 1); // 中点 1 + int leftMax = process(arr, L, mid); + int rightMax = process(arr, mid + 1, R); + return Math.max(leftMax, rightMax); + } + +} diff --git a/体系学习班/class03/HashMapAndSortedMap.java b/体系学习班/class03/HashMapAndSortedMap.java new file mode 100644 index 0000000..bd54a06 --- /dev/null +++ b/体系学习班/class03/HashMapAndSortedMap.java @@ -0,0 +1,128 @@ +package class03; + +import java.util.HashMap; +import java.util.HashSet; +import java.util.TreeMap; + +public class HashMapAndSortedMap { + + public static class Node { + public int value; + + public Node(int v) { + value = v; + } + } + + public static class Zuo { + public int value; + + public Zuo(int v) { + value = v; + } + } + + public static void main(String[] args) { + + HashMap test = new HashMap<>(); + Integer a = 19000000; + Integer b = 19000000; + System.out.println(a == b); + + test.put(a, "我是3"); + System.out.println(test.containsKey(b)); + + Zuo z1 = new Zuo(1); + Zuo z2 = new Zuo(1); + HashMap test2 = new HashMap<>(); + test2.put(z1, "我是z1"); + System.out.println(test2.containsKey(z2)); + + // UnSortedMap + HashMap map = new HashMap<>(); + map.put(1000000, "我是1000000"); + map.put(2, "我是2"); + map.put(3, "我是3"); + map.put(4, "我是4"); + map.put(5, "我是5"); + map.put(6, "我是6"); + map.put(1000000, "我是1000001"); + + System.out.println(map.containsKey(1)); + System.out.println(map.containsKey(10)); + + System.out.println(map.get(4)); + System.out.println(map.get(10)); + + map.put(4, "他是4"); + System.out.println(map.get(4)); + + map.remove(4); + System.out.println(map.get(4)); + + // key + HashSet set = new HashSet<>(); + set.add("abc"); + set.contains("abc"); + set.remove("abc"); + + // 哈希表，增、删、改、查，在使用时，O（1） + + System.out.println("====================="); + + Integer c = 100000; + Integer d = 100000; + System.out.println(c.equals(d)); + + Integer e = 127; // - 128 ~ 127 + Integer f = 127; + System.out.println(e == f); + + HashMap map2 = new HashMap<>(); + Node node1 = new Node(1); + Node node2 = node1; + map2.put(node1, "我是node1"); + map2.put(node2, "我是node1"); + System.out.println(map2.size()); + + System.out.println("======================"); + + // TreeMap 有序表：接口名 + // 红黑树、avl、sb树、跳表 + // O(logN) + System.out.println("有序表测试开始"); + TreeMap treeMap = new TreeMap<>(); + + treeMap.put(3, "我是3"); + treeMap.put(4, "我是4"); + treeMap.put(8, "我是8"); + treeMap.put(5, "我是5"); + treeMap.put(7, "我是7"); + treeMap.put(1, "我是1"); + treeMap.put(2, "我是2"); + + System.out.println(treeMap.containsKey(1)); + System.out.println(treeMap.containsKey(10)); + + System.out.println(treeMap.get(4)); + System.out.println(treeMap.get(10)); + + treeMap.put(4, "他是4"); + System.out.println(treeMap.get(4)); + + // treeMap.remove(4); + System.out.println(treeMap.get(4)); + + System.out.println("新鲜："); + + System.out.println(treeMap.firstKey()); + System.out.println(treeMap.lastKey()); + // <= 4 + System.out.println(treeMap.floorKey(4)); + // >= 4 + System.out.println(treeMap.ceilingKey(4)); + // O(logN) + + } + +} diff --git a/体系学习班/class04/Code01_MergeSort.java b/体系学习班/class04/Code01_MergeSort.java new file mode 100644 index 0000000..c0ddb6b --- /dev/null +++ b/体系学习班/class04/Code01_MergeSort.java @@ -0,0 +1,147 @@ +package class04; + +public class Code01_MergeSort { + + // 递归方法实现 + public static void mergeSort1(int[] arr) { + if (arr == null || arr.length < 2) { + return; + } + process(arr, 0, arr.length - 1); + } + + // 请把arr[L..R]排有序 + // l...r N + // T(N) = 2 * T(N / 2) + O(N) + // O(N * logN) + public static void process(int[] arr, int L, int R) { + if (L == R) { // base case + return; + } + int mid = L + ((R - L) >> 1); + process(arr, L, mid); + process(arr, mid + 1, R); + merge(arr, L, mid, R); + } + + public static void merge(int[] arr, int L, int M, int R) { + int[] help = new int[R - L + 1]; + int i = 0; + int p1 = L; + int p2 = M + 1; + while (p1 <= M && p2 <= R) { + help[i++] = arr[p1] <= arr[p2] ? arr[p1++] : arr[p2++]; + } + // 要么p1越界了，要么p2越界了 + while (p1 <= M) { + help[i++] = arr[p1++]; + } + while (p2 <= R) { + help[i++] = arr[p2++]; + } + for (i = 0; i < help.length; i++) { + arr[L + i] = help[i]; + } + } + + // 非递归方法实现 + public static void mergeSort2(int[] arr) { + if (arr == null || arr.length < 2) { + return; + } + int N = arr.length; + // 步长 + int mergeSize = 1; + while (mergeSize < N) { // log N + // 当前左组的，第一个位置 + int L = 0; + while (L < N) { + if (mergeSize >= N - L) { + break; + } + int M = L + mergeSize - 1; + int R = M + Math.min(mergeSize, N - M - 1); + merge(arr, L, M, R); + L = R + 1; + } + // 防止溢出 + if (mergeSize > N / 2) { + break; + } + mergeSize <<= 1; + } + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + return arr; + } + + // for test + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 100; + int maxValue = 100; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int[] arr1 = generateRandomArray(maxSize, maxValue); + int[] arr2 = copyArray(arr1); + mergeSort1(arr1); + mergeSort2(arr2); + if (!isEqual(arr1, arr2)) { + System.out.println("出错了！"); + printArray(arr1); + printArray(arr2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class04/Code02_SmallSum.java b/体系学习班/class04/Code02_SmallSum.java new file mode 100644 index 0000000..1167a64 --- /dev/null +++ b/体系学习班/class04/Code02_SmallSum.java @@ -0,0 +1,137 @@ +package class04; + +public class Code02_SmallSum { + + public static int smallSum(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + return process(arr, 0, arr.length - 1); + } + + // arr[L..R]既要排好序，也要求小和返回 + // 所有merge时，产生的小和，累加 + // 左 排序 merge + // 右 排序 merge + // merge + public static int process(int[] arr, int l, int r) { + if (l == r) { + return 0; + } + // l < r + int mid = l + ((r - l) >> 1); + return + process(arr, l, mid) + + + process(arr, mid + 1, r) + + + merge(arr, l, mid, r); + } + + public static int merge(int[] arr, int L, int m, int r) { + int[] help = new int[r - L + 1]; + int i = 0; + int p1 = L; + int p2 = m + 1; + int res = 0; + while (p1 <= m && p2 <= r) { + res += arr[p1] < arr[p2] ? (r - p2 + 1) * arr[p1] : 0; + help[i++] = arr[p1] < arr[p2] ? arr[p1++] : arr[p2++]; + } + while (p1 <= m) { + help[i++] = arr[p1++]; + } + while (p2 <= r) { + help[i++] = arr[p2++]; + } + for (i = 0; i < help.length; i++) { + arr[L + i] = help[i]; + } + return res; + } + + // for test + public static int comparator(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int res = 0; + for (int i = 1; i < arr.length; i++) { + for (int j = 0; j < i; j++) { + res += arr[j] < arr[i] ? arr[j] : 0; + } + } + return res; + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + return arr; + } + + // for test + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 100; + int maxValue = 100; + boolean succeed = true; + for (int i = 0; i < testTime; i++) { + int[] arr1 = generateRandomArray(maxSize, maxValue); + int[] arr2 = copyArray(arr1); + if (smallSum(arr1) != comparator(arr2)) { + succeed = false; + printArray(arr1); + printArray(arr2); + break; + } + } + System.out.println(succeed ? "Nice!" : "Fucking fucked!"); + } + +} diff --git a/体系学习班/class04/Code03_ReversePair.java b/体系学习班/class04/Code03_ReversePair.java new file mode 100644 index 0000000..0e6e531 --- /dev/null +++ b/体系学习班/class04/Code03_ReversePair.java @@ -0,0 +1,130 @@ +package class04; + +public class Code03_ReversePair { + + public static int reverPairNumber(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + return process(arr, 0, arr.length - 1); + } + + // arr[L..R]既要排好序，也要求逆序对数量返回 + // 所有merge时，产生的逆序对数量，累加，返回 + // 左 排序 merge并产生逆序对数量 + // 右 排序 merge并产生逆序对数量 + public static int process(int[] arr, int l, int r) { + if (l == r) { + return 0; + } + // l < r + int mid = l + ((r - l) >> 1); + return process(arr, l, mid) + process(arr, mid + 1, r) + merge(arr, l, mid, r); + } + + public static int merge(int[] arr, int L, int m, int r) { + int[] help = new int[r - L + 1]; + int i = help.length - 1; + int p1 = m; + int p2 = r; + int res = 0; + while (p1 >= L && p2 > m) { + res += arr[p1] > arr[p2] ? (p2 - m) : 0; + help[i--] = arr[p1] > arr[p2] ? arr[p1--] : arr[p2--]; + } + while (p1 >= L) { + help[i--] = arr[p1--]; + } + while (p2 > m) { + help[i--] = arr[p2--]; + } + for (i = 0; i < help.length; i++) { + arr[L + i] = help[i]; + } + return res; + } + + // for test + public static int comparator(int[] arr) { + int ans = 0; + for (int i = 0; i < arr.length; i++) { + for (int j = i + 1; j < arr.length; j++) { + if (arr[i] > arr[j]) { + ans++; + } + } + } + return ans; + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + return arr; + } + + // for test + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 100; + int maxValue = 100; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int[] arr1 = generateRandomArray(maxSize, maxValue); + int[] arr2 = copyArray(arr1); + if (reverPairNumber(arr1) != comparator(arr2)) { + System.out.println("Oops!"); + printArray(arr1); + printArray(arr2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class04/Code04_BiggerThanRightTwice.java b/体系学习班/class04/Code04_BiggerThanRightTwice.java new file mode 100644 index 0000000..7eb4f7e --- /dev/null +++ b/体系学习班/class04/Code04_BiggerThanRightTwice.java @@ -0,0 +1,135 @@ +package class04; + +// 本题测试链接 : https://leetcode.com/problems/reverse-pairs/ +public class Code04_BiggerThanRightTwice { + + public static int reversePairs(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + return process(arr, 0, arr.length - 1); + } + + public static int process(int[] arr, int l, int r) { + if (l == r) { + return 0; + } + // l < r + int mid = l + ((r - l) >> 1); + return process(arr, l, mid) + process(arr, mid + 1, r) + merge(arr, l, mid, r); + } + + public static int merge(int[] arr, int L, int m, int r) { + // [L....M] [M+1....R] + int ans = 0; + // 目前囊括进来的数，是从[M+1, windowR) + int windowR = m + 1; + for (int i = L; i <= m; i++) { + while (windowR <= r && (long) arr[i] > (long) arr[windowR] * 2) { + windowR++; + } + ans += windowR - m - 1; + } + int[] help = new int[r - L + 1]; + int i = 0; + int p1 = L; + int p2 = m + 1; + while (p1 <= m && p2 <= r) { + help[i++] = arr[p1] <= arr[p2] ? arr[p1++] : arr[p2++]; + } + while (p1 <= m) { + help[i++] = arr[p1++]; + } + while (p2 <= r) { + help[i++] = arr[p2++]; + } + for (i = 0; i < help.length; i++) { + arr[L + i] = help[i]; + } + return ans; + } + + // for test + public static int comparator(int[] arr) { + int ans = 0; + for (int i = 0; i < arr.length; i++) { + for (int j = i + 1; j < arr.length; j++) { + if (arr[i] > (arr[j] << 1)) { + ans++; + } + } + } + return ans; + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) ((maxValue + 1) * Math.random()); + } + return arr; + } + + // for test + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 100; + int maxValue = 100; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int[] arr1 = generateRandomArray(maxSize, maxValue); + int[] arr2 = copyArray(arr1); + if (reversePairs(arr1) != comparator(arr2)) { + System.out.println("Oops!"); + printArray(arr1); + printArray(arr2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class05/Code01_CountOfRangeSum.java b/体系学习班/class05/Code01_CountOfRangeSum.java new file mode 100644 index 0000000..559a7d8 --- /dev/null +++ b/体系学习班/class05/Code01_CountOfRangeSum.java @@ -0,0 +1,63 @@ +package class05; + +// 这道题直接在leetcode测评： +// https://leetcode.com/problems/count-of-range-sum/ +public class Code01_CountOfRangeSum { + + public static int countRangeSum(int[] nums, int lower, int upper) { + if (nums == null || nums.length == 0) { + return 0; + } + long[] sum = new long[nums.length]; + sum[0] = nums[0]; + for (int i = 1; i < nums.length; i++) { + sum[i] = sum[i - 1] + nums[i]; + } + return process(sum, 0, sum.length - 1, lower, upper); + } + + public static int process(long[] sum, int L, int R, int lower, int upper) { + if (L == R) { + return sum[L] >= lower && sum[L] <= upper ? 1 : 0; + } + int M = L + ((R - L) >> 1); + return process(sum, L, M, lower, upper) + process(sum, M + 1, R, lower, upper) + + merge(sum, L, M, R, lower, upper); + } + + public static int merge(long[] arr, int L, int M, int R, int lower, int upper) { + int ans = 0; + int windowL = L; + int windowR = L; + // [windowL, windowR) + for (int i = M + 1; i <= R; i++) { + long min = arr[i] - upper; + long max = arr[i] - lower; + while (windowR <= M && arr[windowR] <= max) { + windowR++; + } + while (windowL <= M && arr[windowL] < min) { + windowL++; + } + ans += windowR - windowL; + } + long[] help = new long[R - L + 1]; + int i = 0; + int p1 = L; + int p2 = M + 1; + while (p1 <= M && p2 <= R) { + help[i++] = arr[p1] <= arr[p2] ? arr[p1++] : arr[p2++]; + } + while (p1 <= M) { + help[i++] = arr[p1++]; + } + while (p2 <= R) { + help[i++] = arr[p2++]; + } + for (i = 0; i < help.length; i++) { + arr[L + i] = help[i]; + } + return ans; + } + +} diff --git a/体系学习班/class05/Code02_PartitionAndQuickSort.java b/体系学习班/class05/Code02_PartitionAndQuickSort.java new file mode 100644 index 0000000..e8a9d7a --- /dev/null +++ b/体系学习班/class05/Code02_PartitionAndQuickSort.java @@ -0,0 +1,197 @@ +package class05; + +public class Code02_PartitionAndQuickSort { + + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + + // arr[L..R]上，以arr[R]位置的数做划分值 + // <= X > X + // <= X X + public static int partition(int[] arr, int L, int R) { + if (L > R) { + return -1; + } + if (L == R) { + return L; + } + int lessEqual = L - 1; + int index = L; + while (index < R) { + if (arr[index] <= arr[R]) { + swap(arr, index, ++lessEqual); + } + index++; + } + swap(arr, ++lessEqual, R); + return lessEqual; + } + + // arr[L...R] 玩荷兰国旗问题的划分，以arr[R]做划分值 + // arr[R] + public static int[] netherlandsFlag(int[] arr, int L, int R) { + if (L > R) { // L...R L>R + return new int[] { -1, -1 }; + } + if (L == R) { + return new int[] { L, R }; + } + int less = L - 1; // < 区 右边界 + int more = R; // > 区 左边界 + int index = L; + while (index < more) { // 当前位置，不能和 >区的左边界撞上 + if (arr[index] == arr[R]) { + index++; + } else if (arr[index] < arr[R]) { +// swap(arr, less + 1, index); +// less++; +// index++; + swap(arr, index++, ++less); + } else { // > + swap(arr, index, --more); + } + } + swap(arr, more, R); // <[R] =[R] >[R] + return new int[] { less + 1, more }; + } + + public static void quickSort1(int[] arr) { + if (arr == null || arr.length < 2) { + return; + } + process1(arr, 0, arr.length - 1); + } + + public static void process1(int[] arr, int L, int R) { + if (L >= R) { + return; + } + // L..R partition arr[R] [ <=arr[R] arr[R] >arr[R] ] + int M = partition(arr, L, R); + process1(arr, L, M - 1); + process1(arr, M + 1, R); + } + + + + + + + public static void quickSort2(int[] arr) { + if (arr == null || arr.length < 2) { + return; + } + process2(arr, 0, arr.length - 1); + } + + // arr[L...R] 排有序，快排2.0方式 + public static void process2(int[] arr, int L, int R) { + if (L >= R) { + return; + } + // [ equalArea[0] , equalArea[0]] + int[] equalArea = netherlandsFlag(arr, L, R); + process2(arr, L, equalArea[0] - 1); + process2(arr, equalArea[1] + 1, R); + } + + + + + + + + public static void quickSort3(int[] arr) { + if (arr == null || arr.length < 2) { + return; + } + process3(arr, 0, arr.length - 1); + } + + public static void process3(int[] arr, int L, int R) { + if (L >= R) { + return; + } + swap(arr, L + (int) (Math.random() * (R - L + 1)), R); + int[] equalArea = netherlandsFlag(arr, L, R); + process3(arr, L, equalArea[0] - 1); + process3(arr, equalArea[1] + 1, R); + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + return arr; + } + + // for test + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 100; + int maxValue = 100; + boolean succeed = true; + for (int i = 0; i < testTime; i++) { + int[] arr1 = generateRandomArray(maxSize, maxValue); + int[] arr2 = copyArray(arr1); + int[] arr3 = copyArray(arr1); + quickSort1(arr1); + quickSort2(arr2); + quickSort3(arr3); + if (!isEqual(arr1, arr2) || !isEqual(arr2, arr3)) { + succeed = false; + break; + } + } + System.out.println(succeed ? "Nice!" : "Oops!"); + + } + +} diff --git a/体系学习班/class05/Code03_QuickSortRecursiveAndUnrecursive.java b/体系学习班/class05/Code03_QuickSortRecursiveAndUnrecursive.java new file mode 100644 index 0000000..57a7657 --- /dev/null +++ b/体系学习班/class05/Code03_QuickSortRecursiveAndUnrecursive.java @@ -0,0 +1,195 @@ +package class05; + +import java.util.LinkedList; +import java.util.Queue; +import java.util.Stack; + +public class Code03_QuickSortRecursiveAndUnrecursive { + + // 荷兰国旗问题 + public static int[] netherlandsFlag(int[] arr, int L, int R) { + if (L > R) { + return new int[] { -1, -1 }; + } + if (L == R) { + return new int[] { L, R }; + } + int less = L - 1; + int more = R; + int index = L; + while (index < more) { + if (arr[index] == arr[R]) { + index++; + } else if (arr[index] < arr[R]) { + swap(arr, index++, ++less); + } else { + swap(arr, index, --more); + } + } + swap(arr, more, R); + return new int[] { less + 1, more }; + } + + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + + // 快排递归版本 + public static void quickSort1(int[] arr) { + if (arr == null || arr.length < 2) { + return; + } + process(arr, 0, arr.length - 1); + } + + public static void process(int[] arr, int L, int R) { + if (L >= R) { + return; + } + swap(arr, L + (int) (Math.random() * (R - L + 1)), R); + int[] equalArea = netherlandsFlag(arr, L, R); + process(arr, L, equalArea[0] - 1); + process(arr, equalArea[1] + 1, R); + } + + // 快排非递归版本需要的辅助类 + // 要处理的是什么范围上的排序 + public static class Op { + public int l; + public int r; + + public Op(int left, int right) { + l = left; + r = right; + } + } + + // 快排3.0 非递归版本 用栈来执行 + public static void quickSort2(int[] arr) { + if (arr == null || arr.length < 2) { + return; + } + int N = arr.length; + swap(arr, (int) (Math.random() * N), N - 1); + int[] equalArea = netherlandsFlag(arr, 0, N - 1); + int el = equalArea[0]; + int er = equalArea[1]; + Stack stack = new Stack<>(); + stack.push(new Op(0, el - 1)); + stack.push(new Op(er + 1, N - 1)); + while (!stack.isEmpty()) { + Op op = stack.pop(); // op.l ... op.r + if (op.l < op.r) { + swap(arr, op.l + (int) (Math.random() * (op.r - op.l + 1)), op.r); + equalArea = netherlandsFlag(arr, op.l, op.r); + el = equalArea[0]; + er = equalArea[1]; + stack.push(new Op(op.l, el - 1)); + stack.push(new Op(er + 1, op.r)); + } + } + } + + // 快排3.0 非递归版本 用队列来执行 + public static void quickSort3(int[] arr) { + if (arr == null || arr.length < 2) { + return; + } + int N = arr.length; + swap(arr, (int) (Math.random() * N), N - 1); + int[] equalArea = netherlandsFlag(arr, 0, N - 1); + int el = equalArea[0]; + int er = equalArea[1]; + Queue queue = new LinkedList<>(); + queue.offer(new Op(0, el - 1)); + queue.offer(new Op(er + 1, N - 1)); + while (!queue.isEmpty()) { + Op op = queue.poll(); + if (op.l < op.r) { + swap(arr, op.l + (int) (Math.random() * (op.r - op.l + 1)), op.r); + equalArea = netherlandsFlag(arr, op.l, op.r); + el = equalArea[0]; + er = equalArea[1]; + queue.offer(new Op(op.l, el - 1)); + queue.offer(new Op(er + 1, op.r)); + } + } + } + + // 生成随机数组（用于测试） + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + return arr; + } + + // 拷贝数组（用于测试） + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // 对比两个数组（用于测试） + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // 打印数组（用于测试） + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // 跑大样本随机测试（对数器） + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 100; + int maxValue = 100; + boolean succeed = true; + System.out.println("test begin"); + for (int i = 0; i < testTime; i++) { + int[] arr1 = generateRandomArray(maxSize, maxValue); + int[] arr2 = copyArray(arr1); + int[] arr3 = copyArray(arr1); + quickSort1(arr1); + quickSort2(arr2); + quickSort3(arr3); + if (!isEqual(arr1, arr2) || !isEqual(arr1, arr3)) { + succeed = false; + break; + } + } + System.out.println("test end"); + System.out.println("测试" + testTime + "组是否全部通过：" + (succeed ? "是" : "否")); + } + +} diff --git a/体系学习班/class05/Code04_DoubleLinkedListQuickSort.java b/体系学习班/class05/Code04_DoubleLinkedListQuickSort.java new file mode 100644 index 0000000..a3014dc --- /dev/null +++ b/体系学习班/class05/Code04_DoubleLinkedListQuickSort.java @@ -0,0 +1,300 @@ +package class05; + +import java.util.ArrayList; +import java.util.Comparator; + +// 双向链表的随机快速排序 +// 课上没有讲，因为这是群里同学问的问题 +// 作为补充放在这，有需要的同学可以看看 +// 和课上讲的数组的经典快速排序在算法上没有区别 +// 但是coding需要更小心 +public class Code04_DoubleLinkedListQuickSort { + + public static class Node { + public int value; + public Node last; + public Node next; + + public Node(int v) { + value = v; + } + } + + public static Node quickSort(Node h) { + if (h == null) { + return null; + } + int N = 0; + Node c = h; + Node e = null; + while (c != null) { + N++; + e = c; + c = c.next; + } + return process(h, e, N).h; + } + + public static class HeadTail { + public Node h; + public Node t; + + public HeadTail(Node head, Node tail) { + h = head; + t = tail; + } + } + + // L...R是一个双向链表的头和尾, + // L的last指针指向null，R的next指针指向null + // 也就是说L的左边没有，R的右边也没节点 + // 就是一个正常的双向链表，一共有N个节点 + // 将这一段用随机快排的方式排好序 + // 返回排好序之后的双向链表的头和尾(HeadTail) + public static HeadTail process(Node L, Node R, int N) { + if (L == null) { + return null; + } + if (L == R) { + return new HeadTail(L, R); + } + // L..R上不只一个节点 + // 随机得到一个随机下标 + int randomIndex = (int) (Math.random() * N); + // 根据随机下标得到随机节点 + Node randomNode = L; + while (randomIndex-- != 0) { + randomNode = randomNode.next; + } + // 把随机节点从原来的环境里分离出来 + // 比如 a(L) -> b -> c -> d(R), 如果randomNode = c，那么调整之后 + // a(L) -> b -> d(R), c会被挖出来，randomNode = c + if (randomNode == L || randomNode == R) { + if (randomNode == L) { + L = randomNode.next; + L.last = null; + } else { + randomNode.last.next = null; + } + } else { // randomNode一定是中间的节点 + randomNode.last.next = randomNode.next; + randomNode.next.last = randomNode.last; + } + randomNode.last = null; + randomNode.next = null; + Info info = partition(L, randomNode); + // randomNode的部分去排序 + HeadTail rht = process(info.rh, info.rt, info.rs); + // 左部分排好序、右部分排好序 + // 把它们串在一起 + if (lht != null) { + lht.t.next = info.eh; + info.eh.last = lht.t; + } + if (rht != null) { + info.et.next = rht.h; + rht.h.last = info.et; + } + // 返回排好序之后总的头和总的尾 + Node h = lht != null ? lht.h : info.eh; + Node t = rht != null ? rht.t : info.et; + return new HeadTail(h, t); + } + + public static class Info { + public Node lh; + public Node lt; + public int ls; + public Node rh; + public Node rt; + public int rs; + public Node eh; + public Node et; + + public Info(Node lH, Node lT, int lS, Node rH, Node rT, int rS, Node eH, Node eT) { + lh = lH; + lt = lT; + ls = lS; + rh = rH; + rt = rT; + rs = rS; + eh = eH; + et = eT; + } + } + + // (L....一直到空)，是一个双向链表 + // pivot是一个不在(L....一直到空)的独立节点，它作为划分值 + // 根据荷兰国旗问题的划分方式，把(L....一直到空)划分成: + // pivot 三个部分，然后把pivot融进=pivot的部分 + // 比如 4(L)->6->7->1->5->0->9->null pivot=5(这个5和链表中的5，是不同的节点) + // 调整完成后: + // 4->1->0 小于的部分 + // 5->5 等于的部分 + // 6->7->9 大于的部分 + // 三个部分是断开的 + // 然后返回Info： + // 小于部分的头、尾、节点个数 : lh,lt,ls + // 大于部分的头、尾、节点个数 : rh,rt,rs + // 等于部分的头、尾 : eh,et + public static Info partition(Node L, Node pivot) { + Node lh = null; + Node lt = null; + int ls = 0; + Node rh = null; + Node rt = null; + int rs = 0; + Node eh = pivot; + Node et = pivot; + Node tmp = null; + while (L != null) { + tmp = L.next; + L.next = null; + L.last = null; + if (L.value < pivot.value) { + ls++; + if (lh == null) { + lh = L; + lt = L; + } else { + lt.next = L; + L.last = lt; + lt = L; + } + } else if (L.value > pivot.value) { + rs++; + if (rh == null) { + rh = L; + rt = L; + } else { + rt.next = L; + L.last = rt; + rt = L; + } + } else { + et.next = L; + L.last = et; + et = L; + } + L = tmp; + } + return new Info(lh, lt, ls, rh, rt, rs, eh, et); + } + + // 为了测试 + public static class NodeComp implements Comparator { + + @Override + public int compare(Node o1, Node o2) { + return o1.value - o2.value; + } + + } + + // 为了测试 + public static Node sort(Node head) { + if (head == null) { + return null; + } + ArrayList arr = new ArrayList<>(); + while (head != null) { + arr.add(head); + head = head.next; + } + arr.sort(new NodeComp()); + Node h = arr.get(0); + h.last = null; + Node p = h; + for (int i = 1; i < arr.size(); i++) { + Node c = arr.get(i); + p.next = c; + c.last = p; + c.next = null; + p = c; + } + return h; + } + + // 为了测试 + public static Node generateRandomDoubleLinkedList(int n, int v) { + if (n == 0) { + return null; + } + Node[] arr = new Node[n]; + for (int i = 0; i < n; i++) { + arr[i] = new Node((int) (Math.random() * v)); + } + Node head = arr[0]; + Node pre = head; + for (int i = 1; i < n; i++) { + pre.next = arr[i]; + arr[i].last = pre; + pre = arr[i]; + } + return head; + } + + // 为了测试 + public static Node cloneDoubleLinkedList(Node head) { + if (head == null) { + return null; + } + Node h = new Node(head.value); + Node p = h; + head = head.next; + while (head != null) { + Node c = new Node(head.value); + p.next = c; + c.last = p; + p = c; + head = head.next; + } + return h; + } + + // 为了测试 + public static boolean equal(Node h1, Node h2) { + return doubleLinkedListToString(h1).equals(doubleLinkedListToString(h2)); + } + + // 为了测试 + public static String doubleLinkedListToString(Node head) { + Node cur = head; + Node end = null; + StringBuilder builder = new StringBuilder(); + while (cur != null) { + builder.append(cur.value + " "); + end = cur; + cur = cur.next; + } + builder.append("| "); + while (end != null) { + builder.append(end.value + " "); + end = end.last; + } + return builder.toString(); + } + + // 为了测试 + public static void main(String[] args) { + int N = 500; + int V = 500; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int size = (int) (Math.random() * N); + Node head1 = generateRandomDoubleLinkedList(size, V); + Node head2 = cloneDoubleLinkedList(head1); + Node sort1 = quickSort(head1); + Node sort2 = sort(head2); + if (!equal(sort1, sort2)) { + System.out.println("出错了!"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class06/Code01_Comparator.java b/体系学习班/class06/Code01_Comparator.java new file mode 100644 index 0000000..af23b64 --- /dev/null +++ b/体系学习班/class06/Code01_Comparator.java @@ -0,0 +1,168 @@ +package class06; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.Comparator; +import java.util.TreeMap; + +public class Code01_Comparator { + + public static class Student { + public String name; + public int id; + public int age; + + public Student(String name, int id, int age) { + this.name = name; + this.id = id; + this.age = age; + } + } + + // 任何比较器： + // compare方法里，遵循一个统一的规范： + // 返回负数的时候，认为第一个参数应该排在前面 + // 返回正数的时候，认为第二个参数应该排在前面 + // 返回0的时候，认为无所谓谁放前面 + public static class IdShengAgeJiangOrder implements Comparator { + + // 根据id从小到大，但是如果id一样，按照年龄从大到小 + @Override + public int compare(Student o1, Student o2) { + return o1.id != o2.id ? (o1.id - o2.id) : (o2.age - o1.age); + } + + } + + public static class IdAscendingComparator implements Comparator { + + // 返回负数的时候，第一个参数排在前面 + // 返回正数的时候，第二个参数排在前面 + // 返回0的时候，谁在前面无所谓 + @Override + public int compare(Student o1, Student o2) { + return o1.id - o2.id; + } + + } + + public static class IdDescendingComparator implements Comparator { + + @Override + public int compare(Student o1, Student o2) { + return o2.id - o1.id; + } + + } + + // 先按照id排序，id小的，放前面； + // id一样，age大的，前面； + public static class IdInAgeDe implements Comparator { + + @Override + public int compare(Student o1, Student o2) { + return o1.id != o2.id ? o1.id - o2.id : (o2.age - o1.age); + } + + } + + public static void printStudents(Student[] students) { + for (Student student : students) { + System.out.println("Name : " + student.name + ", Id : " + student.id + ", Age : " + student.age); + } + } + + public static void printArray(Integer[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + public static class MyComp implements Comparator { + + @Override + public int compare(Integer o1, Integer o2) { + return o2 - o1; + } + + } + + public static class AComp implements Comparator { + + // 如果返回负数，认为第一个参数应该拍在前面 + // 如果返回正数，认为第二个参数应该拍在前面 + // 如果返回0，认为谁放前面都行 + @Override + public int compare(Integer arg0, Integer arg1) { + + return arg1 - arg0; + +// return 0; + } + + } + + public static void main(String[] args) { + + Integer[] arr = { 5, 4, 3, 2, 7, 9, 1, 0 }; + + Arrays.sort(arr, new AComp()); + + for (int i = 0; i < arr.length; i++) { + System.out.println(arr[i]); + } + + System.out.println("==========================="); + + Student student1 = new Student("A", 4, 40); + Student student2 = new Student("B", 4, 21); + Student student3 = new Student("C", 3, 12); + Student student4 = new Student("D", 3, 62); + Student student5 = new Student("E", 3, 42); + // D E C A B + + Student[] students = new Student[] { student1, student2, student3, student4, student5 }; + System.out.println("第一条打印"); + + Arrays.sort(students, new IdShengAgeJiangOrder()); + for (int i = 0; i < students.length; i++) { + Student s = students[i]; + System.out.println(s.name + "," + s.id + "," + s.age); + } + + System.out.println("第二条打印"); + ArrayList studentList = new ArrayList<>(); + studentList.add(student1); + studentList.add(student2); + studentList.add(student3); + studentList.add(student4); + studentList.add(student5); + studentList.sort(new IdShengAgeJiangOrder()); + for (int i = 0; i < studentList.size(); i++) { + Student s = studentList.get(i); + System.out.println(s.name + "," + s.id + "," + s.age); + } + // N * logN + System.out.println("第三条打印"); + student1 = new Student("A", 4, 40); + student2 = new Student("B", 4, 21); + student3 = new Student("C", 4, 12); + student4 = new Student("D", 4, 62); + student5 = new Student("E", 4, 42); + TreeMap treeMap = new TreeMap<>((a, b) -> (a.id - b.id)); + treeMap.put(student1, "我是学生1，我的名字叫A"); + treeMap.put(student2, "我是学生2，我的名字叫B"); + treeMap.put(student3, "我是学生3，我的名字叫C"); + treeMap.put(student4, "我是学生4，我的名字叫D"); + treeMap.put(student5, "我是学生5，我的名字叫E"); + for (Student s : treeMap.keySet()) { + System.out.println(s.name + "," + s.id + "," + s.age); + } + + } + +} diff --git a/体系学习班/class06/Code02_Heap.java b/体系学习班/class06/Code02_Heap.java new file mode 100644 index 0000000..28140a1 --- /dev/null +++ b/体系学习班/class06/Code02_Heap.java @@ -0,0 +1,192 @@ +package class06; + +import java.util.Comparator; +import java.util.PriorityQueue; + +public class Code02_Heap { + + public static class MyMaxHeap { + private int[] heap; + private final int limit; + private int heapSize; + + public MyMaxHeap(int limit) { + heap = new int[limit]; + this.limit = limit; + heapSize = 0; + } + + public boolean isEmpty() { + return heapSize == 0; + } + + public boolean isFull() { + return heapSize == limit; + } + + public void push(int value) { + if (heapSize == limit) { + throw new RuntimeException("heap is full"); + } + heap[heapSize] = value; + // value heapSize + heapInsert(heap, heapSize++); + } + + // 用户此时，让你返回最大值，并且在大根堆中，把最大值删掉 + // 剩下的数，依然保持大根堆组织 + public int pop() { + int ans = heap[0]; + swap(heap, 0, --heapSize); + heapify(heap, 0, heapSize); + return ans; + } + + // 新加进来的数，现在停在了index位置，请依次往上移动， + // 移动到0位置，或者干不掉自己的父亲了，停！ + private void heapInsert(int[] arr, int index) { + // [index] [index-1]/2 + // index == 0 + while (arr[index] > arr[(index - 1) / 2]) { + swap(arr, index, (index - 1) / 2); + index = (index - 1) / 2; + } + } + + // 从index位置，往下看，不断的下沉 + // 停：较大的孩子都不再比index位置的数大；已经没孩子了 + private void heapify(int[] arr, int index, int heapSize) { + int left = index * 2 + 1; + while (left < heapSize) { // 如果有左孩子，有没有右孩子，可能有可能没有！ + // 把较大孩子的下标，给largest + int largest = left + 1 < heapSize && arr[left + 1] > arr[left] ? left + 1 : left; + largest = arr[largest] > arr[index] ? largest : index; + if (largest == index) { + break; + } + // index和较大孩子，要互换 + swap(arr, largest, index); + index = largest; + left = index * 2 + 1; + } + } + + private void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + + } + + public static class RightMaxHeap { + private int[] arr; + private final int limit; + private int size; + + public RightMaxHeap(int limit) { + arr = new int[limit]; + this.limit = limit; + size = 0; + } + + public boolean isEmpty() { + return size == 0; + } + + public boolean isFull() { + return size == limit; + } + + public void push(int value) { + if (size == limit) { + throw new RuntimeException("heap is full"); + } + arr[size++] = value; + } + + public int pop() { + int maxIndex = 0; + for (int i = 1; i < size; i++) { + if (arr[i] > arr[maxIndex]) { + maxIndex = i; + } + } + int ans = arr[maxIndex]; + arr[maxIndex] = arr[--size]; + return ans; + } + + } + + public static class MyComparator implements Comparator { + + @Override + public int compare(Integer o1, Integer o2) { + return o2 - o1; + } + + } + + public static void main(String[] args) { + + // 小根堆 + PriorityQueue heap = new PriorityQueue<>(new MyComparator()); + heap.add(5); + heap.add(5); + heap.add(5); + heap.add(3); + // 5 , 3 + System.out.println(heap.peek()); + heap.add(7); + heap.add(0); + heap.add(7); + heap.add(0); + heap.add(7); + heap.add(0); + System.out.println(heap.peek()); + while (!heap.isEmpty()) { + System.out.println(heap.poll()); + } + + int value = 1000; + int limit = 100; + int testTimes = 1000000; + for (int i = 0; i < testTimes; i++) { + int curLimit = (int) (Math.random() * limit) + 1; + MyMaxHeap my = new MyMaxHeap(curLimit); + RightMaxHeap test = new RightMaxHeap(curLimit); + int curOpTimes = (int) (Math.random() * limit); + for (int j = 0; j < curOpTimes; j++) { + if (my.isEmpty() != test.isEmpty()) { + System.out.println("Oops!"); + } + if (my.isFull() != test.isFull()) { + System.out.println("Oops!"); + } + if (my.isEmpty()) { + int curValue = (int) (Math.random() * value); + my.push(curValue); + test.push(curValue); + } else if (my.isFull()) { + if (my.pop() != test.pop()) { + System.out.println("Oops!"); + } + } else { + if (Math.random() < 0.5) { + int curValue = (int) (Math.random() * value); + my.push(curValue); + test.push(curValue); + } else { + if (my.pop() != test.pop()) { + System.out.println("Oops!"); + } + } + } + } + } + System.out.println("finish!"); + + } + +} diff --git a/体系学习班/class06/Code03_HeapSort.java b/体系学习班/class06/Code03_HeapSort.java new file mode 100644 index 0000000..49e743a --- /dev/null +++ b/体系学习班/class06/Code03_HeapSort.java @@ -0,0 +1,158 @@ +package class06; + +import java.util.Arrays; +import java.util.PriorityQueue; + +public class Code03_HeapSort { + + // 堆排序额外空间复杂度O(1) + public static void heapSort(int[] arr) { + if (arr == null || arr.length < 2) { + return; + } + // O(N*logN) +// for (int i = 0; i < arr.length; i++) { // O(N) +// heapInsert(arr, i); // O(logN) +// } + // O(N) + for (int i = arr.length - 1; i >= 0; i--) { + heapify(arr, i, arr.length); + } + int heapSize = arr.length; + swap(arr, 0, --heapSize); + // O(N*logN) + while (heapSize > 0) { // O(N) + heapify(arr, 0, heapSize); // O(logN) + swap(arr, 0, --heapSize); // O(1) + } + } + + // arr[index]刚来的数，往上 + public static void heapInsert(int[] arr, int index) { + while (arr[index] > arr[(index - 1) / 2]) { + swap(arr, index, (index - 1) / 2); + index = (index - 1) / 2; + } + } + + // arr[index]位置的数，能否往下移动 + public static void heapify(int[] arr, int index, int heapSize) { + int left = index * 2 + 1; // 左孩子的下标 + while (left < heapSize) { // 下方还有孩子的时候 + // 两个孩子中，谁的值大，把下标给largest + // 1）只有左孩子，left -> largest + // 2) 同时有左孩子和右孩子，右孩子的值<= 左孩子的值，left -> largest + // 3) 同时有左孩子和右孩子并且右孩子的值> 左孩子的值， right -> largest + int largest = left + 1 < heapSize && arr[left + 1] > arr[left] ? left + 1 : left; + // 父和较大的孩子之间，谁的值大，把下标给largest + largest = arr[largest] > arr[index] ? largest : index; + if (largest == index) { + break; + } + swap(arr, largest, index); + index = largest; + left = index * 2 + 1; + } + } + + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + + // for test + public static void comparator(int[] arr) { + Arrays.sort(arr); + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + return arr; + } + + // for test + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + + // 默认小根堆 + PriorityQueue heap = new PriorityQueue<>(); + heap.add(6); + heap.add(8); + heap.add(0); + heap.add(2); + heap.add(9); + heap.add(1); + + while (!heap.isEmpty()) { + System.out.println(heap.poll()); + } + + int testTime = 500000; + int maxSize = 100; + int maxValue = 100; + boolean succeed = true; + for (int i = 0; i < testTime; i++) { + int[] arr1 = generateRandomArray(maxSize, maxValue); + int[] arr2 = copyArray(arr1); + heapSort(arr1); + comparator(arr2); + if (!isEqual(arr1, arr2)) { + succeed = false; + break; + } + } + System.out.println(succeed ? "Nice!" : "Fucking fucked!"); + + int[] arr = generateRandomArray(maxSize, maxValue); + printArray(arr); + heapSort(arr); + printArray(arr); + } + +} diff --git a/体系学习班/class06/Code04_SortArrayDistanceLessK.java b/体系学习班/class06/Code04_SortArrayDistanceLessK.java new file mode 100644 index 0000000..0d1b1ea --- /dev/null +++ b/体系学习班/class06/Code04_SortArrayDistanceLessK.java @@ -0,0 +1,127 @@ +package class06; + +import java.util.Arrays; +import java.util.PriorityQueue; + +public class Code04_SortArrayDistanceLessK { + + public static void sortedArrDistanceLessK(int[] arr, int k) { + if (k == 0) { + return; + } + // 默认小根堆 + PriorityQueue heap = new PriorityQueue<>(); + int index = 0; + // 0...K-1 + for (; index <= Math.min(arr.length - 1, k - 1); index++) { + heap.add(arr[index]); + } + int i = 0; + for (; index < arr.length; i++, index++) { + heap.add(arr[index]); + arr[i] = heap.poll(); + } + while (!heap.isEmpty()) { + arr[i++] = heap.poll(); + } + } + + // for test + public static void comparator(int[] arr, int k) { + Arrays.sort(arr); + } + + // for test + public static int[] randomArrayNoMoveMoreK(int maxSize, int maxValue, int K) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + // 先排个序 + Arrays.sort(arr); + // 然后开始随意交换，但是保证每个数距离不超过K + // swap[i] == true, 表示i位置已经参与过交换 + // swap[i] == false, 表示i位置没有参与过交换 + boolean[] isSwap = new boolean[arr.length]; + for (int i = 0; i < arr.length; i++) { + int j = Math.min(i + (int) (Math.random() * (K + 1)), arr.length - 1); + if (!isSwap[i] && !isSwap[j]) { + isSwap[i] = true; + isSwap[j] = true; + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + } + return arr; + } + + // for test + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + System.out.println("test begin"); + int testTime = 500000; + int maxSize = 100; + int maxValue = 100; + boolean succeed = true; + for (int i = 0; i < testTime; i++) { + int k = (int) (Math.random() * maxSize) + 1; + int[] arr = randomArrayNoMoveMoreK(maxSize, maxValue, k); + int[] arr1 = copyArray(arr); + int[] arr2 = copyArray(arr); + sortedArrDistanceLessK(arr1, k); + comparator(arr2, k); + if (!isEqual(arr1, arr2)) { + succeed = false; + System.out.println("K : " + k); + printArray(arr); + printArray(arr1); + printArray(arr2); + break; + } + } + System.out.println(succeed ? "Nice!" : "Fucking fucked!"); + } + +} \ No newline at end of file diff --git a/体系学习班/class07/Code01_CoverMax.java b/体系学习班/class07/Code01_CoverMax.java new file mode 100644 index 0000000..6260f97 --- /dev/null +++ b/体系学习班/class07/Code01_CoverMax.java @@ -0,0 +1,155 @@ +package class07; + +import java.util.Arrays; +import java.util.Comparator; +import java.util.PriorityQueue; + +public class Code01_CoverMax { + + public static int maxCover1(int[][] lines) { + int min = Integer.MAX_VALUE; + int max = Integer.MIN_VALUE; + for (int i = 0; i < lines.length; i++) { + min = Math.min(min, lines[i][0]); + max = Math.max(max, lines[i][1]); + } + int cover = 0; + for (double p = min + 0.5; p < max; p += 1) { + int cur = 0; + for (int i = 0; i < lines.length; i++) { + if (lines[i][0] < p && lines[i][1] > p) { + cur++; + } + } + cover = Math.max(cover, cur); + } + return cover; + } + + public static int maxCover2(int[][] m) { + Line[] lines = new Line[m.length]; + for (int i = 0; i < m.length; i++) { + lines[i] = new Line(m[i][0], m[i][1]); + } + Arrays.sort(lines, new StartComparator()); + // 小根堆，每一条线段的结尾数值，使用默认的 + PriorityQueue heap = new PriorityQueue<>(); + int max = 0; + for (int i = 0; i < lines.length; i++) { + // lines[i] -> cur 在黑盒中，把<=cur.start 东西都弹出 + while (!heap.isEmpty() && heap.peek() <= lines[i].start) { + heap.poll(); + } + heap.add(lines[i].end); + max = Math.max(max, heap.size()); + } + return max; + } + + public static class Line { + public int start; + public int end; + + public Line(int s, int e) { + start = s; + end = e; + } + } + + public static class EndComparator implements Comparator { + + @Override + public int compare(Line o1, Line o2) { + return o1.end - o2.end; + } + + } + + // 和maxCover2过程是一样的 + // 只是代码更短 + // 不使用类定义的写法 + public static int maxCover3(int[][] m) { + // m是二维数组，可以认为m内部是一个一个的一维数组 + // 每一个一维数组就是一个对象，也就是线段 + // 如下的code，就是根据每一个线段的开始位置排序 + // 比如, m = { {5,7}, {1,4}, {2,6} } 跑完如下的code之后变成：{ {1,4}, {2,6}, {5,7} } + Arrays.sort(m, (a, b) -> (a[0] - b[0])); + // 准备好小根堆，和课堂的说法一样 + PriorityQueue heap = new PriorityQueue<>(); + int max = 0; + for (int[] line : m) { + while (!heap.isEmpty() && heap.peek() <= line[0]) { + heap.poll(); + } + heap.add(line[1]); + max = Math.max(max, heap.size()); + } + return max; + } + + // for test + public static int[][] generateLines(int N, int L, int R) { + int size = (int) (Math.random() * N) + 1; + int[][] ans = new int[size][2]; + for (int i = 0; i < size; i++) { + int a = L + (int) (Math.random() * (R - L + 1)); + int b = L + (int) (Math.random() * (R - L + 1)); + if (a == b) { + b = a + 1; + } + ans[i][0] = Math.min(a, b); + ans[i][1] = Math.max(a, b); + } + return ans; + } + + public static class StartComparator implements Comparator { + + @Override + public int compare(Line o1, Line o2) { + return o1.start - o2.start; + } + + } + + public static void main(String[] args) { + + Line l1 = new Line(4, 9); + Line l2 = new Line(1, 4); + Line l3 = new Line(7, 15); + Line l4 = new Line(2, 4); + Line l5 = new Line(4, 6); + Line l6 = new Line(3, 7); + + // 底层堆结构，heap + PriorityQueue heap = new PriorityQueue<>(new StartComparator()); + heap.add(l1); + heap.add(l2); + heap.add(l3); + heap.add(l4); + heap.add(l5); + heap.add(l6); + + while (!heap.isEmpty()) { + Line cur = heap.poll(); + System.out.println(cur.start + "," + cur.end); + } + + System.out.println("test begin"); + int N = 100; + int L = 0; + int R = 200; + int testTimes = 200000; + for (int i = 0; i < testTimes; i++) { + int[][] lines = generateLines(N, L, R); + int ans1 = maxCover1(lines); + int ans2 = maxCover2(lines); + int ans3 = maxCover3(lines); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("Oops!"); + } + } + System.out.println("test end"); + } + +} diff --git a/体系学习班/class07/Code02_EveryStepShowBoss.java b/体系学习班/class07/Code02_EveryStepShowBoss.java new file mode 100644 index 0000000..a4c123b --- /dev/null +++ b/体系学习班/class07/Code02_EveryStepShowBoss.java @@ -0,0 +1,303 @@ +package class07; + +import java.util.ArrayList; +import java.util.Comparator; +import java.util.HashMap; +import java.util.List; + +public class Code02_EveryStepShowBoss { + + public static class Customer { + public int id; + public int buy; + public int enterTime; + + public Customer(int v, int b, int o) { + id = v; + buy = b; + enterTime = 0; + } + } + + public static class CandidateComparator implements Comparator { + + @Override + public int compare(Customer o1, Customer o2) { + return o1.buy != o2.buy ? (o2.buy - o1.buy) : (o1.enterTime - o2.enterTime); + } + + } + + public static class DaddyComparator implements Comparator { + + @Override + public int compare(Customer o1, Customer o2) { + return o1.buy != o2.buy ? (o1.buy - o2.buy) : (o1.enterTime - o2.enterTime); + } + + } + + public static class WhosYourDaddy { + private HashMap customers; + private HeapGreater candHeap; + private HeapGreater daddyHeap; + private final int daddyLimit; + + public WhosYourDaddy(int limit) { + customers = new HashMap(); + candHeap = new HeapGreater<>(new CandidateComparator()); + daddyHeap = new HeapGreater<>(new DaddyComparator()); + daddyLimit = limit; + } + + // 当前处理i号事件，arr[i] -> id, buyOrRefund + public void operate(int time, int id, boolean buyOrRefund) { + if (!buyOrRefund && !customers.containsKey(id)) { + return; + } + if (!customers.containsKey(id)) { + customers.put(id, new Customer(id, 0, 0)); + } + Customer c = customers.get(id); + if (buyOrRefund) { + c.buy++; + } else { + c.buy--; + } + if (c.buy == 0) { + customers.remove(id); + } + if (!candHeap.contains(c) && !daddyHeap.contains(c)) { + if (daddyHeap.size() < daddyLimit) { + c.enterTime = time; + daddyHeap.push(c); + } else { + c.enterTime = time; + candHeap.push(c); + } + } else if (candHeap.contains(c)) { + if (c.buy == 0) { + candHeap.remove(c); + } else { + candHeap.resign(c); + } + } else { + if (c.buy == 0) { + daddyHeap.remove(c); + } else { + daddyHeap.resign(c); + } + } + daddyMove(time); + } + + public List getDaddies() { + List customers = daddyHeap.getAllElements(); + List ans = new ArrayList<>(); + for (Customer c : customers) { + ans.add(c.id); + } + return ans; + } + + private void daddyMove(int time) { + if (candHeap.isEmpty()) { + return; + } + if (daddyHeap.size() < daddyLimit) { + Customer p = candHeap.pop(); + p.enterTime = time; + daddyHeap.push(p); + } else { + if (candHeap.peek().buy > daddyHeap.peek().buy) { + Customer oldDaddy = daddyHeap.pop(); + Customer newDaddy = candHeap.pop(); + oldDaddy.enterTime = time; + newDaddy.enterTime = time; + daddyHeap.push(newDaddy); + candHeap.push(oldDaddy); + } + } + } + + } + + public static List> topK(int[] arr, boolean[] op, int k) { + List> ans = new ArrayList<>(); + WhosYourDaddy whoDaddies = new WhosYourDaddy(k); + for (int i = 0; i < arr.length; i++) { + whoDaddies.operate(i, arr[i], op[i]); + ans.add(whoDaddies.getDaddies()); + } + return ans; + } + + // 干完所有的事，模拟，不优化 + public static List> compare(int[] arr, boolean[] op, int k) { + HashMap map = new HashMap<>(); + ArrayList cands = new ArrayList<>(); + ArrayList daddy = new ArrayList<>(); + List> ans = new ArrayList<>(); + for (int i = 0; i < arr.length; i++) { + int id = arr[i]; + boolean buyOrRefund = op[i]; + if (!buyOrRefund && !map.containsKey(id)) { + ans.add(getCurAns(daddy)); + continue; + } + // 没有发生：用户购买数为0并且又退货了 + // 用户之前购买数是0，此时买货事件 + // 用户之前购买数>0， 此时买货 + // 用户之前购买数>0, 此时退货 + if (!map.containsKey(id)) { + map.put(id, new Customer(id, 0, 0)); + } + // 买、卖 + Customer c = map.get(id); + if (buyOrRefund) { + c.buy++; + } else { + c.buy--; + } + if (c.buy == 0) { + map.remove(id); + } + // c + // 下面做 + if (!cands.contains(c) && !daddy.contains(c)) { + if (daddy.size() < k) { + c.enterTime = i; + daddy.add(c); + } else { + c.enterTime = i; + cands.add(c); + } + } + cleanZeroBuy(cands); + cleanZeroBuy(daddy); + cands.sort(new CandidateComparator()); + daddy.sort(new DaddyComparator()); + move(cands, daddy, k, i); + ans.add(getCurAns(daddy)); + } + return ans; + } + + public static void move(ArrayList cands, ArrayList daddy, int k, int time) { + if (cands.isEmpty()) { + return; + } + // 候选区不为空 + if (daddy.size() < k) { + Customer c = cands.get(0); + c.enterTime = time; + daddy.add(c); + cands.remove(0); + } else { // 等奖区满了，候选区有东西 + if (cands.get(0).buy > daddy.get(0).buy) { + Customer oldDaddy = daddy.get(0); + daddy.remove(0); + Customer newDaddy = cands.get(0); + cands.remove(0); + newDaddy.enterTime = time; + oldDaddy.enterTime = time; + daddy.add(newDaddy); + cands.add(oldDaddy); + } + } + } + + public static void cleanZeroBuy(ArrayList arr) { + List noZero = new ArrayList(); + for (Customer c : arr) { + if (c.buy != 0) { + noZero.add(c); + } + } + arr.clear(); + for (Customer c : noZero) { + arr.add(c); + } + } + + public static List getCurAns(ArrayList daddy) { + List ans = new ArrayList<>(); + for (Customer c : daddy) { + ans.add(c.id); + } + return ans; + } + + // 为了测试 + public static class Data { + public int[] arr; + public boolean[] op; + + public Data(int[] a, boolean[] o) { + arr = a; + op = o; + } + } + + // 为了测试 + public static Data randomData(int maxValue, int maxLen) { + int len = (int) (Math.random() * maxLen) + 1; + int[] arr = new int[len]; + boolean[] op = new boolean[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * maxValue); + op[i] = Math.random() < 0.5 ? true : false; + } + return new Data(arr, op); + } + + // 为了测试 + public static boolean sameAnswer(List> ans1, List> ans2) { + if (ans1.size() != ans2.size()) { + return false; + } + for (int i = 0; i < ans1.size(); i++) { + List cur1 = ans1.get(i); + List cur2 = ans2.get(i); + if (cur1.size() != cur2.size()) { + return false; + } + cur1.sort((a, b) -> a - b); + cur2.sort((a, b) -> a - b); + for (int j = 0; j < cur1.size(); j++) { + if (!cur1.get(j).equals(cur2.get(j))) { + return false; + } + } + } + return true; + } + + public static void main(String[] args) { + int maxValue = 10; + int maxLen = 100; + int maxK = 6; + int testTimes = 100000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + Data testData = randomData(maxValue, maxLen); + int k = (int) (Math.random() * maxK) + 1; + int[] arr = testData.arr; + boolean[] op = testData.op; + List> ans1 = topK(arr, op, k); + List> ans2 = compare(arr, op, k); + if (!sameAnswer(ans1, ans2)) { + for (int j = 0; j < arr.length; j++) { + System.out.println(arr[j] + " , " + op[j]); + } + System.out.println(k); + System.out.println(ans1); + System.out.println(ans2); + System.out.println("出错了！"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class07/HeapGreater.java b/体系学习班/class07/HeapGreater.java new file mode 100644 index 0000000..e58da47 --- /dev/null +++ b/体系学习班/class07/HeapGreater.java @@ -0,0 +1,112 @@ +package class07; + +import java.util.ArrayList; +import java.util.Comparator; +import java.util.HashMap; +import java.util.List; + +/* + * T一定要是非基础类型，有基础类型需求包一层 + */ +public class HeapGreater { + + private ArrayList heap; + private HashMap indexMap; + private int heapSize; + private Comparator comp; + + public HeapGreater(Comparator c) { + heap = new ArrayList<>(); + indexMap = new HashMap<>(); + heapSize = 0; + comp = c; + } + + public boolean isEmpty() { + return heapSize == 0; + } + + public int size() { + return heapSize; + } + + public boolean contains(T obj) { + return indexMap.containsKey(obj); + } + + public T peek() { + return heap.get(0); + } + + public void push(T obj) { + heap.add(obj); + indexMap.put(obj, heapSize); + heapInsert(heapSize++); + } + + public T pop() { + T ans = heap.get(0); + swap(0, heapSize - 1); + indexMap.remove(ans); + heap.remove(--heapSize); + heapify(0); + return ans; + } + + public void remove(T obj) { + T replace = heap.get(heapSize - 1); + int index = indexMap.get(obj); + indexMap.remove(obj); + heap.remove(--heapSize); + if (obj != replace) { + heap.set(index, replace); + indexMap.put(replace, index); + resign(replace); + } + } + + public void resign(T obj) { + heapInsert(indexMap.get(obj)); + heapify(indexMap.get(obj)); + } + + // 请返回堆上的所有元素 + public List getAllElements() { + List ans = new ArrayList<>(); + for (T c : heap) { + ans.add(c); + } + return ans; + } + + private void heapInsert(int index) { + while (comp.compare(heap.get(index), heap.get((index - 1) / 2)) < 0) { + swap(index, (index - 1) / 2); + index = (index - 1) / 2; + } + } + + private void heapify(int index) { + int left = index * 2 + 1; + while (left < heapSize) { + int best = left + 1 < heapSize && comp.compare(heap.get(left + 1), heap.get(left)) < 0 ? (left + 1) : left; + best = comp.compare(heap.get(best), heap.get(index)) < 0 ? best : index; + if (best == index) { + break; + } + swap(best, index); + index = best; + left = index * 2 + 1; + } + } + + private void swap(int i, int j) { + T o1 = heap.get(i); + T o2 = heap.get(j); + heap.set(i, o2); + heap.set(j, o1); + indexMap.put(o2, i); + indexMap.put(o1, j); + } + +} diff --git a/体系学习班/class07/Inner.java b/体系学习班/class07/Inner.java new file mode 100644 index 0000000..d3d5326 --- /dev/null +++ b/体系学习班/class07/Inner.java @@ -0,0 +1,9 @@ +package class07; + +public class Inner { + public T value; + + public Inner(T v) { + value = v; + } +} diff --git a/体系学习班/class08/Code01_TrieTree.java b/体系学习班/class08/Code01_TrieTree.java new file mode 100644 index 0000000..f7f9cf1 --- /dev/null +++ b/体系学习班/class08/Code01_TrieTree.java @@ -0,0 +1,299 @@ +package class08; + +import java.util.HashMap; + +// 该程序的对数器跑不过，你能发现bug在哪吗？ +public class Code01_TrieTree { + + // 前缀树节点类型 + public static class Node1 { + public int pass; + public int end; + public Node1[] nexts; + + public Node1() { + pass = 0; + end = 0; + nexts = new Node1[26]; + } + } + + public static class Trie1 { + private Node1 root; + + public Trie1() { + root = new Node1(); + } + + public void insert(String word) { + if (word == null) { + return; + } + char[] chs = word.toCharArray(); + Node1 node = root; + node.pass++; + int index = 0; + for (int i = 0; i < chs.length; i++) { // 从左往右遍历字符 + index = chs[i] - 'a'; // 由字符，对应成走向哪条路 + if (node.nexts[index] == null) { + node.nexts[index] = new Node1(); + } + node = node.nexts[index]; + node.pass++; + } + node.end++; + } + + public void delete(String word) { + if (search(word) != 0) { + char[] chs = word.toCharArray(); + Node1 node = root; + node.pass--; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = chs[i] - 'a'; + if (--node.nexts[index].pass == 0) { + node.nexts[index] = null; + return; + } + node = node.nexts[index]; + } + node.end--; + } + } + + // word这个单词之前加入过几次 + public int search(String word) { + if (word == null) { + return 0; + } + char[] chs = word.toCharArray(); + Node1 node = root; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = chs[i] - 'a'; + if (node.nexts[index] == null) { + return 0; + } + node = node.nexts[index]; + } + return node.end; + } + + // 所有加入的字符串中，有几个是以pre这个字符串作为前缀的 + public int prefixNumber(String pre) { + if (pre == null) { + return 0; + } + char[] chs = pre.toCharArray(); + Node1 node = root; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = chs[i] - 'a'; + if (node.nexts[index] == null) { + return 0; + } + node = node.nexts[index]; + } + return node.pass; + } + } + + public static class Node2 { + public int pass; + public int end; + public HashMap nexts; + + public Node2() { + pass = 0; + end = 0; + nexts = new HashMap<>(); + } + } + + public static class Trie2 { + private Node2 root; + + public Trie2() { + root = new Node2(); + } + + public void insert(String word) { + if (word == null) { + return; + } + char[] chs = word.toCharArray(); + Node2 node = root; + node.pass++; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = (int) chs[i]; + if (!node.nexts.containsKey(index)) { + node.nexts.put(index, new Node2()); + } + node = node.nexts.get(index); + node.pass++; + } + node.end++; + } + + public void delete(String word) { + if (search(word) != 0) { + char[] chs = word.toCharArray(); + Node2 node = root; + node.pass--; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = (int) chs[i]; + if (--node.nexts.get(index).pass == 0) { + node.nexts.remove(index); + return; + } + node = node.nexts.get(index); + } + node.end--; + } + } + + // word这个单词之前加入过几次 + public int search(String word) { + if (word == null) { + return 0; + } + char[] chs = word.toCharArray(); + Node2 node = root; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = (int) chs[i]; + if (!node.nexts.containsKey(index)) { + return 0; + } + node = node.nexts.get(index); + } + return node.end; + } + + // 所有加入的字符串中，有几个是以pre这个字符串作为前缀的 + public int prefixNumber(String pre) { + if (pre == null) { + return 0; + } + char[] chs = pre.toCharArray(); + Node2 node = root; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = (int) chs[i]; + if (!node.nexts.containsKey(index)) { + return 0; + } + node = node.nexts.get(index); + } + return node.pass; + } + } + + public static class Right { + + private HashMap box; + + public Right() { + box = new HashMap<>(); + } + + public void insert(String word) { + if (!box.containsKey(word)) { + box.put(word, 1); + } else { + box.put(word, box.get(word) + 1); + } + } + + public void delete(String word) { + if (box.containsKey(word)) { + if (box.get(word) == 1) { + box.remove(word); + } else { + box.put(word, box.get(word) - 1); + } + } + } + + public int search(String word) { + if (!box.containsKey(word)) { + return 0; + } else { + return box.get(word); + } + } + + public int prefixNumber(String pre) { + int count = 0; + for (String cur : box.keySet()) { + if (cur.startsWith(pre)) { + count++; + } + } + return count; + } + } + + // for test + public static String generateRandomString(int strLen) { + char[] ans = new char[(int) (Math.random() * strLen) + 1]; + for (int i = 0; i < ans.length; i++) { + int value = (int) (Math.random() * 6); + ans[i] = (char) (97 + value); + } + return String.valueOf(ans); + } + + // for test + public static String[] generateRandomStringArray(int arrLen, int strLen) { + String[] ans = new String[(int) (Math.random() * arrLen) + 1]; + for (int i = 0; i < ans.length; i++) { + ans[i] = generateRandomString(strLen); + } + return ans; + } + + public static void main(String[] args) { + int arrLen = 100; + int strLen = 20; + int testTimes = 100000; + for (int i = 0; i < testTimes; i++) { + String[] arr = generateRandomStringArray(arrLen, strLen); + Trie1 trie1 = new Trie1(); + Trie2 trie2 = new Trie2(); + Right right = new Right(); + for (int j = 0; j < arr.length; j++) { + double decide = Math.random(); + if (decide < 0.25) { + trie1.insert(arr[j]); + trie2.insert(arr[j]); + right.insert(arr[j]); + } else if (decide < 0.5) { + trie1.delete(arr[j]); + trie2.delete(arr[j]); + right.delete(arr[j]); + } else if (decide < 0.75) { + int ans1 = trie1.search(arr[j]); + int ans2 = trie2.search(arr[j]); + int ans3 = right.search(arr[j]); + if (ans1 != ans2 || ans2 != ans3) { + System.out.println("Oops!"); + } + } else { + int ans1 = trie1.prefixNumber(arr[j]); + int ans2 = trie2.prefixNumber(arr[j]); + int ans3 = right.prefixNumber(arr[j]); + if (ans1 != ans2 || ans2 != ans3) { + System.out.println("Oops!"); + } + } + } + } + System.out.println("finish!"); + + } + +} diff --git a/体系学习班/class08/Code02_TrieTree.java b/体系学习班/class08/Code02_TrieTree.java new file mode 100644 index 0000000..44d35cf --- /dev/null +++ b/体系学习班/class08/Code02_TrieTree.java @@ -0,0 +1,306 @@ +package class08; + +import java.util.HashMap; + +// 该程序完全正确 +public class Code02_TrieTree { + + public static class Node1 { + public int pass; + public int end; + public Node1[] nexts; + + // char tmp = 'b' (tmp - 'a') + public Node1() { + pass = 0; + end = 0; + // 0 a + // 1 b + // 2 c + // .. .. + // 25 z + // nexts[i] == null i方向的路不存在 + // nexts[i] != null i方向的路存在 + nexts = new Node1[26]; + } + } + + public static class Trie1 { + private Node1 root; + + public Trie1() { + root = new Node1(); + } + + public void insert(String word) { + if (word == null) { + return; + } + char[] str = word.toCharArray(); + Node1 node = root; + node.pass++; + int path = 0; + for (int i = 0; i < str.length; i++) { // 从左往右遍历字符 + path = str[i] - 'a'; // 由字符，对应成走向哪条路 + if (node.nexts[path] == null) { + node.nexts[path] = new Node1(); + } + node = node.nexts[path]; + node.pass++; + } + node.end++; + } + + public void delete(String word) { + if (search(word) != 0) { + char[] chs = word.toCharArray(); + Node1 node = root; + node.pass--; + int path = 0; + for (int i = 0; i < chs.length; i++) { + path = chs[i] - 'a'; + if (--node.nexts[path].pass == 0) { + node.nexts[path] = null; + return; + } + node = node.nexts[path]; + } + node.end--; + } + } + + // word这个单词之前加入过几次 + public int search(String word) { + if (word == null) { + return 0; + } + char[] chs = word.toCharArray(); + Node1 node = root; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = chs[i] - 'a'; + if (node.nexts[index] == null) { + return 0; + } + node = node.nexts[index]; + } + return node.end; + } + + // 所有加入的字符串中，有几个是以pre这个字符串作为前缀的 + public int prefixNumber(String pre) { + if (pre == null) { + return 0; + } + char[] chs = pre.toCharArray(); + Node1 node = root; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = chs[i] - 'a'; + if (node.nexts[index] == null) { + return 0; + } + node = node.nexts[index]; + } + return node.pass; + } + } + + public static class Node2 { + public int pass; + public int end; + public HashMap nexts; + + public Node2() { + pass = 0; + end = 0; + nexts = new HashMap<>(); + } + } + + public static class Trie2 { + private Node2 root; + + public Trie2() { + root = new Node2(); + } + + public void insert(String word) { + if (word == null) { + return; + } + char[] chs = word.toCharArray(); + Node2 node = root; + node.pass++; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = (int) chs[i]; + if (!node.nexts.containsKey(index)) { + node.nexts.put(index, new Node2()); + } + node = node.nexts.get(index); + node.pass++; + } + node.end++; + } + + public void delete(String word) { + if (search(word) != 0) { + char[] chs = word.toCharArray(); + Node2 node = root; + node.pass--; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = (int) chs[i]; + if (--node.nexts.get(index).pass == 0) { + node.nexts.remove(index); + return; + } + node = node.nexts.get(index); + } + node.end--; + } + } + + // word这个单词之前加入过几次 + public int search(String word) { + if (word == null) { + return 0; + } + char[] chs = word.toCharArray(); + Node2 node = root; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = (int) chs[i]; + if (!node.nexts.containsKey(index)) { + return 0; + } + node = node.nexts.get(index); + } + return node.end; + } + + // 所有加入的字符串中，有几个是以pre这个字符串作为前缀的 + public int prefixNumber(String pre) { + if (pre == null) { + return 0; + } + char[] chs = pre.toCharArray(); + Node2 node = root; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = (int) chs[i]; + if (!node.nexts.containsKey(index)) { + return 0; + } + node = node.nexts.get(index); + } + return node.pass; + } + } + + public static class Right { + + private HashMap box; + + public Right() { + box = new HashMap<>(); + } + + public void insert(String word) { + if (!box.containsKey(word)) { + box.put(word, 1); + } else { + box.put(word, box.get(word) + 1); + } + } + + public void delete(String word) { + if (box.containsKey(word)) { + if (box.get(word) == 1) { + box.remove(word); + } else { + box.put(word, box.get(word) - 1); + } + } + } + + public int search(String word) { + if (!box.containsKey(word)) { + return 0; + } else { + return box.get(word); + } + } + + public int prefixNumber(String pre) { + int count = 0; + for (String cur : box.keySet()) { + if (cur.startsWith(pre)) { + count += box.get(cur); + } + } + return count; + } + } + + // for test + public static String generateRandomString(int strLen) { + char[] ans = new char[(int) (Math.random() * strLen) + 1]; + for (int i = 0; i < ans.length; i++) { + int value = (int) (Math.random() * 6); + ans[i] = (char) (97 + value); + } + return String.valueOf(ans); + } + + // for test + public static String[] generateRandomStringArray(int arrLen, int strLen) { + String[] ans = new String[(int) (Math.random() * arrLen) + 1]; + for (int i = 0; i < ans.length; i++) { + ans[i] = generateRandomString(strLen); + } + return ans; + } + + public static void main(String[] args) { + int arrLen = 100; + int strLen = 20; + int testTimes = 100000; + for (int i = 0; i < testTimes; i++) { + String[] arr = generateRandomStringArray(arrLen, strLen); + Trie1 trie1 = new Trie1(); + Trie2 trie2 = new Trie2(); + Right right = new Right(); + for (int j = 0; j < arr.length; j++) { + double decide = Math.random(); + if (decide < 0.25) { + trie1.insert(arr[j]); + trie2.insert(arr[j]); + right.insert(arr[j]); + } else if (decide < 0.5) { + trie1.delete(arr[j]); + trie2.delete(arr[j]); + right.delete(arr[j]); + } else if (decide < 0.75) { + int ans1 = trie1.search(arr[j]); + int ans2 = trie2.search(arr[j]); + int ans3 = right.search(arr[j]); + if (ans1 != ans2 || ans2 != ans3) { + System.out.println("Oops!"); + } + } else { + int ans1 = trie1.prefixNumber(arr[j]); + int ans2 = trie2.prefixNumber(arr[j]); + int ans3 = right.prefixNumber(arr[j]); + if (ans1 != ans2 || ans2 != ans3) { + System.out.println("Oops!"); + } + } + } + } + System.out.println("finish!"); + + } + +} diff --git a/体系学习班/class08/Code03_CountSort.java b/体系学习班/class08/Code03_CountSort.java new file mode 100644 index 0000000..d7864ec --- /dev/null +++ b/体系学习班/class08/Code03_CountSort.java @@ -0,0 +1,111 @@ +package class08; + +import java.util.Arrays; + +public class Code03_CountSort { + + // only for 0~200 value + public static void countSort(int[] arr) { + if (arr == null || arr.length < 2) { + return; + } + int max = Integer.MIN_VALUE; + for (int i = 0; i < arr.length; i++) { + max = Math.max(max, arr[i]); + } + int[] bucket = new int[max + 1]; + for (int i = 0; i < arr.length; i++) { + bucket[arr[i]]++; + } + int i = 0; + for (int j = 0; j < bucket.length; j++) { + while (bucket[j]-- > 0) { + arr[i++] = j; + } + } + } + + // for test + public static void comparator(int[] arr) { + Arrays.sort(arr); + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()); + } + return arr; + } + + // for test + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 100; + int maxValue = 150; + boolean succeed = true; + for (int i = 0; i < testTime; i++) { + int[] arr1 = generateRandomArray(maxSize, maxValue); + int[] arr2 = copyArray(arr1); + countSort(arr1); + comparator(arr2); + if (!isEqual(arr1, arr2)) { + succeed = false; + printArray(arr1); + printArray(arr2); + break; + } + } + System.out.println(succeed ? "Nice!" : "Fucking fucked!"); + + int[] arr = generateRandomArray(maxSize, maxValue); + printArray(arr); + countSort(arr); + printArray(arr); + + } + +} diff --git a/体系学习班/class08/Code04_RadixSort.java b/体系学习班/class08/Code04_RadixSort.java new file mode 100644 index 0000000..af35db3 --- /dev/null +++ b/体系学习班/class08/Code04_RadixSort.java @@ -0,0 +1,148 @@ +package class08; + +import java.util.Arrays; + +public class Code04_RadixSort { + + // only for no-negative value + public static void radixSort(int[] arr) { + if (arr == null || arr.length < 2) { + return; + } + radixSort(arr, 0, arr.length - 1, maxbits(arr)); + } + + public static int maxbits(int[] arr) { + int max = Integer.MIN_VALUE; + for (int i = 0; i < arr.length; i++) { + max = Math.max(max, arr[i]); + } + int res = 0; + while (max != 0) { + res++; + max /= 10; + } + return res; + } + + // arr[L..R]排序 , 最大值的十进制位数digit + public static void radixSort(int[] arr, int L, int R, int digit) { + final int radix = 10; + int i = 0, j = 0; + // 有多少个数准备多少个辅助空间 + int[] help = new int[R - L + 1]; + for (int d = 1; d <= digit; d++) { // 有多少位就进出几次 + // 10个空间 + // count[0] 当前位(d位)是0的数字有多少个 + // count[1] 当前位(d位)是(0和1)的数字有多少个 + // count[2] 当前位(d位)是(0、1和2)的数字有多少个 + // count[i] 当前位(d位)是(0~i)的数字有多少个 + int[] count = new int[radix]; // count[0..9] + for (i = L; i <= R; i++) { + // 103 1 3 + // 209 1 9 + j = getDigit(arr[i], d); + count[j]++; + } + for (i = 1; i < radix; i++) { + count[i] = count[i] + count[i - 1]; + } + for (i = R; i >= L; i--) { + j = getDigit(arr[i], d); + help[count[j] - 1] = arr[i]; + count[j]--; + } + for (i = L, j = 0; i <= R; i++, j++) { + arr[i] = help[j]; + } + } + } + + public static int getDigit(int x, int d) { + return ((x / ((int) Math.pow(10, d - 1))) % 10); + } + + // for test + public static void comparator(int[] arr) { + Arrays.sort(arr); + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()); + } + return arr; + } + + // for test + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 100; + int maxValue = 100000; + boolean succeed = true; + for (int i = 0; i < testTime; i++) { + int[] arr1 = generateRandomArray(maxSize, maxValue); + int[] arr2 = copyArray(arr1); + radixSort(arr1); + comparator(arr2); + if (!isEqual(arr1, arr2)) { + succeed = false; + printArray(arr1); + printArray(arr2); + break; + } + } + System.out.println(succeed ? "Nice!" : "Fucking fucked!"); + + int[] arr = generateRandomArray(maxSize, maxValue); + printArray(arr); + radixSort(arr); + printArray(arr); + + } + +} diff --git a/体系学习班/class09/Code01_LinkedListMid.java b/体系学习班/class09/Code01_LinkedListMid.java new file mode 100644 index 0000000..3283443 --- /dev/null +++ b/体系学习班/class09/Code01_LinkedListMid.java @@ -0,0 +1,162 @@ +package class09; + +import java.util.ArrayList; + +public class Code01_LinkedListMid { + + public static class Node { + public int value; + public Node next; + + public Node(int v) { + value = v; + } + } + + // head 头 + public static Node midOrUpMidNode(Node head) { + if (head == null || head.next == null || head.next.next == null) { + return head; + } + // 链表有3个点或以上 + Node slow = head.next; + Node fast = head.next.next; + while (fast.next != null && fast.next.next != null) { + slow = slow.next; + fast = fast.next.next; + } + return slow; + } + + public static Node midOrDownMidNode(Node head) { + if (head == null || head.next == null) { + return head; + } + Node slow = head.next; + Node fast = head.next; + while (fast.next != null && fast.next.next != null) { + slow = slow.next; + fast = fast.next.next; + } + return slow; + } + + public static Node midOrUpMidPreNode(Node head) { + if (head == null || head.next == null || head.next.next == null) { + return null; + } + Node slow = head; + Node fast = head.next.next; + while (fast.next != null && fast.next.next != null) { + slow = slow.next; + fast = fast.next.next; + } + return slow; + } + + public static Node midOrDownMidPreNode(Node head) { + if (head == null || head.next == null) { + return null; + } + if (head.next.next == null) { + return head; + } + Node slow = head; + Node fast = head.next; + while (fast.next != null && fast.next.next != null) { + slow = slow.next; + fast = fast.next.next; + } + return slow; + } + + public static Node right1(Node head) { + if (head == null) { + return null; + } + Node cur = head; + ArrayList arr = new ArrayList<>(); + while (cur != null) { + arr.add(cur); + cur = cur.next; + } + return arr.get((arr.size() - 1) / 2); + } + + public static Node right2(Node head) { + if (head == null) { + return null; + } + Node cur = head; + ArrayList arr = new ArrayList<>(); + while (cur != null) { + arr.add(cur); + cur = cur.next; + } + return arr.get(arr.size() / 2); + } + + public static Node right3(Node head) { + if (head == null || head.next == null || head.next.next == null) { + return null; + } + Node cur = head; + ArrayList arr = new ArrayList<>(); + while (cur != null) { + arr.add(cur); + cur = cur.next; + } + return arr.get((arr.size() - 3) / 2); + } + + public static Node right4(Node head) { + if (head == null || head.next == null) { + return null; + } + Node cur = head; + ArrayList arr = new ArrayList<>(); + while (cur != null) { + arr.add(cur); + cur = cur.next; + } + return arr.get((arr.size() - 2) / 2); + } + + public static void main(String[] args) { + Node test = null; + test = new Node(0); + test.next = new Node(1); + test.next.next = new Node(2); + test.next.next.next = new Node(3); + test.next.next.next.next = new Node(4); + test.next.next.next.next.next = new Node(5); + test.next.next.next.next.next.next = new Node(6); + test.next.next.next.next.next.next.next = new Node(7); + test.next.next.next.next.next.next.next.next = new Node(8); + + Node ans1 = null; + Node ans2 = null; + + ans1 = midOrUpMidNode(test); + ans2 = right1(test); + System.out.println(ans1 != null ? ans1.value : "无"); + System.out.println(ans2 != null ? ans2.value : "无"); + + ans1 = midOrDownMidNode(test); + ans2 = right2(test); + System.out.println(ans1 != null ? ans1.value : "无"); + System.out.println(ans2 != null ? ans2.value : "无"); + + ans1 = midOrUpMidPreNode(test); + ans2 = right3(test); + System.out.println(ans1 != null ? ans1.value : "无"); + System.out.println(ans2 != null ? ans2.value : "无"); + + ans1 = midOrDownMidPreNode(test); + ans2 = right4(test); + System.out.println(ans1 != null ? ans1.value : "无"); + System.out.println(ans2 != null ? ans2.value : "无"); + + } + +} diff --git a/体系学习班/class09/Code02_IsPalindromeList.java b/体系学习班/class09/Code02_IsPalindromeList.java new file mode 100644 index 0000000..63d6eed --- /dev/null +++ b/体系学习班/class09/Code02_IsPalindromeList.java @@ -0,0 +1,204 @@ +package class09; + +import java.util.Stack; + +public class Code02_IsPalindromeList { + + public static class Node { + public int value; + public Node next; + + public Node(int data) { + this.value = data; + } + } + + // need n extra space + public static boolean isPalindrome1(Node head) { + Stack stack = new Stack(); + Node cur = head; + while (cur != null) { + stack.push(cur); + cur = cur.next; + } + while (head != null) { + if (head.value != stack.pop().value) { + return false; + } + head = head.next; + } + return true; + } + + // need n/2 extra space + public static boolean isPalindrome2(Node head) { + if (head == null || head.next == null) { + return true; + } + Node right = head.next; + Node cur = head; + while (cur.next != null && cur.next.next != null) { + right = right.next; + cur = cur.next.next; + } + Stack stack = new Stack(); + while (right != null) { + stack.push(right); + right = right.next; + } + while (!stack.isEmpty()) { + if (head.value != stack.pop().value) { + return false; + } + head = head.next; + } + return true; + } + + // need O(1) extra space + public static boolean isPalindrome3(Node head) { + if (head == null || head.next == null) { + return true; + } + Node n1 = head; + Node n2 = head; + while (n2.next != null && n2.next.next != null) { // find mid node + n1 = n1.next; // n1 -> mid + n2 = n2.next.next; // n2 -> end + } + // n1 中点 + + + n2 = n1.next; // n2 -> right part first node + n1.next = null; // mid.next -> null + Node n3 = null; + while (n2 != null) { // right part convert + n3 = n2.next; // n3 -> save next node + n2.next = n1; // next of right node convert + n1 = n2; // n1 move + n2 = n3; // n2 move + } + n3 = n1; // n3 -> save last node + n2 = head;// n2 -> left first node + boolean res = true; + while (n1 != null && n2 != null) { // check palindrome + if (n1.value != n2.value) { + res = false; + break; + } + n1 = n1.next; // left to mid + n2 = n2.next; // right to mid + } + n1 = n3.next; + n3.next = null; + while (n1 != null) { // recover list + n2 = n1.next; + n1.next = n3; + n3 = n1; + n1 = n2; + } + return res; + } + + public static void printLinkedList(Node node) { + System.out.print("Linked List: "); + while (node != null) { + System.out.print(node.value + " "); + node = node.next; + } + System.out.println(); + } + + public static void main(String[] args) { + + Node head = null; + printLinkedList(head); + System.out.print(isPalindrome1(head) + " | "); + System.out.print(isPalindrome2(head) + " | "); + System.out.println(isPalindrome3(head) + " | "); + printLinkedList(head); + System.out.println("========================="); + + head = new Node(1); + printLinkedList(head); + System.out.print(isPalindrome1(head) + " | "); + System.out.print(isPalindrome2(head) + " | "); + System.out.println(isPalindrome3(head) + " | "); + printLinkedList(head); + System.out.println("========================="); + + head = new Node(1); + head.next = new Node(2); + printLinkedList(head); + System.out.print(isPalindrome1(head) + " | "); + System.out.print(isPalindrome2(head) + " | "); + System.out.println(isPalindrome3(head) + " | "); + printLinkedList(head); + System.out.println("========================="); + + head = new Node(1); + head.next = new Node(1); + printLinkedList(head); + System.out.print(isPalindrome1(head) + " | "); + System.out.print(isPalindrome2(head) + " | "); + System.out.println(isPalindrome3(head) + " | "); + printLinkedList(head); + System.out.println("========================="); + + head = new Node(1); + head.next = new Node(2); + head.next.next = new Node(3); + printLinkedList(head); + System.out.print(isPalindrome1(head) + " | "); + System.out.print(isPalindrome2(head) + " | "); + System.out.println(isPalindrome3(head) + " | "); + printLinkedList(head); + System.out.println("========================="); + + head = new Node(1); + head.next = new Node(2); + head.next.next = new Node(1); + printLinkedList(head); + System.out.print(isPalindrome1(head) + " | "); + System.out.print(isPalindrome2(head) + " | "); + System.out.println(isPalindrome3(head) + " | "); + printLinkedList(head); + System.out.println("========================="); + + head = new Node(1); + head.next = new Node(2); + head.next.next = new Node(3); + head.next.next.next = new Node(1); + printLinkedList(head); + System.out.print(isPalindrome1(head) + " | "); + System.out.print(isPalindrome2(head) + " | "); + System.out.println(isPalindrome3(head) + " | "); + printLinkedList(head); + System.out.println("========================="); + + head = new Node(1); + head.next = new Node(2); + head.next.next = new Node(2); + head.next.next.next = new Node(1); + printLinkedList(head); + System.out.print(isPalindrome1(head) + " | "); + System.out.print(isPalindrome2(head) + " | "); + System.out.println(isPalindrome3(head) + " | "); + printLinkedList(head); + System.out.println("========================="); + + head = new Node(1); + head.next = new Node(2); + head.next.next = new Node(3); + head.next.next.next = new Node(2); + head.next.next.next.next = new Node(1); + printLinkedList(head); + System.out.print(isPalindrome1(head) + " | "); + System.out.print(isPalindrome2(head) + " | "); + System.out.println(isPalindrome3(head) + " | "); + printLinkedList(head); + System.out.println("========================="); + + } + +} diff --git a/体系学习班/class09/Code03_SmallerEqualBigger.java b/体系学习班/class09/Code03_SmallerEqualBigger.java new file mode 100644 index 0000000..33203e4 --- /dev/null +++ b/体系学习班/class09/Code03_SmallerEqualBigger.java @@ -0,0 +1,138 @@ +package class09; + +public class Code03_SmallerEqualBigger { + + public static class Node { + public int value; + public Node next; + + public Node(int data) { + this.value = data; + } + } + + public static Node listPartition1(Node head, int pivot) { + if (head == null) { + return head; + } + Node cur = head; + int i = 0; + while (cur != null) { + i++; + cur = cur.next; + } + Node[] nodeArr = new Node[i]; + i = 0; + cur = head; + for (i = 0; i != nodeArr.length; i++) { + nodeArr[i] = cur; + cur = cur.next; + } + arrPartition(nodeArr, pivot); + for (i = 1; i != nodeArr.length; i++) { + nodeArr[i - 1].next = nodeArr[i]; + } + nodeArr[i - 1].next = null; + return nodeArr[0]; + } + + public static void arrPartition(Node[] nodeArr, int pivot) { + int small = -1; + int big = nodeArr.length; + int index = 0; + while (index != big) { + if (nodeArr[index].value < pivot) { + swap(nodeArr, ++small, index++); + } else if (nodeArr[index].value == pivot) { + index++; + } else { + swap(nodeArr, --big, index); + } + } + } + + public static void swap(Node[] nodeArr, int a, int b) { + Node tmp = nodeArr[a]; + nodeArr[a] = nodeArr[b]; + nodeArr[b] = tmp; + } + + public static Node listPartition2(Node head, int pivot) { + Node sH = null; // small head + Node sT = null; // small tail + Node eH = null; // equal head + Node eT = null; // equal tail + Node mH = null; // big head + Node mT = null; // big tail + Node next = null; // save next node + // every node distributed to three lists + while (head != null) { + next = head.next; + head.next = null; + if (head.value < pivot) { + if (sH == null) { + sH = head; + sT = head; + } else { + sT.next = head; + sT = head; + } + } else if (head.value == pivot) { + if (eH == null) { + eH = head; + eT = head; + } else { + eT.next = head; + eT = head; + } + } else { + if (mH == null) { + mH = head; + mT = head; + } else { + mT.next = head; + mT = head; + } + } + head = next; + } + // 小于区域的尾巴，连等于区域的头，等于区域的尾巴连大于区域的头 + if (sT != null) { // 如果有小于区域 + sT.next = eH; + eT = eT == null ? sT : eT; // 下一步，谁去连大于区域的头，谁就变成eT + } + // 下一步，一定是需要用eT 去接 大于区域的头 + // 有等于区域，eT -> 等于区域的尾结点 + // 无等于区域，eT -> 小于区域的尾结点 + // eT 尽量不为空的尾巴节点 + if (eT != null) { // 如果小于区域和等于区域，不是都没有 + eT.next = mH; + } + return sH != null ? sH : (eH != null ? eH : mH); + } + + public static void printLinkedList(Node node) { + System.out.print("Linked List: "); + while (node != null) { + System.out.print(node.value + " "); + node = node.next; + } + System.out.println(); + } + + public static void main(String[] args) { + Node head1 = new Node(7); + head1.next = new Node(9); + head1.next.next = new Node(1); + head1.next.next.next = new Node(8); + head1.next.next.next.next = new Node(5); + head1.next.next.next.next.next = new Node(2); + head1.next.next.next.next.next.next = new Node(5); + printLinkedList(head1); + // head1 = listPartition1(head1, 4); + head1 = listPartition2(head1, 5); + printLinkedList(head1); + + } + +} diff --git a/体系学习班/class09/Code04_CopyListWithRandom.java b/体系学习班/class09/Code04_CopyListWithRandom.java new file mode 100644 index 0000000..f42454f --- /dev/null +++ b/体系学习班/class09/Code04_CopyListWithRandom.java @@ -0,0 +1,79 @@ +package class09; + +import java.util.HashMap; + +// 测试链接 : https://leetcode.com/problems/copy-list-with-random-pointer/ +public class Code04_CopyListWithRandom { + + public static class Node { + int val; + Node next; + Node random; + + public Node(int val) { + this.val = val; + this.next = null; + this.random = null; + } + } + + public static Node copyRandomList1(Node head) { + // key 老节点 + // value 新节点 + HashMap map = new HashMap(); + Node cur = head; + while (cur != null) { + map.put(cur, new Node(cur.val)); + cur = cur.next; + } + cur = head; + while (cur != null) { + // cur 老 + // map.get(cur) 新 + // 新.next -> cur.next克隆节点找到 + map.get(cur).next = map.get(cur.next); + map.get(cur).random = map.get(cur.random); + cur = cur.next; + } + return map.get(head); + } + + public static Node copyRandomList2(Node head) { + if (head == null) { + return null; + } + Node cur = head; + Node next = null; + // 1 -> 2 -> 3 -> null + // 1 -> 1' -> 2 -> 2' -> 3 -> 3' + while (cur != null) { + next = cur.next; + cur.next = new Node(cur.val); + cur.next.next = next; + cur = next; + } + cur = head; + Node copy = null; + // 1 1' 2 2' 3 3' + // 依次设置 1' 2' 3' random指针 + while (cur != null) { + next = cur.next.next; + copy = cur.next; + copy.random = cur.random != null ? cur.random.next : null; + cur = next; + } + Node res = head.next; + cur = head; + // 老 新 混在一起，next方向上，random正确 + // next方向上，把新老链表分离 + while (cur != null) { + next = cur.next.next; + copy = cur.next; + cur.next = next; + copy.next = next != null ? next.next : null; + cur = next; + } + return res; + } + +} diff --git a/体系学习班/class10/Code01_FindFirstIntersectNode.java b/体系学习班/class10/Code01_FindFirstIntersectNode.java new file mode 100644 index 0000000..8cec65d --- /dev/null +++ b/体系学习班/class10/Code01_FindFirstIntersectNode.java @@ -0,0 +1,170 @@ +package class10; + +public class Code01_FindFirstIntersectNode { + + public static class Node { + public int value; + public Node next; + + public Node(int data) { + this.value = data; + } + } + + public static Node getIntersectNode(Node head1, Node head2) { + if (head1 == null || head2 == null) { + return null; + } + Node loop1 = getLoopNode(head1); + Node loop2 = getLoopNode(head2); + if (loop1 == null && loop2 == null) { + return noLoop(head1, head2); + } + if (loop1 != null && loop2 != null) { + return bothLoop(head1, loop1, head2, loop2); + } + return null; + } + + // 找到链表第一个入环节点，如果无环，返回null + public static Node getLoopNode(Node head) { + if (head == null || head.next == null || head.next.next == null) { + return null; + } + // n1 慢 n2 快 + Node slow = head.next; // n1 -> slow + Node fast = head.next.next; // n2 -> fast + while (slow != fast) { + if (fast.next == null || fast.next.next == null) { + return null; + } + fast = fast.next.next; + slow = slow.next; + } + // slow fast 相遇 + fast = head; // n2 -> walk again from head + while (slow != fast) { + slow = slow.next; + fast = fast.next; + } + return slow; + } + + // 如果两个链表都无环，返回第一个相交节点，如果不想交，返回null + public static Node noLoop(Node head1, Node head2) { + if (head1 == null || head2 == null) { + return null; + } + Node cur1 = head1; + Node cur2 = head2; + int n = 0; + while (cur1.next != null) { + n++; + cur1 = cur1.next; + } + while (cur2.next != null) { + n--; + cur2 = cur2.next; + } + if (cur1 != cur2) { + return null; + } + // n : 链表1长度减去链表2长度的值 + cur1 = n > 0 ? head1 : head2; // 谁长，谁的头变成cur1 + cur2 = cur1 == head1 ? head2 : head1; // 谁短，谁的头变成cur2 + n = Math.abs(n); + while (n != 0) { + n--; + cur1 = cur1.next; + } + while (cur1 != cur2) { + cur1 = cur1.next; + cur2 = cur2.next; + } + return cur1; + } + + // 两个有环链表，返回第一个相交节点，如果不想交返回null + public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) { + Node cur1 = null; + Node cur2 = null; + if (loop1 == loop2) { + cur1 = head1; + cur2 = head2; + int n = 0; + while (cur1 != loop1) { + n++; + cur1 = cur1.next; + } + while (cur2 != loop2) { + n--; + cur2 = cur2.next; + } + cur1 = n > 0 ? head1 : head2; + cur2 = cur1 == head1 ? head2 : head1; + n = Math.abs(n); + while (n != 0) { + n--; + cur1 = cur1.next; + } + while (cur1 != cur2) { + cur1 = cur1.next; + cur2 = cur2.next; + } + return cur1; + } else { + cur1 = loop1.next; + while (cur1 != loop1) { + if (cur1 == loop2) { + return loop1; + } + cur1 = cur1.next; + } + return null; + } + } + + public static void main(String[] args) { + // 1->2->3->4->5->6->7->null + Node head1 = new Node(1); + head1.next = new Node(2); + head1.next.next = new Node(3); + head1.next.next.next = new Node(4); + head1.next.next.next.next = new Node(5); + head1.next.next.next.next.next = new Node(6); + head1.next.next.next.next.next.next = new Node(7); + + // 0->9->8->6->7->null + Node head2 = new Node(0); + head2.next = new Node(9); + head2.next.next = new Node(8); + head2.next.next.next = head1.next.next.next.next.next; // 8->6 + System.out.println(getIntersectNode(head1, head2).value); + + // 1->2->3->4->5->6->7->4... + head1 = new Node(1); + head1.next = new Node(2); + head1.next.next = new Node(3); + head1.next.next.next = new Node(4); + head1.next.next.next.next = new Node(5); + head1.next.next.next.next.next = new Node(6); + head1.next.next.next.next.next.next = new Node(7); + head1.next.next.next.next.next.next = head1.next.next.next; // 7->4 + + // 0->9->8->2... + head2 = new Node(0); + head2.next = new Node(9); + head2.next.next = new Node(8); + head2.next.next.next = head1.next; // 8->2 + System.out.println(getIntersectNode(head1, head2).value); + + // 0->9->8->6->4->5->6.. + head2 = new Node(0); + head2.next = new Node(9); + head2.next.next = new Node(8); + head2.next.next.next = head1.next.next.next.next.next; // 8->6 + System.out.println(getIntersectNode(head1, head2).value); + + } + +} diff --git a/体系学习班/class10/Code02_RecursiveTraversalBT.java b/体系学习班/class10/Code02_RecursiveTraversalBT.java new file mode 100644 index 0000000..c7f2b2c --- /dev/null +++ b/体系学习班/class10/Code02_RecursiveTraversalBT.java @@ -0,0 +1,72 @@ +package class10; + +public class Code02_RecursiveTraversalBT { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int v) { + value = v; + } + } + + public static void f(Node head) { + if (head == null) { + return; + } + // 1 + f(head.left); + // 2 + f(head.right); + // 3 + } + + // 先序打印所有节点 + public static void pre(Node head) { + if (head == null) { + return; + } + System.out.println(head.value); + pre(head.left); + pre(head.right); + } + + public static void in(Node head) { + if (head == null) { + return; + } + in(head.left); + System.out.println(head.value); + in(head.right); + } + + public static void pos(Node head) { + if (head == null) { + return; + } + pos(head.left); + pos(head.right); + System.out.println(head.value); + } + + public static void main(String[] args) { + Node head = new Node(1); + head.left = new Node(2); + head.right = new Node(3); + head.left.left = new Node(4); + head.left.right = new Node(5); + head.right.left = new Node(6); + head.right.right = new Node(7); + + pre(head); + System.out.println("========"); + in(head); + System.out.println("========"); + pos(head); + System.out.println("========"); + + } + +} diff --git a/体系学习班/class10/Code03_UnRecursiveTraversalBT.java b/体系学习班/class10/Code03_UnRecursiveTraversalBT.java new file mode 100644 index 0000000..96ef25e --- /dev/null +++ b/体系学习班/class10/Code03_UnRecursiveTraversalBT.java @@ -0,0 +1,118 @@ +package class10; + +import java.util.Stack; + +public class Code03_UnRecursiveTraversalBT { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int v) { + value = v; + } + } + + public static void pre(Node head) { + System.out.print("pre-order: "); + if (head != null) { + Stack stack = new Stack(); + stack.add(head); + while (!stack.isEmpty()) { + head = stack.pop(); + System.out.print(head.value + " "); + if (head.right != null) { + stack.push(head.right); + } + if (head.left != null) { + stack.push(head.left); + } + } + } + System.out.println(); + } + + public static void in(Node cur) { + System.out.print("in-order: "); + if (cur != null) { + Stack stack = new Stack(); + while (!stack.isEmpty() || cur != null) { + if (cur != null) { + stack.push(cur); + cur = cur.left; + } else { + cur = stack.pop(); + System.out.print(cur.value + " "); + cur = cur.right; + } + } + } + System.out.println(); + } + + public static void pos1(Node head) { + System.out.print("pos-order: "); + if (head != null) { + Stack s1 = new Stack(); + Stack s2 = new Stack(); + s1.push(head); + while (!s1.isEmpty()) { + head = s1.pop(); // 头 右 左 + s2.push(head); + if (head.left != null) { + s1.push(head.left); + } + if (head.right != null) { + s1.push(head.right); + } + } + // 左 右 头 + while (!s2.isEmpty()) { + System.out.print(s2.pop().value + " "); + } + } + System.out.println(); + } + + public static void pos2(Node h) { + System.out.print("pos-order: "); + if (h != null) { + Stack stack = new Stack(); + stack.push(h); + Node c = null; + while (!stack.isEmpty()) { + c = stack.peek(); + if (c.left != null && h != c.left && h != c.right) { + stack.push(c.left); + } else if (c.right != null && h != c.right) { + stack.push(c.right); + } else { + System.out.print(stack.pop().value + " "); + h = c; + } + } + } + System.out.println(); + } + + public static void main(String[] args) { + Node head = new Node(1); + head.left = new Node(2); + head.right = new Node(3); + head.left.left = new Node(4); + head.left.right = new Node(5); + head.right.left = new Node(6); + head.right.right = new Node(7); + + pre(head); + System.out.println("========"); + in(head); + System.out.println("========"); + pos1(head); + System.out.println("========"); + pos2(head); + System.out.println("========"); + } + +} diff --git a/体系学习班/class11/Code01_LevelTraversalBT.java b/体系学习班/class11/Code01_LevelTraversalBT.java new file mode 100644 index 0000000..12e8ef0 --- /dev/null +++ b/体系学习班/class11/Code01_LevelTraversalBT.java @@ -0,0 +1,49 @@ +package class11; + +import java.util.LinkedList; +import java.util.Queue; + +public class Code01_LevelTraversalBT { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int v) { + value = v; + } + } + + public static void level(Node head) { + if (head == null) { + return; + } + Queue queue = new LinkedList<>(); + queue.add(head); + while (!queue.isEmpty()) { + Node cur = queue.poll(); + System.out.println(cur.value); + if (cur.left != null) { + queue.add(cur.left); + } + if (cur.right != null) { + queue.add(cur.right); + } + } + } + + public static void main(String[] args) { + Node head = new Node(1); + head.left = new Node(2); + head.right = new Node(3); + head.left.left = new Node(4); + head.left.right = new Node(5); + head.right.left = new Node(6); + head.right.right = new Node(7); + + level(head); + System.out.println("========"); + } + +} diff --git a/体系学习班/class11/Code02_SerializeAndReconstructTree.java b/体系学习班/class11/Code02_SerializeAndReconstructTree.java new file mode 100644 index 0000000..6e6da7e --- /dev/null +++ b/体系学习班/class11/Code02_SerializeAndReconstructTree.java @@ -0,0 +1,264 @@ +package class11; + +import java.util.LinkedList; +import java.util.Queue; +import java.util.Stack; + +public class Code02_SerializeAndReconstructTree { + /* + * 二叉树可以通过先序、后序或者按层遍历的方式序列化和反序列化， + * 以下代码全部实现了。 + * 但是，二叉树无法通过中序遍历的方式实现序列化和反序列化 + * 因为不同的两棵树，可能得到同样的中序序列，即便补了空位置也可能一样。 + * 比如如下两棵树 + * __2 + * / + * 1 + * 和 + * 1__ + * \ + * 2 + * 补足空位置的中序遍历结果都是{ null, 1, null, 2, null} + * + * */ + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int data) { + this.value = data; + } + } + + public static Queue preSerial(Node head) { + Queue ans = new LinkedList<>(); + pres(head, ans); + return ans; + } + + public static void pres(Node head, Queue ans) { + if (head == null) { + ans.add(null); + } else { + ans.add(String.valueOf(head.value)); + pres(head.left, ans); + pres(head.right, ans); + } + } + + public static Queue inSerial(Node head) { + Queue ans = new LinkedList<>(); + ins(head, ans); + return ans; + } + + public static void ins(Node head, Queue ans) { + if (head == null) { + ans.add(null); + } else { + ins(head.left, ans); + ans.add(String.valueOf(head.value)); + ins(head.right, ans); + } + } + + public static Queue posSerial(Node head) { + Queue ans = new LinkedList<>(); + poss(head, ans); + return ans; + } + + public static void poss(Node head, Queue ans) { + if (head == null) { + ans.add(null); + } else { + poss(head.left, ans); + poss(head.right, ans); + ans.add(String.valueOf(head.value)); + } + } + + public static Node buildByPreQueue(Queue prelist) { + if (prelist == null || prelist.size() == 0) { + return null; + } + return preb(prelist); + } + + public static Node preb(Queue prelist) { + String value = prelist.poll(); + if (value == null) { + return null; + } + Node head = new Node(Integer.valueOf(value)); + head.left = preb(prelist); + head.right = preb(prelist); + return head; + } + + public static Node buildByPosQueue(Queue poslist) { + if (poslist == null || poslist.size() == 0) { + return null; + } + // 左右中 -> stack(中右左) + Stack stack = new Stack<>(); + while (!poslist.isEmpty()) { + stack.push(poslist.poll()); + } + return posb(stack); + } + + public static Node posb(Stack posstack) { + String value = posstack.pop(); + if (value == null) { + return null; + } + Node head = new Node(Integer.valueOf(value)); + head.right = posb(posstack); + head.left = posb(posstack); + return head; + } + + public static Queue levelSerial(Node head) { + Queue ans = new LinkedList<>(); + if (head == null) { + ans.add(null); + } else { + ans.add(String.valueOf(head.value)); + Queue queue = new LinkedList(); + queue.add(head); + while (!queue.isEmpty()) { + head = queue.poll(); // head 父 子 + if (head.left != null) { + ans.add(String.valueOf(head.left.value)); + queue.add(head.left); + } else { + ans.add(null); + } + if (head.right != null) { + ans.add(String.valueOf(head.right.value)); + queue.add(head.right); + } else { + ans.add(null); + } + } + } + return ans; + } + + public static Node buildByLevelQueue(Queue levelList) { + if (levelList == null || levelList.size() == 0) { + return null; + } + Node head = generateNode(levelList.poll()); + Queue queue = new LinkedList(); + if (head != null) { + queue.add(head); + } + Node node = null; + while (!queue.isEmpty()) { + node = queue.poll(); + node.left = generateNode(levelList.poll()); + node.right = generateNode(levelList.poll()); + if (node.left != null) { + queue.add(node.left); + } + if (node.right != null) { + queue.add(node.right); + } + } + return head; + } + + public static Node generateNode(String val) { + if (val == null) { + return null; + } + return new Node(Integer.valueOf(val)); + } + + // for test + public static Node generateRandomBST(int maxLevel, int maxValue) { + return generate(1, maxLevel, maxValue); + } + + // for test + public static Node generate(int level, int maxLevel, int maxValue) { + if (level > maxLevel || Math.random() < 0.5) { + return null; + } + Node head = new Node((int) (Math.random() * maxValue)); + head.left = generate(level + 1, maxLevel, maxValue); + head.right = generate(level + 1, maxLevel, maxValue); + return head; + } + + // for test + public static boolean isSameValueStructure(Node head1, Node head2) { + if (head1 == null && head2 != null) { + return false; + } + if (head1 != null && head2 == null) { + return false; + } + if (head1 == null && head2 == null) { + return true; + } + if (head1.value != head2.value) { + return false; + } + return isSameValueStructure(head1.left, head2.left) && isSameValueStructure(head1.right, head2.right); + } + + // for test + public static void printTree(Node head) { + System.out.println("Binary Tree:"); + printInOrder(head, 0, "H", 17); + System.out.println(); + } + + public static void printInOrder(Node head, int height, String to, int len) { + if (head == null) { + return; + } + printInOrder(head.right, height + 1, "v", len); + String val = to + head.value + to; + int lenM = val.length(); + int lenL = (len - lenM) / 2; + int lenR = len - lenM - lenL; + val = getSpace(lenL) + val + getSpace(lenR); + System.out.println(getSpace(height * len) + val); + printInOrder(head.left, height + 1, "^", len); + } + + public static String getSpace(int num) { + String space = " "; + StringBuffer buf = new StringBuffer(""); + for (int i = 0; i < num; i++) { + buf.append(space); + } + return buf.toString(); + } + + public static void main(String[] args) { + int maxLevel = 5; + int maxValue = 100; + int testTimes = 1000000; + System.out.println("test begin"); + for (int i = 0; i < testTimes; i++) { + Node head = generateRandomBST(maxLevel, maxValue); + Queue pre = preSerial(head); + Queue pos = posSerial(head); + Queue level = levelSerial(head); + Node preBuild = buildByPreQueue(pre); + Node posBuild = buildByPosQueue(pos); + Node levelBuild = buildByLevelQueue(level); + if (!isSameValueStructure(preBuild, posBuild) || !isSameValueStructure(posBuild, levelBuild)) { + System.out.println("Oops!"); + } + } + System.out.println("test finish!"); + + } +} diff --git a/体系学习班/class11/Code03_EncodeNaryTreeToBinaryTree.java b/体系学习班/class11/Code03_EncodeNaryTreeToBinaryTree.java new file mode 100644 index 0000000..aa9f2ab --- /dev/null +++ b/体系学习班/class11/Code03_EncodeNaryTreeToBinaryTree.java @@ -0,0 +1,86 @@ +package class11; + +import java.util.ArrayList; +import java.util.List; + +// 本题测试链接：https://leetcode.com/problems/encode-n-ary-tree-to-binary-tree +public class Code03_EncodeNaryTreeToBinaryTree { + + // 提交时不要提交这个类 + public static class Node { + public int val; + public List children; + + public Node() { + } + + public Node(int _val) { + val = _val; + } + + public Node(int _val, List _children) { + val = _val; + children = _children; + } + }; + + // 提交时不要提交这个类 + public static class TreeNode { + int val; + TreeNode left; + TreeNode right; + + TreeNode(int x) { + val = x; + } + } + + // 只提交这个类即可 + class Codec { + // Encodes an n-ary tree to a binary tree. + public TreeNode encode(Node root) { + if (root == null) { + return null; + } + TreeNode head = new TreeNode(root.val); + head.left = en(root.children); + return head; + } + + private TreeNode en(List children) { + TreeNode head = null; + TreeNode cur = null; + for (Node child : children) { + TreeNode tNode = new TreeNode(child.val); + if (head == null) { + head = tNode; + } else { + cur.right = tNode; + } + cur = tNode; + cur.left = en(child.children); + } + return head; + } + + // Decodes your binary tree to an n-ary tree. + public Node decode(TreeNode root) { + if (root == null) { + return null; + } + return new Node(root.val, de(root.left)); + } + + public List de(TreeNode root) { + List children = new ArrayList<>(); + while (root != null) { + Node cur = new Node(root.val, de(root.left)); + children.add(cur); + root = root.right; + } + return children; + } + + } + +} diff --git a/体系学习班/class11/Code04_PrintBinaryTree.java b/体系学习班/class11/Code04_PrintBinaryTree.java new file mode 100644 index 0000000..75125f7 --- /dev/null +++ b/体系学习班/class11/Code04_PrintBinaryTree.java @@ -0,0 +1,74 @@ +package class11; + +public class Code04_PrintBinaryTree { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int data) { + this.value = data; + } + } + + public static void printTree(Node head) { + System.out.println("Binary Tree:"); + printInOrder(head, 0, "H", 17); + System.out.println(); + } + + public static void printInOrder(Node head, int height, String to, int len) { + if (head == null) { + return; + } + printInOrder(head.right, height + 1, "v", len); + String val = to + head.value + to; + int lenM = val.length(); + int lenL = (len - lenM) / 2; + int lenR = len - lenM - lenL; + val = getSpace(lenL) + val + getSpace(lenR); + System.out.println(getSpace(height * len) + val); + printInOrder(head.left, height + 1, "^", len); + } + + public static String getSpace(int num) { + String space = " "; + StringBuffer buf = new StringBuffer(""); + for (int i = 0; i < num; i++) { + buf.append(space); + } + return buf.toString(); + } + + public static void main(String[] args) { + Node head = new Node(1); + head.left = new Node(-222222222); + head.right = new Node(3); + head.left.left = new Node(Integer.MIN_VALUE); + head.right.left = new Node(55555555); + head.right.right = new Node(66); + head.left.left.right = new Node(777); + printTree(head); + + head = new Node(1); + head.left = new Node(2); + head.right = new Node(3); + head.left.left = new Node(4); + head.right.left = new Node(5); + head.right.right = new Node(6); + head.left.left.right = new Node(7); + printTree(head); + + head = new Node(1); + head.left = new Node(1); + head.right = new Node(1); + head.left.left = new Node(1); + head.right.left = new Node(1); + head.right.right = new Node(1); + head.left.left.right = new Node(1); + printTree(head); + + } + +} diff --git a/体系学习班/class11/Code05_TreeMaxWidth.java b/体系学习班/class11/Code05_TreeMaxWidth.java new file mode 100644 index 0000000..787bd01 --- /dev/null +++ b/体系学习班/class11/Code05_TreeMaxWidth.java @@ -0,0 +1,114 @@ +package class11; + +import java.util.HashMap; +import java.util.LinkedList; +import java.util.Queue; + +public class Code05_TreeMaxWidth { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int data) { + this.value = data; + } + } + + public static int maxWidthUseMap(Node head) { + if (head == null) { + return 0; + } + Queue queue = new LinkedList<>(); + queue.add(head); + // key 在 哪一层，value + HashMap levelMap = new HashMap<>(); + levelMap.put(head, 1); + int curLevel = 1; // 当前你正在统计哪一层的宽度 + int curLevelNodes = 0; // 当前层curLevel层，宽度目前是多少 + int max = 0; + while (!queue.isEmpty()) { + Node cur = queue.poll(); + int curNodeLevel = levelMap.get(cur); + if (cur.left != null) { + levelMap.put(cur.left, curNodeLevel + 1); + queue.add(cur.left); + } + if (cur.right != null) { + levelMap.put(cur.right, curNodeLevel + 1); + queue.add(cur.right); + } + if (curNodeLevel == curLevel) { + curLevelNodes++; + } else { + max = Math.max(max, curLevelNodes); + curLevel++; + curLevelNodes = 1; + } + } + max = Math.max(max, curLevelNodes); + return max; + } + + public static int maxWidthNoMap(Node head) { + if (head == null) { + return 0; + } + Queue queue = new LinkedList<>(); + queue.add(head); + Node curEnd = head; // 当前层，最右节点是谁 + Node nextEnd = null; // 下一层，最右节点是谁 + int max = 0; + int curLevelNodes = 0; // 当前层的节点数 + while (!queue.isEmpty()) { + Node cur = queue.poll(); + if (cur.left != null) { + queue.add(cur.left); + nextEnd = cur.left; + } + if (cur.right != null) { + queue.add(cur.right); + nextEnd = cur.right; + } + curLevelNodes++; + if (cur == curEnd) { + max = Math.max(max, curLevelNodes); + curLevelNodes = 0; + curEnd = nextEnd; + } + } + return max; + } + + // for test + public static Node generateRandomBST(int maxLevel, int maxValue) { + return generate(1, maxLevel, maxValue); + } + + // for test + public static Node generate(int level, int maxLevel, int maxValue) { + if (level > maxLevel || Math.random() < 0.5) { + return null; + } + Node head = new Node((int) (Math.random() * maxValue)); + head.left = generate(level + 1, maxLevel, maxValue); + head.right = generate(level + 1, maxLevel, maxValue); + return head; + } + + public static void main(String[] args) { + int maxLevel = 10; + int maxValue = 100; + int testTimes = 1000000; + for (int i = 0; i < testTimes; i++) { + Node head = generateRandomBST(maxLevel, maxValue); + if (maxWidthUseMap(head) != maxWidthNoMap(head)) { + System.out.println("Oops!"); + } + } + System.out.println("finish!"); + + } + +} diff --git a/体系学习班/class11/Code06_SuccessorNode.java b/体系学习班/class11/Code06_SuccessorNode.java new file mode 100644 index 0000000..b33d4b5 --- /dev/null +++ b/体系学习班/class11/Code06_SuccessorNode.java @@ -0,0 +1,86 @@ +package class11; + +public class Code06_SuccessorNode { + + public static class Node { + public int value; + public Node left; + public Node right; + public Node parent; + + public Node(int data) { + this.value = data; + } + } + + public static Node getSuccessorNode(Node node) { + if (node == null) { + return node; + } + if (node.right != null) { + return getLeftMost(node.right); + } else { // 无右子树 + Node parent = node.parent; + while (parent != null && parent.right == node) { // 当前节点是其父亲节点右孩子 + node = parent; + parent = node.parent; + } + return parent; + } + } + + public static Node getLeftMost(Node node) { + if (node == null) { + return node; + } + while (node.left != null) { + node = node.left; + } + return node; + } + + public static void main(String[] args) { + Node head = new Node(6); + head.parent = null; + head.left = new Node(3); + head.left.parent = head; + head.left.left = new Node(1); + head.left.left.parent = head.left; + head.left.left.right = new Node(2); + head.left.left.right.parent = head.left.left; + head.left.right = new Node(4); + head.left.right.parent = head.left; + head.left.right.right = new Node(5); + head.left.right.right.parent = head.left.right; + head.right = new Node(9); + head.right.parent = head; + head.right.left = new Node(8); + head.right.left.parent = head.right; + head.right.left.left = new Node(7); + head.right.left.left.parent = head.right.left; + head.right.right = new Node(10); + head.right.right.parent = head.right; + + Node test = head.left.left; + System.out.println(test.value + " next: " + getSuccessorNode(test).value); + test = head.left.left.right; + System.out.println(test.value + " next: " + getSuccessorNode(test).value); + test = head.left; + System.out.println(test.value + " next: " + getSuccessorNode(test).value); + test = head.left.right; + System.out.println(test.value + " next: " + getSuccessorNode(test).value); + test = head.left.right.right; + System.out.println(test.value + " next: " + getSuccessorNode(test).value); + test = head; + System.out.println(test.value + " next: " + getSuccessorNode(test).value); + test = head.right.left.left; + System.out.println(test.value + " next: " + getSuccessorNode(test).value); + test = head.right.left; + System.out.println(test.value + " next: " + getSuccessorNode(test).value); + test = head.right; + System.out.println(test.value + " next: " + getSuccessorNode(test).value); + test = head.right.right; // 10's next is null + System.out.println(test.value + " next: " + getSuccessorNode(test)); + } + +} diff --git a/体系学习班/class11/Code07_PaperFolding.java b/体系学习班/class11/Code07_PaperFolding.java new file mode 100644 index 0000000..3f3b0db --- /dev/null +++ b/体系学习班/class11/Code07_PaperFolding.java @@ -0,0 +1,28 @@ +package class11; + +public class Code07_PaperFolding { + + public static void printAllFolds(int N) { + process(1, N, true); + System.out.println(); + } + + // 当前你来了一个节点，脑海中想象的！ + // 这个节点在第i层，一共有N层，N固定不变的 + // 这个节点如果是凹的话，down = T + // 这个节点如果是凸的话，down = F + // 函数的功能：中序打印以你想象的节点为头的整棵树！ + public static void process(int i, int N, boolean down) { + if (i > N) { + return; + } + process(i + 1, N, true); + System.out.print(down ? "凹 " : "凸 "); + process(i + 1, N, false); + } + + public static void main(String[] args) { + int N = 4; + printAllFolds(N); + } +} diff --git a/体系学习班/class12/Code01_IsCBT.java b/体系学习班/class12/Code01_IsCBT.java new file mode 100644 index 0000000..e0b9cab --- /dev/null +++ b/体系学习班/class12/Code01_IsCBT.java @@ -0,0 +1,149 @@ +package class12; + +import java.util.LinkedList; + +public class Code01_IsCBT { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int data) { + this.value = data; + } + } + + public static boolean isCBT1(Node head) { + if (head == null) { + return true; + } + LinkedList queue = new LinkedList<>(); + // 是否遇到过左右两个孩子不双全的节点 + boolean leaf = false; + Node l = null; + Node r = null; + queue.add(head); + while (!queue.isEmpty()) { + head = queue.poll(); + l = head.left; + r = head.right; + if ( + // 如果遇到了不双全的节点之后，又发现当前节点不是叶节点 + (leaf && (l != null || r != null)) + || + (l == null && r != null) + + ) { + return false; + } + if (l != null) { + queue.add(l); + } + if (r != null) { + queue.add(r); + } + if (l == null || r == null) { + leaf = true; + } + } + return true; + } + + public static boolean isCBT2(Node head) { + if (head == null) { + return true; + } + return process(head).isCBT; + } + + // 对每一棵子树，是否是满二叉树、是否是完全二叉树、高度 + public static class Info { + public boolean isFull; + public boolean isCBT; + public int height; + + public Info(boolean full, boolean cbt, int h) { + isFull = full; + isCBT = cbt; + height = h; + } + } + + public static Info process(Node X) { + if (X == null) { + return new Info(true, true, 0); + } + Info leftInfo = process(X.left); + Info rightInfo = process(X.right); + + + + int height = Math.max(leftInfo.height, rightInfo.height) + 1; + + + boolean isFull = leftInfo.isFull + && + rightInfo.isFull + && leftInfo.height == rightInfo.height; + + + boolean isCBT = false; + if (isFull) { + isCBT = true; + } else { // 以x为头整棵树，不满 + if (leftInfo.isCBT && rightInfo.isCBT) { + + + if (leftInfo.isCBT + && rightInfo.isFull + && leftInfo.height == rightInfo.height + 1) { + isCBT = true; + } + if (leftInfo.isFull + && + rightInfo.isFull + && leftInfo.height == rightInfo.height + 1) { + isCBT = true; + } + if (leftInfo.isFull + && rightInfo.isCBT && leftInfo.height == rightInfo.height) { + isCBT = true; + } + + + } + } + return new Info(isFull, isCBT, height); + } + + // for test + public static Node generateRandomBST(int maxLevel, int maxValue) { + return generate(1, maxLevel, maxValue); + } + + // for test + public static Node generate(int level, int maxLevel, int maxValue) { + if (level > maxLevel || Math.random() < 0.5) { + return null; + } + Node head = new Node((int) (Math.random() * maxValue)); + head.left = generate(level + 1, maxLevel, maxValue); + head.right = generate(level + 1, maxLevel, maxValue); + return head; + } + + public static void main(String[] args) { + int maxLevel = 5; + int maxValue = 100; + int testTimes = 1000000; + for (int i = 0; i < testTimes; i++) { + Node head = generateRandomBST(maxLevel, maxValue); + if (isCBT1(head) != isCBT2(head)) { + System.out.println("Oops!"); + } + } + System.out.println("finish!"); + } + +} diff --git a/体系学习班/class12/Code02_IsBST.java b/体系学习班/class12/Code02_IsBST.java new file mode 100644 index 0000000..30d99dc --- /dev/null +++ b/体系学习班/class12/Code02_IsBST.java @@ -0,0 +1,125 @@ +package class12; + +import java.util.ArrayList; + +public class Code02_IsBST { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int data) { + this.value = data; + } + } + + public static boolean isBST1(Node head) { + if (head == null) { + return true; + } + ArrayList arr = new ArrayList<>(); + in(head, arr); + for (int i = 1; i < arr.size(); i++) { + if (arr.get(i).value <= arr.get(i - 1).value) { + return false; + } + } + return true; + } + + public static void in(Node head, ArrayList arr) { + if (head == null) { + return; + } + in(head.left, arr); + arr.add(head); + in(head.right, arr); + } + + public static boolean isBST2(Node head) { + if (head == null) { + return true; + } + return process(head).isBST; + } + + public static class Info { + public boolean isBST; + public int max; + public int min; + + public Info(boolean i, int ma, int mi) { + isBST = i; + max = ma; + min = mi; + } + + } + + public static Info process(Node x) { + if (x == null) { + return null; + } + Info leftInfo = process(x.left); + Info rightInfo = process(x.right); + int max = x.value; + if (leftInfo != null) { + max = Math.max(max, leftInfo.max); + } + if (rightInfo != null) { + max = Math.max(max, rightInfo.max); + } + int min = x.value; + if (leftInfo != null) { + min = Math.min(min, leftInfo.min); + } + if (rightInfo != null) { + min = Math.min(min, rightInfo.min); + } + boolean isBST = true; + if (leftInfo != null && !leftInfo.isBST) { + isBST = false; + } + if (rightInfo != null && !rightInfo.isBST) { + isBST = false; + } + if (leftInfo != null && leftInfo.max >= x.value) { + isBST = false; + } + if (rightInfo != null && rightInfo.min <= x.value) { + isBST = false; + } + return new Info(isBST, max, min); + } + + // for test + public static Node generateRandomBST(int maxLevel, int maxValue) { + return generate(1, maxLevel, maxValue); + } + + // for test + public static Node generate(int level, int maxLevel, int maxValue) { + if (level > maxLevel || Math.random() < 0.5) { + return null; + } + Node head = new Node((int) (Math.random() * maxValue)); + head.left = generate(level + 1, maxLevel, maxValue); + head.right = generate(level + 1, maxLevel, maxValue); + return head; + } + + public static void main(String[] args) { + int maxLevel = 4; + int maxValue = 100; + int testTimes = 1000000; + for (int i = 0; i < testTimes; i++) { + Node head = generateRandomBST(maxLevel, maxValue); + if (isBST1(head) != isBST2(head)) { + System.out.println("Oops!"); + } + } + System.out.println("finish!"); + } + +} diff --git a/体系学习班/class12/Code03_IsBalanced.java b/体系学习班/class12/Code03_IsBalanced.java new file mode 100644 index 0000000..e8a6c0f --- /dev/null +++ b/体系学习班/class12/Code03_IsBalanced.java @@ -0,0 +1,102 @@ +package class12; + +public class Code03_IsBalanced { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int data) { + this.value = data; + } + } + + public static boolean isBalanced1(Node head) { + boolean[] ans = new boolean[1]; + ans[0] = true; + process1(head, ans); + return ans[0]; + } + + public static int process1(Node head, boolean[] ans) { + if (!ans[0] || head == null) { + return -1; + } + int leftHeight = process1(head.left, ans); + int rightHeight = process1(head.right, ans); + if (Math.abs(leftHeight - rightHeight) > 1) { + ans[0] = false; + } + return Math.max(leftHeight, rightHeight) + 1; + } + + public static boolean isBalanced2(Node head) { + return process(head).isBalanced; + } + + public static class Info{ + public boolean isBalanced; + public int height; + + public Info(boolean i, int h) { + isBalanced = i; + height = h; + } + } + + public static Info process(Node x) { + if(x == null) { + return new Info(true, 0); + } + Info leftInfo = process(x.left); + Info rightInfo = process(x.right); + int height = Math.max(leftInfo.height, rightInfo.height) + 1; + boolean isBalanced = true; + if(!leftInfo.isBalanced) { + isBalanced = false; + } + if(!rightInfo.isBalanced) { + isBalanced = false; + } + if(Math.abs(leftInfo.height - rightInfo.height) > 1) { + isBalanced = false; + } + return new Info(isBalanced, height); + } + + + + + + + // for test + public static Node generateRandomBST(int maxLevel, int maxValue) { + return generate(1, maxLevel, maxValue); + } + + // for test + public static Node generate(int level, int maxLevel, int maxValue) { + if (level > maxLevel || Math.random() < 0.5) { + return null; + } + Node head = new Node((int) (Math.random() * maxValue)); + head.left = generate(level + 1, maxLevel, maxValue); + head.right = generate(level + 1, maxLevel, maxValue); + return head; + } + + public static void main(String[] args) { + int maxLevel = 5; + int maxValue = 100; + int testTimes = 1000000; + for (int i = 0; i < testTimes; i++) { + Node head = generateRandomBST(maxLevel, maxValue); + if (isBalanced1(head) != isBalanced2(head)) { + System.out.println("Oops!"); + } + } + System.out.println("finish!"); + } + +} diff --git a/体系学习班/class12/Code04_IsFull.java b/体系学习班/class12/Code04_IsFull.java new file mode 100644 index 0000000..a83c952 --- /dev/null +++ b/体系学习班/class12/Code04_IsFull.java @@ -0,0 +1,109 @@ +package class12; + +public class Code04_IsFull { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int data) { + this.value = data; + } + } + + // 第一种方法 + // 收集整棵树的高度h，和节点数n + // 只有满二叉树满足 : 2 ^ h - 1 == n + public static boolean isFull1(Node head) { + if (head == null) { + return true; + } + Info1 all = process1(head); + return (1 << all.height) - 1 == all.nodes; + } + + public static class Info1 { + public int height; + public int nodes; + + public Info1(int h, int n) { + height = h; + nodes = n; + } + } + + public static Info1 process1(Node head) { + if (head == null) { + return new Info1(0, 0); + } + Info1 leftInfo = process1(head.left); + Info1 rightInfo = process1(head.right); + int height = Math.max(leftInfo.height, rightInfo.height) + 1; + int nodes = leftInfo.nodes + rightInfo.nodes + 1; + return new Info1(height, nodes); + } + + // 第二种方法 + // 收集子树是否是满二叉树 + // 收集子树的高度 + // 左树满 && 右树满 && 左右树高度一样 -> 整棵树是满的 + public static boolean isFull2(Node head) { + if (head == null) { + return true; + } + return process2(head).isFull; + } + + public static class Info2 { + public boolean isFull; + public int height; + + public Info2(boolean f, int h) { + isFull = f; + height = h; + } + } + + public static Info2 process2(Node h) { + if (h == null) { + return new Info2(true, 0); + } + Info2 leftInfo = process2(h.left); + Info2 rightInfo = process2(h.right); + boolean isFull = leftInfo.isFull && rightInfo.isFull && leftInfo.height == rightInfo.height; + int height = Math.max(leftInfo.height, rightInfo.height) + 1; + return new Info2(isFull, height); + } + + // for test + public static Node generateRandomBST(int maxLevel, int maxValue) { + return generate(1, maxLevel, maxValue); + } + + // for test + public static Node generate(int level, int maxLevel, int maxValue) { + if (level > maxLevel || Math.random() < 0.5) { + return null; + } + Node head = new Node((int) (Math.random() * maxValue)); + head.left = generate(level + 1, maxLevel, maxValue); + head.right = generate(level + 1, maxLevel, maxValue); + return head; + } + + public static void main(String[] args) { + int maxLevel = 5; + int maxValue = 100; + int testTimes = 1000000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + Node head = generateRandomBST(maxLevel, maxValue); + if (isFull1(head) != isFull2(head)) { + System.out.println("出错了!"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class12/Code05_MaxSubBSTSize.java b/体系学习班/class12/Code05_MaxSubBSTSize.java new file mode 100644 index 0000000..85ee0cd --- /dev/null +++ b/体系学习班/class12/Code05_MaxSubBSTSize.java @@ -0,0 +1,157 @@ +package class12; + +import java.util.ArrayList; + +// 在线测试链接 : https://leetcode.com/problems/largest-bst-subtree +public class Code05_MaxSubBSTSize { + + // 提交时不要提交这个类 + public static class TreeNode { + public int val; + public TreeNode left; + public TreeNode right; + + public TreeNode(int value) { + val = value; + } + } + + // 提交如下的largestBSTSubtree方法，可以直接通过 + public static int largestBSTSubtree(TreeNode head) { + if (head == null) { + return 0; + } + return process(head).maxBSTSubtreeSize; + } + + public static class Info { + public int maxBSTSubtreeSize; + public int allSize; + public int max; + public int min; + + public Info(int m, int a, int ma, int mi) { + maxBSTSubtreeSize = m; + allSize = a; + max = ma; + min = mi; + } + } + + public static Info process(TreeNode x) { + if (x == null) { + return null; + } + Info leftInfo = process(x.left); + Info rightInfo = process(x.right); + int max = x.val; + int min = x.val; + int allSize = 1; + if (leftInfo != null) { + max = Math.max(leftInfo.max, max); + min = Math.min(leftInfo.min, min); + allSize += leftInfo.allSize; + } + if (rightInfo != null) { + max = Math.max(rightInfo.max, max); + min = Math.min(rightInfo.min, min); + allSize += rightInfo.allSize; + } + int p1 = -1; + if (leftInfo != null) { + p1 = leftInfo.maxBSTSubtreeSize; + } + int p2 = -1; + if (rightInfo != null) { + p2 = rightInfo.maxBSTSubtreeSize; + } + int p3 = -1; + boolean leftBST = leftInfo == null ? true : (leftInfo.maxBSTSubtreeSize == leftInfo.allSize); + boolean rightBST = rightInfo == null ? true : (rightInfo.maxBSTSubtreeSize == rightInfo.allSize); + if (leftBST && rightBST) { + boolean leftMaxLessX = leftInfo == null ? true : (leftInfo.max < x.val); + boolean rightMinMoreX = rightInfo == null ? true : (x.val < rightInfo.min); + if (leftMaxLessX && rightMinMoreX) { + int leftSize = leftInfo == null ? 0 : leftInfo.allSize; + int rightSize = rightInfo == null ? 0 : rightInfo.allSize; + p3 = leftSize + rightSize + 1; + } + } + return new Info(Math.max(p1, Math.max(p2, p3)), allSize, max, min); + } + + // 为了验证 + // 对数器方法 + public static int right(TreeNode head) { + if (head == null) { + return 0; + } + int h = getBSTSize(head); + if (h != 0) { + return h; + } + return Math.max(right(head.left), right(head.right)); + } + + // 为了验证 + // 对数器方法 + public static int getBSTSize(TreeNode head) { + if (head == null) { + return 0; + } + ArrayList arr = new ArrayList<>(); + in(head, arr); + for (int i = 1; i < arr.size(); i++) { + if (arr.get(i).val <= arr.get(i - 1).val) { + return 0; + } + } + return arr.size(); + } + + // 为了验证 + // 对数器方法 + public static void in(TreeNode head, ArrayList arr) { + if (head == null) { + return; + } + in(head.left, arr); + arr.add(head); + in(head.right, arr); + } + + // 为了验证 + // 对数器方法 + public static TreeNode generateRandomBST(int maxLevel, int maxValue) { + return generate(1, maxLevel, maxValue); + } + + // 为了验证 + // 对数器方法 + public static TreeNode generate(int level, int maxLevel, int maxValue) { + if (level > maxLevel || Math.random() < 0.5) { + return null; + } + TreeNode head = new TreeNode((int) (Math.random() * maxValue)); + head.left = generate(level + 1, maxLevel, maxValue); + head.right = generate(level + 1, maxLevel, maxValue); + return head; + } + + // 为了验证 + // 对数器方法 + public static void main(String[] args) { + int maxLevel = 4; + int maxValue = 100; + int testTimes = 1000000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + TreeNode head = generateRandomBST(maxLevel, maxValue); + if (largestBSTSubtree(head) != right(head)) { + System.out.println("出错了！"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class12/Code06_MaxDistance.java b/体系学习班/class12/Code06_MaxDistance.java new file mode 100644 index 0000000..8c82c0e --- /dev/null +++ b/体系学习班/class12/Code06_MaxDistance.java @@ -0,0 +1,180 @@ +package class12; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.HashSet; + +public class Code06_MaxDistance { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int data) { + this.value = data; + } + } + + public static int maxDistance1(Node head) { + if (head == null) { + return 0; + } + ArrayList arr = getPrelist(head); + HashMap parentMap = getParentMap(head); + int max = 0; + for (int i = 0; i < arr.size(); i++) { + for (int j = i; j < arr.size(); j++) { + max = Math.max(max, distance(parentMap, arr.get(i), arr.get(j))); + } + } + return max; + } + + public static ArrayList getPrelist(Node head) { + ArrayList arr = new ArrayList<>(); + fillPrelist(head, arr); + return arr; + } + + public static void fillPrelist(Node head, ArrayList arr) { + if (head == null) { + return; + } + arr.add(head); + fillPrelist(head.left, arr); + fillPrelist(head.right, arr); + } + + public static HashMap getParentMap(Node head) { + HashMap map = new HashMap<>(); + map.put(head, null); + fillParentMap(head, map); + return map; + } + + public static void fillParentMap(Node head, HashMap parentMap) { + if (head.left != null) { + parentMap.put(head.left, head); + fillParentMap(head.left, parentMap); + } + if (head.right != null) { + parentMap.put(head.right, head); + fillParentMap(head.right, parentMap); + } + } + + public static int distance(HashMap parentMap, Node o1, Node o2) { + HashSet o1Set = new HashSet<>(); + Node cur = o1; + o1Set.add(cur); + while (parentMap.get(cur) != null) { + cur = parentMap.get(cur); + o1Set.add(cur); + } + cur = o2; + while (!o1Set.contains(cur)) { + cur = parentMap.get(cur); + } + Node lowestAncestor = cur; + cur = o1; + int distance1 = 1; + while (cur != lowestAncestor) { + cur = parentMap.get(cur); + distance1++; + } + cur = o2; + int distance2 = 1; + while (cur != lowestAncestor) { + cur = parentMap.get(cur); + distance2++; + } + return distance1 + distance2 - 1; + } + +// public static int maxDistance2(Node head) { +// return process(head).maxDistance; +// } +// +// public static class Info { +// public int maxDistance; +// public int height; +// +// public Info(int dis, int h) { +// maxDistance = dis; +// height = h; +// } +// } +// +// public static Info process(Node X) { +// if (X == null) { +// return new Info(0, 0); +// } +// Info leftInfo = process(X.left); +// Info rightInfo = process(X.right); +// int height = Math.max(leftInfo.height, rightInfo.height) + 1; +// int maxDistance = Math.max( +// Math.max(leftInfo.maxDistance, rightInfo.maxDistance), +// leftInfo.height + rightInfo.height + 1); +// return new Info(maxDistance, height); +// } + + public static int maxDistance2(Node head) { + return process(head).maxDistance; + } + + public static class Info { + public int maxDistance; + public int height; + + public Info(int m, int h) { + maxDistance = m; + height = h; + } + + } + + public static Info process(Node x) { + if (x == null) { + return new Info(0, 0); + } + Info leftInfo = process(x.left); + Info rightInfo = process(x.right); + int height = Math.max(leftInfo.height, rightInfo.height) + 1; + int p1 = leftInfo.maxDistance; + int p2 = rightInfo.maxDistance; + int p3 = leftInfo.height + rightInfo.height + 1; + int maxDistance = Math.max(Math.max(p1, p2), p3); + return new Info(maxDistance, height); + } + + // for test + public static Node generateRandomBST(int maxLevel, int maxValue) { + return generate(1, maxLevel, maxValue); + } + + // for test + public static Node generate(int level, int maxLevel, int maxValue) { + if (level > maxLevel || Math.random() < 0.5) { + return null; + } + Node head = new Node((int) (Math.random() * maxValue)); + head.left = generate(level + 1, maxLevel, maxValue); + head.right = generate(level + 1, maxLevel, maxValue); + return head; + } + + public static void main(String[] args) { + int maxLevel = 4; + int maxValue = 100; + int testTimes = 1000000; + for (int i = 0; i < testTimes; i++) { + Node head = generateRandomBST(maxLevel, maxValue); + if (maxDistance1(head) != maxDistance2(head)) { + System.out.println("Oops!"); + } + } + System.out.println("finish!"); + } + +} diff --git a/体系学习班/class13/Code01_IsCBT.java b/体系学习班/class13/Code01_IsCBT.java new file mode 100644 index 0000000..c991c8e --- /dev/null +++ b/体系学习班/class13/Code01_IsCBT.java @@ -0,0 +1,120 @@ +package class13; + +import java.util.LinkedList; + +// 测试链接 : https://leetcode.com/problems/check-completeness-of-a-binary-tree/ + +public class Code01_IsCBT { + + // 不要提交这个类 + public static class TreeNode { + public int val; + public TreeNode left; + public TreeNode right; + + public TreeNode(int v) { + val = v; + } + } + + public static boolean isCompleteTree1(TreeNode head) { + if (head == null) { + return true; + } + LinkedList queue = new LinkedList<>(); + // 是否遇到过左右两个孩子不双全的节点 + boolean leaf = false; + TreeNode l = null; + TreeNode r = null; + queue.add(head); + while (!queue.isEmpty()) { + head = queue.poll(); + l = head.left; + r = head.right; + if ( + // 如果遇到了不双全的节点之后，又发现当前节点不是叶节点 + (leaf && (l != null || r != null)) || (l == null && r != null) + + ) { + return false; + } + if (l != null) { + queue.add(l); + } + if (r != null) { + queue.add(r); + } + if (l == null || r == null) { + leaf = true; + } + } + return true; + } + + public static boolean isCompleteTree2(TreeNode head) { + return process(head).isCBT; + } + + public static class Info { + public boolean isFull; + public boolean isCBT; + public int height; + + public Info(boolean full, boolean cbt, int h) { + isFull = full; + isCBT = cbt; + height = h; + } + } + + public static Info process(TreeNode x) { + if (x == null) { + return new Info(true, true, 0); + } + Info leftInfo = process(x.left); + Info rightInfo = process(x.right); + int height = Math.max(leftInfo.height, rightInfo.height) + 1; + boolean isFull = leftInfo.isFull && rightInfo.isFull && leftInfo.height == rightInfo.height; + boolean isCBT = false; + if (leftInfo.isFull && rightInfo.isFull && leftInfo.height == rightInfo.height) { + isCBT = true; + } else if (leftInfo.isCBT && rightInfo.isFull && leftInfo.height == rightInfo.height + 1) { + isCBT = true; + } else if (leftInfo.isFull && rightInfo.isFull && leftInfo.height == rightInfo.height + 1) { + isCBT = true; + } else if (leftInfo.isFull && rightInfo.isCBT && leftInfo.height == rightInfo.height) { + isCBT = true; + } + return new Info(isFull, isCBT, height); + } + + // for test + public static TreeNode generateRandomBST(int maxLevel, int maxValue) { + return generate(1, maxLevel, maxValue); + } + + // for test + public static TreeNode generate(int level, int maxLevel, int maxValue) { + if (level > maxLevel || Math.random() < 0.5) { + return null; + } + TreeNode head = new TreeNode((int) (Math.random() * maxValue)); + head.left = generate(level + 1, maxLevel, maxValue); + head.right = generate(level + 1, maxLevel, maxValue); + return head; + } + + public static void main(String[] args) { + int maxLevel = 5; + int maxValue = 100; + int testTimes = 1000000; + for (int i = 0; i < testTimes; i++) { + TreeNode head = generateRandomBST(maxLevel, maxValue); + if (isCompleteTree1(head) != isCompleteTree2(head)) { + System.out.println("Oops!"); + } + } + System.out.println("finish!"); + } + +} diff --git a/体系学习班/class13/Code02_MaxSubBSTHead.java b/体系学习班/class13/Code02_MaxSubBSTHead.java new file mode 100644 index 0000000..e64c37a --- /dev/null +++ b/体系学习班/class13/Code02_MaxSubBSTHead.java @@ -0,0 +1,136 @@ +package class13; + +import java.util.ArrayList; + +public class Code02_MaxSubBSTHead { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int data) { + this.value = data; + } + } + + public static int getBSTSize(Node head) { + if (head == null) { + return 0; + } + ArrayList arr = new ArrayList<>(); + in(head, arr); + for (int i = 1; i < arr.size(); i++) { + if (arr.get(i).value <= arr.get(i - 1).value) { + return 0; + } + } + return arr.size(); + } + + public static void in(Node head, ArrayList arr) { + if (head == null) { + return; + } + in(head.left, arr); + arr.add(head); + in(head.right, arr); + } + + public static Node maxSubBSTHead1(Node head) { + if (head == null) { + return null; + } + if (getBSTSize(head) != 0) { + return head; + } + Node leftAns = maxSubBSTHead1(head.left); + Node rightAns = maxSubBSTHead1(head.right); + return getBSTSize(leftAns) >= getBSTSize(rightAns) ? leftAns : rightAns; + } + + public static Node maxSubBSTHead2(Node head) { + if (head == null) { + return null; + } + return process(head).maxSubBSTHead; + } + + // 每一棵子树 + public static class Info { + public Node maxSubBSTHead; + public int maxSubBSTSize; + public int min; + public int max; + + public Info(Node h, int size, int mi, int ma) { + maxSubBSTHead = h; + maxSubBSTSize = size; + min = mi; + max = ma; + } + } + + public static Info process(Node X) { + if (X == null) { + return null; + } + Info leftInfo = process(X.left); + Info rightInfo = process(X.right); + int min = X.value; + int max = X.value; + Node maxSubBSTHead = null; + int maxSubBSTSize = 0; + if (leftInfo != null) { + min = Math.min(min, leftInfo.min); + max = Math.max(max, leftInfo.max); + maxSubBSTHead = leftInfo.maxSubBSTHead; + maxSubBSTSize = leftInfo.maxSubBSTSize; + } + if (rightInfo != null) { + min = Math.min(min, rightInfo.min); + max = Math.max(max, rightInfo.max); + if (rightInfo.maxSubBSTSize > maxSubBSTSize) { + maxSubBSTHead = rightInfo.maxSubBSTHead; + maxSubBSTSize = rightInfo.maxSubBSTSize; + } + } + if ((leftInfo == null ? true : (leftInfo.maxSubBSTHead == X.left && leftInfo.max < X.value)) + && (rightInfo == null ? true : (rightInfo.maxSubBSTHead == X.right && rightInfo.min > X.value))) { + maxSubBSTHead = X; + maxSubBSTSize = (leftInfo == null ? 0 : leftInfo.maxSubBSTSize) + + (rightInfo == null ? 0 : rightInfo.maxSubBSTSize) + 1; + } + return new Info(maxSubBSTHead, maxSubBSTSize, min, max); + } + + // for test + public static Node generateRandomBST(int maxLevel, int maxValue) { + return generate(1, maxLevel, maxValue); + } + + // for test + public static Node generate(int level, int maxLevel, int maxValue) { + if (level > maxLevel || Math.random() < 0.5) { + return null; + } + Node head = new Node((int) (Math.random() * maxValue)); + head.left = generate(level + 1, maxLevel, maxValue); + head.right = generate(level + 1, maxLevel, maxValue); + return head; + } + + public static void main(String[] args) { + int maxLevel = 4; + int maxValue = 100; + int testTimes = 1000000; + for (int i = 0; i < testTimes; i++) { + Node head = generateRandomBST(maxLevel, maxValue); + if (maxSubBSTHead1(head) != maxSubBSTHead2(head)) { + System.out.println("Oops!"); + } + } + System.out.println("finish!"); + } + +} diff --git a/体系学习班/class13/Code03_lowestAncestor.java b/体系学习班/class13/Code03_lowestAncestor.java new file mode 100644 index 0000000..e2a9d5f --- /dev/null +++ b/体系学习班/class13/Code03_lowestAncestor.java @@ -0,0 +1,141 @@ +package class13; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.HashSet; + +public class Code03_lowestAncestor { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int data) { + this.value = data; + } + } + + public static Node lowestAncestor1(Node head, Node o1, Node o2) { + if (head == null) { + return null; + } + // key的父节点是value + HashMap parentMap = new HashMap<>(); + parentMap.put(head, null); + fillParentMap(head, parentMap); + HashSet o1Set = new HashSet<>(); + Node cur = o1; + o1Set.add(cur); + while (parentMap.get(cur) != null) { + cur = parentMap.get(cur); + o1Set.add(cur); + } + cur = o2; + while (!o1Set.contains(cur)) { + cur = parentMap.get(cur); + } + return cur; + } + + public static void fillParentMap(Node head, HashMap parentMap) { + if (head.left != null) { + parentMap.put(head.left, head); + fillParentMap(head.left, parentMap); + } + if (head.right != null) { + parentMap.put(head.right, head); + fillParentMap(head.right, parentMap); + } + } + + public static Node lowestAncestor2(Node head, Node a, Node b) { + return process(head, a, b).ans; + } + + public static class Info { + public boolean findA; + public boolean findB; + public Node ans; + + public Info(boolean fA, boolean fB, Node an) { + findA = fA; + findB = fB; + ans = an; + } + } + + public static Info process(Node x, Node a, Node b) { + if (x == null) { + return new Info(false, false, null); + } + Info leftInfo = process(x.left, a, b); + Info rightInfo = process(x.right, a, b); + boolean findA = (x == a) || leftInfo.findA || rightInfo.findA; + boolean findB = (x == b) || leftInfo.findB || rightInfo.findB; + Node ans = null; + if (leftInfo.ans != null) { + ans = leftInfo.ans; + } else if (rightInfo.ans != null) { + ans = rightInfo.ans; + } else { + if (findA && findB) { + ans = x; + } + } + return new Info(findA, findB, ans); + } + + // for test + public static Node generateRandomBST(int maxLevel, int maxValue) { + return generate(1, maxLevel, maxValue); + } + + // for test + public static Node generate(int level, int maxLevel, int maxValue) { + if (level > maxLevel || Math.random() < 0.5) { + return null; + } + Node head = new Node((int) (Math.random() * maxValue)); + head.left = generate(level + 1, maxLevel, maxValue); + head.right = generate(level + 1, maxLevel, maxValue); + return head; + } + + // for test + public static Node pickRandomOne(Node head) { + if (head == null) { + return null; + } + ArrayList arr = new ArrayList<>(); + fillPrelist(head, arr); + int randomIndex = (int) (Math.random() * arr.size()); + return arr.get(randomIndex); + } + + // for test + public static void fillPrelist(Node head, ArrayList arr) { + if (head == null) { + return; + } + arr.add(head); + fillPrelist(head.left, arr); + fillPrelist(head.right, arr); + } + + public static void main(String[] args) { + int maxLevel = 4; + int maxValue = 100; + int testTimes = 1000000; + for (int i = 0; i < testTimes; i++) { + Node head = generateRandomBST(maxLevel, maxValue); + Node o1 = pickRandomOne(head); + Node o2 = pickRandomOne(head); + if (lowestAncestor1(head, o1, o2) != lowestAncestor2(head, o1, o2)) { + System.out.println("Oops!"); + } + } + System.out.println("finish!"); + } + +} diff --git a/体系学习班/class13/Code04_MaxHappy.java b/体系学习班/class13/Code04_MaxHappy.java new file mode 100644 index 0000000..3cb21e7 --- /dev/null +++ b/体系学习班/class13/Code04_MaxHappy.java @@ -0,0 +1,116 @@ +package class13; + +import java.util.ArrayList; +import java.util.List; + +public class Code04_MaxHappy { + + public static class Employee { + public int happy; + public List nexts; + + public Employee(int h) { + happy = h; + nexts = new ArrayList<>(); + } + + } + + public static int maxHappy1(Employee boss) { + if (boss == null) { + return 0; + } + return process1(boss, false); + } + + // 当前来到的节点叫cur， + // up表示cur的上级是否来， + // 该函数含义： + // 如果up为true，表示在cur上级已经确定来，的情况下，cur整棵树能够提供最大的快乐值是多少？ + // 如果up为false，表示在cur上级已经确定不来，的情况下，cur整棵树能够提供最大的快乐值是多少？ + public static int process1(Employee cur, boolean up) { + if (up) { // 如果cur的上级来的话，cur没得选，只能不来 + int ans = 0; + for (Employee next : cur.nexts) { + ans += process1(next, false); + } + return ans; + } else { // 如果cur的上级不来的话，cur可以选，可以来也可以不来 + int p1 = cur.happy; + int p2 = 0; + for (Employee next : cur.nexts) { + p1 += process1(next, true); + p2 += process1(next, false); + } + return Math.max(p1, p2); + } + } + + public static int maxHappy2(Employee head) { + Info allInfo = process(head); + return Math.max(allInfo.no, allInfo.yes); + } + + public static class Info { + public int no; + public int yes; + + public Info(int n, int y) { + no = n; + yes = y; + } + } + + public static Info process(Employee x) { + if (x == null) { + return new Info(0, 0); + } + int no = 0; + int yes = x.happy; + for (Employee next : x.nexts) { + Info nextInfo = process(next); + no += Math.max(nextInfo.no, nextInfo.yes); + yes += nextInfo.no; + + } + return new Info(no, yes); + } + + // for test + public static Employee genarateBoss(int maxLevel, int maxNexts, int maxHappy) { + if (Math.random() < 0.02) { + return null; + } + Employee boss = new Employee((int) (Math.random() * (maxHappy + 1))); + genarateNexts(boss, 1, maxLevel, maxNexts, maxHappy); + return boss; + } + + // for test + public static void genarateNexts(Employee e, int level, int maxLevel, int maxNexts, int maxHappy) { + if (level > maxLevel) { + return; + } + int nextsSize = (int) (Math.random() * (maxNexts + 1)); + for (int i = 0; i < nextsSize; i++) { + Employee next = new Employee((int) (Math.random() * (maxHappy + 1))); + e.nexts.add(next); + genarateNexts(next, level + 1, maxLevel, maxNexts, maxHappy); + } + } + + public static void main(String[] args) { + int maxLevel = 4; + int maxNexts = 7; + int maxHappy = 100; + int testTimes = 100000; + for (int i = 0; i < testTimes; i++) { + Employee boss = genarateBoss(maxLevel, maxNexts, maxHappy); + if (maxHappy1(boss) != maxHappy2(boss)) { + System.out.println("Oops!"); + } + } + System.out.println("finish!"); + } + +} diff --git a/体系学习班/class13/Code05_LowestLexicography.java b/体系学习班/class13/Code05_LowestLexicography.java new file mode 100644 index 0000000..13ceedd --- /dev/null +++ b/体系学习班/class13/Code05_LowestLexicography.java @@ -0,0 +1,116 @@ +package class13; + +import java.util.Arrays; +import java.util.Comparator; +import java.util.TreeSet; + +public class Code05_LowestLexicography { + + public static String lowestString1(String[] strs) { + if (strs == null || strs.length == 0) { + return ""; + } + TreeSet ans = process(strs); + return ans.size() == 0 ? "" : ans.first(); + } + + // strs中所有字符串全排列，返回所有可能的结果 + public static TreeSet process(String[] strs) { + TreeSet ans = new TreeSet<>(); + if (strs.length == 0) { + ans.add(""); + return ans; + } + for (int i = 0; i < strs.length; i++) { + String first = strs[i]; + String[] nexts = removeIndexString(strs, i); + TreeSet next = process(nexts); + for (String cur : next) { + ans.add(first + cur); + } + } + return ans; + } + + // {"abc", "cks", "bct"} + // 0 1 2 + // removeIndexString(arr , 1) -> {"abc", "bct"} + public static String[] removeIndexString(String[] arr, int index) { + int N = arr.length; + String[] ans = new String[N - 1]; + int ansIndex = 0; + for (int i = 0; i < N; i++) { + if (i != index) { + ans[ansIndex++] = arr[i]; + } + } + return ans; + } + + public static class MyComparator implements Comparator { + @Override + public int compare(String a, String b) { + return (a + b).compareTo(b + a); + } + } + + public static String lowestString2(String[] strs) { + if (strs == null || strs.length == 0) { + return ""; + } + Arrays.sort(strs, new MyComparator()); + String res = ""; + for (int i = 0; i < strs.length; i++) { + res += strs[i]; + } + return res; + } + + // for test + public static String generateRandomString(int strLen) { + char[] ans = new char[(int) (Math.random() * strLen) + 1]; + for (int i = 0; i < ans.length; i++) { + int value = (int) (Math.random() * 5); + ans[i] = (Math.random() <= 0.5) ? (char) (65 + value) : (char) (97 + value); + } + return String.valueOf(ans); + } + + // for test + public static String[] generateRandomStringArray(int arrLen, int strLen) { + String[] ans = new String[(int) (Math.random() * arrLen) + 1]; + for (int i = 0; i < ans.length; i++) { + ans[i] = generateRandomString(strLen); + } + return ans; + } + + // for test + public static String[] copyStringArray(String[] arr) { + String[] ans = new String[arr.length]; + for (int i = 0; i < ans.length; i++) { + ans[i] = String.valueOf(arr[i]); + } + return ans; + } + + public static void main(String[] args) { + int arrLen = 6; + int strLen = 5; + int testTimes = 10000; + System.out.println("test begin"); + for (int i = 0; i < testTimes; i++) { + String[] arr1 = generateRandomStringArray(arrLen, strLen); + String[] arr2 = copyStringArray(arr1); + if (!lowestString1(arr1).equals(lowestString2(arr2))) { + for (String str : arr1) { + System.out.print(str + ","); + } + System.out.println(); + System.out.println("Oops!"); + } + } + System.out.println("finish!"); + } + +} diff --git a/体系学习班/class14/Code01_Light.java b/体系学习班/class14/Code01_Light.java new file mode 100644 index 0000000..7cbfde2 --- /dev/null +++ b/体系学习班/class14/Code01_Light.java @@ -0,0 +1,105 @@ +package class14; + +import java.util.HashSet; + +public class Code01_Light { + + public static int minLight1(String road) { + if (road == null || road.length() == 0) { + return 0; + } + return process(road.toCharArray(), 0, new HashSet<>()); + } + + // str[index....]位置，自由选择放灯还是不放灯 + // str[0..index-1]位置呢？已经做完决定了，那些放了灯的位置，存在lights里 + // 要求选出能照亮所有.的方案，并且在这些有效的方案中，返回最少需要几个灯 + public static int process(char[] str, int index, HashSet lights) { + if (index == str.length) { // 结束的时候 + for (int i = 0; i < str.length; i++) { + if (str[i] != 'X') { // 当前位置是点的话 + if (!lights.contains(i - 1) && !lights.contains(i) && !lights.contains(i + 1)) { + return Integer.MAX_VALUE; + } + } + } + return lights.size(); + } else { // str还没结束 + // i X . + int no = process(str, index + 1, lights); + int yes = Integer.MAX_VALUE; + if (str[index] == '.') { + lights.add(index); + yes = process(str, index + 1, lights); + lights.remove(index); + } + return Math.min(no, yes); + } + } + + public static int minLight2(String road) { + char[] str = road.toCharArray(); + int i = 0; + int light = 0; + while (i < str.length) { + if (str[i] == 'X') { + i++; + } else { + light++; + if (i + 1 == str.length) { + break; + } else { // 有i位置 i+ 1 X . + if (str[i + 1] == 'X') { + i = i + 2; + } else { + i = i + 3; + } + } + } + } + return light; + } + + // 更简洁的解法 + // 两个X之间，数一下.的数量，然后除以3，向上取整 + // 把灯数累加 + public static int minLight3(String road) { + char[] str = road.toCharArray(); + int cur = 0; + int light = 0; + for (char c : str) { + if (c == 'X') { + light += (cur + 2) / 3; + cur = 0; + } else { + cur++; + } + } + light += (cur + 2) / 3; + return light; + } + + // for test + public static String randomString(int len) { + char[] res = new char[(int) (Math.random() * len) + 1]; + for (int i = 0; i < res.length; i++) { + res[i] = Math.random() < 0.5 ? 'X' : '.'; + } + return String.valueOf(res); + } + + public static void main(String[] args) { + int len = 20; + int testTime = 100000; + for (int i = 0; i < testTime; i++) { + String test = randomString(len); + int ans1 = minLight1(test); + int ans2 = minLight2(test); + int ans3 = minLight3(test); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("oops!"); + } + } + System.out.println("finish!"); + } +} diff --git a/体系学习班/class14/Code02_LessMoneySplitGold.java b/体系学习班/class14/Code02_LessMoneySplitGold.java new file mode 100644 index 0000000..682fa12 --- /dev/null +++ b/体系学习班/class14/Code02_LessMoneySplitGold.java @@ -0,0 +1,79 @@ +package class14; + +import java.util.PriorityQueue; + +public class Code02_LessMoneySplitGold { + + // 纯暴力！ + public static int lessMoney1(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + return process(arr, 0); + } + + // 等待合并的数都在arr里，pre之前的合并行为产生了多少总代价 + // arr中只剩一个数字的时候，停止合并，返回最小的总代价 + public static int process(int[] arr, int pre) { + if (arr.length == 1) { + return pre; + } + int ans = Integer.MAX_VALUE; + for (int i = 0; i < arr.length; i++) { + for (int j = i + 1; j < arr.length; j++) { + ans = Math.min(ans, process(copyAndMergeTwo(arr, i, j), pre + arr[i] + arr[j])); + } + } + return ans; + } + + public static int[] copyAndMergeTwo(int[] arr, int i, int j) { + int[] ans = new int[arr.length - 1]; + int ansi = 0; + for (int arri = 0; arri < arr.length; arri++) { + if (arri != i && arri != j) { + ans[ansi++] = arr[arri]; + } + } + ans[ansi] = arr[i] + arr[j]; + return ans; + } + + public static int lessMoney2(int[] arr) { + PriorityQueue pQ = new PriorityQueue<>(); + for (int i = 0; i < arr.length; i++) { + pQ.add(arr[i]); + } + int sum = 0; + int cur = 0; + while (pQ.size() > 1) { + cur = pQ.poll() + pQ.poll(); + sum += cur; + pQ.add(cur); + } + return sum; + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) (Math.random() * (maxValue + 1)); + } + return arr; + } + + public static void main(String[] args) { + int testTime = 100000; + int maxSize = 6; + int maxValue = 1000; + for (int i = 0; i < testTime; i++) { + int[] arr = generateRandomArray(maxSize, maxValue); + if (lessMoney1(arr) != lessMoney2(arr)) { + System.out.println("Oops!"); + } + } + System.out.println("finish!"); + } + +} diff --git a/体系学习班/class14/Code03_BestArrange.java b/体系学习班/class14/Code03_BestArrange.java new file mode 100644 index 0000000..c8a79ff --- /dev/null +++ b/体系学习班/class14/Code03_BestArrange.java @@ -0,0 +1,111 @@ +package class14; + +import java.util.Arrays; +import java.util.Comparator; + +public class Code03_BestArrange { + + public static class Program { + public int start; + public int end; + + public Program(int start, int end) { + this.start = start; + this.end = end; + } + } + + // 暴力！所有情况都尝试！ + public static int bestArrange1(Program[] programs) { + if (programs == null || programs.length == 0) { + return 0; + } + return process(programs, 0, 0); + } + + // 还剩下的会议都放在programs里 + // done之前已经安排了多少会议的数量 + // timeLine目前来到的时间点是什么 + + // 目前来到timeLine的时间点，已经安排了done多的会议，剩下的会议programs可以自由安排 + // 返回能安排的最多会议数量 + public static int process(Program[] programs, int done, int timeLine) { + if (programs.length == 0) { + return done; + } + // 还剩下会议 + int max = done; + // 当前安排的会议是什么会，每一个都枚举 + for (int i = 0; i < programs.length; i++) { + if (programs[i].start >= timeLine) { + Program[] next = copyButExcept(programs, i); + max = Math.max(max, process(next, done + 1, programs[i].end)); + } + } + return max; + } + + public static Program[] copyButExcept(Program[] programs, int i) { + Program[] ans = new Program[programs.length - 1]; + int index = 0; + for (int k = 0; k < programs.length; k++) { + if (k != i) { + ans[index++] = programs[k]; + } + } + return ans; + } + + // 会议的开始时间和结束时间，都是数值，不会 < 0 + public static int bestArrange2(Program[] programs) { + Arrays.sort(programs, new ProgramComparator()); + int timeLine = 0; + int result = 0; + // 依次遍历每一个会议，结束时间早的会议先遍历 + for (int i = 0; i < programs.length; i++) { + if (timeLine <= programs[i].start) { + result++; + timeLine = programs[i].end; + } + } + return result; + } + + public static class ProgramComparator implements Comparator { + + @Override + public int compare(Program o1, Program o2) { + return o1.end - o2.end; + } + + } + + // for test + public static Program[] generatePrograms(int programSize, int timeMax) { + Program[] ans = new Program[(int) (Math.random() * (programSize + 1))]; + for (int i = 0; i < ans.length; i++) { + int r1 = (int) (Math.random() * (timeMax + 1)); + int r2 = (int) (Math.random() * (timeMax + 1)); + if (r1 == r2) { + ans[i] = new Program(r1, r1 + 1); + } else { + ans[i] = new Program(Math.min(r1, r2), Math.max(r1, r2)); + } + } + return ans; + } + + public static void main(String[] args) { + int programSize = 12; + int timeMax = 20; + int timeTimes = 1000000; + for (int i = 0; i < timeTimes; i++) { + Program[] programs = generatePrograms(programSize, timeMax); + if (bestArrange1(programs) != bestArrange2(programs)) { + System.out.println("Oops!"); + } + } + System.out.println("finish!"); + } + +} diff --git a/体系学习班/class14/Code04_IPO.java b/体系学习班/class14/Code04_IPO.java new file mode 100644 index 0000000..0ac3b85 --- /dev/null +++ b/体系学习班/class14/Code04_IPO.java @@ -0,0 +1,58 @@ +package class14; + +import java.util.Comparator; +import java.util.PriorityQueue; + +public class Code04_IPO { + + // 最多K个项目 + // W是初始资金 + // Profits[] Capital[] 一定等长 + // 返回最终最大的资金 + public static int findMaximizedCapital(int K, int W, int[] Profits, int[] Capital) { + PriorityQueue minCostQ = new PriorityQueue<>(new MinCostComparator()); + PriorityQueue maxProfitQ = new PriorityQueue<>(new MaxProfitComparator()); + for (int i = 0; i < Profits.length; i++) { + minCostQ.add(new Program(Profits[i], Capital[i])); + } + for (int i = 0; i < K; i++) { + while (!minCostQ.isEmpty() && minCostQ.peek().c <= W) { + maxProfitQ.add(minCostQ.poll()); + } + if (maxProfitQ.isEmpty()) { + return W; + } + W += maxProfitQ.poll().p; + } + return W; + } + + public static class Program { + public int p; + public int c; + + public Program(int p, int c) { + this.p = p; + this.c = c; + } + } + + public static class MinCostComparator implements Comparator { + + @Override + public int compare(Program o1, Program o2) { + return o1.c - o2.c; + } + + } + + public static class MaxProfitComparator implements Comparator { + + @Override + public int compare(Program o1, Program o2) { + return o2.p - o1.p; + } + + } + +} diff --git a/体系学习班/class14/Code05_UnionFind.java b/体系学习班/class14/Code05_UnionFind.java new file mode 100644 index 0000000..0d9e246 --- /dev/null +++ b/体系学习班/class14/Code05_UnionFind.java @@ -0,0 +1,70 @@ +package class14; + +import java.util.HashMap; +import java.util.List; +import java.util.Stack; + +public class Code05_UnionFind { + + public static class Node { + V value; + + public Node(V v) { + value = v; + } + } + + public static class UnionFind { + public HashMap> nodes; + public HashMap, Node> parents; + public HashMap, Integer> sizeMap; + + public UnionFind(List values) { + nodes = new HashMap<>(); + parents = new HashMap<>(); + sizeMap = new HashMap<>(); + for (V cur : values) { + Node node = new Node<>(cur); + nodes.put(cur, node); + parents.put(node, node); + sizeMap.put(node, 1); + } + } + + // 给你一个节点，请你往上到不能再往上，把代表返回 + public Node findFather(Node cur) { + Stack> path = new Stack<>(); + while (cur != parents.get(cur)) { + path.push(cur); + cur = parents.get(cur); + } + while (!path.isEmpty()) { + parents.put(path.pop(), cur); + } + return cur; + } + + public boolean isSameSet(V a, V b) { + return findFather(nodes.get(a)) == findFather(nodes.get(b)); + } + + public void union(V a, V b) { + Node aHead = findFather(nodes.get(a)); + Node bHead = findFather(nodes.get(b)); + if (aHead != bHead) { + int aSetSize = sizeMap.get(aHead); + int bSetSize = sizeMap.get(bHead); + Node big = aSetSize >= bSetSize ? aHead : bHead; + Node small = big == aHead ? bHead : aHead; + parents.put(small, big); + sizeMap.put(big, aSetSize + bSetSize); + sizeMap.remove(small); + } + } + + public int sets() { + return sizeMap.size(); + } + + } +} diff --git a/体系学习班/class15/Code01_FriendCircles.java b/体系学习班/class15/Code01_FriendCircles.java new file mode 100644 index 0000000..dd4253d --- /dev/null +++ b/体系学习班/class15/Code01_FriendCircles.java @@ -0,0 +1,78 @@ +package class15; + +// 本题为leetcode原题 +// 测试链接：https://leetcode.com/problems/friend-circles/ +// 可以直接通过 +public class Code01_FriendCircles { + + public static int findCircleNum(int[][] M) { + int N = M.length; + // {0} {1} {2} {N-1} + UnionFind unionFind = new UnionFind(N); + for (int i = 0; i < N; i++) { + for (int j = i + 1; j < N; j++) { + if (M[i][j] == 1) { // i和j互相认识 + unionFind.union(i, j); + } + } + } + return unionFind.sets(); + } + + public static class UnionFind { + // parent[i] = k ： i的父亲是k + private int[] parent; + // size[i] = k ： 如果i是代表节点，size[i]才有意义，否则无意义 + // i所在的集合大小是多少 + private int[] size; + // 辅助结构 + private int[] help; + // 一共有多少个集合 + private int sets; + + public UnionFind(int N) { + parent = new int[N]; + size = new int[N]; + help = new int[N]; + sets = N; + for (int i = 0; i < N; i++) { + parent[i] = i; + size[i] = 1; + } + } + + // 从i开始一直往上，往上到不能再往上，代表节点，返回 + // 这个过程要做路径压缩 + private int find(int i) { + int hi = 0; + while (i != parent[i]) { + help[hi++] = i; + i = parent[i]; + } + for (hi--; hi >= 0; hi--) { + parent[help[hi]] = i; + } + return i; + } + + public void union(int i, int j) { + int f1 = find(i); + int f2 = find(j); + if (f1 != f2) { + if (size[f1] >= size[f2]) { + size[f1] += size[f2]; + parent[f2] = f1; + } else { + size[f2] += size[f1]; + parent[f1] = f2; + } + sets--; + } + } + + public int sets() { + return sets; + } + } + +} diff --git a/体系学习班/class15/Code02_NumberOfIslands.java b/体系学习班/class15/Code02_NumberOfIslands.java new file mode 100644 index 0000000..6065d42 --- /dev/null +++ b/体系学习班/class15/Code02_NumberOfIslands.java @@ -0,0 +1,317 @@ +package class15; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.List; +import java.util.Stack; + +// 本题为leetcode原题 +// 测试链接：https://leetcode.com/problems/number-of-islands/ +// 所有方法都可以直接通过 +public class Code02_NumberOfIslands { + + public static int numIslands3(char[][] board) { + int islands = 0; + for (int i = 0; i < board.length; i++) { + for (int j = 0; j < board[0].length; j++) { + if (board[i][j] == '1') { + islands++; + infect(board, i, j); + } + } + } + return islands; + } + + // 从(i,j)这个位置出发，把所有练成一片的'1'字符，变成0 + public static void infect(char[][] board, int i, int j) { + if (i < 0 || i == board.length || j < 0 || j == board[0].length || board[i][j] != '1') { + return; + } + board[i][j] = 0; + infect(board, i - 1, j); + infect(board, i + 1, j); + infect(board, i, j - 1); + infect(board, i, j + 1); + } + + public static int numIslands1(char[][] board) { + int row = board.length; + int col = board[0].length; + Dot[][] dots = new Dot[row][col]; + List dotList = new ArrayList<>(); + for (int i = 0; i < row; i++) { + for (int j = 0; j < col; j++) { + if (board[i][j] == '1') { + dots[i][j] = new Dot(); + dotList.add(dots[i][j]); + } + } + } + UnionFind1 uf = new UnionFind1<>(dotList); + for (int j = 1; j < col; j++) { + // (0,j) (0,0)跳过了 (0,1) (0,2) (0,3) + if (board[0][j - 1] == '1' && board[0][j] == '1') { + uf.union(dots[0][j - 1], dots[0][j]); + } + } + for (int i = 1; i < row; i++) { + if (board[i - 1][0] == '1' && board[i][0] == '1') { + uf.union(dots[i - 1][0], dots[i][0]); + } + } + for (int i = 1; i < row; i++) { + for (int j = 1; j < col; j++) { + if (board[i][j] == '1') { + if (board[i][j - 1] == '1') { + uf.union(dots[i][j - 1], dots[i][j]); + } + if (board[i - 1][j] == '1') { + uf.union(dots[i - 1][j], dots[i][j]); + } + } + } + } + return uf.sets(); + } + + public static class Dot { + + } + + public static class Node { + + V value; + + public Node(V v) { + value = v; + } + + } + + public static class UnionFind1 { + public HashMap> nodes; + public HashMap, Node> parents; + public HashMap, Integer> sizeMap; + + public UnionFind1(List values) { + nodes = new HashMap<>(); + parents = new HashMap<>(); + sizeMap = new HashMap<>(); + for (V cur : values) { + Node node = new Node<>(cur); + nodes.put(cur, node); + parents.put(node, node); + sizeMap.put(node, 1); + } + } + + public Node findFather(Node cur) { + Stack> path = new Stack<>(); + while (cur != parents.get(cur)) { + path.push(cur); + cur = parents.get(cur); + } + while (!path.isEmpty()) { + parents.put(path.pop(), cur); + } + return cur; + } + + public void union(V a, V b) { + Node aHead = findFather(nodes.get(a)); + Node bHead = findFather(nodes.get(b)); + if (aHead != bHead) { + int aSetSize = sizeMap.get(aHead); + int bSetSize = sizeMap.get(bHead); + Node big = aSetSize >= bSetSize ? aHead : bHead; + Node small = big == aHead ? bHead : aHead; + parents.put(small, big); + sizeMap.put(big, aSetSize + bSetSize); + sizeMap.remove(small); + } + } + + public int sets() { + return sizeMap.size(); + } + + } + + public static int numIslands2(char[][] board) { + int row = board.length; + int col = board[0].length; + UnionFind2 uf = new UnionFind2(board); + for (int j = 1; j < col; j++) { + if (board[0][j - 1] == '1' && board[0][j] == '1') { + uf.union(0, j - 1, 0, j); + } + } + for (int i = 1; i < row; i++) { + if (board[i - 1][0] == '1' && board[i][0] == '1') { + uf.union(i - 1, 0, i, 0); + } + } + for (int i = 1; i < row; i++) { + for (int j = 1; j < col; j++) { + if (board[i][j] == '1') { + if (board[i][j - 1] == '1') { + uf.union(i, j - 1, i, j); + } + if (board[i - 1][j] == '1') { + uf.union(i - 1, j, i, j); + } + } + } + } + return uf.sets(); + } + + public static class UnionFind2 { + private int[] parent; + private int[] size; + private int[] help; + private int col; + private int sets; + + public UnionFind2(char[][] board) { + col = board[0].length; + sets = 0; + int row = board.length; + int len = row * col; + parent = new int[len]; + size = new int[len]; + help = new int[len]; + for (int r = 0; r < row; r++) { + for (int c = 0; c < col; c++) { + if (board[r][c] == '1') { + int i = index(r, c); + parent[i] = i; + size[i] = 1; + sets++; + } + } + } + } + + // (r,c) -> i + private int index(int r, int c) { + return r * col + c; + } + + // 原始位置 -> 下标 + private int find(int i) { + int hi = 0; + while (i != parent[i]) { + help[hi++] = i; + i = parent[i]; + } + for (hi--; hi >= 0; hi--) { + parent[help[hi]] = i; + } + return i; + } + + public void union(int r1, int c1, int r2, int c2) { + int i1 = index(r1, c1); + int i2 = index(r2, c2); + int f1 = find(i1); + int f2 = find(i2); + if (f1 != f2) { + if (size[f1] >= size[f2]) { + size[f1] += size[f2]; + parent[f2] = f1; + } else { + size[f2] += size[f1]; + parent[f1] = f2; + } + sets--; + } + } + + public int sets() { + return sets; + } + + } + + // 为了测试 + public static char[][] generateRandomMatrix(int row, int col) { + char[][] board = new char[row][col]; + for (int i = 0; i < row; i++) { + for (int j = 0; j < col; j++) { + board[i][j] = Math.random() < 0.5 ? '1' : '0'; + } + } + return board; + } + + // 为了测试 + public static char[][] copy(char[][] board) { + int row = board.length; + int col = board[0].length; + char[][] ans = new char[row][col]; + for (int i = 0; i < row; i++) { + for (int j = 0; j < col; j++) { + ans[i][j] = board[i][j]; + } + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int row = 0; + int col = 0; + char[][] board1 = null; + char[][] board2 = null; + char[][] board3 = null; + long start = 0; + long end = 0; + + row = 1000; + col = 1000; + board1 = generateRandomMatrix(row, col); + board2 = copy(board1); + board3 = copy(board1); + + System.out.println("感染方法、并查集(map实现)、并查集(数组实现)的运行结果和运行时间"); + System.out.println("随机生成的二维矩阵规模 : " + row + " * " + col); + + start = System.currentTimeMillis(); + System.out.println("感染方法的运行结果: " + numIslands3(board1)); + end = System.currentTimeMillis(); + System.out.println("感染方法的运行时间: " + (end - start) + " ms"); + + start = System.currentTimeMillis(); + System.out.println("并查集(map实现)的运行结果: " + numIslands1(board2)); + end = System.currentTimeMillis(); + System.out.println("并查集(map实现)的运行时间: " + (end - start) + " ms"); + + start = System.currentTimeMillis(); + System.out.println("并查集(数组实现)的运行结果: " + numIslands2(board3)); + end = System.currentTimeMillis(); + System.out.println("并查集(数组实现)的运行时间: " + (end - start) + " ms"); + + System.out.println(); + + row = 10000; + col = 10000; + board1 = generateRandomMatrix(row, col); + board3 = copy(board1); + System.out.println("感染方法、并查集(数组实现)的运行结果和运行时间"); + System.out.println("随机生成的二维矩阵规模 : " + row + " * " + col); + + start = System.currentTimeMillis(); + System.out.println("感染方法的运行结果: " + numIslands3(board1)); + end = System.currentTimeMillis(); + System.out.println("感染方法的运行时间: " + (end - start) + " ms"); + + start = System.currentTimeMillis(); + System.out.println("并查集(数组实现)的运行结果: " + numIslands2(board3)); + end = System.currentTimeMillis(); + System.out.println("并查集(数组实现)的运行时间: " + (end - start) + " ms"); + + } + +} diff --git a/体系学习班/class15/Code03_NumberOfIslandsII.java b/体系学习班/class15/Code03_NumberOfIslandsII.java new file mode 100644 index 0000000..d109276 --- /dev/null +++ b/体系学习班/class15/Code03_NumberOfIslandsII.java @@ -0,0 +1,165 @@ +package class15; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.List; + +// 本题为leetcode原题 +// 测试链接：https://leetcode.com/problems/number-of-islands-ii/ +// 所有方法都可以直接通过 +public class Code03_NumberOfIslandsII { + + public static List numIslands21(int m, int n, int[][] positions) { + UnionFind1 uf = new UnionFind1(m, n); + List ans = new ArrayList<>(); + for (int[] position : positions) { + ans.add(uf.connect(position[0], position[1])); + } + return ans; + } + + public static class UnionFind1 { + private int[] parent; + private int[] size; + private int[] help; + private final int row; + private final int col; + private int sets; + + public UnionFind1(int m, int n) { + row = m; + col = n; + sets = 0; + int len = row * col; + parent = new int[len]; + size = new int[len]; + help = new int[len]; + } + + private int index(int r, int c) { + return r * col + c; + } + + private int find(int i) { + int hi = 0; + while (i != parent[i]) { + help[hi++] = i; + i = parent[i]; + } + for (hi--; hi >= 0; hi--) { + parent[help[hi]] = i; + } + return i; + } + + private void union(int r1, int c1, int r2, int c2) { + if (r1 < 0 || r1 == row || r2 < 0 || r2 == row || c1 < 0 || c1 == col || c2 < 0 || c2 == col) { + return; + } + int i1 = index(r1, c1); + int i2 = index(r2, c2); + if (size[i1] == 0 || size[i2] == 0) { + return; + } + int f1 = find(i1); + int f2 = find(i2); + if (f1 != f2) { + if (size[f1] >= size[f2]) { + size[f1] += size[f2]; + parent[f2] = f1; + } else { + size[f2] += size[f1]; + parent[f1] = f2; + } + sets--; + } + } + + public int connect(int r, int c) { + int index = index(r, c); + if (size[index] == 0) { + parent[index] = index; + size[index] = 1; + sets++; + union(r - 1, c, r, c); + union(r + 1, c, r, c); + union(r, c - 1, r, c); + union(r, c + 1, r, c); + } + return sets; + } + + } + + // 课上讲的如果m*n比较大，会经历很重的初始化，而k比较小，怎么优化的方法 + public static List numIslands22(int m, int n, int[][] positions) { + UnionFind2 uf = new UnionFind2(); + List ans = new ArrayList<>(); + for (int[] position : positions) { + ans.add(uf.connect(position[0], position[1])); + } + return ans; + } + + public static class UnionFind2 { + private HashMap parent; + private HashMap size; + private ArrayList help; + private int sets; + + public UnionFind2() { + parent = new HashMap<>(); + size = new HashMap<>(); + help = new ArrayList<>(); + sets = 0; + } + + private String find(String cur) { + while (!cur.equals(parent.get(cur))) { + help.add(cur); + cur = parent.get(cur); + } + for (String str : help) { + parent.put(str, cur); + } + help.clear(); + return cur; + } + + private void union(String s1, String s2) { + if (parent.containsKey(s1) && parent.containsKey(s2)) { + String f1 = find(s1); + String f2 = find(s2); + if (!f1.equals(f2)) { + int size1 = size.get(f1); + int size2 = size.get(f2); + String big = size1 >= size2 ? f1 : f2; + String small = big == f1 ? f2 : f1; + parent.put(small, big); + size.put(big, size1 + size2); + sets--; + } + } + } + + public int connect(int r, int c) { + String key = String.valueOf(r) + "_" + String.valueOf(c); + if (!parent.containsKey(key)) { + parent.put(key, key); + size.put(key, 1); + sets++; + String up = String.valueOf(r - 1) + "_" + String.valueOf(c); + String down = String.valueOf(r + 1) + "_" + String.valueOf(c); + String left = String.valueOf(r) + "_" + String.valueOf(c - 1); + String right = String.valueOf(r) + "_" + String.valueOf(c + 1); + union(up, key); + union(down, key); + union(left, key); + union(right, key); + } + return sets; + } + + } + +} diff --git a/体系学习班/class16/Code01_BFS.java b/体系学习班/class16/Code01_BFS.java new file mode 100644 index 0000000..c2d4d9f --- /dev/null +++ b/体系学习班/class16/Code01_BFS.java @@ -0,0 +1,30 @@ +package class16; + +import java.util.HashSet; +import java.util.LinkedList; +import java.util.Queue; + +public class Code01_BFS { + + // 从node出发，进行宽度优先遍历 + public static void bfs(Node start) { + if (start == null) { + return; + } + Queue queue = new LinkedList<>(); + HashSet set = new HashSet<>(); + queue.add(start); + set.add(start); + while (!queue.isEmpty()) { + Node cur = queue.poll(); + System.out.println(cur.value); + for (Node next : cur.nexts) { + if (!set.contains(next)) { + set.add(next); + queue.add(next); + } + } + } + } + +} diff --git a/体系学习班/class16/Code02_DFS.java b/体系学习班/class16/Code02_DFS.java new file mode 100644 index 0000000..fa85ca4 --- /dev/null +++ b/体系学习班/class16/Code02_DFS.java @@ -0,0 +1,34 @@ +package class16; + +import java.util.HashSet; +import java.util.Stack; + +public class Code02_DFS { + + public static void dfs(Node node) { + if (node == null) { + return; + } + Stack stack = new Stack<>(); + HashSet set = new HashSet<>(); + stack.add(node); + set.add(node); + System.out.println(node.value); + while (!stack.isEmpty()) { + Node cur = stack.pop(); + for (Node next : cur.nexts) { + if (!set.contains(next)) { + stack.push(cur); + stack.push(next); + set.add(next); + System.out.println(next.value); + break; + } + } + } + } + + + + +} diff --git a/体系学习班/class16/Code03_TopologicalOrderBFS.java b/体系学习班/class16/Code03_TopologicalOrderBFS.java new file mode 100644 index 0000000..5e54581 --- /dev/null +++ b/体系学习班/class16/Code03_TopologicalOrderBFS.java @@ -0,0 +1,53 @@ +package class16; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.LinkedList; +import java.util.Queue; + +// OJ链接：https://www.lintcode.com/problem/topological-sorting +public class Code03_TopologicalOrderBFS { + + // 不要提交这个类 + public static class DirectedGraphNode { + public int label; + public ArrayList neighbors; + + public DirectedGraphNode(int x) { + label = x; + neighbors = new ArrayList(); + } + } + + // 提交下面的 + public static ArrayList topSort(ArrayList graph) { + HashMap indegreeMap = new HashMap<>(); + for (DirectedGraphNode cur : graph) { + indegreeMap.put(cur, 0); + } + for (DirectedGraphNode cur : graph) { + for (DirectedGraphNode next : cur.neighbors) { + indegreeMap.put(next, indegreeMap.get(next) + 1); + } + } + Queue zeroQueue = new LinkedList<>(); + for (DirectedGraphNode cur : indegreeMap.keySet()) { + if (indegreeMap.get(cur) == 0) { + zeroQueue.add(cur); + } + } + ArrayList ans = new ArrayList<>(); + while (!zeroQueue.isEmpty()) { + DirectedGraphNode cur = zeroQueue.poll(); + ans.add(cur); + for (DirectedGraphNode next : cur.neighbors) { + indegreeMap.put(next, indegreeMap.get(next) - 1); + if (indegreeMap.get(next) == 0) { + zeroQueue.offer(next); + } + } + } + return ans; + } + +} diff --git a/体系学习班/class16/Code03_TopologicalOrderDFS1.java b/体系学习班/class16/Code03_TopologicalOrderDFS1.java new file mode 100644 index 0000000..57f1bb7 --- /dev/null +++ b/体系学习班/class16/Code03_TopologicalOrderDFS1.java @@ -0,0 +1,70 @@ +package class16; + +import java.util.ArrayList; +import java.util.Comparator; +import java.util.HashMap; + +// OJ链接：https://www.lintcode.com/problem/topological-sorting +public class Code03_TopologicalOrderDFS1 { + + // 不要提交这个类 + public static class DirectedGraphNode { + public int label; + public ArrayList neighbors; + + public DirectedGraphNode(int x) { + label = x; + neighbors = new ArrayList(); + } + } + + // 提交下面的 + public static class Record { + public DirectedGraphNode node; + public int deep; + + public Record(DirectedGraphNode n, int o) { + node = n; + deep = o; + } + } + + public static class MyComparator implements Comparator { + + @Override + public int compare(Record o1, Record o2) { + return o2.deep - o1.deep; + } + } + + public static ArrayList topSort(ArrayList graph) { + HashMap order = new HashMap<>(); + for (DirectedGraphNode cur : graph) { + f(cur, order); + } + ArrayList recordArr = new ArrayList<>(); + for (Record r : order.values()) { + recordArr.add(r); + } + recordArr.sort(new MyComparator()); + ArrayList ans = new ArrayList(); + for (Record r : recordArr) { + ans.add(r.node); + } + return ans; + } + + public static Record f(DirectedGraphNode cur, HashMap order) { + if (order.containsKey(cur)) { + return order.get(cur); + } + int follow = 0; + for (DirectedGraphNode next : cur.neighbors) { + follow = Math.max(follow, f(next, order).deep); + } + Record ans = new Record(cur, follow + 1); + order.put(cur, ans); + return ans; + } + +} diff --git a/体系学习班/class16/Code03_TopologicalOrderDFS2.java b/体系学习班/class16/Code03_TopologicalOrderDFS2.java new file mode 100644 index 0000000..4c90d1d --- /dev/null +++ b/体系学习班/class16/Code03_TopologicalOrderDFS2.java @@ -0,0 +1,76 @@ +package class16; + +import java.util.ArrayList; +import java.util.Comparator; +import java.util.HashMap; + +// OJ链接：https://www.lintcode.com/problem/topological-sorting +public class Code03_TopologicalOrderDFS2 { + + // 不要提交这个类 + public static class DirectedGraphNode { + public int label; + public ArrayList neighbors; + + public DirectedGraphNode(int x) { + label = x; + neighbors = new ArrayList(); + } + } + + // 提交下面的 + public static class Record { + public DirectedGraphNode node; + public long nodes; + + public Record(DirectedGraphNode n, long o) { + node = n; + nodes = o; + } + } + + public static class MyComparator implements Comparator { + + @Override + public int compare(Record o1, Record o2) { + return o1.nodes == o2.nodes ? 0 : (o1.nodes > o2.nodes ? -1 : 1); + } + } + + public static ArrayList topSort(ArrayList graph) { + HashMap order = new HashMap<>(); + for (DirectedGraphNode cur : graph) { + f(cur, order); + } + ArrayList recordArr = new ArrayList<>(); + for (Record r : order.values()) { + recordArr.add(r); + } + recordArr.sort(new MyComparator()); + ArrayList ans = new ArrayList(); + for (Record r : recordArr) { + ans.add(r.node); + } + return ans; + } + + // 当前来到cur点，请返回cur点所到之处，所有的点次！ + // 返回（cur，点次） + // 缓存！！！！！order + // key : 某一个点的点次，之前算过了！ + // value : 点次是多少 + public static Record f(DirectedGraphNode cur, HashMap order) { + if (order.containsKey(cur)) { + return order.get(cur); + } + // cur的点次之前没算过！ + long nodes = 0; + for (DirectedGraphNode next : cur.neighbors) { + nodes += f(next, order).nodes; + } + Record ans = new Record(cur, nodes + 1); + order.put(cur, ans); + return ans; + } + +} diff --git a/体系学习班/class16/Code03_TopologySort.java b/体系学习班/class16/Code03_TopologySort.java new file mode 100644 index 0000000..735dca0 --- /dev/null +++ b/体系学习班/class16/Code03_TopologySort.java @@ -0,0 +1,36 @@ +package class16; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.LinkedList; +import java.util.List; +import java.util.Queue; + +public class Code03_TopologySort { + + // directed graph and no loop + public static List sortedTopology(Graph graph) { + // key 某个节点 value 剩余的入度 + HashMap inMap = new HashMap<>(); + // 只有剩余入度为0的点，才进入这个队列 + Queue zeroInQueue = new LinkedList<>(); + for (Node node : graph.nodes.values()) { + inMap.put(node, node.in); + if (node.in == 0) { + zeroInQueue.add(node); + } + } + List result = new ArrayList<>(); + while (!zeroInQueue.isEmpty()) { + Node cur = zeroInQueue.poll(); + result.add(cur); + for (Node next : cur.nexts) { + inMap.put(next, inMap.get(next) - 1); + if (inMap.get(next) == 0) { + zeroInQueue.add(next); + } + } + } + return result; + } +} diff --git a/体系学习班/class16/Code04_Kruskal.java b/体系学习班/class16/Code04_Kruskal.java new file mode 100644 index 0000000..cb0ea62 --- /dev/null +++ b/体系学习班/class16/Code04_Kruskal.java @@ -0,0 +1,101 @@ +package class16; + +import java.util.Collection; +import java.util.Comparator; +import java.util.HashMap; +import java.util.HashSet; +import java.util.PriorityQueue; +import java.util.Set; +import java.util.Stack; + +//undirected graph only +public class Code04_Kruskal { + + // Union-Find Set + public static class UnionFind { + // key 某一个节点， value key节点往上的节点 + private HashMap fatherMap; + // key 某一个集合的代表节点, value key所在集合的节点个数 + private HashMap sizeMap; + + public UnionFind() { + fatherMap = new HashMap(); + sizeMap = new HashMap(); + } + + public void makeSets(Collection nodes) { + fatherMap.clear(); + sizeMap.clear(); + for (Node node : nodes) { + fatherMap.put(node, node); + sizeMap.put(node, 1); + } + } + + private Node findFather(Node n) { + Stack path = new Stack<>(); + while(n != fatherMap.get(n)) { + path.add(n); + n = fatherMap.get(n); + } + while(!path.isEmpty()) { + fatherMap.put(path.pop(), n); + } + return n; + } + + public boolean isSameSet(Node a, Node b) { + return findFather(a) == findFather(b); + } + + public void union(Node a, Node b) { + if (a == null || b == null) { + return; + } + Node aDai = findFather(a); + Node bDai = findFather(b); + if (aDai != bDai) { + int aSetSize = sizeMap.get(aDai); + int bSetSize = sizeMap.get(bDai); + if (aSetSize <= bSetSize) { + fatherMap.put(aDai, bDai); + sizeMap.put(bDai, aSetSize + bSetSize); + sizeMap.remove(aDai); + } else { + fatherMap.put(bDai, aDai); + sizeMap.put(aDai, aSetSize + bSetSize); + sizeMap.remove(bDai); + } + } + } + } + + + public static class EdgeComparator implements Comparator { + + @Override + public int compare(Edge o1, Edge o2) { + return o1.weight - o2.weight; + } + + } + + public static Set kruskalMST(Graph graph) { + UnionFind unionFind = new UnionFind(); + unionFind.makeSets(graph.nodes.values()); + // 从小的边到大的边，依次弹出，小根堆！ + PriorityQueue priorityQueue = new PriorityQueue<>(new EdgeComparator()); + for (Edge edge : graph.edges) { // M 条边 + priorityQueue.add(edge); // O(logM) + } + Set result = new HashSet<>(); + while (!priorityQueue.isEmpty()) { // M 条边 + Edge edge = priorityQueue.poll(); // O(logM) + if (!unionFind.isSameSet(edge.from, edge.to)) { // O(1) + result.add(edge); + unionFind.union(edge.from, edge.to); + } + } + return result; + } +} diff --git a/体系学习班/class16/Code05_Prim.java b/体系学习班/class16/Code05_Prim.java new file mode 100644 index 0000000..d00dc75 --- /dev/null +++ b/体系学习班/class16/Code05_Prim.java @@ -0,0 +1,94 @@ +package class16; + +import java.util.Comparator; +import java.util.HashSet; +import java.util.PriorityQueue; +import java.util.Set; + +// undirected graph only +public class Code05_Prim { + + public static class EdgeComparator implements Comparator { + + @Override + public int compare(Edge o1, Edge o2) { + return o1.weight - o2.weight; + } + + } + + public static Set primMST(Graph graph) { + // 解锁的边进入小根堆 + PriorityQueue priorityQueue = new PriorityQueue<>(new EdgeComparator()); + + // 哪些点被解锁出来了 + HashSet nodeSet = new HashSet<>(); + + + + Set result = new HashSet<>(); // 依次挑选的的边在result里 + + for (Node node : graph.nodes.values()) { // 随便挑了一个点 + // node 是开始点 + if (!nodeSet.contains(node)) { + nodeSet.add(node); + for (Edge edge : node.edges) { // 由一个点，解锁所有相连的边 + priorityQueue.add(edge); + } + while (!priorityQueue.isEmpty()) { + Edge edge = priorityQueue.poll(); // 弹出解锁的边中，最小的边 + Node toNode = edge.to; // 可能的一个新的点 + if (!nodeSet.contains(toNode)) { // 不含有的时候，就是新的点 + nodeSet.add(toNode); + result.add(edge); + for (Edge nextEdge : toNode.edges) { + priorityQueue.add(nextEdge); + } + } + } + } + // break; + } + return result; + } + + // 请保证graph是连通图 + // graph[i][j]表示点i到点j的距离，如果是系统最大值代表无路 + // 返回值是最小连通图的路径之和 + public static int prim(int[][] graph) { + int size = graph.length; + int[] distances = new int[size]; + boolean[] visit = new boolean[size]; + visit[0] = true; + for (int i = 0; i < size; i++) { + distances[i] = graph[0][i]; + } + int sum = 0; + for (int i = 1; i < size; i++) { + int minPath = Integer.MAX_VALUE; + int minIndex = -1; + for (int j = 0; j < size; j++) { + if (!visit[j] && distances[j] < minPath) { + minPath = distances[j]; + minIndex = j; + } + } + if (minIndex == -1) { + return sum; + } + visit[minIndex] = true; + sum += minPath; + for (int j = 0; j < size; j++) { + if (!visit[j] && distances[j] > graph[minIndex][j]) { + distances[j] = graph[minIndex][j]; + } + } + } + return sum; + } + + public static void main(String[] args) { + System.out.println("hello world!"); + } + +} diff --git a/体系学习班/class16/Code06_Dijkstra.java b/体系学习班/class16/Code06_Dijkstra.java new file mode 100644 index 0000000..2b819c6 --- /dev/null +++ b/体系学习班/class16/Code06_Dijkstra.java @@ -0,0 +1,160 @@ +package class16; + +import java.util.HashMap; +import java.util.HashSet; +import java.util.Map.Entry; + +// no negative weight +public class Code06_Dijkstra { + + public static HashMap dijkstra1(Node from) { + HashMap distanceMap = new HashMap<>(); + distanceMap.put(from, 0); + // 打过对号的点 + HashSet selectedNodes = new HashSet<>(); + Node minNode = getMinDistanceAndUnselectedNode(distanceMap, selectedNodes); + while (minNode != null) { + // 原始点 -> minNode(跳转点) 最小距离distance + int distance = distanceMap.get(minNode); + for (Edge edge : minNode.edges) { + Node toNode = edge.to; + if (!distanceMap.containsKey(toNode)) { + distanceMap.put(toNode, distance + edge.weight); + } else { // toNode + distanceMap.put(edge.to, Math.min(distanceMap.get(toNode), distance + edge.weight)); + } + } + selectedNodes.add(minNode); + minNode = getMinDistanceAndUnselectedNode(distanceMap, selectedNodes); + } + return distanceMap; + } + + public static Node getMinDistanceAndUnselectedNode(HashMap distanceMap, HashSet touchedNodes) { + Node minNode = null; + int minDistance = Integer.MAX_VALUE; + for (Entry entry : distanceMap.entrySet()) { + Node node = entry.getKey(); + int distance = entry.getValue(); + if (!touchedNodes.contains(node) && distance < minDistance) { + minNode = node; + minDistance = distance; + } + } + return minNode; + } + + public static class NodeRecord { + public Node node; + public int distance; + + public NodeRecord(Node node, int distance) { + this.node = node; + this.distance = distance; + } + } + + public static class NodeHeap { + private Node[] nodes; // 实际的堆结构 + // key 某一个node， value 上面堆中的位置 + private HashMap heapIndexMap; + // key 某一个节点， value 从源节点出发到该节点的目前最小距离 + private HashMap distanceMap; + private int size; // 堆上有多少个点 + + public NodeHeap(int size) { + nodes = new Node[size]; + heapIndexMap = new HashMap<>(); + distanceMap = new HashMap<>(); + size = 0; + } + + public boolean isEmpty() { + return size == 0; + } + + // 有一个点叫node，现在发现了一个从源节点出发到达node的距离为distance + // 判断要不要更新，如果需要的话，就更新 + public void addOrUpdateOrIgnore(Node node, int distance) { + if (inHeap(node)) { + distanceMap.put(node, Math.min(distanceMap.get(node), distance)); + insertHeapify(heapIndexMap.get(node)); + } + if (!isEntered(node)) { + nodes[size] = node; + heapIndexMap.put(node, size); + distanceMap.put(node, distance); + insertHeapify(size++); + } + } + + public NodeRecord pop() { + NodeRecord nodeRecord = new NodeRecord(nodes[0], distanceMap.get(nodes[0])); + swap(0, size - 1); + heapIndexMap.put(nodes[size - 1], -1); + distanceMap.remove(nodes[size - 1]); + // free C++同学还要把原本堆顶节点析构，对java同学不必 + nodes[size - 1] = null; + heapify(0, --size); + return nodeRecord; + } + + private void insertHeapify(int index) { + while (distanceMap.get(nodes[index]) < distanceMap.get(nodes[(index - 1) / 2])) { + swap(index, (index - 1) / 2); + index = (index - 1) / 2; + } + } + + private void heapify(int index, int size) { + int left = index * 2 + 1; + while (left < size) { + int smallest = left + 1 < size && distanceMap.get(nodes[left + 1]) < distanceMap.get(nodes[left]) + ? left + 1 + : left; + smallest = distanceMap.get(nodes[smallest]) < distanceMap.get(nodes[index]) ? smallest : index; + if (smallest == index) { + break; + } + swap(smallest, index); + index = smallest; + left = index * 2 + 1; + } + } + + private boolean isEntered(Node node) { + return heapIndexMap.containsKey(node); + } + + private boolean inHeap(Node node) { + return isEntered(node) && heapIndexMap.get(node) != -1; + } + + private void swap(int index1, int index2) { + heapIndexMap.put(nodes[index1], index2); + heapIndexMap.put(nodes[index2], index1); + Node tmp = nodes[index1]; + nodes[index1] = nodes[index2]; + nodes[index2] = tmp; + } + } + + // 改进后的dijkstra算法 + // 从head出发，所有head能到达的节点，生成到达每个节点的最小路径记录并返回 + public static HashMap dijkstra2(Node head, int size) { + NodeHeap nodeHeap = new NodeHeap(size); + nodeHeap.addOrUpdateOrIgnore(head, 0); + HashMap result = new HashMap<>(); + while (!nodeHeap.isEmpty()) { + NodeRecord record = nodeHeap.pop(); + Node cur = record.node; + int distance = record.distance; + for (Edge edge : cur.edges) { + nodeHeap.addOrUpdateOrIgnore(edge.to, edge.weight + distance); + } + result.put(cur, distance); + } + return result; + } + +} diff --git a/体系学习班/class16/Code06_NetworkDelayTime.java b/体系学习班/class16/Code06_NetworkDelayTime.java new file mode 100644 index 0000000..0be520a --- /dev/null +++ b/体系学习班/class16/Code06_NetworkDelayTime.java @@ -0,0 +1,151 @@ +package class16; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.PriorityQueue; + +// 课上没有讲这个题，这是我给同学们找的练习题 +// leetcode 743题，可以用这道题来练习Dijkstra算法 +// 测试链接 : https://leetcode.com/problems/network-delay-time +public class Code06_NetworkDelayTime { + + // 方法一 : 普通堆 + 屏蔽已经计算过的点 + public static int networkDelayTime1(int[][] times, int n, int k) { + ArrayList> nexts = new ArrayList<>(); + for (int i = 0; i <= n; i++) { + nexts.add(new ArrayList<>()); + } + for (int[] delay : times) { + nexts.get(delay[0]).add(new int[] { delay[1], delay[2] }); + } + PriorityQueue heap = new PriorityQueue<>((a, b) -> a[1] - b[1]); + heap.add(new int[] { k, 0 }); + boolean[] used = new boolean[n + 1]; + int num = 0; + int max = 0; + while (!heap.isEmpty() && num < n) { + int[] record = heap.poll(); + int cur = record[0]; + int delay = record[1]; + if (used[cur]) { + continue; + } + used[cur] = true; + num++; + max = Math.max(max, delay); + for (int[] next : nexts.get(cur)) { + heap.add(new int[] { next[0], delay + next[1] }); + } + } + return num < n ? -1 : max; + } + + // 方法二 : 加强堆的解法 + public static int networkDelayTime2(int[][] times, int n, int k) { + ArrayList> nexts = new ArrayList<>(); + for (int i = 0; i <= n; i++) { + nexts.add(new ArrayList<>()); + } + for (int[] delay : times) { + nexts.get(delay[0]).add(new int[] { delay[1], delay[2] }); + } + Heap heap = new Heap(n); + heap.add(k, 0); + int num = 0; + int max = 0; + while (!heap.isEmpty()) { + int[] record = heap.poll(); + int cur = record[0]; + int delay = record[1]; + num++; + max = Math.max(max, delay); + for (int[] next : nexts.get(cur)) { + heap.add(next[0], delay + next[1]); + } + } + return num < n ? -1 : max; + } + + // 加强堆 + public static class Heap { + public boolean[] used; + public int[][] heap; + public int[] hIndex; + public int size; + + public Heap(int n) { + used = new boolean[n + 1]; + heap = new int[n + 1][2]; + hIndex = new int[n + 1]; + Arrays.fill(hIndex, -1); + size = 0; + } + + public void add(int cur, int delay) { + if (used[cur]) { + return; + } + if (hIndex[cur] == -1) { + heap[size][0] = cur; + heap[size][1] = delay; + hIndex[cur] = size; + heapInsert(size++); + } else { + int hi = hIndex[cur]; + if (delay <= heap[hi][1]) { + heap[hi][1] = delay; + heapInsert(hi); + } + } + } + + public int[] poll() { + int[] ans = heap[0]; + swap(0, --size); + heapify(0); + used[ans[0]] = true; + hIndex[ans[0]] = -1; + return ans; + } + + public boolean isEmpty() { + return size == 0; + } + + private void heapInsert(int i) { + int parent = (i - 1) / 2; + while (heap[i][1] < heap[parent][1]) { + swap(i, parent); + i = parent; + parent = (i - 1) / 2; + } + } + + private void heapify(int i) { + int l = (i * 2) + 1; + while (l < size) { + int smallest = l + 1 < size && heap[l + 1][1] < heap[l][1] ? (l + 1) : l; + smallest = heap[smallest][1] < heap[i][1] ? smallest : i; + if (smallest == i) { + break; + } + swap(smallest, i); + i = smallest; + l = (i * 2) + 1; + } + } + + private void swap(int i, int j) { + int[] o1 = heap[i]; + int[] o2 = heap[j]; + int o1hi = hIndex[o1[0]]; + int o2hi = hIndex[o2[0]]; + heap[i] = o2; + heap[j] = o1; + hIndex[o1[0]] = o2hi; + hIndex[o2[0]] = o1hi; + } + + } + +} diff --git a/体系学习班/class16/Edge.java b/体系学习班/class16/Edge.java new file mode 100644 index 0000000..7c56891 --- /dev/null +++ b/体系学习班/class16/Edge.java @@ -0,0 +1,14 @@ +package class16; + +public class Edge { + public int weight; + public Node from; + public Node to; + + public Edge(int weight, Node from, Node to) { + this.weight = weight; + this.from = from; + this.to = to; + } + +} diff --git a/体系学习班/class16/Graph.java b/体系学习班/class16/Graph.java new file mode 100644 index 0000000..933e43b --- /dev/null +++ b/体系学习班/class16/Graph.java @@ -0,0 +1,14 @@ +package class16; + +import java.util.HashMap; +import java.util.HashSet; + +public class Graph { + public HashMap nodes; + public HashSet edges; + + public Graph() { + nodes = new HashMap<>(); + edges = new HashSet<>(); + } +} diff --git a/体系学习班/class16/GraphGenerator.java b/体系学习班/class16/GraphGenerator.java new file mode 100644 index 0000000..fe3387d --- /dev/null +++ b/体系学习班/class16/GraphGenerator.java @@ -0,0 +1,37 @@ +package class16; + +public class GraphGenerator { + + // matrix 所有的边 + // N*3 的矩阵 + // [weight, from节点上面的值，to节点上面的值] + // + // [ 5 , 0 , 7] + // [ 3 , 0, 1] + // + public static Graph createGraph(int[][] matrix) { + Graph graph = new Graph(); + for (int i = 0; i < matrix.length; i++) { + // 拿到每一条边， matrix[i] + int weight = matrix[i][0]; + int from = matrix[i][1]; + int to = matrix[i][2]; + if (!graph.nodes.containsKey(from)) { + graph.nodes.put(from, new Node(from)); + } + if (!graph.nodes.containsKey(to)) { + graph.nodes.put(to, new Node(to)); + } + Node fromNode = graph.nodes.get(from); + Node toNode = graph.nodes.get(to); + Edge newEdge = new Edge(weight, fromNode, toNode); + fromNode.nexts.add(toNode); + fromNode.out++; + toNode.in++; + fromNode.edges.add(newEdge); + graph.edges.add(newEdge); + } + return graph; + } + +} diff --git a/体系学习班/class16/Node.java b/体系学习班/class16/Node.java new file mode 100644 index 0000000..a468aaf --- /dev/null +++ b/体系学习班/class16/Node.java @@ -0,0 +1,20 @@ +package class16; + +import java.util.ArrayList; + +// 点结构的描述 +public class Node { + public int value; + public int in; + public int out; + public ArrayList nexts; + public ArrayList edges; + + public Node(int value) { + this.value = value; + in = 0; + out = 0; + nexts = new ArrayList<>(); + edges = new ArrayList<>(); + } +} diff --git a/体系学习班/class17/Code01_Dijkstra.java b/体系学习班/class17/Code01_Dijkstra.java new file mode 100644 index 0000000..0b39615 --- /dev/null +++ b/体系学习班/class17/Code01_Dijkstra.java @@ -0,0 +1,162 @@ +package class17; + +import java.util.HashMap; +import java.util.HashSet; +import java.util.Map.Entry; + +// no negative weight +public class Code01_Dijkstra { + + public static HashMap dijkstra1(Node from) { + HashMap distanceMap = new HashMap<>(); + distanceMap.put(from, 0); + // 打过对号的点 + HashSet selectedNodes = new HashSet<>(); + Node minNode = getMinDistanceAndUnselectedNode(distanceMap, selectedNodes); + while (minNode != null) { + // 原始点 -> minNode(跳转点) 最小距离distance + int distance = distanceMap.get(minNode); + for (Edge edge : minNode.edges) { + Node toNode = edge.to; + if (!distanceMap.containsKey(toNode)) { + distanceMap.put(toNode, distance + edge.weight); + } else { // toNode + distanceMap.put(edge.to, Math.min(distanceMap.get(toNode), distance + edge.weight)); + } + } + selectedNodes.add(minNode); + minNode = getMinDistanceAndUnselectedNode(distanceMap, selectedNodes); + } + return distanceMap; + } + + public static Node getMinDistanceAndUnselectedNode(HashMap distanceMap, HashSet touchedNodes) { + Node minNode = null; + int minDistance = Integer.MAX_VALUE; + for (Entry entry : distanceMap.entrySet()) { + Node node = entry.getKey(); + int distance = entry.getValue(); + if (!touchedNodes.contains(node) && distance < minDistance) { + minNode = node; + minDistance = distance; + } + } + return minNode; + } + + public static class NodeRecord { + public Node node; + public int distance; + + public NodeRecord(Node node, int distance) { + this.node = node; + this.distance = distance; + } + } + + public static class NodeHeap { + // 堆！ + private Node[] nodes; + // node -> 堆上的什么位置？ + + private HashMap heapIndexMap; + private HashMap distanceMap; + private int size; + + public NodeHeap(int size) { + nodes = new Node[size]; + heapIndexMap = new HashMap<>(); + distanceMap = new HashMap<>(); + size = 0; + } + + public boolean isEmpty() { + return size == 0; + } + + // 有一个点叫node，现在发现了一个从源节点出发到达node的距离为distance + // 判断要不要更新，如果需要的话，就更新 + public void addOrUpdateOrIgnore(Node node, int distance) { + if (inHeap(node)) { // update + distanceMap.put(node, Math.min(distanceMap.get(node), distance)); + insertHeapify(node, heapIndexMap.get(node)); + } + if (!isEntered(node)) { // add + nodes[size] = node; + heapIndexMap.put(node, size); + distanceMap.put(node, distance); + insertHeapify(node, size++); + } + // ignore + } + + public NodeRecord pop() { + NodeRecord nodeRecord = new NodeRecord(nodes[0], distanceMap.get(nodes[0])); + swap(0, size - 1); // 0 > size - 1 size - 1 > 0 + heapIndexMap.put(nodes[size - 1], -1); + distanceMap.remove(nodes[size - 1]); + // free C++同学还要把原本堆顶节点析构，对java同学不必 + nodes[size - 1] = null; + heapify(0, --size); + return nodeRecord; + } + + private void insertHeapify(Node node, int index) { + while (distanceMap.get(nodes[index]) < distanceMap.get(nodes[(index - 1) / 2])) { + swap(index, (index - 1) / 2); + index = (index - 1) / 2; + } + } + + private void heapify(int index, int size) { + int left = index * 2 + 1; + while (left < size) { + int smallest = left + 1 < size && distanceMap.get(nodes[left + 1]) < distanceMap.get(nodes[left]) + ? left + 1 + : left; + smallest = distanceMap.get(nodes[smallest]) < distanceMap.get(nodes[index]) ? smallest : index; + if (smallest == index) { + break; + } + swap(smallest, index); + index = smallest; + left = index * 2 + 1; + } + } + + private boolean isEntered(Node node) { + return heapIndexMap.containsKey(node); + } + + private boolean inHeap(Node node) { + return isEntered(node) && heapIndexMap.get(node) != -1; + } + + private void swap(int index1, int index2) { + heapIndexMap.put(nodes[index1], index2); + heapIndexMap.put(nodes[index2], index1); + Node tmp = nodes[index1]; + nodes[index1] = nodes[index2]; + nodes[index2] = tmp; + } + } + + // 改进后的dijkstra算法 + // 从head出发，所有head能到达的节点，生成到达每个节点的最小路径记录并返回 + public static HashMap dijkstra2(Node head, int size) { + NodeHeap nodeHeap = new NodeHeap(size); + nodeHeap.addOrUpdateOrIgnore(head, 0); + HashMap result = new HashMap<>(); + while (!nodeHeap.isEmpty()) { + NodeRecord record = nodeHeap.pop(); + Node cur = record.node; + int distance = record.distance; + for (Edge edge : cur.edges) { + nodeHeap.addOrUpdateOrIgnore(edge.to, edge.weight + distance); + } + result.put(cur, distance); + } + return result; + } + +} diff --git a/体系学习班/class17/Code02_Hanoi.java b/体系学习班/class17/Code02_Hanoi.java new file mode 100644 index 0000000..3d9ba6e --- /dev/null +++ b/体系学习班/class17/Code02_Hanoi.java @@ -0,0 +1,139 @@ +package class17; + +import java.util.Stack; + +public class Code02_Hanoi { + + public static void hanoi1(int n) { + leftToRight(n); + } + + // 请把1~N层圆盘 从左 -> 右 + public static void leftToRight(int n) { + if (n == 1) { // base case + System.out.println("Move 1 from left to right"); + return; + } + leftToMid(n - 1); + System.out.println("Move " + n + " from left to right"); + midToRight(n - 1); + } + + // 请把1~N层圆盘 从左 -> 中 + public static void leftToMid(int n) { + if (n == 1) { + System.out.println("Move 1 from left to mid"); + return; + } + leftToRight(n - 1); + System.out.println("Move " + n + " from left to mid"); + rightToMid(n - 1); + } + + public static void rightToMid(int n) { + if (n == 1) { + System.out.println("Move 1 from right to mid"); + return; + } + rightToLeft(n - 1); + System.out.println("Move " + n + " from right to mid"); + leftToMid(n - 1); + } + + public static void midToRight(int n) { + if (n == 1) { + System.out.println("Move 1 from mid to right"); + return; + } + midToLeft(n - 1); + System.out.println("Move " + n + " from mid to right"); + leftToRight(n - 1); + } + + public static void midToLeft(int n) { + if (n == 1) { + System.out.println("Move 1 from mid to left"); + return; + } + midToRight(n - 1); + System.out.println("Move " + n + " from mid to left"); + rightToLeft(n - 1); + } + + public static void rightToLeft(int n) { + if (n == 1) { + System.out.println("Move 1 from right to left"); + return; + } + rightToMid(n - 1); + System.out.println("Move " + n + " from right to left"); + midToLeft(n - 1); + } + + public static void hanoi2(int n) { + if (n > 0) { + func(n, "left", "right", "mid"); + } + } + + public static void func(int N, String from, String to, String other) { + if (N == 1) { // base + System.out.println("Move 1 from " + from + " to " + to); + } else { + func(N - 1, from, other, to); + System.out.println("Move " + N + " from " + from + " to " + to); + func(N - 1, other, to, from); + } + } + + public static class Record { + public boolean finish1; + public int base; + public String from; + public String to; + public String other; + + public Record(boolean f1, int b, String f, String t, String o) { + finish1 = false; + base = b; + from = f; + to = t; + other = o; + } + } + + public static void hanoi3(int N) { + if (N < 1) { + return; + } + Stack stack = new Stack<>(); + stack.add(new Record(false, N, "left", "right", "mid")); + while (!stack.isEmpty()) { + Record cur = stack.pop(); + if (cur.base == 1) { + System.out.println("Move 1 from " + cur.from + " to " + cur.to); + if (!stack.isEmpty()) { + stack.peek().finish1 = true; + } + } else { + if (!cur.finish1) { + stack.push(cur); + stack.push(new Record(false, cur.base - 1, cur.from, cur.other, cur.to)); + } else { + System.out.println("Move " + cur.base + " from " + cur.from + " to " + cur.to); + stack.push(new Record(false, cur.base - 1, cur.other, cur.to, cur.from)); + } + } + } + } + + public static void main(String[] args) { + int n = 3; + hanoi1(n); + System.out.println("============"); + hanoi2(n); +// System.out.println("============"); +// hanoi3(n); + } + +} diff --git a/体系学习班/class17/Code03_PrintAllSubsquences.java b/体系学习班/class17/Code03_PrintAllSubsquences.java new file mode 100644 index 0000000..05b6a03 --- /dev/null +++ b/体系学习班/class17/Code03_PrintAllSubsquences.java @@ -0,0 +1,74 @@ +package class17; + +import java.util.ArrayList; +import java.util.HashSet; +import java.util.List; + +public class Code03_PrintAllSubsquences { + + // s -> "abc" -> + public static List subs(String s) { + char[] str = s.toCharArray(); + String path = ""; + List ans = new ArrayList<>(); + process1(str, 0, ans, path); + return ans; + } + + // str 固定参数 + // 来到了str[index]字符，index是位置 + // str[0..index-1]已经走过了！之前的决定，都在path上 + // 之前的决定已经不能改变了，就是path + // str[index....]还能决定，之前已经确定，而后面还能自由选择的话， + // 把所有生成的子序列，放入到ans里去 + public static void process1(char[] str, int index, List ans, String path) { + if (index == str.length) { + ans.add(path); + return; + } + // 没有要index位置的字符 + process1(str, index + 1, ans, path); + // 要了index位置的字符 + process1(str, index + 1, ans, path + String.valueOf(str[index])); + } + + public static List subsNoRepeat(String s) { + char[] str = s.toCharArray(); + String path = ""; + HashSet set = new HashSet<>(); + process2(str, 0, set, path); + List ans = new ArrayList<>(); + for (String cur : set) { + ans.add(cur); + } + return ans; + } + + public static void process2(char[] str, int index, HashSet set, String path) { + if (index == str.length) { + set.add(path); + return; + } + String no = path; + process2(str, index + 1, set, no); + String yes = path + String.valueOf(str[index]); + process2(str, index + 1, set, yes); + } + + public static void main(String[] args) { + String test = "acccc"; + List ans1 = subs(test); + List ans2 = subsNoRepeat(test); + + for (String str : ans1) { + System.out.println(str); + } + System.out.println("================="); + for (String str : ans2) { + System.out.println(str); + } + System.out.println("================="); + + } + +} diff --git a/体系学习班/class17/Code04_PrintAllPermutations.java b/体系学习班/class17/Code04_PrintAllPermutations.java new file mode 100644 index 0000000..0337a95 --- /dev/null +++ b/体系学习班/class17/Code04_PrintAllPermutations.java @@ -0,0 +1,110 @@ +package class17; + +import java.util.ArrayList; +import java.util.List; + +public class Code04_PrintAllPermutations { + + public static List permutation1(String s) { + List ans = new ArrayList<>(); + if (s == null || s.length() == 0) { + return ans; + } + char[] str = s.toCharArray(); + ArrayList rest = new ArrayList(); + for (char cha : str) { + rest.add(cha); + } + String path = ""; + f(rest, path, ans); + return ans; + } + + public static void f(ArrayList rest, String path, List ans) { + if (rest.isEmpty()) { + ans.add(path); + } else { + int N = rest.size(); + for (int i = 0; i < N; i++) { + char cur = rest.get(i); + rest.remove(i); + f(rest, path + cur, ans); + rest.add(i, cur); + } + } + } + + public static List permutation2(String s) { + List ans = new ArrayList<>(); + if (s == null || s.length() == 0) { + return ans; + } + char[] str = s.toCharArray(); + g1(str, 0, ans); + return ans; + } + + public static void g1(char[] str, int index, List ans) { + if (index == str.length) { + ans.add(String.valueOf(str)); + } else { + for (int i = index; i < str.length; i++) { + swap(str, index, i); + g1(str, index + 1, ans); + swap(str, index, i); + } + } + } + + public static List permutation3(String s) { + List ans = new ArrayList<>(); + if (s == null || s.length() == 0) { + return ans; + } + char[] str = s.toCharArray(); + g2(str, 0, ans); + return ans; + } + + public static void g2(char[] str, int index, List ans) { + if (index == str.length) { + ans.add(String.valueOf(str)); + } else { + boolean[] visited = new boolean[256]; + for (int i = index; i < str.length; i++) { + if (!visited[str[i]]) { + visited[str[i]] = true; + swap(str, index, i); + g2(str, index + 1, ans); + swap(str, index, i); + } + } + } + } + + public static void swap(char[] chs, int i, int j) { + char tmp = chs[i]; + chs[i] = chs[j]; + chs[j] = tmp; + } + + public static void main(String[] args) { + String s = "acc"; + List ans1 = permutation1(s); + for (String str : ans1) { + System.out.println(str); + } + System.out.println("======="); + List ans2 = permutation2(s); + for (String str : ans2) { + System.out.println(str); + } + System.out.println("======="); + List ans3 = permutation3(s); + for (String str : ans3) { + System.out.println(str); + } + + } + +} diff --git a/体系学习班/class17/Code05_ReverseStackUsingRecursive.java b/体系学习班/class17/Code05_ReverseStackUsingRecursive.java new file mode 100644 index 0000000..731056a --- /dev/null +++ b/体系学习班/class17/Code05_ReverseStackUsingRecursive.java @@ -0,0 +1,44 @@ +package class17; + +import java.util.Stack; + +public class Code05_ReverseStackUsingRecursive { + + public static void reverse(Stack stack) { + if (stack.isEmpty()) { + return; + } + int i = f(stack); + reverse(stack); + stack.push(i); + } + + // 栈底元素移除掉 + // 上面的元素盖下来 + // 返回移除掉的栈底元素 + public static int f(Stack stack) { + int result = stack.pop(); + if (stack.isEmpty()) { + return result; + } else { + int last = f(stack); + stack.push(result); + return last; + } + } + + public static void main(String[] args) { + Stack test = new Stack(); + test.push(1); + test.push(2); + test.push(3); + test.push(4); + test.push(5); + reverse(test); + while (!test.isEmpty()) { + System.out.println(test.pop()); + } + + } + +} diff --git a/体系学习班/class17/Edge.java b/体系学习班/class17/Edge.java new file mode 100644 index 0000000..eacd58b --- /dev/null +++ b/体系学习班/class17/Edge.java @@ -0,0 +1,14 @@ +package class17; + +public class Edge { + public int weight; + public Node from; + public Node to; + + public Edge(int weight, Node from, Node to) { + this.weight = weight; + this.from = from; + this.to = to; + } + +} diff --git a/体系学习班/class17/Graph.java b/体系学习班/class17/Graph.java new file mode 100644 index 0000000..f9314a2 --- /dev/null +++ b/体系学习班/class17/Graph.java @@ -0,0 +1,14 @@ +package class17; + +import java.util.HashMap; +import java.util.HashSet; + +public class Graph { + public HashMap nodes; + public HashSet edges; + + public Graph() { + nodes = new HashMap<>(); + edges = new HashSet<>(); + } +} diff --git a/体系学习班/class17/Node.java b/体系学习班/class17/Node.java new file mode 100644 index 0000000..1056ead --- /dev/null +++ b/体系学习班/class17/Node.java @@ -0,0 +1,20 @@ +package class17; + +import java.util.ArrayList; + +// 点结构的描述 +public class Node { + public int value; + public int in; + public int out; + public ArrayList nexts; + public ArrayList edges; + + public Node(int value) { + this.value = value; + in = 0; + out = 0; + nexts = new ArrayList<>(); + edges = new ArrayList<>(); + } +} diff --git a/体系学习班/class18/Code01_RobotWalk.java b/体系学习班/class18/Code01_RobotWalk.java new file mode 100644 index 0000000..f9ee3f2 --- /dev/null +++ b/体系学习班/class18/Code01_RobotWalk.java @@ -0,0 +1,94 @@ +package class18; + +public class Code01_RobotWalk { + + public static int ways1(int N, int start, int aim, int K) { + if (N < 2 || start < 1 || start > N || aim < 1 || aim > N || K < 1) { + return -1; + } + return process1(start, K, aim, N); + } + + // 机器人当前来到的位置是cur， + // 机器人还有rest步需要去走， + // 最终的目标是aim， + // 有哪些位置？1~N + // 返回：机器人从cur出发，走过rest步之后，最终停在aim的方法数，是多少？ + public static int process1(int cur, int rest, int aim, int N) { + if (rest == 0) { // 如果已经不需要走了，走完了！ + return cur == aim ? 1 : 0; + } + // (cur, rest) + if (cur == 1) { // 1 -> 2 + return process1(2, rest - 1, aim, N); + } + // (cur, rest) + if (cur == N) { // N-1 <- N + return process1(N - 1, rest - 1, aim, N); + } + // (cur, rest) + return process1(cur - 1, rest - 1, aim, N) + process1(cur + 1, rest - 1, aim, N); + } + + public static int ways2(int N, int start, int aim, int K) { + if (N < 2 || start < 1 || start > N || aim < 1 || aim > N || K < 1) { + return -1; + } + int[][] dp = new int[N + 1][K + 1]; + for (int i = 0; i <= N; i++) { + for (int j = 0; j <= K; j++) { + dp[i][j] = -1; + } + } + // dp就是缓存表 + // dp[cur][rest] == -1 -> process1(cur, rest)之前没算过！ + // dp[cur][rest] != -1 -> process1(cur, rest)之前算过！返回值，dp[cur][rest] + // N+1 * K+1 + return process2(start, K, aim, N, dp); + } + + // cur 范: 1 ~ N + // rest 范：0 ~ K + public static int process2(int cur, int rest, int aim, int N, int[][] dp) { + if (dp[cur][rest] != -1) { + return dp[cur][rest]; + } + // 之前没算过！ + int ans = 0; + if (rest == 0) { + ans = cur == aim ? 1 : 0; + } else if (cur == 1) { + ans = process2(2, rest - 1, aim, N, dp); + } else if (cur == N) { + ans = process2(N - 1, rest - 1, aim, N, dp); + } else { + ans = process2(cur - 1, rest - 1, aim, N, dp) + process2(cur + 1, rest - 1, aim, N, dp); + } + dp[cur][rest] = ans; + return ans; + + } + + public static int ways3(int N, int start, int aim, int K) { + if (N < 2 || start < 1 || start > N || aim < 1 || aim > N || K < 1) { + return -1; + } + int[][] dp = new int[N + 1][K + 1]; + dp[aim][0] = 1; + for (int rest = 1; rest <= K; rest++) { + dp[1][rest] = dp[2][rest - 1]; + for (int cur = 2; cur < N; cur++) { + dp[cur][rest] = dp[cur - 1][rest - 1] + dp[cur + 1][rest - 1]; + } + dp[N][rest] = dp[N - 1][rest - 1]; + } + return dp[start][K]; + } + + public static void main(String[] args) { + System.out.println(ways1(5, 2, 4, 6)); + System.out.println(ways2(5, 2, 4, 6)); + System.out.println(ways3(5, 2, 4, 6)); + } + +} diff --git a/体系学习班/class18/Code02_CardsInLine.java b/体系学习班/class18/Code02_CardsInLine.java new file mode 100644 index 0000000..919a4cb --- /dev/null +++ b/体系学习班/class18/Code02_CardsInLine.java @@ -0,0 +1,116 @@ +package class18; + +public class Code02_CardsInLine { + + // 根据规则，返回获胜者的分数 + public static int win1(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int first = f1(arr, 0, arr.length - 1); + int second = g1(arr, 0, arr.length - 1); + return Math.max(first, second); + } + + // arr[L..R]，先手获得的最好分数返回 + public static int f1(int[] arr, int L, int R) { + if (L == R) { + return arr[L]; + } + int p1 = arr[L] + g1(arr, L + 1, R); + int p2 = arr[R] + g1(arr, L, R - 1); + return Math.max(p1, p2); + } + + // // arr[L..R]，后手获得的最好分数返回 + public static int g1(int[] arr, int L, int R) { + if (L == R) { + return 0; + } + int p1 = f1(arr, L + 1, R); // 对手拿走了L位置的数 + int p2 = f1(arr, L, R - 1); // 对手拿走了R位置的数 + return Math.min(p1, p2); + } + + public static int win2(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int N = arr.length; + int[][] fmap = new int[N][N]; + int[][] gmap = new int[N][N]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < N; j++) { + fmap[i][j] = -1; + gmap[i][j] = -1; + } + } + int first = f2(arr, 0, arr.length - 1, fmap, gmap); + int second = g2(arr, 0, arr.length - 1, fmap, gmap); + return Math.max(first, second); + } + + // arr[L..R]，先手获得的最好分数返回 + public static int f2(int[] arr, int L, int R, int[][] fmap, int[][] gmap) { + if (fmap[L][R] != -1) { + return fmap[L][R]; + } + int ans = 0; + if (L == R) { + ans = arr[L]; + } else { + int p1 = arr[L] + g2(arr, L + 1, R, fmap, gmap); + int p2 = arr[R] + g2(arr, L, R - 1, fmap, gmap); + ans = Math.max(p1, p2); + } + fmap[L][R] = ans; + return ans; + } + + // // arr[L..R]，后手获得的最好分数返回 + public static int g2(int[] arr, int L, int R, int[][] fmap, int[][] gmap) { + if (gmap[L][R] != -1) { + return gmap[L][R]; + } + int ans = 0; + if (L != R) { + int p1 = f2(arr, L + 1, R, fmap, gmap); // 对手拿走了L位置的数 + int p2 = f2(arr, L, R - 1, fmap, gmap); // 对手拿走了R位置的数 + ans = Math.min(p1, p2); + } + gmap[L][R] = ans; + return ans; + } + + public static int win3(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int N = arr.length; + int[][] fmap = new int[N][N]; + int[][] gmap = new int[N][N]; + for (int i = 0; i < N; i++) { + fmap[i][i] = arr[i]; + } + for (int startCol = 1; startCol < N; startCol++) { + int L = 0; + int R = startCol; + while (R < N) { + fmap[L][R] = Math.max(arr[L] + gmap[L + 1][R], arr[R] + gmap[L][R - 1]); + gmap[L][R] = Math.min(fmap[L + 1][R], fmap[L][R - 1]); + L++; + R++; + } + } + return Math.max(fmap[0][N - 1], gmap[0][N - 1]); + } + + public static void main(String[] args) { + int[] arr = { 5, 7, 4, 5, 8, 1, 6, 0, 3, 4, 6, 1, 7 }; + System.out.println(win1(arr)); + System.out.println(win2(arr)); + System.out.println(win3(arr)); + + } + +} diff --git a/体系学习班/class19/Code01_Knapsack.java b/体系学习班/class19/Code01_Knapsack.java new file mode 100644 index 0000000..ab355a2 --- /dev/null +++ b/体系学习班/class19/Code01_Knapsack.java @@ -0,0 +1,63 @@ +package class19; + +public class Code01_Knapsack { + + // 所有的货，重量和价值，都在w和v数组里 + // 为了方便，其中没有负数 + // bag背包容量，不能超过这个载重 + // 返回：不超重的情况下，能够得到的最大价值 + public static int maxValue(int[] w, int[] v, int bag) { + if (w == null || v == null || w.length != v.length || w.length == 0) { + return 0; + } + // 尝试函数！ + return process(w, v, 0, bag); + } + + // index 0~N + // rest 负~bag + public static int process(int[] w, int[] v, int index, int rest) { + if (rest < 0) { + return -1; + } + if (index == w.length) { + return 0; + } + int p1 = process(w, v, index + 1, rest); + int p2 = 0; + int next = process(w, v, index + 1, rest - w[index]); + if (next != -1) { + p2 = v[index] + next; + } + return Math.max(p1, p2); + } + + public static int dp(int[] w, int[] v, int bag) { + if (w == null || v == null || w.length != v.length || w.length == 0) { + return 0; + } + int N = w.length; + int[][] dp = new int[N + 1][bag + 1]; + for (int index = N - 1; index >= 0; index--) { + for (int rest = 0; rest <= bag; rest++) { + int p1 = dp[index + 1][rest]; + int p2 = 0; + int next = rest - w[index] < 0 ? -1 : dp[index + 1][rest - w[index]]; + if (next != -1) { + p2 = v[index] + next; + } + dp[index][rest] = Math.max(p1, p2); + } + } + return dp[0][bag]; + } + + public static void main(String[] args) { + int[] weights = { 3, 2, 4, 7, 3, 1, 7 }; + int[] values = { 5, 6, 3, 19, 12, 4, 2 }; + int bag = 15; + System.out.println(maxValue(weights, values, bag)); + System.out.println(dp(weights, values, bag)); + } + +} diff --git a/体系学习班/class19/Code02_ConvertToLetterString.java b/体系学习班/class19/Code02_ConvertToLetterString.java new file mode 100644 index 0000000..b780ca7 --- /dev/null +++ b/体系学习班/class19/Code02_ConvertToLetterString.java @@ -0,0 +1,123 @@ +package class19; + +public class Code02_ConvertToLetterString { + + // str只含有数字字符0~9 + // 返回多少种转化方案 + public static int number(String str) { + if (str == null || str.length() == 0) { + return 0; + } + return process(str.toCharArray(), 0); + } + + // str[0..i-1]转化无需过问 + // str[i.....]去转化，返回有多少种转化方法 + public static int process(char[] str, int i) { + if (i == str.length) { + return 1; + } + // i没到最后，说明有字符 + if (str[i] == '0') { // 之前的决定有问题 + return 0; + } + // str[i] != '0' + // 可能性一，i单转 + int ways = process(str, i + 1); + if (i + 1 < str.length && (str[i] - '0') * 10 + str[i + 1] - '0' < 27) { + ways += process(str, i + 2); + } + return ways; + } + + // 从右往左的动态规划 + // 就是上面方法的动态规划版本 + // dp[i]表示：str[i...]有多少种转化方式 + public static int dp1(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + int N = str.length; + int[] dp = new int[N + 1]; + dp[N] = 1; + for (int i = N - 1; i >= 0; i--) { + if (str[i] != '0') { + int ways = dp[i + 1]; + if (i + 1 < str.length && (str[i] - '0') * 10 + str[i + 1] - '0' < 27) { + ways += dp[i + 2]; + } + dp[i] = ways; + } + } + return dp[0]; + } + + // 从左往右的动态规划 + // dp[i]表示：str[0...i]有多少种转化方式 + public static int dp2(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + int N = str.length; + if (str[0] == '0') { + return 0; + } + int[] dp = new int[N]; + dp[0] = 1; + for (int i = 1; i < N; i++) { + if (str[i] == '0') { + // 如果此时str[i]=='0'，那么他是一定要拉前一个字符(i-1的字符)一起拼的， + // 那么就要求前一个字符，不能也是‘0’，否则拼不了。 + // 前一个字符不是‘0’就够了嘛？不够，还得要求拼完了要么是10，要么是20，如果更大的话，拼不了。 + // 这就够了嘛？还不够，你们拼完了，还得要求str[0...i-2]真的可以被分解！ + // 如果str[0...i-2]都不存在分解方案，那i和i-1拼成了也不行，因为之前的搞定不了。 + if (str[i - 1] == '0' || str[i - 1] > '2' || (i - 2 >= 0 && dp[i - 2] == 0)) { + return 0; + } else { + dp[i] = i - 2 >= 0 ? dp[i - 2] : 1; + } + } else { + dp[i] = dp[i - 1]; + if (str[i - 1] != '0' && (str[i - 1] - '0') * 10 + str[i] - '0' <= 26) { + dp[i] += i - 2 >= 0 ? dp[i - 2] : 1; + } + } + } + return dp[N - 1]; + } + + // 为了测试 + public static String randomString(int len) { + char[] str = new char[len]; + for (int i = 0; i < len; i++) { + str[i] = (char) ((int) (Math.random() * 10) + '0'); + } + return String.valueOf(str); + } + + // 为了测试 + public static void main(String[] args) { + int N = 30; + int testTime = 1000000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * N); + String s = randomString(len); + int ans0 = number(s); + int ans1 = dp1(s); + int ans2 = dp2(s); + if (ans0 != ans1 || ans0 != ans2) { + System.out.println(s); + System.out.println(ans0); + System.out.println(ans1); + System.out.println(ans2); + System.out.println("Oops!"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class19/Code03_StickersToSpellWord.java b/体系学习班/class19/Code03_StickersToSpellWord.java new file mode 100644 index 0000000..90b3a4d --- /dev/null +++ b/体系学习班/class19/Code03_StickersToSpellWord.java @@ -0,0 +1,151 @@ +package class19; + +import java.util.HashMap; + +// 本题测试链接：https://leetcode.com/problems/stickers-to-spell-word +public class Code03_StickersToSpellWord { + + public static int minStickers1(String[] stickers, String target) { + int ans = process1(stickers, target); + return ans == Integer.MAX_VALUE ? -1 : ans; + } + + // 所有贴纸stickers，每一种贴纸都有无穷张 + // target + // 最少张数 + public static int process1(String[] stickers, String target) { + if (target.length() == 0) { + return 0; + } + int min = Integer.MAX_VALUE; + for (String first : stickers) { + String rest = minus(target, first); + if (rest.length() != target.length()) { + min = Math.min(min, process1(stickers, rest)); + } + } + return min + (min == Integer.MAX_VALUE ? 0 : 1); + } + + public static String minus(String s1, String s2) { + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + int[] count = new int[26]; + for (char cha : str1) { + count[cha - 'a']++; + } + for (char cha : str2) { + count[cha - 'a']--; + } + StringBuilder builder = new StringBuilder(); + for (int i = 0; i < 26; i++) { + if (count[i] > 0) { + for (int j = 0; j < count[i]; j++) { + builder.append((char) (i + 'a')); + } + } + } + return builder.toString(); + } + + public static int minStickers2(String[] stickers, String target) { + int N = stickers.length; + // 关键优化(用词频表替代贴纸数组) + int[][] counts = new int[N][26]; + for (int i = 0; i < N; i++) { + char[] str = stickers[i].toCharArray(); + for (char cha : str) { + counts[i][cha - 'a']++; + } + } + int ans = process2(counts, target); + return ans == Integer.MAX_VALUE ? -1 : ans; + } + + // stickers[i] 数组，当初i号贴纸的字符统计 int[][] stickers -> 所有的贴纸 + // 每一种贴纸都有无穷张 + // 返回搞定target的最少张数 + // 最少张数 + public static int process2(int[][] stickers, String t) { + if (t.length() == 0) { + return 0; + } + // target做出词频统计 + // target aabbc 2 2 1.. + // 0 1 2.. + char[] target = t.toCharArray(); + int[] tcounts = new int[26]; + for (char cha : target) { + tcounts[cha - 'a']++; + } + int N = stickers.length; + int min = Integer.MAX_VALUE; + for (int i = 0; i < N; i++) { + // 尝试第一张贴纸是谁 + int[] sticker = stickers[i]; + // 最关键的优化(重要的剪枝!这一步也是贪心!) + if (sticker[target[0] - 'a'] > 0) { + StringBuilder builder = new StringBuilder(); + for (int j = 0; j < 26; j++) { + if (tcounts[j] > 0) { + int nums = tcounts[j] - sticker[j]; + for (int k = 0; k < nums; k++) { + builder.append((char) (j + 'a')); + } + } + } + String rest = builder.toString(); + min = Math.min(min, process2(stickers, rest)); + } + } + return min + (min == Integer.MAX_VALUE ? 0 : 1); + } + + public static int minStickers3(String[] stickers, String target) { + int N = stickers.length; + int[][] counts = new int[N][26]; + for (int i = 0; i < N; i++) { + char[] str = stickers[i].toCharArray(); + for (char cha : str) { + counts[i][cha - 'a']++; + } + } + HashMap dp = new HashMap<>(); + dp.put("", 0); + int ans = process3(counts, target, dp); + return ans == Integer.MAX_VALUE ? -1 : ans; + } + + public static int process3(int[][] stickers, String t, HashMap dp) { + if (dp.containsKey(t)) { + return dp.get(t); + } + char[] target = t.toCharArray(); + int[] tcounts = new int[26]; + for (char cha : target) { + tcounts[cha - 'a']++; + } + int N = stickers.length; + int min = Integer.MAX_VALUE; + for (int i = 0; i < N; i++) { + int[] sticker = stickers[i]; + if (sticker[target[0] - 'a'] > 0) { + StringBuilder builder = new StringBuilder(); + for (int j = 0; j < 26; j++) { + if (tcounts[j] > 0) { + int nums = tcounts[j] - sticker[j]; + for (int k = 0; k < nums; k++) { + builder.append((char) (j + 'a')); + } + } + } + String rest = builder.toString(); + min = Math.min(min, process3(stickers, rest, dp)); + } + } + int ans = min + (min == Integer.MAX_VALUE ? 0 : 1); + dp.put(t, ans); + return ans; + } + +} diff --git a/体系学习班/class19/Code04_LongestCommonSubsequence.java b/体系学习班/class19/Code04_LongestCommonSubsequence.java new file mode 100644 index 0000000..5e49551 --- /dev/null +++ b/体系学习班/class19/Code04_LongestCommonSubsequence.java @@ -0,0 +1,124 @@ +package class19; + +// 这个问题leetcode上可以直接测 +// 链接：https://leetcode.com/problems/longest-common-subsequence/ +public class Code04_LongestCommonSubsequence { + + public static int longestCommonSubsequence1(String s1, String s2) { + if (s1 == null || s2 == null || s1.length() == 0 || s2.length() == 0) { + return 0; + } + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + // 尝试 + return process1(str1, str2, str1.length - 1, str2.length - 1); + } + + // str1[0...i]和str2[0...j]，这个范围上最长公共子序列长度是多少？ + // 可能性分类: + // a) 最长公共子序列，一定不以str1[i]字符结尾、也一定不以str2[j]字符结尾 + // b) 最长公共子序列，可能以str1[i]字符结尾、但是一定不以str2[j]字符结尾 + // c) 最长公共子序列，一定不以str1[i]字符结尾、但是可能以str2[j]字符结尾 + // d) 最长公共子序列，必须以str1[i]字符结尾、也必须以str2[j]字符结尾 + // 注意：a)、b)、c)、d)并不是完全互斥的，他们可能会有重叠的情况 + // 但是可以肯定，答案不会超过这四种可能性的范围 + // 那么我们分别来看一下，这几种可能性怎么调用后续的递归。 + // a) 最长公共子序列，一定不以str1[i]字符结尾、也一定不以str2[j]字符结尾 + // 如果是这种情况，那么有没有str1[i]和str2[j]就根本不重要了，因为这两个字符一定没用啊 + // 所以砍掉这两个字符，最长公共子序列 = str1[0...i-1]与str2[0...j-1]的最长公共子序列长度(后续递归) + // b) 最长公共子序列，可能以str1[i]字符结尾、但是一定不以str2[j]字符结尾 + // 如果是这种情况，那么我们可以确定str2[j]一定没有用，要砍掉；但是str1[i]可能有用，所以要保留 + // 所以，最长公共子序列 = str1[0...i]与str2[0...j-1]的最长公共子序列长度(后续递归) + // c) 最长公共子序列，一定不以str1[i]字符结尾、但是可能以str2[j]字符结尾 + // 跟上面分析过程类似，最长公共子序列 = str1[0...i-1]与str2[0...j]的最长公共子序列长度(后续递归) + // d) 最长公共子序列，必须以str1[i]字符结尾、也必须以str2[j]字符结尾 + // 同时可以看到，可能性d)存在的条件，一定是在str1[i] == str2[j]的情况下，才成立的 + // 所以，最长公共子序列总长度 = str1[0...i-1]与str2[0...j-1]的最长公共子序列长度(后续递归) + 1(共同的结尾) + // 综上，四种情况已经穷尽了所有可能性。四种情况中取最大即可 + // 其中b)、c)一定参与最大值的比较， + // 当str1[i] == str2[j]时，a)一定比d)小，所以d)参与 + // 当str1[i] != str2[j]时，d)压根不存在，所以a)参与 + // 但是再次注意了！ + // a)是：str1[0...i-1]与str2[0...j-1]的最长公共子序列长度 + // b)是：str1[0...i]与str2[0...j-1]的最长公共子序列长度 + // c)是：str1[0...i-1]与str2[0...j]的最长公共子序列长度 + // a)中str1的范围 < b)中str1的范围，a)中str2的范围 == b)中str2的范围 + // 所以a)不用求也知道，它比不过b)啊，因为有一个样本的范围比b)小啊！ + // a)中str1的范围 == c)中str1的范围，a)中str2的范围 < c)中str2的范围 + // 所以a)不用求也知道，它比不过c)啊，因为有一个样本的范围比c)小啊！ + // 至此，可以知道，a)就是个垃圾，有它没它，都不影响最大值的决策 + // 所以，当str1[i] == str2[j]时，b)、c)、d)中选出最大值 + // 当str1[i] != str2[j]时，b)、c)中选出最大值 + public static int process1(char[] str1, char[] str2, int i, int j) { + if (i == 0 && j == 0) { + // str1[0..0]和str2[0..0]，都只剩一个字符了 + // 那如果字符相等，公共子序列长度就是1，不相等就是0 + // 这显而易见 + return str1[i] == str2[j] ? 1 : 0; + } else if (i == 0) { + // 这里的情况为： + // str1[0...0]和str2[0...j]，str1只剩1个字符了，但是str2不只一个字符 + // 因为str1只剩一个字符了，所以str1[0...0]和str2[0...j]公共子序列最多长度为1 + // 如果str1[0] == str2[j]，那么此时相等已经找到了！公共子序列长度就是1，也不可能更大了 + // 如果str1[0] != str2[j]，只是此时不相等而已， + // 那么str2[0...j-1]上有没有字符等于str1[0]呢？不知道，所以递归继续找 + if (str1[i] == str2[j]) { + return 1; + } else { + return process1(str1, str2, i, j - 1); + } + } else if (j == 0) { + // 和上面的else if同理 + // str1[0...i]和str2[0...0]，str2只剩1个字符了，但是str1不只一个字符 + // 因为str2只剩一个字符了，所以str1[0...i]和str2[0...0]公共子序列最多长度为1 + // 如果str1[i] == str2[0]，那么此时相等已经找到了！公共子序列长度就是1，也不可能更大了 + // 如果str1[i] != str2[0]，只是此时不相等而已， + // 那么str1[0...i-1]上有没有字符等于str2[0]呢？不知道，所以递归继续找 + if (str1[i] == str2[j]) { + return 1; + } else { + return process1(str1, str2, i - 1, j); + } + } else { // i != 0 && j != 0 + // 这里的情况为： + // str1[0...i]和str2[0...i]，str1和str2都不只一个字符 + // 看函数开始之前的注释部分 + // p1就是可能性c) + int p1 = process1(str1, str2, i - 1, j); + // p2就是可能性b) + int p2 = process1(str1, str2, i, j - 1); + // p3就是可能性d)，如果可能性d)存在，即str1[i] == str2[j]，那么p3就求出来，参与pk + // 如果可能性d)不存在，即str1[i] != str2[j]，那么让p3等于0，然后去参与pk，反正不影响 + int p3 = str1[i] == str2[j] ? (1 + process1(str1, str2, i - 1, j - 1)) : 0; + return Math.max(p1, Math.max(p2, p3)); + } + } + + public static int longestCommonSubsequence2(String s1, String s2) { + if (s1 == null || s2 == null || s1.length() == 0 || s2.length() == 0) { + return 0; + } + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + int N = str1.length; + int M = str2.length; + int[][] dp = new int[N][M]; + dp[0][0] = str1[0] == str2[0] ? 1 : 0; + for (int j = 1; j < M; j++) { + dp[0][j] = str1[0] == str2[j] ? 1 : dp[0][j - 1]; + } + for (int i = 1; i < N; i++) { + dp[i][0] = str1[i] == str2[0] ? 1 : dp[i - 1][0]; + } + for (int i = 1; i < N; i++) { + for (int j = 1; j < M; j++) { + int p1 = dp[i - 1][j]; + int p2 = dp[i][j - 1]; + int p3 = str1[i] == str2[j] ? (1 + dp[i - 1][j - 1]) : 0; + dp[i][j] = Math.max(p1, Math.max(p2, p3)); + } + } + return dp[N - 1][M - 1]; + } + +} diff --git a/体系学习班/class20/Code01_PalindromeSubsequence.java b/体系学习班/class20/Code01_PalindromeSubsequence.java new file mode 100644 index 0000000..dd38429 --- /dev/null +++ b/体系学习班/class20/Code01_PalindromeSubsequence.java @@ -0,0 +1,121 @@ +package class20; + +// 测试链接：https://leetcode.com/problems/longest-palindromic-subsequence/ +public class Code01_PalindromeSubsequence { + + public static int lpsl1(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + return f(str, 0, str.length - 1); + } + + // str[L..R]最长回文子序列长度返回 + public static int f(char[] str, int L, int R) { + if (L == R) { + return 1; + } + if (L == R - 1) { + return str[L] == str[R] ? 2 : 1; + } + int p1 = f(str, L + 1, R - 1); + int p2 = f(str, L, R - 1); + int p3 = f(str, L + 1, R); + int p4 = str[L] != str[R] ? 0 : (2 + f(str, L + 1, R - 1)); + return Math.max(Math.max(p1, p2), Math.max(p3, p4)); + } + + public static int lpsl2(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + int N = str.length; + int[][] dp = new int[N][N]; + dp[N - 1][N - 1] = 1; + for (int i = 0; i < N - 1; i++) { + dp[i][i] = 1; + dp[i][i + 1] = str[i] == str[i + 1] ? 2 : 1; + } + for (int L = N - 3; L >= 0; L--) { + for (int R = L + 2; R < N; R++) { + dp[L][R] = Math.max(dp[L][R - 1], dp[L + 1][R]); + if (str[L] == str[R]) { + dp[L][R] = Math.max(dp[L][R], 2 + dp[L + 1][R - 1]); + } + } + } + return dp[0][N - 1]; + } + + public static int longestPalindromeSubseq1(String s) { + if (s == null || s.length() == 0) { + return 0; + } + if (s.length() == 1) { + return 1; + } + char[] str = s.toCharArray(); + char[] reverse = reverse(str); + return longestCommonSubsequence(str, reverse); + } + + public static char[] reverse(char[] str) { + int N = str.length; + char[] reverse = new char[str.length]; + for (int i = 0; i < str.length; i++) { + reverse[--N] = str[i]; + } + return reverse; + } + + public static int longestCommonSubsequence(char[] str1, char[] str2) { + int N = str1.length; + int M = str2.length; + int[][] dp = new int[N][M]; + dp[0][0] = str1[0] == str2[0] ? 1 : 0; + for (int i = 1; i < N; i++) { + dp[i][0] = str1[i] == str2[0] ? 1 : dp[i - 1][0]; + } + for (int j = 1; j < M; j++) { + dp[0][j] = str1[0] == str2[j] ? 1 : dp[0][j - 1]; + } + for (int i = 1; i < N; i++) { + for (int j = 1; j < M; j++) { + dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]); + if (str1[i] == str2[j]) { + dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - 1] + 1); + } + } + } + return dp[N - 1][M - 1]; + } + + public static int longestPalindromeSubseq2(String s) { + if (s == null || s.length() == 0) { + return 0; + } + if (s.length() == 1) { + return 1; + } + char[] str = s.toCharArray(); + int N = str.length; + int[][] dp = new int[N][N]; + dp[N - 1][N - 1] = 1; + for (int i = 0; i < N - 1; i++) { + dp[i][i] = 1; + dp[i][i + 1] = str[i] == str[i + 1] ? 2 : 1; + } + for (int i = N - 3; i >= 0; i--) { + for (int j = i + 2; j < N; j++) { + dp[i][j] = Math.max(dp[i][j - 1], dp[i + 1][j]); + if (str[i] == str[j]) { + dp[i][j] = Math.max(dp[i][j], dp[i + 1][j - 1] + 2); + } + } + } + return dp[0][N - 1]; + } + +} diff --git a/体系学习班/class20/Code02_HorseJump.java b/体系学习班/class20/Code02_HorseJump.java new file mode 100644 index 0000000..f96f4a1 --- /dev/null +++ b/体系学习班/class20/Code02_HorseJump.java @@ -0,0 +1,109 @@ +package class20; + +public class Code02_HorseJump { + + // 当前来到的位置是（x,y） + // 还剩下rest步需要跳 + // 跳完rest步，正好跳到a，b的方法数是多少？ + // 10 * 9 + public static int jump(int a, int b, int k) { + return process(0, 0, k, a, b); + } + + public static int process(int x, int y, int rest, int a, int b) { + if (x < 0 || x > 9 || y < 0 || y > 8) { + return 0; + } + if (rest == 0) { + return (x == a && y == b) ? 1 : 0; + } + int ways = process(x + 2, y + 1, rest - 1, a, b); + ways += process(x + 1, y + 2, rest - 1, a, b); + ways += process(x - 1, y + 2, rest - 1, a, b); + ways += process(x - 2, y + 1, rest - 1, a, b); + ways += process(x - 2, y - 1, rest - 1, a, b); + ways += process(x - 1, y - 2, rest - 1, a, b); + ways += process(x + 1, y - 2, rest - 1, a, b); + ways += process(x + 2, y - 1, rest - 1, a, b); + return ways; + } + + public static int dp(int a, int b, int k) { + int[][][] dp = new int[10][9][k + 1]; + dp[a][b][0] = 1; + for (int rest = 1; rest <= k; rest++) { + for (int x = 0; x < 10; x++) { + for (int y = 0; y < 9; y++) { + int ways = pick(dp, x + 2, y + 1, rest - 1); + ways += pick(dp, x + 1, y + 2, rest - 1); + ways += pick(dp, x - 1, y + 2, rest - 1); + ways += pick(dp, x - 2, y + 1, rest - 1); + ways += pick(dp, x - 2, y - 1, rest - 1); + ways += pick(dp, x - 1, y - 2, rest - 1); + ways += pick(dp, x + 1, y - 2, rest - 1); + ways += pick(dp, x + 2, y - 1, rest - 1); + dp[x][y][rest] = ways; + } + } + } + return dp[0][0][k]; + } + + public static int pick(int[][][] dp, int x, int y, int rest) { + if (x < 0 || x > 9 || y < 0 || y > 8) { + return 0; + } + return dp[x][y][rest]; + } + + public static int ways(int a, int b, int step) { + return f(0, 0, step, a, b); + } + + public static int f(int i, int j, int step, int a, int b) { + if (i < 0 || i > 9 || j < 0 || j > 8) { + return 0; + } + if (step == 0) { + return (i == a && j == b) ? 1 : 0; + } + return f(i - 2, j + 1, step - 1, a, b) + f(i - 1, j + 2, step - 1, a, b) + f(i + 1, j + 2, step - 1, a, b) + + f(i + 2, j + 1, step - 1, a, b) + f(i + 2, j - 1, step - 1, a, b) + f(i + 1, j - 2, step - 1, a, b) + + f(i - 1, j - 2, step - 1, a, b) + f(i - 2, j - 1, step - 1, a, b); + + } + + public static int waysdp(int a, int b, int s) { + int[][][] dp = new int[10][9][s + 1]; + dp[a][b][0] = 1; + for (int step = 1; step <= s; step++) { // 按层来 + for (int i = 0; i < 10; i++) { + for (int j = 0; j < 9; j++) { + dp[i][j][step] = getValue(dp, i - 2, j + 1, step - 1) + getValue(dp, i - 1, j + 2, step - 1) + + getValue(dp, i + 1, j + 2, step - 1) + getValue(dp, i + 2, j + 1, step - 1) + + getValue(dp, i + 2, j - 1, step - 1) + getValue(dp, i + 1, j - 2, step - 1) + + getValue(dp, i - 1, j - 2, step - 1) + getValue(dp, i - 2, j - 1, step - 1); + } + } + } + return dp[0][0][s]; + } + + // 在dp表中，得到dp[i][j][step]的值，但如果(i，j)位置越界的话，返回0； + public static int getValue(int[][][] dp, int i, int j, int step) { + if (i < 0 || i > 9 || j < 0 || j > 8) { + return 0; + } + return dp[i][j][step]; + } + + public static void main(String[] args) { + int x = 7; + int y = 7; + int step = 10; + System.out.println(ways(x, y, step)); + System.out.println(dp(x, y, step)); + + System.out.println(jump(x, y, step)); + } +} diff --git a/体系学习班/class20/Code03_Coffee.java b/体系学习班/class20/Code03_Coffee.java new file mode 100644 index 0000000..eb2b620 --- /dev/null +++ b/体系学习班/class20/Code03_Coffee.java @@ -0,0 +1,204 @@ +package class20; + +import java.util.Arrays; +import java.util.Comparator; +import java.util.PriorityQueue; + +// 题目 +// 数组arr代表每一个咖啡机冲一杯咖啡的时间，每个咖啡机只能串行的制造咖啡。 +// 现在有n个人需要喝咖啡，只能用咖啡机来制造咖啡。 +// 认为每个人喝咖啡的时间非常短，冲好的时间即是喝完的时间。 +// 每个人喝完之后咖啡杯可以选择洗或者自然挥发干净，只有一台洗咖啡杯的机器，只能串行的洗咖啡杯。 +// 洗杯子的机器洗完一个杯子时间为a，任何一个杯子自然挥发干净的时间为b。 +// 四个参数：arr, n, a, b +// 假设时间点从0开始，返回所有人喝完咖啡并洗完咖啡杯的全部过程结束后，至少来到什么时间点。 +public class Code03_Coffee { + + // 验证的方法 + // 彻底的暴力 + // 很慢但是绝对正确 + public static int right(int[] arr, int n, int a, int b) { + int[] times = new int[arr.length]; + int[] drink = new int[n]; + return forceMake(arr, times, 0, drink, n, a, b); + } + + // 每个人暴力尝试用每一个咖啡机给自己做咖啡 + public static int forceMake(int[] arr, int[] times, int kth, int[] drink, int n, int a, int b) { + if (kth == n) { + int[] drinkSorted = Arrays.copyOf(drink, kth); + Arrays.sort(drinkSorted); + return forceWash(drinkSorted, a, b, 0, 0, 0); + } + int time = Integer.MAX_VALUE; + for (int i = 0; i < arr.length; i++) { + int work = arr[i]; + int pre = times[i]; + drink[kth] = pre + work; + times[i] = pre + work; + time = Math.min(time, forceMake(arr, times, kth + 1, drink, n, a, b)); + drink[kth] = 0; + times[i] = pre; + } + return time; + } + + public static int forceWash(int[] drinks, int a, int b, int index, int washLine, int time) { + if (index == drinks.length) { + return time; + } + // 选择一：当前index号咖啡杯，选择用洗咖啡机刷干净 + int wash = Math.max(drinks[index], washLine) + a; + int ans1 = forceWash(drinks, a, b, index + 1, wash, Math.max(wash, time)); + + // 选择二：当前index号咖啡杯，选择自然挥发 + int dry = drinks[index] + b; + int ans2 = forceWash(drinks, a, b, index + 1, washLine, Math.max(dry, time)); + return Math.min(ans1, ans2); + } + + // 以下为贪心+优良暴力 + public static class Machine { + public int timePoint; + public int workTime; + + public Machine(int t, int w) { + timePoint = t; + workTime = w; + } + } + + public static class MachineComparator implements Comparator { + + @Override + public int compare(Machine o1, Machine o2) { + return (o1.timePoint + o1.workTime) - (o2.timePoint + o2.workTime); + } + + } + + // 优良一点的暴力尝试的方法 + public static int minTime1(int[] arr, int n, int a, int b) { + PriorityQueue heap = new PriorityQueue(new MachineComparator()); + for (int i = 0; i < arr.length; i++) { + heap.add(new Machine(0, arr[i])); + } + int[] drinks = new int[n]; + for (int i = 0; i < n; i++) { + Machine cur = heap.poll(); + cur.timePoint += cur.workTime; + drinks[i] = cur.timePoint; + heap.add(cur); + } + return bestTime(drinks, a, b, 0, 0); + } + + // drinks 所有杯子可以开始洗的时间 + // wash 单杯洗干净的时间（串行） + // air 挥发干净的时间(并行) + // free 洗的机器什么时候可用 + // drinks[index.....]都变干净，最早的结束时间（返回） + public static int bestTime(int[] drinks, int wash, int air, int index, int free) { + if (index == drinks.length) { + return 0; + } + // index号杯子 决定洗 + int selfClean1 = Math.max(drinks[index], free) + wash; + int restClean1 = bestTime(drinks, wash, air, index + 1, selfClean1); + int p1 = Math.max(selfClean1, restClean1); + + // index号杯子 决定挥发 + int selfClean2 = drinks[index] + air; + int restClean2 = bestTime(drinks, wash, air, index + 1, free); + int p2 = Math.max(selfClean2, restClean2); + return Math.min(p1, p2); + } + + // 贪心+优良尝试改成动态规划 + public static int minTime2(int[] arr, int n, int a, int b) { + PriorityQueue heap = new PriorityQueue(new MachineComparator()); + for (int i = 0; i < arr.length; i++) { + heap.add(new Machine(0, arr[i])); + } + int[] drinks = new int[n]; + for (int i = 0; i < n; i++) { + Machine cur = heap.poll(); + cur.timePoint += cur.workTime; + drinks[i] = cur.timePoint; + heap.add(cur); + } + return bestTimeDp(drinks, a, b); + } + + public static int bestTimeDp(int[] drinks, int wash, int air) { + int N = drinks.length; + int maxFree = 0; + for (int i = 0; i < drinks.length; i++) { + maxFree = Math.max(maxFree, drinks[i]) + wash; + } + int[][] dp = new int[N + 1][maxFree + 1]; + for (int index = N - 1; index >= 0; index--) { + for (int free = 0; free <= maxFree; free++) { + int selfClean1 = Math.max(drinks[index], free) + wash; + if (selfClean1 > maxFree) { + break; // 因为后面的也都不用填了 + } + // index号杯子 决定洗 + int restClean1 = dp[index + 1][selfClean1]; + int p1 = Math.max(selfClean1, restClean1); + // index号杯子 决定挥发 + int selfClean2 = drinks[index] + air; + int restClean2 = dp[index + 1][free]; + int p2 = Math.max(selfClean2, restClean2); + dp[index][free] = Math.min(p1, p2); + } + } + return dp[0][0]; + } + + // for test + public static int[] randomArray(int len, int max) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * max) + 1; + } + return arr; + } + + // for test + public static void printArray(int[] arr) { + System.out.print("arr : "); + for (int j = 0; j < arr.length; j++) { + System.out.print(arr[j] + ", "); + } + System.out.println(); + } + + public static void main(String[] args) { + int len = 10; + int max = 10; + int testTime = 10; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int[] arr = randomArray(len, max); + int n = (int) (Math.random() * 7) + 1; + int a = (int) (Math.random() * 7) + 1; + int b = (int) (Math.random() * 10) + 1; + int ans1 = right(arr, n, a, b); + int ans2 = minTime1(arr, n, a, b); + int ans3 = minTime2(arr, n, a, b); + if (ans1 != ans2 || ans2 != ans3) { + printArray(arr); + System.out.println("n : " + n); + System.out.println("a : " + a); + System.out.println("b : " + b); + System.out.println(ans1 + " , " + ans2 + " , " + ans3); + System.out.println("==============="); + break; + } + } + System.out.println("测试结束"); + + } + +} diff --git a/体系学习班/class21/Code01_MinPathSum.java b/体系学习班/class21/Code01_MinPathSum.java new file mode 100644 index 0000000..c537438 --- /dev/null +++ b/体系学习班/class21/Code01_MinPathSum.java @@ -0,0 +1,79 @@ +package class21; + +public class Code01_MinPathSum { + + public static int minPathSum1(int[][] m) { + if (m == null || m.length == 0 || m[0] == null || m[0].length == 0) { + return 0; + } + int row = m.length; + int col = m[0].length; + int[][] dp = new int[row][col]; + dp[0][0] = m[0][0]; + for (int i = 1; i < row; i++) { + dp[i][0] = dp[i - 1][0] + m[i][0]; + } + for (int j = 1; j < col; j++) { + dp[0][j] = dp[0][j - 1] + m[0][j]; + } + for (int i = 1; i < row; i++) { + for (int j = 1; j < col; j++) { + dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + m[i][j]; + } + } + return dp[row - 1][col - 1]; + } + + public static int minPathSum2(int[][] m) { + if (m == null || m.length == 0 || m[0] == null || m[0].length == 0) { + return 0; + } + int row = m.length; + int col = m[0].length; + int[] dp = new int[col]; + dp[0] = m[0][0]; + for (int j = 1; j < col; j++) { + dp[j] = dp[j - 1] + m[0][j]; + } + for (int i = 1; i < row; i++) { + dp[0] += m[i][0]; + for (int j = 1; j < col; j++) { + dp[j] = Math.min(dp[j - 1], dp[j]) + m[i][j]; + } + } + return dp[col - 1]; + } + + // for test + public static int[][] generateRandomMatrix(int rowSize, int colSize) { + if (rowSize < 0 || colSize < 0) { + return null; + } + int[][] result = new int[rowSize][colSize]; + for (int i = 0; i != result.length; i++) { + for (int j = 0; j != result[0].length; j++) { + result[i][j] = (int) (Math.random() * 100); + } + } + return result; + } + + // for test + public static void printMatrix(int[][] matrix) { + for (int i = 0; i != matrix.length; i++) { + for (int j = 0; j != matrix[0].length; j++) { + System.out.print(matrix[i][j] + " "); + } + System.out.println(); + } + } + + public static void main(String[] args) { + int rowSize = 10; + int colSize = 10; + int[][] m = generateRandomMatrix(rowSize, colSize); + System.out.println(minPathSum1(m)); + System.out.println(minPathSum2(m)); + + } +} diff --git a/体系学习班/class21/Code02_CoinsWayEveryPaperDifferent.java b/体系学习班/class21/Code02_CoinsWayEveryPaperDifferent.java new file mode 100644 index 0000000..a162b06 --- /dev/null +++ b/体系学习班/class21/Code02_CoinsWayEveryPaperDifferent.java @@ -0,0 +1,77 @@ +package class21; + +public class Code02_CoinsWayEveryPaperDifferent { + + public static int coinWays(int[] arr, int aim) { + return process(arr, 0, aim); + } + + // arr[index....] 组成正好rest这么多的钱，有几种方法 + public static int process(int[] arr, int index, int rest) { + if (rest < 0) { + return 0; + } + if (index == arr.length) { // 没钱了！ + return rest == 0 ? 1 : 0; + } else { + return process(arr, index + 1, rest) + process(arr, index + 1, rest - arr[index]); + } + } + + public static int dp(int[] arr, int aim) { + if (aim == 0) { + return 1; + } + int N = arr.length; + int[][] dp = new int[N + 1][aim + 1]; + dp[N][0] = 1; + for (int index = N - 1; index >= 0; index--) { + for (int rest = 0; rest <= aim; rest++) { + dp[index][rest] = dp[index + 1][rest] + (rest - arr[index] >= 0 ? dp[index + 1][rest - arr[index]] : 0); + } + } + return dp[0][aim]; + } + + // 为了测试 + public static int[] randomArray(int maxLen, int maxValue) { + int N = (int) (Math.random() * maxLen); + int[] arr = new int[N]; + for (int i = 0; i < N; i++) { + arr[i] = (int) (Math.random() * maxValue) + 1; + } + return arr; + } + + // 为了测试 + public static void printArray(int[] arr) { + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // 为了测试 + public static void main(String[] args) { + int maxLen = 20; + int maxValue = 30; + int testTime = 1000000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int[] arr = randomArray(maxLen, maxValue); + int aim = (int) (Math.random() * maxValue); + int ans1 = coinWays(arr, aim); + int ans2 = dp(arr, aim); + if (ans1 != ans2) { + System.out.println("Oops!"); + printArray(arr); + System.out.println(aim); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class21/Code03_CoinsWayNoLimit.java b/体系学习班/class21/Code03_CoinsWayNoLimit.java new file mode 100644 index 0000000..7759d9c --- /dev/null +++ b/体系学习班/class21/Code03_CoinsWayNoLimit.java @@ -0,0 +1,108 @@ +package class21; + +public class Code03_CoinsWayNoLimit { + + public static int coinsWay(int[] arr, int aim) { + if (arr == null || arr.length == 0 || aim < 0) { + return 0; + } + return process(arr, 0, aim); + } + + // arr[index....] 所有的面值，每一个面值都可以任意选择张数，组成正好rest这么多钱，方法数多少？ + public static int process(int[] arr, int index, int rest) { + if (index == arr.length) { // 没钱了 + return rest == 0 ? 1 : 0; + } + int ways = 0; + for (int zhang = 0; zhang * arr[index] <= rest; zhang++) { + ways += process(arr, index + 1, rest - (zhang * arr[index])); + } + return ways; + } + + public static int dp1(int[] arr, int aim) { + if (arr == null || arr.length == 0 || aim < 0) { + return 0; + } + int N = arr.length; + int[][] dp = new int[N + 1][aim + 1]; + dp[N][0] = 1; + for (int index = N - 1; index >= 0; index--) { + for (int rest = 0; rest <= aim; rest++) { + int ways = 0; + for (int zhang = 0; zhang * arr[index] <= rest; zhang++) { + ways += dp[index + 1][rest - (zhang * arr[index])]; + } + dp[index][rest] = ways; + } + } + return dp[0][aim]; + } + + public static int dp2(int[] arr, int aim) { + if (arr == null || arr.length == 0 || aim < 0) { + return 0; + } + int N = arr.length; + int[][] dp = new int[N + 1][aim + 1]; + dp[N][0] = 1; + for (int index = N - 1; index >= 0; index--) { + for (int rest = 0; rest <= aim; rest++) { + dp[index][rest] = dp[index + 1][rest]; + if (rest - arr[index] >= 0) { + dp[index][rest] += dp[index][rest - arr[index]]; + } + } + } + return dp[0][aim]; + } + + // 为了测试 + public static int[] randomArray(int maxLen, int maxValue) { + int N = (int) (Math.random() * maxLen); + int[] arr = new int[N]; + boolean[] has = new boolean[maxValue + 1]; + for (int i = 0; i < N; i++) { + do { + arr[i] = (int) (Math.random() * maxValue) + 1; + } while (has[arr[i]]); + has[arr[i]] = true; + } + return arr; + } + + // 为了测试 + public static void printArray(int[] arr) { + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // 为了测试 + public static void main(String[] args) { + int maxLen = 10; + int maxValue = 30; + int testTime = 1000000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int[] arr = randomArray(maxLen, maxValue); + int aim = (int) (Math.random() * maxValue); + int ans1 = coinsWay(arr, aim); + int ans2 = dp1(arr, aim); + int ans3 = dp2(arr, aim); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("Oops!"); + printArray(arr); + System.out.println(aim); + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class21/Code04_CoinsWaySameValueSamePapper.java b/体系学习班/class21/Code04_CoinsWaySameValueSamePapper.java new file mode 100644 index 0000000..f45c919 --- /dev/null +++ b/体系学习班/class21/Code04_CoinsWaySameValueSamePapper.java @@ -0,0 +1,148 @@ +package class21; + +import java.util.HashMap; +import java.util.Map.Entry; + +public class Code04_CoinsWaySameValueSamePapper { + + public static class Info { + public int[] coins; + public int[] zhangs; + + public Info(int[] c, int[] z) { + coins = c; + zhangs = z; + } + } + + public static Info getInfo(int[] arr) { + HashMap counts = new HashMap<>(); + for (int value : arr) { + if (!counts.containsKey(value)) { + counts.put(value, 1); + } else { + counts.put(value, counts.get(value) + 1); + } + } + int N = counts.size(); + int[] coins = new int[N]; + int[] zhangs = new int[N]; + int index = 0; + for (Entry entry : counts.entrySet()) { + coins[index] = entry.getKey(); + zhangs[index++] = entry.getValue(); + } + return new Info(coins, zhangs); + } + + public static int coinsWay(int[] arr, int aim) { + if (arr == null || arr.length == 0 || aim < 0) { + return 0; + } + Info info = getInfo(arr); + return process(info.coins, info.zhangs, 0, aim); + } + + // coins 面值数组，正数且去重 + // zhangs 每种面值对应的张数 + public static int process(int[] coins, int[] zhangs, int index, int rest) { + if (index == coins.length) { + return rest == 0 ? 1 : 0; + } + int ways = 0; + for (int zhang = 0; zhang * coins[index] <= rest && zhang <= zhangs[index]; zhang++) { + ways += process(coins, zhangs, index + 1, rest - (zhang * coins[index])); + } + return ways; + } + + public static int dp1(int[] arr, int aim) { + if (arr == null || arr.length == 0 || aim < 0) { + return 0; + } + Info info = getInfo(arr); + int[] coins = info.coins; + int[] zhangs = info.zhangs; + int N = coins.length; + int[][] dp = new int[N + 1][aim + 1]; + dp[N][0] = 1; + for (int index = N - 1; index >= 0; index--) { + for (int rest = 0; rest <= aim; rest++) { + int ways = 0; + for (int zhang = 0; zhang * coins[index] <= rest && zhang <= zhangs[index]; zhang++) { + ways += dp[index + 1][rest - (zhang * coins[index])]; + } + dp[index][rest] = ways; + } + } + return dp[0][aim]; + } + + public static int dp2(int[] arr, int aim) { + if (arr == null || arr.length == 0 || aim < 0) { + return 0; + } + Info info = getInfo(arr); + int[] coins = info.coins; + int[] zhangs = info.zhangs; + int N = coins.length; + int[][] dp = new int[N + 1][aim + 1]; + dp[N][0] = 1; + for (int index = N - 1; index >= 0; index--) { + for (int rest = 0; rest <= aim; rest++) { + dp[index][rest] = dp[index + 1][rest]; + if (rest - coins[index] >= 0) { + dp[index][rest] += dp[index][rest - coins[index]]; + } + if (rest - coins[index] * (zhangs[index] + 1) >= 0) { + dp[index][rest] -= dp[index + 1][rest - coins[index] * (zhangs[index] + 1)]; + } + } + } + return dp[0][aim]; + } + + // 为了测试 + public static int[] randomArray(int maxLen, int maxValue) { + int N = (int) (Math.random() * maxLen); + int[] arr = new int[N]; + for (int i = 0; i < N; i++) { + arr[i] = (int) (Math.random() * maxValue) + 1; + } + return arr; + } + + // 为了测试 + public static void printArray(int[] arr) { + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // 为了测试 + public static void main(String[] args) { + int maxLen = 10; + int maxValue = 20; + int testTime = 1000000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int[] arr = randomArray(maxLen, maxValue); + int aim = (int) (Math.random() * maxValue); + int ans1 = coinsWay(arr, aim); + int ans2 = dp1(arr, aim); + int ans3 = dp2(arr, aim); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("Oops!"); + printArray(arr); + System.out.println(aim); + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class21/Code05_BobDie.java b/体系学习班/class21/Code05_BobDie.java new file mode 100644 index 0000000..439fffe --- /dev/null +++ b/体系学习班/class21/Code05_BobDie.java @@ -0,0 +1,58 @@ +package class21; + +public class Code05_BobDie { + + public static double livePosibility1(int row, int col, int k, int N, int M) { + return (double) process(row, col, k, N, M) / Math.pow(4, k); + } + + // 目前在row，col位置，还有rest步要走，走完了如果还在棋盘中就获得1个生存点，返回总的生存点数 + public static long process(int row, int col, int rest, int N, int M) { + if (row < 0 || row == N || col < 0 || col == M) { + return 0; + } + // 还在棋盘中！ + if (rest == 0) { + return 1; + } + // 还在棋盘中！还有步数要走 + long up = process(row - 1, col, rest - 1, N, M); + long down = process(row + 1, col, rest - 1, N, M); + long left = process(row, col - 1, rest - 1, N, M); + long right = process(row, col + 1, rest - 1, N, M); + return up + down + left + right; + } + + public static double livePosibility2(int row, int col, int k, int N, int M) { + long[][][] dp = new long[N][M][k + 1]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + dp[i][j][0] = 1; + } + } + for (int rest = 1; rest <= k; rest++) { + for (int r = 0; r < N; r++) { + for (int c = 0; c < M; c++) { + dp[r][c][rest] = pick(dp, N, M, r - 1, c, rest - 1); + dp[r][c][rest] += pick(dp, N, M, r + 1, c, rest - 1); + dp[r][c][rest] += pick(dp, N, M, r, c - 1, rest - 1); + dp[r][c][rest] += pick(dp, N, M, r, c + 1, rest - 1); + } + } + } + return (double) dp[row][col][k] / Math.pow(4, k); + } + + public static long pick(long[][][] dp, int N, int M, int r, int c, int rest) { + if (r < 0 || r == N || c < 0 || c == M) { + return 0; + } + return dp[r][c][rest]; + } + + public static void main(String[] args) { + System.out.println(livePosibility1(6, 6, 10, 50, 50)); + System.out.println(livePosibility2(6, 6, 10, 50, 50)); + } + +} diff --git a/体系学习班/class22/Code01_KillMonster.java b/体系学习班/class22/Code01_KillMonster.java new file mode 100644 index 0000000..77fae7b --- /dev/null +++ b/体系学习班/class22/Code01_KillMonster.java @@ -0,0 +1,100 @@ +package class22; + +public class Code01_KillMonster { + + public static double right(int N, int M, int K) { + if (N < 1 || M < 1 || K < 1) { + return 0; + } + long all = (long) Math.pow(M + 1, K); + long kill = process(K, M, N); + return (double) ((double) kill / (double) all); + } + + // 怪兽还剩hp点血 + // 每次的伤害在[0~M]范围上 + // 还有times次可以砍 + // 返回砍死的情况数！ + public static long process(int times, int M, int hp) { + if (times == 0) { + return hp <= 0 ? 1 : 0; + } + if (hp <= 0) { + return (long) Math.pow(M + 1, times); + } + long ways = 0; + for (int i = 0; i <= M; i++) { + ways += process(times - 1, M, hp - i); + } + return ways; + } + + public static double dp1(int N, int M, int K) { + if (N < 1 || M < 1 || K < 1) { + return 0; + } + long all = (long) Math.pow(M + 1, K); + long[][] dp = new long[K + 1][N + 1]; + dp[0][0] = 1; + for (int times = 1; times <= K; times++) { + dp[times][0] = (long) Math.pow(M + 1, times); + for (int hp = 1; hp <= N; hp++) { + long ways = 0; + for (int i = 0; i <= M; i++) { + if (hp - i >= 0) { + ways += dp[times - 1][hp - i]; + } else { + ways += (long) Math.pow(M + 1, times - 1); + } + } + dp[times][hp] = ways; + } + } + long kill = dp[K][N]; + return (double) ((double) kill / (double) all); + } + + public static double dp2(int N, int M, int K) { + if (N < 1 || M < 1 || K < 1) { + return 0; + } + long all = (long) Math.pow(M + 1, K); + long[][] dp = new long[K + 1][N + 1]; + dp[0][0] = 1; + for (int times = 1; times <= K; times++) { + dp[times][0] = (long) Math.pow(M + 1, times); + for (int hp = 1; hp <= N; hp++) { + dp[times][hp] = dp[times][hp - 1] + dp[times - 1][hp]; + if (hp - 1 - M >= 0) { + dp[times][hp] -= dp[times - 1][hp - 1 - M]; + } else { + dp[times][hp] -= Math.pow(M + 1, times - 1); + } + } + } + long kill = dp[K][N]; + return (double) ((double) kill / (double) all); + } + + public static void main(String[] args) { + int NMax = 10; + int MMax = 10; + int KMax = 10; + int testTime = 200; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int N = (int) (Math.random() * NMax); + int M = (int) (Math.random() * MMax); + int K = (int) (Math.random() * KMax); + double ans1 = right(N, M, K); + double ans2 = dp1(N, M, K); + double ans3 = dp2(N, M, K); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("Oops!"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class22/Code02_MinCoinsNoLimit.java b/体系学习班/class22/Code02_MinCoinsNoLimit.java new file mode 100644 index 0000000..06e4843 --- /dev/null +++ b/体系学习班/class22/Code02_MinCoinsNoLimit.java @@ -0,0 +1,121 @@ +package class22; + +public class Code02_MinCoinsNoLimit { + + public static int minCoins(int[] arr, int aim) { + return process(arr, 0, aim); + } + + // arr[index...]面值，每种面值张数自由选择， + // 搞出rest正好这么多钱，返回最小张数 + // 拿Integer.MAX_VALUE标记怎么都搞定不了 + public static int process(int[] arr, int index, int rest) { + if (index == arr.length) { + return rest == 0 ? 0 : Integer.MAX_VALUE; + } else { + int ans = Integer.MAX_VALUE; + for (int zhang = 0; zhang * arr[index] <= rest; zhang++) { + int next = process(arr, index + 1, rest - zhang * arr[index]); + if (next != Integer.MAX_VALUE) { + ans = Math.min(ans, zhang + next); + } + } + return ans; + } + } + + public static int dp1(int[] arr, int aim) { + if (aim == 0) { + return 0; + } + int N = arr.length; + int[][] dp = new int[N + 1][aim + 1]; + dp[N][0] = 0; + for (int j = 1; j <= aim; j++) { + dp[N][j] = Integer.MAX_VALUE; + } + for (int index = N - 1; index >= 0; index--) { + for (int rest = 0; rest <= aim; rest++) { + int ans = Integer.MAX_VALUE; + for (int zhang = 0; zhang * arr[index] <= rest; zhang++) { + int next = dp[index + 1][rest - zhang * arr[index]]; + if (next != Integer.MAX_VALUE) { + ans = Math.min(ans, zhang + next); + } + } + dp[index][rest] = ans; + } + } + return dp[0][aim]; + } + + public static int dp2(int[] arr, int aim) { + if (aim == 0) { + return 0; + } + int N = arr.length; + int[][] dp = new int[N + 1][aim + 1]; + dp[N][0] = 0; + for (int j = 1; j <= aim; j++) { + dp[N][j] = Integer.MAX_VALUE; + } + for (int index = N - 1; index >= 0; index--) { + for (int rest = 0; rest <= aim; rest++) { + dp[index][rest] = dp[index + 1][rest]; + if (rest - arr[index] >= 0 + && dp[index][rest - arr[index]] != Integer.MAX_VALUE) { + dp[index][rest] = Math.min(dp[index][rest], dp[index][rest - arr[index]] + 1); + } + } + } + return dp[0][aim]; + } + + // 为了测试 + public static int[] randomArray(int maxLen, int maxValue) { + int N = (int) (Math.random() * maxLen); + int[] arr = new int[N]; + boolean[] has = new boolean[maxValue + 1]; + for (int i = 0; i < N; i++) { + do { + arr[i] = (int) (Math.random() * maxValue) + 1; + } while (has[arr[i]]); + has[arr[i]] = true; + } + return arr; + } + + // 为了测试 + public static void printArray(int[] arr) { + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // 为了测试 + public static void main(String[] args) { + int maxLen = 20; + int maxValue = 30; + int testTime = 300000; + System.out.println("功能测试开始"); + for (int i = 0; i < testTime; i++) { + int N = (int) (Math.random() * maxLen); + int[] arr = randomArray(N, maxValue); + int aim = (int) (Math.random() * maxValue); + int ans1 = minCoins(arr, aim); + int ans2 = dp1(arr, aim); + int ans3 = dp2(arr, aim); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("Oops!"); + printArray(arr); + System.out.println(aim); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("功能测试结束"); + } + +} diff --git a/体系学习班/class22/Code03_SplitNumber.java b/体系学习班/class22/Code03_SplitNumber.java new file mode 100644 index 0000000..23193bc --- /dev/null +++ b/体系学习班/class22/Code03_SplitNumber.java @@ -0,0 +1,85 @@ +package class22; + +public class Code03_SplitNumber { + + // n为正数 + public static int ways(int n) { + if (n < 0) { + return 0; + } + if (n == 1) { + return 1; + } + return process(1, n); + } + + // 上一个拆出来的数是pre + // 还剩rest需要去拆 + // 返回拆解的方法数 + public static int process(int pre, int rest) { + if (rest == 0) { + return 1; + } + if (pre > rest) { + return 0; + } + int ways = 0; + for (int first = pre; first <= rest; first++) { + ways += process(first, rest - first); + } + return ways; + } + + public static int dp1(int n) { + if (n < 0) { + return 0; + } + if (n == 1) { + return 1; + } + int[][] dp = new int[n + 1][n + 1]; + for (int pre = 1; pre <= n; pre++) { + dp[pre][0] = 1; + dp[pre][pre] = 1; + } + for (int pre = n - 1; pre >= 1; pre--) { + for (int rest = pre + 1; rest <= n; rest++) { + int ways = 0; + for (int first = pre; first <= rest; first++) { + ways += dp[first][rest - first]; + } + dp[pre][rest] = ways; + } + } + return dp[1][n]; + } + + public static int dp2(int n) { + if (n < 0) { + return 0; + } + if (n == 1) { + return 1; + } + int[][] dp = new int[n + 1][n + 1]; + for (int pre = 1; pre <= n; pre++) { + dp[pre][0] = 1; + dp[pre][pre] = 1; + } + for (int pre = n - 1; pre >= 1; pre--) { + for (int rest = pre + 1; rest <= n; rest++) { + dp[pre][rest] = dp[pre + 1][rest]; + dp[pre][rest] += dp[pre][rest - pre]; + } + } + return dp[1][n]; + } + + public static void main(String[] args) { + int test = 39; + System.out.println(ways(test)); + System.out.println(dp1(test)); + System.out.println(dp2(test)); + } + +} diff --git a/体系学习班/class23/Code01_SplitSumClosed.java b/体系学习班/class23/Code01_SplitSumClosed.java new file mode 100644 index 0000000..6f82c49 --- /dev/null +++ b/体系学习班/class23/Code01_SplitSumClosed.java @@ -0,0 +1,94 @@ +package class23; + +public class Code01_SplitSumClosed { + + public static int right(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int sum = 0; + for (int num : arr) { + sum += num; + } + return process(arr, 0, sum / 2); + } + + // arr[i...]可以自由选择，请返回累加和尽量接近rest，但不能超过rest的情况下，最接近的累加和是多少？ + public static int process(int[] arr, int i, int rest) { + if (i == arr.length) { + return 0; + } else { // 还有数，arr[i]这个数 + // 可能性1，不使用arr[i] + int p1 = process(arr, i + 1, rest); + // 可能性2，要使用arr[i] + int p2 = 0; + if (arr[i] <= rest) { + p2 = arr[i] + process(arr, i + 1, rest - arr[i]); + } + return Math.max(p1, p2); + } + } + + public static int dp(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int sum = 0; + for (int num : arr) { + sum += num; + } + sum /= 2; + int N = arr.length; + int[][] dp = new int[N + 1][sum + 1]; + for (int i = N - 1; i >= 0; i--) { + for (int rest = 0; rest <= sum; rest++) { + // 可能性1，不使用arr[i] + int p1 = dp[i + 1][rest]; + // 可能性2，要使用arr[i] + int p2 = 0; + if (arr[i] <= rest) { + p2 = arr[i] + dp[i + 1][rest - arr[i]]; + } + dp[i][rest] = Math.max(p1, p2); + } + } + return dp[0][sum]; + } + + public static int[] randomArray(int len, int value) { + int[] arr = new int[len]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) (Math.random() * value); + } + return arr; + } + + public static void printArray(int[] arr) { + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + } + + public static void main(String[] args) { + int maxLen = 20; + int maxValue = 50; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * maxLen); + int[] arr = randomArray(len, maxValue); + int ans1 = right(arr); + int ans2 = dp(arr); + if (ans1 != ans2) { + printArray(arr); + System.out.println(ans1); + System.out.println(ans2); + System.out.println("Oops!"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class23/Code02_SplitSumClosedSizeHalf.java b/体系学习班/class23/Code02_SplitSumClosedSizeHalf.java new file mode 100644 index 0000000..0da40fb --- /dev/null +++ b/体系学习班/class23/Code02_SplitSumClosedSizeHalf.java @@ -0,0 +1,243 @@ +package class23; + +public class Code02_SplitSumClosedSizeHalf { + + public static int right(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int sum = 0; + for (int num : arr) { + sum += num; + } + if ((arr.length & 1) == 0) { + return process(arr, 0, arr.length / 2, sum / 2); + } else { + return Math.max(process(arr, 0, arr.length / 2, sum / 2), process(arr, 0, arr.length / 2 + 1, sum / 2)); + } + } + + // arr[i....]自由选择，挑选的个数一定要是picks个，累加和<=rest, 离rest最近的返回 + public static int process(int[] arr, int i, int picks, int rest) { + if (i == arr.length) { + return picks == 0 ? 0 : -1; + } else { + int p1 = process(arr, i + 1, picks, rest); + // 就是要使用arr[i]这个数 + int p2 = -1; + int next = -1; + if (arr[i] <= rest) { + next = process(arr, i + 1, picks - 1, rest - arr[i]); + } + if (next != -1) { + p2 = arr[i] + next; + } + return Math.max(p1, p2); + } + } + + public static int dp(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int sum = 0; + for (int num : arr) { + sum += num; + } + sum /= 2; + int N = arr.length; + int M = (N + 1) / 2; + int[][][] dp = new int[N + 1][M + 1][sum + 1]; + for (int i = 0; i <= N; i++) { + for (int j = 0; j <= M; j++) { + for (int k = 0; k <= sum; k++) { + dp[i][j][k] = -1; + } + } + } + for (int rest = 0; rest <= sum; rest++) { + dp[N][0][rest] = 0; + } + for (int i = N - 1; i >= 0; i--) { + for (int picks = 0; picks <= M; picks++) { + for (int rest = 0; rest <= sum; rest++) { + int p1 = dp[i + 1][picks][rest]; + // 就是要使用arr[i]这个数 + int p2 = -1; + int next = -1; + if (picks - 1 >= 0 && arr[i] <= rest) { + next = dp[i + 1][picks - 1][rest - arr[i]]; + } + if (next != -1) { + p2 = arr[i] + next; + } + dp[i][picks][rest] = Math.max(p1, p2); + } + } + } + if ((arr.length & 1) == 0) { + return dp[0][arr.length / 2][sum]; + } else { + return Math.max(dp[0][arr.length / 2][sum], dp[0][(arr.length / 2) + 1][sum]); + } + } + +// public static int right(int[] arr) { +// if (arr == null || arr.length < 2) { +// return 0; +// } +// int sum = 0; +// for (int num : arr) { +// sum += num; +// } +// return process(arr, 0, 0, sum >> 1); +// } +// +// public static int process(int[] arr, int i, int picks, int rest) { +// if (i == arr.length) { +// if ((arr.length & 1) == 0) { +// return picks == (arr.length >> 1) ? 0 : -1; +// } else { +// return (picks == (arr.length >> 1) || picks == (arr.length >> 1) + 1) ? 0 : -1; +// } +// } +// int p1 = process(arr, i + 1, picks, rest); +// int p2 = -1; +// int next2 = -1; +// if (arr[i] <= rest) { +// next2 = process(arr, i + 1, picks + 1, rest - arr[i]); +// } +// if (next2 != -1) { +// p2 = arr[i] + next2; +// } +// return Math.max(p1, p2); +// } +// +// public static int dp1(int[] arr) { +// if (arr == null || arr.length < 2) { +// return 0; +// } +// int sum = 0; +// for (int num : arr) { +// sum += num; +// } +// sum >>= 1; +// int N = arr.length; +// int M = (arr.length + 1) >> 1; +// int[][][] dp = new int[N + 1][M + 1][sum + 1]; +// for (int i = 0; i <= N; i++) { +// for (int j = 0; j <= M; j++) { +// for (int k = 0; k <= sum; k++) { +// dp[i][j][k] = -1; +// } +// } +// } +// for (int k = 0; k <= sum; k++) { +// dp[N][M][k] = 0; +// } +// if ((arr.length & 1) != 0) { +// for (int k = 0; k <= sum; k++) { +// dp[N][M - 1][k] = 0; +// } +// } +// for (int i = N - 1; i >= 0; i--) { +// for (int picks = 0; picks <= M; picks++) { +// for (int rest = 0; rest <= sum; rest++) { +// int p1 = dp[i + 1][picks][rest]; +// int p2 = -1; +// int next2 = -1; +// if (picks + 1 <= M && arr[i] <= rest) { +// next2 = dp[i + 1][picks + 1][rest - arr[i]]; +// } +// if (next2 != -1) { +// p2 = arr[i] + next2; +// } +// dp[i][picks][rest] = Math.max(p1, p2); +// } +// } +// } +// return dp[0][0][sum]; +// } + + public static int dp2(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int sum = 0; + for (int num : arr) { + sum += num; + } + sum >>= 1; + int N = arr.length; + int M = (arr.length + 1) >> 1; + int[][][] dp = new int[N][M + 1][sum + 1]; + for (int i = 0; i < N; i++) { + for (int j = 0; j <= M; j++) { + for (int k = 0; k <= sum; k++) { + dp[i][j][k] = Integer.MIN_VALUE; + } + } + } + for (int i = 0; i < N; i++) { + for (int k = 0; k <= sum; k++) { + dp[i][0][k] = 0; + } + } + for (int k = 0; k <= sum; k++) { + dp[0][1][k] = arr[0] <= k ? arr[0] : Integer.MIN_VALUE; + } + for (int i = 1; i < N; i++) { + for (int j = 1; j <= Math.min(i + 1, M); j++) { + for (int k = 0; k <= sum; k++) { + dp[i][j][k] = dp[i - 1][j][k]; + if (k - arr[i] >= 0) { + dp[i][j][k] = Math.max(dp[i][j][k], dp[i - 1][j - 1][k - arr[i]] + arr[i]); + } + } + } + } + return Math.max(dp[N - 1][M][sum], dp[N - 1][N - M][sum]); + } + + // for test + public static int[] randomArray(int len, int value) { + int[] arr = new int[len]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) (Math.random() * value); + } + return arr; + } + + // for test + public static void printArray(int[] arr) { + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + int maxLen = 10; + int maxValue = 50; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * maxLen); + int[] arr = randomArray(len, maxValue); + int ans1 = right(arr); + int ans2 = dp(arr); + int ans3 = dp2(arr); + if (ans1 != ans2 || ans1 != ans3) { + printArray(arr); + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + System.out.println("Oops!"); + break; + } + } + System.out.println("测试结束"); + } + +} \ No newline at end of file diff --git a/体系学习班/class23/Code03_NQueens.java b/体系学习班/class23/Code03_NQueens.java new file mode 100644 index 0000000..a9e454a --- /dev/null +++ b/体系学习班/class23/Code03_NQueens.java @@ -0,0 +1,89 @@ +package class23; + +public class Code03_NQueens { + + public static int num1(int n) { + if (n < 1) { + return 0; + } + int[] record = new int[n]; + return process1(0, record, n); + } + + // 当前来到i行，一共是0~N-1行 + // 在i行上放皇后，所有列都尝试 + // 必须要保证跟之前所有的皇后不打架 + // int[] record record[x] = y 之前的第x行的皇后，放在了y列上 + // 返回：不关心i以上发生了什么，i.... 后续有多少合法的方法数 + public static int process1(int i, int[] record, int n) { + if (i == n) { + return 1; + } + int res = 0; + // i行的皇后，放哪一列呢？j列， + for (int j = 0; j < n; j++) { + if (isValid(record, i, j)) { + record[i] = j; + res += process1(i + 1, record, n); + } + } + return res; + } + + public static boolean isValid(int[] record, int i, int j) { + // 0..i-1 + for (int k = 0; k < i; k++) { + if (j == record[k] || Math.abs(record[k] - j) == Math.abs(i - k)) { + return false; + } + } + return true; + } + + // 请不要超过32皇后问题 + public static int num2(int n) { + if (n < 1 || n > 32) { + return 0; + } + // 如果你是13皇后问题，limit 最右13个1，其他都是0 + int limit = n == 32 ? -1 : (1 << n) - 1; + return process2(limit, 0, 0, 0); + } + + // 7皇后问题 + // limit : 0....0 1 1 1 1 1 1 1 + // 之前皇后的列影响：colLim + // 之前皇后的左下对角线影响：leftDiaLim + // 之前皇后的右下对角线影响：rightDiaLim + public static int process2(int limit, int colLim, int leftDiaLim, int rightDiaLim) { + if (colLim == limit) { + return 1; + } + // pos中所有是1的位置，是你可以去尝试皇后的位置 + int pos = limit & (~(colLim | leftDiaLim | rightDiaLim)); + int mostRightOne = 0; + int res = 0; + while (pos != 0) { + mostRightOne = pos & (~pos + 1); + pos = pos - mostRightOne; + res += process2(limit, colLim | mostRightOne, (leftDiaLim | mostRightOne) << 1, + (rightDiaLim | mostRightOne) >>> 1); + } + return res; + } + + public static void main(String[] args) { + int n = 15; + + long start = System.currentTimeMillis(); + System.out.println(num2(n)); + long end = System.currentTimeMillis(); + System.out.println("cost time: " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + System.out.println(num1(n)); + end = System.currentTimeMillis(); + System.out.println("cost time: " + (end - start) + "ms"); + + } +} diff --git a/体系学习班/class24/Code01_SlidingWindowMaxArray.java b/体系学习班/class24/Code01_SlidingWindowMaxArray.java new file mode 100644 index 0000000..d115951 --- /dev/null +++ b/体系学习班/class24/Code01_SlidingWindowMaxArray.java @@ -0,0 +1,99 @@ +package class24; + +import java.util.LinkedList; + +public class Code01_SlidingWindowMaxArray { + + // 暴力的对数器方法 + public static int[] right(int[] arr, int w) { + if (arr == null || w < 1 || arr.length < w) { + return null; + } + int N = arr.length; + int[] res = new int[N - w + 1]; + int index = 0; + int L = 0; + int R = w - 1; + while (R < N) { + int max = arr[L]; + for (int i = L + 1; i <= R; i++) { + max = Math.max(max, arr[i]); + + } + res[index++] = max; + L++; + R++; + } + return res; + } + + public static int[] getMaxWindow(int[] arr, int w) { + if (arr == null || w < 1 || arr.length < w) { + return null; + } + // qmax 窗口最大值的更新结构 + // 放下标 + LinkedList qmax = new LinkedList(); + int[] res = new int[arr.length - w + 1]; + int index = 0; + for (int R = 0; R < arr.length; R++) { + while (!qmax.isEmpty() && arr[qmax.peekLast()] <= arr[R]) { + qmax.pollLast(); + } + qmax.addLast(R); + if (qmax.peekFirst() == R - w) { + qmax.pollFirst(); + } + if (R >= w - 1) { + res[index++] = arr[qmax.peekFirst()]; + } + } + return res; + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) (Math.random() * (maxValue + 1)); + } + return arr; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + public static void main(String[] args) { + int testTime = 100000; + int maxSize = 100; + int maxValue = 100; + System.out.println("test begin"); + for (int i = 0; i < testTime; i++) { + int[] arr = generateRandomArray(maxSize, maxValue); + int w = (int) (Math.random() * (arr.length + 1)); + int[] ans1 = getMaxWindow(arr, w); + int[] ans2 = right(arr, w); + if (!isEqual(ans1, ans2)) { + System.out.println("Oops!"); + } + } + System.out.println("test finish"); + } + +} diff --git a/体系学习班/class24/Code02_AllLessNumSubArray.java b/体系学习班/class24/Code02_AllLessNumSubArray.java new file mode 100644 index 0000000..7bbd5de --- /dev/null +++ b/体系学习班/class24/Code02_AllLessNumSubArray.java @@ -0,0 +1,109 @@ +package class24; + +import java.util.LinkedList; + +public class Code02_AllLessNumSubArray { + + // 暴力的对数器方法 + public static int right(int[] arr, int sum) { + if (arr == null || arr.length == 0 || sum < 0) { + return 0; + } + int N = arr.length; + int count = 0; + for (int L = 0; L < N; L++) { + for (int R = L; R < N; R++) { + int max = arr[L]; + int min = arr[L]; + for (int i = L + 1; i <= R; i++) { + max = Math.max(max, arr[i]); + min = Math.min(min, arr[i]); + } + if (max - min <= sum) { + count++; + } + } + } + return count; + } + + public static int num(int[] arr, int sum) { + if (arr == null || arr.length == 0 || sum < 0) { + return 0; + } + int N = arr.length; + int count = 0; + LinkedList maxWindow = new LinkedList<>(); + LinkedList minWindow = new LinkedList<>(); + int R = 0; + for (int L = 0; L < N; L++) { + while (R < N) { + while (!maxWindow.isEmpty() && arr[maxWindow.peekLast()] <= arr[R]) { + maxWindow.pollLast(); + } + maxWindow.addLast(R); + while (!minWindow.isEmpty() && arr[minWindow.peekLast()] >= arr[R]) { + minWindow.pollLast(); + } + minWindow.addLast(R); + if (arr[maxWindow.peekFirst()] - arr[minWindow.peekFirst()] > sum) { + break; + } else { + R++; + } + } + count += R - L; + if (maxWindow.peekFirst() == L) { + maxWindow.pollFirst(); + } + if (minWindow.peekFirst() == L) { + minWindow.pollFirst(); + } + } + return count; + } + + // for test + public static int[] generateRandomArray(int maxLen, int maxValue) { + int len = (int) (Math.random() * (maxLen + 1)); + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * (maxValue + 1)) - (int) (Math.random() * (maxValue + 1)); + } + return arr; + } + + // for test + public static void printArray(int[] arr) { + if (arr != null) { + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + } + + public static void main(String[] args) { + int maxLen = 100; + int maxValue = 200; + int testTime = 100000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int[] arr = generateRandomArray(maxLen, maxValue); + int sum = (int) (Math.random() * (maxValue + 1)); + int ans1 = right(arr, sum); + int ans2 = num(arr, sum); + if (ans1 != ans2) { + System.out.println("Oops!"); + printArray(arr); + System.out.println(sum); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + + } + +} diff --git a/体系学习班/class24/Code03_GasStation.java b/体系学习班/class24/Code03_GasStation.java new file mode 100644 index 0000000..b80906f --- /dev/null +++ b/体系学习班/class24/Code03_GasStation.java @@ -0,0 +1,53 @@ +package class24; + +import java.util.LinkedList; + +// 测试链接：https://leetcode.com/problems/gas-station +public class Code03_GasStation { + + // 这个方法的时间复杂度O(N)，额外空间复杂度O(N) + public static int canCompleteCircuit(int[] gas, int[] cost) { + boolean[] good = goodArray(gas, cost); + for (int i = 0; i < gas.length; i++) { + if (good[i]) { + return i; + } + } + return -1; + } + + public static boolean[] goodArray(int[] g, int[] c) { + int N = g.length; + int M = N << 1; + int[] arr = new int[M]; + for (int i = 0; i < N; i++) { + arr[i] = g[i] - c[i]; + arr[i + N] = g[i] - c[i]; + } + for (int i = 1; i < M; i++) { + arr[i] += arr[i - 1]; + } + LinkedList w = new LinkedList<>(); + for (int i = 0; i < N; i++) { + while (!w.isEmpty() && arr[w.peekLast()] >= arr[i]) { + w.pollLast(); + } + w.addLast(i); + } + boolean[] ans = new boolean[N]; + for (int offset = 0, i = 0, j = N; j < M; offset = arr[i++], j++) { + if (arr[w.peekFirst()] - offset >= 0) { + ans[i] = true; + } + if (w.peekFirst() == i) { + w.pollFirst(); + } + while (!w.isEmpty() && arr[w.peekLast()] >= arr[j]) { + w.pollLast(); + } + w.addLast(j); + } + return ans; + } + +} diff --git a/体系学习班/class24/Code04_MinCoinsOnePaper.java b/体系学习班/class24/Code04_MinCoinsOnePaper.java new file mode 100644 index 0000000..cfda95a --- /dev/null +++ b/体系学习班/class24/Code04_MinCoinsOnePaper.java @@ -0,0 +1,250 @@ +package class24; + +import java.util.HashMap; +import java.util.Map.Entry; +import java.util.LinkedList; + +public class Code04_MinCoinsOnePaper { + + public static int minCoins(int[] arr, int aim) { + return process(arr, 0, aim); + } + + public static int process(int[] arr, int index, int rest) { + if (rest < 0) { + return Integer.MAX_VALUE; + } + if (index == arr.length) { + return rest == 0 ? 0 : Integer.MAX_VALUE; + } else { + int p1 = process(arr, index + 1, rest); + int p2 = process(arr, index + 1, rest - arr[index]); + if (p2 != Integer.MAX_VALUE) { + p2++; + } + return Math.min(p1, p2); + } + } + + // dp1时间复杂度为：O(arr长度 * aim) + public static int dp1(int[] arr, int aim) { + if (aim == 0) { + return 0; + } + int N = arr.length; + int[][] dp = new int[N + 1][aim + 1]; + dp[N][0] = 0; + for (int j = 1; j <= aim; j++) { + dp[N][j] = Integer.MAX_VALUE; + } + for (int index = N - 1; index >= 0; index--) { + for (int rest = 0; rest <= aim; rest++) { + int p1 = dp[index + 1][rest]; + int p2 = rest - arr[index] >= 0 ? dp[index + 1][rest - arr[index]] : Integer.MAX_VALUE; + if (p2 != Integer.MAX_VALUE) { + p2++; + } + dp[index][rest] = Math.min(p1, p2); + } + } + return dp[0][aim]; + } + + public static class Info { + public int[] coins; + public int[] zhangs; + + public Info(int[] c, int[] z) { + coins = c; + zhangs = z; + } + } + + public static Info getInfo(int[] arr) { + HashMap counts = new HashMap<>(); + for (int value : arr) { + if (!counts.containsKey(value)) { + counts.put(value, 1); + } else { + counts.put(value, counts.get(value) + 1); + } + } + int N = counts.size(); + int[] coins = new int[N]; + int[] zhangs = new int[N]; + int index = 0; + for (Entry entry : counts.entrySet()) { + coins[index] = entry.getKey(); + zhangs[index++] = entry.getValue(); + } + return new Info(coins, zhangs); + } + + // dp2时间复杂度为：O(arr长度) + O(货币种数 * aim * 每种货币的平均张数) + public static int dp2(int[] arr, int aim) { + if (aim == 0) { + return 0; + } + // 得到info时间复杂度O(arr长度) + Info info = getInfo(arr); + int[] coins = info.coins; + int[] zhangs = info.zhangs; + int N = coins.length; + int[][] dp = new int[N + 1][aim + 1]; + dp[N][0] = 0; + for (int j = 1; j <= aim; j++) { + dp[N][j] = Integer.MAX_VALUE; + } + // 这三层for循环，时间复杂度为O(货币种数 * aim * 每种货币的平均张数) + for (int index = N - 1; index >= 0; index--) { + for (int rest = 0; rest <= aim; rest++) { + dp[index][rest] = dp[index + 1][rest]; + for (int zhang = 1; zhang * coins[index] <= aim && zhang <= zhangs[index]; zhang++) { + if (rest - zhang * coins[index] >= 0 + && dp[index + 1][rest - zhang * coins[index]] != Integer.MAX_VALUE) { + dp[index][rest] = Math.min(dp[index][rest], zhang + dp[index + 1][rest - zhang * coins[index]]); + } + } + } + } + return dp[0][aim]; + } + + // dp3时间复杂度为：O(arr长度) + O(货币种数 * aim) + // 优化需要用到窗口内最小值的更新结构 + public static int dp3(int[] arr, int aim) { + if (aim == 0) { + return 0; + } + // 得到info时间复杂度O(arr长度) + Info info = getInfo(arr); + int[] c = info.coins; + int[] z = info.zhangs; + int N = c.length; + int[][] dp = new int[N + 1][aim + 1]; + dp[N][0] = 0; + for (int j = 1; j <= aim; j++) { + dp[N][j] = Integer.MAX_VALUE; + } + // 虽然是嵌套了很多循环，但是时间复杂度为O(货币种数 * aim) + // 因为用了窗口内最小值的更新结构 + for (int i = N - 1; i >= 0; i--) { + for (int mod = 0; mod < Math.min(aim + 1, c[i]); mod++) { + // 当前面值 X + // mod mod + x mod + 2*x mod + 3 * x + LinkedList w = new LinkedList<>(); + w.add(mod); + dp[i][mod] = dp[i + 1][mod]; + for (int r = mod + c[i]; r <= aim; r += c[i]) { + while (!w.isEmpty() && (dp[i + 1][w.peekLast()] == Integer.MAX_VALUE + || dp[i + 1][w.peekLast()] + compensate(w.peekLast(), r, c[i]) >= dp[i + 1][r])) { + w.pollLast(); + } + w.addLast(r); + int overdue = r - c[i] * (z[i] + 1); + if (w.peekFirst() == overdue) { + w.pollFirst(); + } + dp[i][r] = dp[i + 1][w.peekFirst()] + compensate(w.peekFirst(), r, c[i]); + } + } + } + return dp[0][aim]; + } + + public static int compensate(int pre, int cur, int coin) { + return (cur - pre) / coin; + } + + // 为了测试 + public static int[] randomArray(int N, int maxValue) { + int[] arr = new int[N]; + for (int i = 0; i < N; i++) { + arr[i] = (int) (Math.random() * maxValue) + 1; + } + return arr; + } + + // 为了测试 + public static void printArray(int[] arr) { + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // 为了测试 + public static void main(String[] args) { + int maxLen = 20; + int maxValue = 30; + int testTime = 300000; + System.out.println("功能测试开始"); + for (int i = 0; i < testTime; i++) { + int N = (int) (Math.random() * maxLen); + int[] arr = randomArray(N, maxValue); + int aim = (int) (Math.random() * maxValue); + int ans1 = minCoins(arr, aim); + int ans2 = dp1(arr, aim); + int ans3 = dp2(arr, aim); + int ans4 = dp3(arr, aim); + if (ans1 != ans2 || ans3 != ans4 || ans1 != ans3) { + System.out.println("Oops!"); + printArray(arr); + System.out.println(aim); + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + System.out.println(ans4); + break; + } + } + System.out.println("功能测试结束"); + + System.out.println("=========="); + + int aim = 0; + int[] arr = null; + long start; + long end; + int ans2; + int ans3; + + System.out.println("性能测试开始"); + maxLen = 30000; + maxValue = 20; + aim = 60000; + arr = randomArray(maxLen, maxValue); + + start = System.currentTimeMillis(); + ans2 = dp2(arr, aim); + end = System.currentTimeMillis(); + System.out.println("dp2答案 : " + ans2 + ", dp2运行时间 : " + (end - start) + " ms"); + + start = System.currentTimeMillis(); + ans3 = dp3(arr, aim); + end = System.currentTimeMillis(); + System.out.println("dp3答案 : " + ans3 + ", dp3运行时间 : " + (end - start) + " ms"); + System.out.println("性能测试结束"); + + System.out.println("==========="); + + System.out.println("货币大量重复出现情况下，"); + System.out.println("大数据量测试dp3开始"); + maxLen = 20000000; + aim = 10000; + maxValue = 10000; + arr = randomArray(maxLen, maxValue); + start = System.currentTimeMillis(); + ans3 = dp3(arr, aim); + end = System.currentTimeMillis(); + System.out.println("dp3运行时间 : " + (end - start) + " ms"); + System.out.println("大数据量测试dp3结束"); + + System.out.println("==========="); + + System.out.println("当货币很少出现重复，dp2比dp3有常数时间优势"); + System.out.println("当货币大量出现重复，dp3时间复杂度明显优于dp2"); + System.out.println("dp3的优化用到了窗口内最小值的更新结构"); + } + +} diff --git a/体系学习班/class25/Code01_MonotonousStack.java b/体系学习班/class25/Code01_MonotonousStack.java new file mode 100644 index 0000000..b823a30 --- /dev/null +++ b/体系学习班/class25/Code01_MonotonousStack.java @@ -0,0 +1,165 @@ +package class25; + +import java.util.List; +import java.util.ArrayList; +import java.util.Stack; + +public class Code01_MonotonousStack { + + // arr = [ 3, 1, 2, 3] + // 0 1 2 3 + // [ + // 0 : [-1, 1] + // 1 : [-1, -1] + // 2 : [ 1, -1] + // 3 : [ 2, -1] + // ] + public static int[][] getNearLessNoRepeat(int[] arr) { + int[][] res = new int[arr.length][2]; + // 只存位置！ + Stack stack = new Stack<>(); + for (int i = 0; i < arr.length; i++) { // 当遍历到i位置的数，arr[i] + while (!stack.isEmpty() && arr[stack.peek()] > arr[i]) { + int j = stack.pop(); + int leftLessIndex = stack.isEmpty() ? -1 : stack.peek(); + res[j][0] = leftLessIndex; + res[j][1] = i; + } + stack.push(i); + } + while (!stack.isEmpty()) { + int j = stack.pop(); + int leftLessIndex = stack.isEmpty() ? -1 : stack.peek(); + res[j][0] = leftLessIndex; + res[j][1] = -1; + } + return res; + } + + public static int[][] getNearLess(int[] arr) { + int[][] res = new int[arr.length][2]; + Stack> stack = new Stack<>(); + for (int i = 0; i < arr.length; i++) { // i -> arr[i] 进栈 + while (!stack.isEmpty() && arr[stack.peek().get(0)] > arr[i]) { + List popIs = stack.pop(); + int leftLessIndex = stack.isEmpty() ? -1 : stack.peek().get(stack.peek().size() - 1); + for (Integer popi : popIs) { + res[popi][0] = leftLessIndex; + res[popi][1] = i; + } + } + if (!stack.isEmpty() && arr[stack.peek().get(0)] == arr[i]) { + stack.peek().add(Integer.valueOf(i)); + } else { + ArrayList list = new ArrayList<>(); + list.add(i); + stack.push(list); + } + } + while (!stack.isEmpty()) { + List popIs = stack.pop(); + int leftLessIndex = stack.isEmpty() ? -1 : stack.peek().get(stack.peek().size() - 1); + for (Integer popi : popIs) { + res[popi][0] = leftLessIndex; + res[popi][1] = -1; + } + } + return res; + } + + // for test + public static int[] getRandomArrayNoRepeat(int size) { + int[] arr = new int[(int) (Math.random() * size) + 1]; + for (int i = 0; i < arr.length; i++) { + arr[i] = i; + } + for (int i = 0; i < arr.length; i++) { + int swapIndex = (int) (Math.random() * arr.length); + int tmp = arr[swapIndex]; + arr[swapIndex] = arr[i]; + arr[i] = tmp; + } + return arr; + } + + // for test + public static int[] getRandomArray(int size, int max) { + int[] arr = new int[(int) (Math.random() * size) + 1]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) (Math.random() * max) - (int) (Math.random() * max); + } + return arr; + } + + // for test + public static int[][] rightWay(int[] arr) { + int[][] res = new int[arr.length][2]; + for (int i = 0; i < arr.length; i++) { + int leftLessIndex = -1; + int rightLessIndex = -1; + int cur = i - 1; + while (cur >= 0) { + if (arr[cur] < arr[i]) { + leftLessIndex = cur; + break; + } + cur--; + } + cur = i + 1; + while (cur < arr.length) { + if (arr[cur] < arr[i]) { + rightLessIndex = cur; + break; + } + cur++; + } + res[i][0] = leftLessIndex; + res[i][1] = rightLessIndex; + } + return res; + } + + // for test + public static boolean isEqual(int[][] res1, int[][] res2) { + if (res1.length != res2.length) { + return false; + } + for (int i = 0; i < res1.length; i++) { + if (res1[i][0] != res2[i][0] || res1[i][1] != res2[i][1]) { + return false; + } + } + + return true; + } + + // for test + public static void printArray(int[] arr) { + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + public static void main(String[] args) { + int size = 10; + int max = 20; + int testTimes = 2000000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + int[] arr1 = getRandomArrayNoRepeat(size); + int[] arr2 = getRandomArray(size, max); + if (!isEqual(getNearLessNoRepeat(arr1), rightWay(arr1))) { + System.out.println("Oops!"); + printArray(arr1); + break; + } + if (!isEqual(getNearLess(arr2), rightWay(arr2))) { + System.out.println("Oops!"); + printArray(arr2); + break; + } + } + System.out.println("测试结束"); + } +} diff --git a/体系学习班/class25/Code01_MonotonousStackForNowcoder.java b/体系学习班/class25/Code01_MonotonousStackForNowcoder.java new file mode 100644 index 0000000..501b65b --- /dev/null +++ b/体系学习班/class25/Code01_MonotonousStackForNowcoder.java @@ -0,0 +1,71 @@ +package class25; + +// 测试链接 : https://www.nowcoder.com/practice/2a2c00e7a88a498693568cef63a4b7bb +// 如果在牛客上做题，可以用如下的方式来做 +// 提交如下的代码，并把主类名改成"Main" +// 可以直接通过 +import java.io.BufferedReader; +import java.io.IOException; +import java.io.InputStreamReader; + +public class Code01_MonotonousStackForNowcoder { + + public static int[] arr = new int[1000000]; + public static int[][] ans = new int[1000000][2]; + // stack1 : 相等值的位置也放 + // stack2 : 只放不相等值的最后一个位置 + // 比如 : arr = { 3, 3, 3, 4, 4, 6, 6, 6} + // 位置 0 1 2 3 4 5 6 7 + // 如果位置依次压栈， + // stack1中的记录是（位置） : 0 1 2 3 4 5 6 7 + // stack1中的记录是（位置） : 2 4 7 + public static int[] stack1 = new int[1000000]; + public static int[] stack2 = new int[1000000]; + + public static void main(String[] args) throws IOException { + BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); + int n = Integer.parseInt(br.readLine()); + String[] strs = br.readLine().split(" "); + for (int i = 0; i < n; i++) { + arr[i] = Integer.parseInt(strs[i]); + } + getNearLess(n); + StringBuilder builder = new StringBuilder(); + for (int i = 0; i < n; i++) { + builder.append(ans[i][0] + " " + ans[i][1] + "\n"); + } + System.out.println(builder.toString()); + } + + public static void getNearLess(int n) { + int stackSize1 = 0; + int stackSize2 = 0; + for (int i = 0; i < n; i++) { + while (stackSize1 > 0 && arr[stack1[stackSize1 - 1]] > arr[i]) { + int curIndex = stack1[--stackSize1]; + int left = stackSize2 < 2 ? -1 : stack2[stackSize2 - 2]; + ans[curIndex][0] = left; + ans[curIndex][1] = i; + if (stackSize1 == 0 || arr[stack1[stackSize1 - 1]] != arr[curIndex]) { + stackSize2--; + } + } + if (stackSize1 != 0 && arr[stack1[stackSize1 - 1]] == arr[i]) { + stack2[stackSize2 - 1] = i; + } else { + stack2[stackSize2++] = i; + } + stack1[stackSize1++] = i; + } + while (stackSize1 != 0) { + int curIndex = stack1[--stackSize1]; + int left = stackSize2 < 2 ? -1 : stack2[stackSize2 - 2]; + ans[curIndex][0] = left; + ans[curIndex][1] = -1; + if (stackSize1 == 0 || arr[stack1[stackSize1 - 1]] != arr[curIndex]) { + stackSize2--; + } + } + } + +} diff --git a/体系学习班/class25/Code02_AllTimesMinToMax.java b/体系学习班/class25/Code02_AllTimesMinToMax.java new file mode 100644 index 0000000..2f26ac1 --- /dev/null +++ b/体系学习班/class25/Code02_AllTimesMinToMax.java @@ -0,0 +1,98 @@ +package class25; + +import java.util.Stack; + +public class Code02_AllTimesMinToMax { + + public static int max1(int[] arr) { + int max = Integer.MIN_VALUE; + for (int i = 0; i < arr.length; i++) { + for (int j = i; j < arr.length; j++) { + int minNum = Integer.MAX_VALUE; + int sum = 0; + for (int k = i; k <= j; k++) { + sum += arr[k]; + minNum = Math.min(minNum, arr[k]); + } + max = Math.max(max, minNum * sum); + } + } + return max; + } + + public static int max2(int[] arr) { + int size = arr.length; + int[] sums = new int[size]; + sums[0] = arr[0]; + for (int i = 1; i < size; i++) { + sums[i] = sums[i - 1] + arr[i]; + } + int max = Integer.MIN_VALUE; + Stack stack = new Stack(); + for (int i = 0; i < size; i++) { + while (!stack.isEmpty() && arr[stack.peek()] >= arr[i]) { + int j = stack.pop(); + max = Math.max(max, (stack.isEmpty() ? sums[i - 1] : (sums[i - 1] - sums[stack.peek()])) * arr[j]); + } + stack.push(i); + } + while (!stack.isEmpty()) { + int j = stack.pop(); + max = Math.max(max, (stack.isEmpty() ? sums[size - 1] : (sums[size - 1] - sums[stack.peek()])) * arr[j]); + } + return max; + } + + public static int[] gerenareRondomArray() { + int[] arr = new int[(int) (Math.random() * 20) + 10]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) (Math.random() * 101); + } + return arr; + } + + public static void main(String[] args) { + int testTimes = 2000000; + System.out.println("test begin"); + for (int i = 0; i < testTimes; i++) { + int[] arr = gerenareRondomArray(); + if (max1(arr) != max2(arr)) { + System.out.println("FUCK!"); + break; + } + } + System.out.println("test finish"); + } + + // 本题可以在leetcode上找到原题 + // 测试链接 : https://leetcode.com/problems/maximum-subarray-min-product/ + // 注意测试题目数量大，要取模，但是思路和课上讲的是完全一样的 + // 注意溢出的处理即可，也就是用long类型来表示累加和 + // 还有优化就是，你可以用自己手写的数组栈，来替代系统实现的栈，也会快很多 + public static int maxSumMinProduct(int[] arr) { + int size = arr.length; + long[] sums = new long[size]; + sums[0] = arr[0]; + for (int i = 1; i < size; i++) { + sums[i] = sums[i - 1] + arr[i]; + } + long max = Long.MIN_VALUE; + int[] stack = new int[size]; + int stackSize = 0; + for (int i = 0; i < size; i++) { + while (stackSize != 0 && arr[stack[stackSize - 1]] >= arr[i]) { + int j = stack[--stackSize]; + max = Math.max(max, + (stackSize == 0 ? sums[i - 1] : (sums[i - 1] - sums[stack[stackSize - 1]])) * arr[j]); + } + stack[stackSize++] = i; + } + while (stackSize != 0) { + int j = stack[--stackSize]; + max = Math.max(max, + (stackSize == 0 ? sums[size - 1] : (sums[size - 1] - sums[stack[stackSize - 1]])) * arr[j]); + } + return (int) (max % 1000000007); + } + +} diff --git a/体系学习班/class25/Code03_LargestRectangleInHistogram.java b/体系学习班/class25/Code03_LargestRectangleInHistogram.java new file mode 100644 index 0000000..d2d1ce5 --- /dev/null +++ b/体系学习班/class25/Code03_LargestRectangleInHistogram.java @@ -0,0 +1,58 @@ +package class25; + +import java.util.Stack; + +// 测试链接：https://leetcode.com/problems/largest-rectangle-in-histogram +public class Code03_LargestRectangleInHistogram { + + public static int largestRectangleArea1(int[] height) { + if (height == null || height.length == 0) { + return 0; + } + int maxArea = 0; + Stack stack = new Stack(); + for (int i = 0; i < height.length; i++) { + while (!stack.isEmpty() && height[i] <= height[stack.peek()]) { + int j = stack.pop(); + int k = stack.isEmpty() ? -1 : stack.peek(); + int curArea = (i - k - 1) * height[j]; + maxArea = Math.max(maxArea, curArea); + } + stack.push(i); + } + while (!stack.isEmpty()) { + int j = stack.pop(); + int k = stack.isEmpty() ? -1 : stack.peek(); + int curArea = (height.length - k - 1) * height[j]; + maxArea = Math.max(maxArea, curArea); + } + return maxArea; + } + + public static int largestRectangleArea2(int[] height) { + if (height == null || height.length == 0) { + return 0; + } + int N = height.length; + int[] stack = new int[N]; + int si = -1; + int maxArea = 0; + for (int i = 0; i < height.length; i++) { + while (si != -1 && height[i] <= height[stack[si]]) { + int j = stack[si--]; + int k = si == -1 ? -1 : stack[si]; + int curArea = (i - k - 1) * height[j]; + maxArea = Math.max(maxArea, curArea); + } + stack[++si] = i; + } + while (si != -1) { + int j = stack[si--]; + int k = si == -1 ? -1 : stack[si]; + int curArea = (height.length - k - 1) * height[j]; + maxArea = Math.max(maxArea, curArea); + } + return maxArea; + } + +} diff --git a/体系学习班/class25/Code04_MaximalRectangle.java b/体系学习班/class25/Code04_MaximalRectangle.java new file mode 100644 index 0000000..e2b9d37 --- /dev/null +++ b/体系学习班/class25/Code04_MaximalRectangle.java @@ -0,0 +1,48 @@ +package class25; + +import java.util.Stack; + +// 测试链接：https://leetcode.com/problems/maximal-rectangle/ +public class Code04_MaximalRectangle { + + public static int maximalRectangle(char[][] map) { + if (map == null || map.length == 0 || map[0].length == 0) { + return 0; + } + int maxArea = 0; + int[] height = new int[map[0].length]; + for (int i = 0; i < map.length; i++) { + for (int j = 0; j < map[0].length; j++) { + height[j] = map[i][j] == '0' ? 0 : height[j] + 1; + } + maxArea = Math.max(maxRecFromBottom(height), maxArea); + } + return maxArea; + } + + // height是正方图数组 + public static int maxRecFromBottom(int[] height) { + if (height == null || height.length == 0) { + return 0; + } + int maxArea = 0; + Stack stack = new Stack(); + for (int i = 0; i < height.length; i++) { + while (!stack.isEmpty() && height[i] <= height[stack.peek()]) { + int j = stack.pop(); + int k = stack.isEmpty() ? -1 : stack.peek(); + int curArea = (i - k - 1) * height[j]; + maxArea = Math.max(maxArea, curArea); + } + stack.push(i); + } + while (!stack.isEmpty()) { + int j = stack.pop(); + int k = stack.isEmpty() ? -1 : stack.peek(); + int curArea = (height.length - k - 1) * height[j]; + maxArea = Math.max(maxArea, curArea); + } + return maxArea; + } + +} diff --git a/体系学习班/class25/Code05_CountSubmatricesWithAllOnes.java b/体系学习班/class25/Code05_CountSubmatricesWithAllOnes.java new file mode 100644 index 0000000..f24d02e --- /dev/null +++ b/体系学习班/class25/Code05_CountSubmatricesWithAllOnes.java @@ -0,0 +1,81 @@ +package class25; + +// 测试链接：https://leetcode.com/problems/count-submatrices-with-all-ones +public class Code05_CountSubmatricesWithAllOnes { + + public static int numSubmat(int[][] mat) { + if (mat == null || mat.length == 0 || mat[0].length == 0) { + return 0; + } + int nums = 0; + int[] height = new int[mat[0].length]; + for (int i = 0; i < mat.length; i++) { + for (int j = 0; j < mat[0].length; j++) { + height[j] = mat[i][j] == 0 ? 0 : height[j] + 1; + } + nums += countFromBottom(height); + } + return nums; + + } + + // 比如 + // 1 + // 1 + // 1 1 + // 1 1 1 + // 1 1 1 + // 1 1 1 + // + // 2 .... 6 .... 9 + // 如上图，假设在6位置，1的高度为6 + // 在6位置的左边，离6位置最近、且小于高度6的位置是2，2位置的高度是3 + // 在6位置的右边，离6位置最近、且小于高度6的位置是9，9位置的高度是4 + // 此时我们求什么？ + // 1) 求在3~8范围上，必须以高度6作为高的矩形，有几个？ + // 2) 求在3~8范围上，必须以高度5作为高的矩形，有几个？ + // 也就是说，<=4的高度，一律不求 + // 那么，1) 求必须以位置6的高度6作为高的矩形，有几个？ + // 3..3 3..4 3..5 3..6 3..7 3..8 + // 4..4 4..5 4..6 4..7 4..8 + // 5..5 5..6 5..7 5..8 + // 6..6 6..7 6..8 + // 7..7 7..8 + // 8..8 + // 这么多！= 21 = (9 - 2 - 1) * (9 - 2) / 2 + // 这就是任何一个数字从栈里弹出的时候，计算矩形数量的方式 + public static int countFromBottom(int[] height) { + if (height == null || height.length == 0) { + return 0; + } + int nums = 0; + int[] stack = new int[height.length]; + int si = -1; + for (int i = 0; i < height.length; i++) { + while (si != -1 && height[stack[si]] >= height[i]) { + int cur = stack[si--]; + if (height[cur] > height[i]) { + int left = si == -1 ? -1 : stack[si]; + int n = i - left - 1; + int down = Math.max(left == -1 ? 0 : height[left], height[i]); + nums += (height[cur] - down) * num(n); + } + + } + stack[++si] = i; + } + while (si != -1) { + int cur = stack[si--]; + int left = si == -1 ? -1 : stack[si]; + int n = height.length - left - 1; + int down = left == -1 ? 0 : height[left]; + nums += (height[cur] - down) * num(n); + } + return nums; + } + + public static int num(int n) { + return ((n * (1 + n)) >> 1); + } + +} diff --git a/体系学习班/class26/Code01_SumOfSubarrayMinimums.java b/体系学习班/class26/Code01_SumOfSubarrayMinimums.java new file mode 100644 index 0000000..343247b --- /dev/null +++ b/体系学习班/class26/Code01_SumOfSubarrayMinimums.java @@ -0,0 +1,156 @@ +package class26; + +// 测试链接：https://leetcode.com/problems/sum-of-subarray-minimums/ +// subArrayMinSum1是暴力解 +// subArrayMinSum2是最优解的思路 +// sumSubarrayMins是最优解思路下的单调栈优化 +// Leetcode上不要提交subArrayMinSum1、subArrayMinSum2方法，因为没有考虑取摸 +// Leetcode上只提交sumSubarrayMins方法，时间复杂度O(N)，可以直接通过 +public class Code01_SumOfSubarrayMinimums { + + public static int subArrayMinSum1(int[] arr) { + int ans = 0; + for (int i = 0; i < arr.length; i++) { + for (int j = i; j < arr.length; j++) { + int min = arr[i]; + for (int k = i + 1; k <= j; k++) { + min = Math.min(min, arr[k]); + } + ans += min; + } + } + return ans; + } + + // 没有用单调栈 + public static int subArrayMinSum2(int[] arr) { + // left[i] = x : arr[i]左边，离arr[i]最近，<=arr[i]，位置在x + int[] left = leftNearLessEqual2(arr); + // right[i] = y : arr[i]右边，离arr[i]最近，< arr[i],的数，位置在y + int[] right = rightNearLess2(arr); + int ans = 0; + for (int i = 0; i < arr.length; i++) { + int start = i - left[i]; + int end = right[i] - i; + ans += start * end * arr[i]; + } + return ans; + } + + public static int[] leftNearLessEqual2(int[] arr) { + int N = arr.length; + int[] left = new int[N]; + for (int i = 0; i < N; i++) { + int ans = -1; + for (int j = i - 1; j >= 0; j--) { + if (arr[j] <= arr[i]) { + ans = j; + break; + } + } + left[i] = ans; + } + return left; + } + + public static int[] rightNearLess2(int[] arr) { + int N = arr.length; + int[] right = new int[N]; + for (int i = 0; i < N; i++) { + int ans = N; + for (int j = i + 1; j < N; j++) { + if (arr[i] > arr[j]) { + ans = j; + break; + } + } + right[i] = ans; + } + return right; + } + + public static int sumSubarrayMins(int[] arr) { + int[] stack = new int[arr.length]; + int[] left = nearLessEqualLeft(arr, stack); + int[] right = nearLessRight(arr, stack); + long ans = 0; + for (int i = 0; i < arr.length; i++) { + long start = i - left[i]; + long end = right[i] - i; + ans += start * end * (long) arr[i]; + ans %= 1000000007; + } + return (int) ans; + } + + public static int[] nearLessEqualLeft(int[] arr, int[] stack) { + int N = arr.length; + int[] left = new int[N]; + int size = 0; + for (int i = N - 1; i >= 0; i--) { + while (size != 0 && arr[i] <= arr[stack[size - 1]]) { + left[stack[--size]] = i; + } + stack[size++] = i; + } + while (size != 0) { + left[stack[--size]] = -1; + } + return left; + } + + public static int[] nearLessRight(int[] arr, int[] stack) { + int N = arr.length; + int[] right = new int[N]; + int size = 0; + for (int i = 0; i < N; i++) { + while (size != 0 && arr[stack[size - 1]] > arr[i]) { + right[stack[--size]] = i; + } + stack[size++] = i; + } + while (size != 0) { + right[stack[--size]] = N; + } + return right; + } + + public static int[] randomArray(int len, int maxValue) { + int[] ans = new int[len]; + for (int i = 0; i < len; i++) { + ans[i] = (int) (Math.random() * maxValue) + 1; + } + return ans; + } + + public static void printArray(int[] arr) { + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + public static void main(String[] args) { + int maxLen = 100; + int maxValue = 50; + int testTime = 100000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * maxLen); + int[] arr = randomArray(len, maxValue); + int ans1 = subArrayMinSum1(arr); + int ans2 = subArrayMinSum2(arr); + int ans3 = sumSubarrayMins(arr); + if (ans1 != ans2 || ans1 != ans3) { + printArray(arr); + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + System.out.println("出错了！"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class26/Code02_FibonacciProblem.java b/体系学习班/class26/Code02_FibonacciProblem.java new file mode 100644 index 0000000..7f50420 --- /dev/null +++ b/体系学习班/class26/Code02_FibonacciProblem.java @@ -0,0 +1,189 @@ +package class26; + +public class Code02_FibonacciProblem { + + public static int f1(int n) { + if (n < 1) { + return 0; + } + if (n == 1 || n == 2) { + return 1; + } + return f1(n - 1) + f1(n - 2); + } + + public static int f2(int n) { + if (n < 1) { + return 0; + } + if (n == 1 || n == 2) { + return 1; + } + int res = 1; + int pre = 1; + int tmp = 0; + for (int i = 3; i <= n; i++) { + tmp = res; + res = res + pre; + pre = tmp; + } + return res; + } + + // O(logN) + public static int f3(int n) { + if (n < 1) { + return 0; + } + if (n == 1 || n == 2) { + return 1; + } + // [ 1 ,1 ] + // [ 1, 0 ] + int[][] base = { + { 1, 1 }, + { 1, 0 } + }; + int[][] res = matrixPower(base, n - 2); + return res[0][0] + res[1][0]; + } + + public static int[][] matrixPower(int[][] m, int p) { + int[][] res = new int[m.length][m[0].length]; + for (int i = 0; i < res.length; i++) { + res[i][i] = 1; + } + // res = 矩阵中的1 + int[][] t = m;// 矩阵1次方 + for (; p != 0; p >>= 1) { + if ((p & 1) != 0) { + res = product(res, t); + } + t = product(t, t); + } + return res; + } + + // 两个矩阵乘完之后的结果返回 + public static int[][] product(int[][] a, int[][] b) { + int n = a.length; + int m = b[0].length; + int k = a[0].length; // a的列数同时也是b的行数 + int[][] ans = new int[n][m]; + for(int i = 0 ; i < n; i++) { + for(int j = 0 ; j < m;j++) { + for(int c = 0; c < k; c++) { + ans[i][j] += a[i][c] * b[c][j]; + } + } + } + return ans; + } + + public static int s1(int n) { + if (n < 1) { + return 0; + } + if (n == 1 || n == 2) { + return n; + } + return s1(n - 1) + s1(n - 2); + } + + public static int s2(int n) { + if (n < 1) { + return 0; + } + if (n == 1 || n == 2) { + return n; + } + int res = 2; + int pre = 1; + int tmp = 0; + for (int i = 3; i <= n; i++) { + tmp = res; + res = res + pre; + pre = tmp; + } + return res; + } + + public static int s3(int n) { + if (n < 1) { + return 0; + } + if (n == 1 || n == 2) { + return n; + } + int[][] base = { { 1, 1 }, { 1, 0 } }; + int[][] res = matrixPower(base, n - 2); + return 2 * res[0][0] + res[1][0]; + } + + public static int c1(int n) { + if (n < 1) { + return 0; + } + if (n == 1 || n == 2 || n == 3) { + return n; + } + return c1(n - 1) + c1(n - 3); + } + + public static int c2(int n) { + if (n < 1) { + return 0; + } + if (n == 1 || n == 2 || n == 3) { + return n; + } + int res = 3; + int pre = 2; + int prepre = 1; + int tmp1 = 0; + int tmp2 = 0; + for (int i = 4; i <= n; i++) { + tmp1 = res; + tmp2 = pre; + res = res + prepre; + pre = tmp1; + prepre = tmp2; + } + return res; + } + + public static int c3(int n) { + if (n < 1) { + return 0; + } + if (n == 1 || n == 2 || n == 3) { + return n; + } + int[][] base = { + { 1, 1, 0 }, + { 0, 0, 1 }, + { 1, 0, 0 } }; + int[][] res = matrixPower(base, n - 3); + return 3 * res[0][0] + 2 * res[1][0] + res[2][0]; + } + + public static void main(String[] args) { + int n = 19; + System.out.println(f1(n)); + System.out.println(f2(n)); + System.out.println(f3(n)); + System.out.println("==="); + + System.out.println(s1(n)); + System.out.println(s2(n)); + System.out.println(s3(n)); + System.out.println("==="); + + System.out.println(c1(n)); + System.out.println(c2(n)); + System.out.println(c3(n)); + System.out.println("==="); + + } + +} diff --git a/体系学习班/class26/Code03_ZeroLeftOneStringNumber.java b/体系学习班/class26/Code03_ZeroLeftOneStringNumber.java new file mode 100644 index 0000000..8cc3ce2 --- /dev/null +++ b/体系学习班/class26/Code03_ZeroLeftOneStringNumber.java @@ -0,0 +1,113 @@ +package class26; + +public class Code03_ZeroLeftOneStringNumber { + + public static int getNum1(int n) { + if (n < 1) { + return 0; + } + return process(1, n); + } + + public static int process(int i, int n) { + if (i == n - 1) { + return 2; + } + if (i == n) { + return 1; + } + return process(i + 1, n) + process(i + 2, n); + } + + public static int getNum2(int n) { + if (n < 1) { + return 0; + } + if (n == 1) { + return 1; + } + int pre = 1; + int cur = 1; + int tmp = 0; + for (int i = 2; i < n + 1; i++) { + tmp = cur; + cur += pre; + pre = tmp; + } + return cur; + } + + public static int getNum3(int n) { + if (n < 1) { + return 0; + } + if (n == 1 || n == 2) { + return n; + } + int[][] base = { { 1, 1 }, { 1, 0 } }; + int[][] res = matrixPower(base, n - 2); + return 2 * res[0][0] + res[1][0]; + } + + + + + + + public static int fi(int n) { + if (n < 1) { + return 0; + } + if (n == 1 || n == 2) { + return 1; + } + int[][] base = { { 1, 1 }, + { 1, 0 } }; + int[][] res = matrixPower(base, n - 2); + return res[0][0] + res[1][0]; + } + + + + + public static int[][] matrixPower(int[][] m, int p) { + int[][] res = new int[m.length][m[0].length]; + for (int i = 0; i < res.length; i++) { + res[i][i] = 1; + } + int[][] tmp = m; + for (; p != 0; p >>= 1) { + if ((p & 1) != 0) { + res = product(res, tmp); + } + tmp = product(tmp, tmp); + } + return res; + } + + // 两个矩阵乘完之后的结果返回 + public static int[][] product(int[][] a, int[][] b) { + int n = a.length; + int m = b[0].length; + int k = a[0].length; // a的列数同时也是b的行数 + int[][] ans = new int[n][m]; + for(int i = 0 ; i < n; i++) { + for(int j = 0 ; j < m;j++) { + for(int c = 0; c < k; c++) { + ans[i][j] += a[i][c] * b[c][j]; + } + } + } + return ans; + } + + public static void main(String[] args) { + for (int i = 0; i != 20; i++) { + System.out.println(getNum1(i)); + System.out.println(getNum2(i)); + System.out.println(getNum3(i)); + System.out.println("==================="); + } + + } +} diff --git a/体系学习班/class27/Code01_KMP.java b/体系学习班/class27/Code01_KMP.java new file mode 100644 index 0000000..68a51de --- /dev/null +++ b/体系学习班/class27/Code01_KMP.java @@ -0,0 +1,75 @@ +package class27; + +public class Code01_KMP { + + public static int getIndexOf(String s1, String s2) { + if (s1 == null || s2 == null || s2.length() < 1 || s1.length() < s2.length()) { + return -1; + } + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + int x = 0; + int y = 0; + // O(M) m <= n + int[] next = getNextArray(str2); + // O(N) + while (x < str1.length && y < str2.length) { + if (str1[x] == str2[y]) { + x++; + y++; + } else if (next[y] == -1) { // y == 0 + x++; + } else { + y = next[y]; + } + } + return y == str2.length ? x - y : -1; + } + + public static int[] getNextArray(char[] str2) { + if (str2.length == 1) { + return new int[] { -1 }; + } + int[] next = new int[str2.length]; + next[0] = -1; + next[1] = 0; + int i = 2; // 目前在哪个位置上求next数组的值 + int cn = 0; // 当前是哪个位置的值再和i-1位置的字符比较 + while (i < next.length) { + if (str2[i - 1] == str2[cn]) { // 配成功的时候 + next[i++] = ++cn; + } else if (cn > 0) { + cn = next[cn]; + } else { + next[i++] = 0; + } + } + return next; + } + + // for test + public static String getRandomString(int possibilities, int size) { + char[] ans = new char[(int) (Math.random() * size) + 1]; + for (int i = 0; i < ans.length; i++) { + ans[i] = (char) ((int) (Math.random() * possibilities) + 'a'); + } + return String.valueOf(ans); + } + + public static void main(String[] args) { + int possibilities = 5; + int strSize = 20; + int matchSize = 5; + int testTimes = 5000000; + System.out.println("test begin"); + for (int i = 0; i < testTimes; i++) { + String str = getRandomString(possibilities, strSize); + String match = getRandomString(possibilities, matchSize); + if (getIndexOf(str, match) != str.indexOf(match)) { + System.out.println("Oops!"); + } + } + System.out.println("test finish"); + } + +} diff --git a/体系学习班/class27/Code02_TreeEqual.java b/体系学习班/class27/Code02_TreeEqual.java new file mode 100644 index 0000000..657373a --- /dev/null +++ b/体系学习班/class27/Code02_TreeEqual.java @@ -0,0 +1,172 @@ +package class27; + +import java.util.ArrayList; + +public class Code02_TreeEqual { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int v) { + value = v; + } + } + + public static boolean containsTree1(Node big, Node small) { + if (small == null) { + return true; + } + if (big == null) { + return false; + } + if (isSameValueStructure(big, small)) { + return true; + } + return containsTree1(big.left, small) || containsTree1(big.right, small); + } + + public static boolean isSameValueStructure(Node head1, Node head2) { + if (head1 == null && head2 != null) { + return false; + } + if (head1 != null && head2 == null) { + return false; + } + if (head1 == null && head2 == null) { + return true; + } + if (head1.value != head2.value) { + return false; + } + return isSameValueStructure(head1.left, head2.left) + && isSameValueStructure(head1.right, head2.right); + } + + public static boolean containsTree2(Node big, Node small) { + if (small == null) { + return true; + } + if (big == null) { + return false; + } + ArrayList b = preSerial(big); + ArrayList s = preSerial(small); + String[] str = new String[b.size()]; + for (int i = 0; i < str.length; i++) { + str[i] = b.get(i); + } + + String[] match = new String[s.size()]; + for (int i = 0; i < match.length; i++) { + match[i] = s.get(i); + } + return getIndexOf(str, match) != -1; + } + + public static ArrayList preSerial(Node head) { + ArrayList ans = new ArrayList<>(); + pres(head, ans); + return ans; + } + + public static void pres(Node head, ArrayList ans) { + if (head == null) { + ans.add(null); + } else { + ans.add(String.valueOf(head.value)); + pres(head.left, ans); + pres(head.right, ans); + } + } + + public static int getIndexOf(String[] str1, String[] str2) { + if (str1 == null || str2 == null || str1.length < 1 || str1.length < str2.length) { + return -1; + } + int x = 0; + int y = 0; + int[] next = getNextArray(str2); + while (x < str1.length && y < str2.length) { + if (isEqual(str1[x], str2[y])) { + x++; + y++; + } else if (next[y] == -1) { + x++; + } else { + y = next[y]; + } + } + return y == str2.length ? x - y : -1; + } + + public static int[] getNextArray(String[] ms) { + if (ms.length == 1) { + return new int[] { -1 }; + } + int[] next = new int[ms.length]; + next[0] = -1; + next[1] = 0; + int i = 2; + int cn = 0; + while (i < next.length) { + if (isEqual(ms[i - 1], ms[cn])) { + next[i++] = ++cn; + } else if (cn > 0) { + cn = next[cn]; + } else { + next[i++] = 0; + } + } + return next; + } + + public static boolean isEqual(String a, String b) { + if (a == null && b == null) { + return true; + } else { + if (a == null || b == null) { + return false; + } else { + return a.equals(b); + } + } + } + + // for test + public static Node generateRandomBST(int maxLevel, int maxValue) { + return generate(1, maxLevel, maxValue); + } + + // for test + public static Node generate(int level, int maxLevel, int maxValue) { + if (level > maxLevel || Math.random() < 0.5) { + return null; + } + Node head = new Node((int) (Math.random() * maxValue)); + head.left = generate(level + 1, maxLevel, maxValue); + head.right = generate(level + 1, maxLevel, maxValue); + return head; + } + + public static void main(String[] args) { + int bigTreeLevel = 7; + int smallTreeLevel = 4; + int nodeMaxValue = 5; + int testTimes = 100000; + System.out.println("test begin"); + for (int i = 0; i < testTimes; i++) { + Node big = generateRandomBST(bigTreeLevel, nodeMaxValue); + Node small = generateRandomBST(smallTreeLevel, nodeMaxValue); + boolean ans1 = containsTree1(big, small); + boolean ans2 = containsTree2(big, small); + if (ans1 != ans2) { + System.out.println("Oops!"); + } + } + System.out.println("test finish!"); + + } + +} diff --git a/体系学习班/class27/Code03_IsRotation.java b/体系学习班/class27/Code03_IsRotation.java new file mode 100644 index 0000000..605cb81 --- /dev/null +++ b/体系学习班/class27/Code03_IsRotation.java @@ -0,0 +1,64 @@ +package class27; + +public class Code03_IsRotation { + + public static boolean isRotation(String a, String b) { + if (a == null || b == null || a.length() != b.length()) { + return false; + } + String b2 = b + b; + return getIndexOf(b2, a) != -1; + } + + // KMP Algorithm + public static int getIndexOf(String s, String m) { + if (s.length() < m.length()) { + return -1; + } + char[] ss = s.toCharArray(); + char[] ms = m.toCharArray(); + int si = 0; + int mi = 0; + int[] next = getNextArray(ms); + while (si < ss.length && mi < ms.length) { + if (ss[si] == ms[mi]) { + si++; + mi++; + } else if (next[mi] == -1) { + si++; + } else { + mi = next[mi]; + } + } + return mi == ms.length ? si - mi : -1; + } + + public static int[] getNextArray(char[] ms) { + if (ms.length == 1) { + return new int[] { -1 }; + } + int[] next = new int[ms.length]; + next[0] = -1; + next[1] = 0; + int pos = 2; + int cn = 0; + while (pos < next.length) { + if (ms[pos - 1] == ms[cn]) { + next[pos++] = ++cn; + } else if (cn > 0) { + cn = next[cn]; + } else { + next[pos++] = 0; + } + } + return next; + } + + public static void main(String[] args) { + String str1 = "yunzuocheng"; + String str2 = "zuochengyun"; + System.out.println(isRotation(str1, str2)); + + } + +} diff --git a/体系学习班/class28/Code01_Manacher.java b/体系学习班/class28/Code01_Manacher.java new file mode 100644 index 0000000..c75975f --- /dev/null +++ b/体系学习班/class28/Code01_Manacher.java @@ -0,0 +1,90 @@ +package class28; + +public class Code01_Manacher { + + public static int manacher(String s) { + if (s == null || s.length() == 0) { + return 0; + } + // "12132" -> "#1#2#1#3#2#" + char[] str = manacherString(s); + // 回文半径的大小 + int[] pArr = new int[str.length]; + int C = -1; + // 讲述中：R代表最右的扩成功的位置 + // coding：最右的扩成功位置的，再下一个位置 + int R = -1; + int max = Integer.MIN_VALUE; + for (int i = 0; i < str.length; i++) { // 0 1 2 + // R第一个违规的位置，i>= R + // i位置扩出来的答案，i位置扩的区域，至少是多大。 + pArr[i] = R > i ? Math.min(pArr[2 * C - i], R - i) : 1; + while (i + pArr[i] < str.length && i - pArr[i] > -1) { + if (str[i + pArr[i]] == str[i - pArr[i]]) + pArr[i]++; + else { + break; + } + } + if (i + pArr[i] > R) { + R = i + pArr[i]; + C = i; + } + max = Math.max(max, pArr[i]); + } + return max - 1; + } + + public static char[] manacherString(String str) { + char[] charArr = str.toCharArray(); + char[] res = new char[str.length() * 2 + 1]; + int index = 0; + for (int i = 0; i != res.length; i++) { + res[i] = (i & 1) == 0 ? '#' : charArr[index++]; + } + return res; + } + + // for test + public static int right(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = manacherString(s); + int max = 0; + for (int i = 0; i < str.length; i++) { + int L = i - 1; + int R = i + 1; + while (L >= 0 && R < str.length && str[L] == str[R]) { + L--; + R++; + } + max = Math.max(max, R - L - 1); + } + return max / 2; + } + + // for test + public static String getRandomString(int possibilities, int size) { + char[] ans = new char[(int) (Math.random() * size) + 1]; + for (int i = 0; i < ans.length; i++) { + ans[i] = (char) ((int) (Math.random() * possibilities) + 'a'); + } + return String.valueOf(ans); + } + + public static void main(String[] args) { + int possibilities = 5; + int strSize = 20; + int testTimes = 5000000; + System.out.println("test begin"); + for (int i = 0; i < testTimes; i++) { + String str = getRandomString(possibilities, strSize); + if (manacher(str) != right(str)) { + System.out.println("Oops!"); + } + } + System.out.println("test finish"); + } + +} diff --git a/体系学习班/class28/Code02_AddShortestEnd.java b/体系学习班/class28/Code02_AddShortestEnd.java new file mode 100644 index 0000000..00d961c --- /dev/null +++ b/体系学习班/class28/Code02_AddShortestEnd.java @@ -0,0 +1,54 @@ +package class28; + +public class Code02_AddShortestEnd { + + public static String shortestEnd(String s) { + if (s == null || s.length() == 0) { + return null; + } + char[] str = manacherString(s); + int[] pArr = new int[str.length]; + int C = -1; + int R = -1; + int maxContainsEnd = -1; + for (int i = 0; i != str.length; i++) { + pArr[i] = R > i ? Math.min(pArr[2 * C - i], R - i) : 1; + while (i + pArr[i] < str.length && i - pArr[i] > -1) { + if (str[i + pArr[i]] == str[i - pArr[i]]) + pArr[i]++; + else { + break; + } + } + if (i + pArr[i] > R) { + R = i + pArr[i]; + C = i; + } + if (R == str.length) { + maxContainsEnd = pArr[i]; + break; + } + } + char[] res = new char[s.length() - maxContainsEnd + 1]; + for (int i = 0; i < res.length; i++) { + res[res.length - 1 - i] = str[i * 2 + 1]; + } + return String.valueOf(res); + } + + public static char[] manacherString(String str) { + char[] charArr = str.toCharArray(); + char[] res = new char[str.length() * 2 + 1]; + int index = 0; + for (int i = 0; i != res.length; i++) { + res[i] = (i & 1) == 0 ? '#' : charArr[index++]; + } + return res; + } + + public static void main(String[] args) { + String str1 = "abcd123321"; + System.out.println(shortestEnd(str1)); + } + +} diff --git a/体系学习班/class29/Code01_FindMinKth.java b/体系学习班/class29/Code01_FindMinKth.java new file mode 100644 index 0000000..e317550 --- /dev/null +++ b/体系学习班/class29/Code01_FindMinKth.java @@ -0,0 +1,175 @@ +package class29; + +import java.util.Comparator; +import java.util.PriorityQueue; + +public class Code01_FindMinKth { + + public static class MaxHeapComparator implements Comparator { + + @Override + public int compare(Integer o1, Integer o2) { + return o2 - o1; + } + + } + + // 利用大根堆，时间复杂度O(N*logK) + public static int minKth1(int[] arr, int k) { + PriorityQueue maxHeap = new PriorityQueue<>(new MaxHeapComparator()); + for (int i = 0; i < k; i++) { + maxHeap.add(arr[i]); + } + for (int i = k; i < arr.length; i++) { + if (arr[i] < maxHeap.peek()) { + maxHeap.poll(); + maxHeap.add(arr[i]); + } + } + return maxHeap.peek(); + } + + // 改写快排，时间复杂度O(N) + // k >= 1 + public static int minKth2(int[] array, int k) { + int[] arr = copyArray(array); + return process2(arr, 0, arr.length - 1, k - 1); + } + + public static int[] copyArray(int[] arr) { + int[] ans = new int[arr.length]; + for (int i = 0; i != ans.length; i++) { + ans[i] = arr[i]; + } + return ans; + } + + // arr 第k小的数 + // process2(arr, 0, N-1, k-1) + // arr[L..R] 范围上，如果排序的话(不是真的去排序)，找位于index的数 + // index [L..R] + public static int process2(int[] arr, int L, int R, int index) { + if (L == R) { // L = =R ==INDEX + return arr[L]; + } + // 不止一个数 L + [0, R -L] + int pivot = arr[L + (int) (Math.random() * (R - L + 1))]; + int[] range = partition(arr, L, R, pivot); + if (index >= range[0] && index <= range[1]) { + return arr[index]; + } else if (index < range[0]) { + return process2(arr, L, range[0] - 1, index); + } else { + return process2(arr, range[1] + 1, R, index); + } + } + + public static int[] partition(int[] arr, int L, int R, int pivot) { + int less = L - 1; + int more = R + 1; + int cur = L; + while (cur < more) { + if (arr[cur] < pivot) { + swap(arr, ++less, cur++); + } else if (arr[cur] > pivot) { + swap(arr, cur, --more); + } else { + cur++; + } + } + return new int[] { less + 1, more - 1 }; + } + + public static void swap(int[] arr, int i1, int i2) { + int tmp = arr[i1]; + arr[i1] = arr[i2]; + arr[i2] = tmp; + } + + // 利用bfprt算法，时间复杂度O(N) + public static int minKth3(int[] array, int k) { + int[] arr = copyArray(array); + return bfprt(arr, 0, arr.length - 1, k - 1); + } + + // arr[L..R] 如果排序的话，位于index位置的数，是什么，返回 + public static int bfprt(int[] arr, int L, int R, int index) { + if (L == R) { + return arr[L]; + } + // L...R 每五个数一组 + // 每一个小组内部排好序 + // 小组的中位数组成新数组 + // 这个新数组的中位数返回 + int pivot = medianOfMedians(arr, L, R); + int[] range = partition(arr, L, R, pivot); + if (index >= range[0] && index <= range[1]) { + return arr[index]; + } else if (index < range[0]) { + return bfprt(arr, L, range[0] - 1, index); + } else { + return bfprt(arr, range[1] + 1, R, index); + } + } + + // arr[L...R] 五个数一组 + // 每个小组内部排序 + // 每个小组中位数领出来，组成marr + // marr中的中位数，返回 + public static int medianOfMedians(int[] arr, int L, int R) { + int size = R - L + 1; + int offset = size % 5 == 0 ? 0 : 1; + int[] mArr = new int[size / 5 + offset]; + for (int team = 0; team < mArr.length; team++) { + int teamFirst = L + team * 5; + // L ... L + 4 + // L +5 ... L +9 + // L +10....L+14 + mArr[team] = getMedian(arr, teamFirst, Math.min(R, teamFirst + 4)); + } + // marr中，找到中位数 + // marr(0, marr.len - 1, mArr.length / 2 ) + return bfprt(mArr, 0, mArr.length - 1, mArr.length / 2); + } + + public static int getMedian(int[] arr, int L, int R) { + insertionSort(arr, L, R); + return arr[(L + R) / 2]; + } + + public static void insertionSort(int[] arr, int L, int R) { + for (int i = L + 1; i <= R; i++) { + for (int j = i - 1; j >= L && arr[j] > arr[j + 1]; j--) { + swap(arr, j, j + 1); + } + } + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) (Math.random() * maxSize) + 1]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) (Math.random() * (maxValue + 1)); + } + return arr; + } + + public static void main(String[] args) { + int testTime = 1000000; + int maxSize = 100; + int maxValue = 100; + System.out.println("test begin"); + for (int i = 0; i < testTime; i++) { + int[] arr = generateRandomArray(maxSize, maxValue); + int k = (int) (Math.random() * arr.length) + 1; + int ans1 = minKth1(arr, k); + int ans2 = minKth2(arr, k); + int ans3 = minKth3(arr, k); + if (ans1 != ans2 || ans2 != ans3) { + System.out.println("Oops!"); + } + } + System.out.println("test finish"); + } + +} diff --git a/体系学习班/class29/Code02_MaxTopK.java b/体系学习班/class29/Code02_MaxTopK.java new file mode 100644 index 0000000..b300171 --- /dev/null +++ b/体系学习班/class29/Code02_MaxTopK.java @@ -0,0 +1,223 @@ +package class29; + +import java.util.Arrays; + +public class Code02_MaxTopK { + + // 时间复杂度O(N*logN) + // 排序+收集 + public static int[] maxTopK1(int[] arr, int k) { + if (arr == null || arr.length == 0) { + return new int[0]; + } + int N = arr.length; + k = Math.min(N, k); + Arrays.sort(arr); + int[] ans = new int[k]; + for (int i = N - 1, j = 0; j < k; i--, j++) { + ans[j] = arr[i]; + } + return ans; + } + + // 方法二，时间复杂度O(N + K*logN) + // 解释：堆 + public static int[] maxTopK2(int[] arr, int k) { + if (arr == null || arr.length == 0) { + return new int[0]; + } + int N = arr.length; + k = Math.min(N, k); + // 从底向上建堆，时间复杂度O(N) + for (int i = N - 1; i >= 0; i--) { + heapify(arr, i, N); + } + // 只把前K个数放在arr末尾，然后收集，O(K*logN) + int heapSize = N; + swap(arr, 0, --heapSize); + int count = 1; + while (heapSize > 0 && count < k) { + heapify(arr, 0, heapSize); + swap(arr, 0, --heapSize); + count++; + } + int[] ans = new int[k]; + for (int i = N - 1, j = 0; j < k; i--, j++) { + ans[j] = arr[i]; + } + return ans; + } + + public static void heapInsert(int[] arr, int index) { + while (arr[index] > arr[(index - 1) / 2]) { + swap(arr, index, (index - 1) / 2); + index = (index - 1) / 2; + } + } + + public static void heapify(int[] arr, int index, int heapSize) { + int left = index * 2 + 1; + while (left < heapSize) { + int largest = left + 1 < heapSize && arr[left + 1] > arr[left] ? left + 1 : left; + largest = arr[largest] > arr[index] ? largest : index; + if (largest == index) { + break; + } + swap(arr, largest, index); + index = largest; + left = index * 2 + 1; + } + } + + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + + // 方法三，时间复杂度O(n + k * logk) + public static int[] maxTopK3(int[] arr, int k) { + if (arr == null || arr.length == 0) { + return new int[0]; + } + int N = arr.length; + k = Math.min(N, k); + // O(N) + int num = minKth(arr, N - k); + int[] ans = new int[k]; + int index = 0; + for (int i = 0; i < N; i++) { + if (arr[i] > num) { + ans[index++] = arr[i]; + } + } + for (; index < k; index++) { + ans[index] = num; + } + // O(k*logk) + Arrays.sort(ans); + for (int L = 0, R = k - 1; L < R; L++, R--) { + swap(ans, L, R); + } + return ans; + } + + // 时间复杂度O(N) + public static int minKth(int[] arr, int index) { + int L = 0; + int R = arr.length - 1; + int pivot = 0; + int[] range = null; + while (L < R) { + pivot = arr[L + (int) (Math.random() * (R - L + 1))]; + range = partition(arr, L, R, pivot); + if (index < range[0]) { + R = range[0] - 1; + } else if (index > range[1]) { + L = range[1] + 1; + } else { + return pivot; + } + } + return arr[L]; + } + + public static int[] partition(int[] arr, int L, int R, int pivot) { + int less = L - 1; + int more = R + 1; + int cur = L; + while (cur < more) { + if (arr[cur] < pivot) { + swap(arr, ++less, cur++); + } else if (arr[cur] > pivot) { + swap(arr, cur, --more); + } else { + cur++; + } + } + return new int[] { less + 1, more - 1 }; + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + // [-? , +?] + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + return arr; + } + + // for test + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // 生成随机数组测试 + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 100; + int maxValue = 100; + boolean pass = true; + System.out.println("测试开始，没有打印出错信息说明测试通过"); + for (int i = 0; i < testTime; i++) { + int k = (int) (Math.random() * maxSize) + 1; + int[] arr = generateRandomArray(maxSize, maxValue); + + int[] arr1 = copyArray(arr); + int[] arr2 = copyArray(arr); + int[] arr3 = copyArray(arr); + + int[] ans1 = maxTopK1(arr1, k); + int[] ans2 = maxTopK2(arr2, k); + int[] ans3 = maxTopK3(arr3, k); + if (!isEqual(ans1, ans2) || !isEqual(ans1, ans3)) { + pass = false; + System.out.println("出错了！"); + printArray(ans1); + printArray(ans2); + printArray(ans3); + break; + } + } + System.out.println("测试结束了，测试了" + testTime + "组，是否所有测试用例都通过？" + (pass ? "是" : "否")); + } + +} diff --git a/体系学习班/class29/Code03_ReservoirSampling.java b/体系学习班/class29/Code03_ReservoirSampling.java new file mode 100644 index 0000000..c2ce9e3 --- /dev/null +++ b/体系学习班/class29/Code03_ReservoirSampling.java @@ -0,0 +1,94 @@ +package class29; + +public class Code03_ReservoirSampling { + + public static class RandomBox { + private int[] bag; + private int N; + private int count; + + public RandomBox(int capacity) { + bag = new int[capacity]; + N = capacity; + count = 0; + } + + private int rand(int max) { + return (int) (Math.random() * max) + 1; + } + + public void add(int num) { + count++; + if (count <= N) { + bag[count - 1] = num; + } else { + if (rand(count) <= N) { + bag[rand(N) - 1] = num; + } + } + } + + public int[] choices() { + int[] ans = new int[N]; + for (int i = 0; i < N; i++) { + ans[i] = bag[i]; + } + return ans; + } + + } + + // 请等概率返回1~i中的一个数字 + public static int random(int i) { + return (int) (Math.random() * i) + 1; + } + + public static void main(String[] args) { + System.out.println("hello"); + int test = 10000; + int ballNum = 17; + int[] count = new int[ballNum + 1]; + for (int i = 0; i < test; i++) { + int[] bag = new int[10]; + int bagi = 0; + for (int num = 1; num <= ballNum; num++) { + if (num <= 10) { + bag[bagi++] = num; + } else { // num > 10 + if (random(num) <= 10) { // 一定要把num球入袋子 + bagi = (int) (Math.random() * 10); + bag[bagi] = num; + } + } + + } + for (int num : bag) { + count[num]++; + } + } + for (int i = 0; i <= ballNum; i++) { + System.out.println(count[i]); + } + + System.out.println("hello"); + int all = 100; + int choose = 10; + int testTimes = 50000; + int[] counts = new int[all + 1]; + for (int i = 0; i < testTimes; i++) { + RandomBox box = new RandomBox(choose); + for (int num = 1; num <= all; num++) { + box.add(num); + } + int[] ans = box.choices(); + for (int j = 0; j < ans.length; j++) { + counts[ans[j]]++; + } + } + + for (int i = 0; i < counts.length; i++) { + System.out.println(i + " times : " + counts[i]); + } + + } +} diff --git a/体系学习班/class30/Code01_MorrisTraversal.java b/体系学习班/class30/Code01_MorrisTraversal.java new file mode 100644 index 0000000..2e3a4a4 --- /dev/null +++ b/体系学习班/class30/Code01_MorrisTraversal.java @@ -0,0 +1,230 @@ +package class30; + +public class Code01_MorrisTraversal { + + public static class Node { + public int value; + Node left; + Node right; + + public Node(int data) { + this.value = data; + } + } + + public static void process(Node root) { + if (root == null) { + return; + } + // 1 + process(root.left); + // 2 + process(root.right); + // 3 + } + + public static void morris(Node head) { + if (head == null) { + return; + } + Node cur = head; + Node mostRight = null; + while (cur != null) { + mostRight = cur.left; + if (mostRight != null) { + while (mostRight.right != null && mostRight.right != cur) { + mostRight = mostRight.right; + } + if (mostRight.right == null) { + mostRight.right = cur; + cur = cur.left; + continue; + } else { + mostRight.right = null; + } + } + cur = cur.right; + } + } + + public static void morrisPre(Node head) { + if (head == null) { + return; + } + Node cur = head; + Node mostRight = null; + while (cur != null) { + mostRight = cur.left; + if (mostRight != null) { + while (mostRight.right != null && mostRight.right != cur) { + mostRight = mostRight.right; + } + if (mostRight.right == null) { + System.out.print(cur.value + " "); + mostRight.right = cur; + cur = cur.left; + continue; + } else { + mostRight.right = null; + } + } else { + System.out.print(cur.value + " "); + } + cur = cur.right; + } + System.out.println(); + } + + public static void morrisIn(Node head) { + if (head == null) { + return; + } + Node cur = head; + Node mostRight = null; + while (cur != null) { + mostRight = cur.left; + if (mostRight != null) { + while (mostRight.right != null && mostRight.right != cur) { + mostRight = mostRight.right; + } + if (mostRight.right == null) { + mostRight.right = cur; + cur = cur.left; + continue; + } else { + mostRight.right = null; + } + } + System.out.print(cur.value + " "); + cur = cur.right; + } + System.out.println(); + } + + public static void morrisPos(Node head) { + if (head == null) { + return; + } + Node cur = head; + Node mostRight = null; + while (cur != null) { + mostRight = cur.left; + if (mostRight != null) { + while (mostRight.right != null && mostRight.right != cur) { + mostRight = mostRight.right; + } + if (mostRight.right == null) { + mostRight.right = cur; + cur = cur.left; + continue; + } else { + mostRight.right = null; + printEdge(cur.left); + } + } + cur = cur.right; + } + printEdge(head); + System.out.println(); + } + + public static void printEdge(Node head) { + Node tail = reverseEdge(head); + Node cur = tail; + while (cur != null) { + System.out.print(cur.value + " "); + cur = cur.right; + } + reverseEdge(tail); + } + + public static Node reverseEdge(Node from) { + Node pre = null; + Node next = null; + while (from != null) { + next = from.right; + from.right = pre; + pre = from; + from = next; + } + return pre; + } + + // for test -- print tree + public static void printTree(Node head) { + System.out.println("Binary Tree:"); + printInOrder(head, 0, "H", 17); + System.out.println(); + } + + public static void printInOrder(Node head, int height, String to, int len) { + if (head == null) { + return; + } + printInOrder(head.right, height + 1, "v", len); + String val = to + head.value + to; + int lenM = val.length(); + int lenL = (len - lenM) / 2; + int lenR = len - lenM - lenL; + val = getSpace(lenL) + val + getSpace(lenR); + System.out.println(getSpace(height * len) + val); + printInOrder(head.left, height + 1, "^", len); + } + + public static String getSpace(int num) { + String space = " "; + StringBuffer buf = new StringBuffer(""); + for (int i = 0; i < num; i++) { + buf.append(space); + } + return buf.toString(); + } + + public static boolean isBST(Node head) { + if (head == null) { + return true; + } + Node cur = head; + Node mostRight = null; + Integer pre = null; + boolean ans = true; + while (cur != null) { + mostRight = cur.left; + if (mostRight != null) { + while (mostRight.right != null && mostRight.right != cur) { + mostRight = mostRight.right; + } + if (mostRight.right == null) { + mostRight.right = cur; + cur = cur.left; + continue; + } else { + mostRight.right = null; + } + } + if (pre != null && pre >= cur.value) { + ans = false; + } + pre = cur.value; + cur = cur.right; + } + return ans; + } + + public static void main(String[] args) { + Node head = new Node(4); + head.left = new Node(2); + head.right = new Node(6); + head.left.left = new Node(1); + head.left.right = new Node(3); + head.right.left = new Node(5); + head.right.right = new Node(7); + printTree(head); + morrisIn(head); + morrisPre(head); + morrisPos(head); + printTree(head); + + } + +} diff --git a/体系学习班/class30/Code02_MinDepth.java b/体系学习班/class30/Code02_MinDepth.java new file mode 100644 index 0000000..fa2160c --- /dev/null +++ b/体系学习班/class30/Code02_MinDepth.java @@ -0,0 +1,88 @@ +package class30; + +// 本题测试链接 : https://leetcode-cn.com/problems/minimum-depth-of-binary-tree/ +public class Code02_MinDepth { + + // 不提交这个类 + public static class TreeNode { + public int val; + public TreeNode left; + public TreeNode right; + + public TreeNode(int x) { + val = x; + } + } + + // 下面的方法是一般解 + public static int minDepth1(TreeNode head) { + if (head == null) { + return 0; + } + return p(head); + } + + // 返回x为头的树，最小深度是多少 + public static int p(TreeNode x) { + if (x.left == null && x.right == null) { + return 1; + } + // 左右子树起码有一个不为空 + int leftH = Integer.MAX_VALUE; + if (x.left != null) { + leftH = p(x.left); + } + int rightH = Integer.MAX_VALUE; + if (x.right != null) { + rightH = p(x.right); + } + return 1 + Math.min(leftH, rightH); + } + + // 下面的方法是morris遍历的解 + public static int minDepth2(TreeNode head) { + if (head == null) { + return 0; + } + TreeNode cur = head; + TreeNode mostRight = null; + int curLevel = 0; + int minHeight = Integer.MAX_VALUE; + while (cur != null) { + mostRight = cur.left; + if (mostRight != null) { + int rightBoardSize = 1; + while (mostRight.right != null && mostRight.right != cur) { + rightBoardSize++; + mostRight = mostRight.right; + } + if (mostRight.right == null) { // 第一次到达 + curLevel++; + mostRight.right = cur; + cur = cur.left; + continue; + } else { // 第二次到达 + if (mostRight.left == null) { + minHeight = Math.min(minHeight, curLevel); + } + curLevel -= rightBoardSize; + mostRight.right = null; + } + } else { // 只有一次到达 + curLevel++; + } + cur = cur.right; + } + int finalRight = 1; + cur = head; + while (cur.right != null) { + finalRight++; + cur = cur.right; + } + if (cur.left == null && cur.right == null) { + minHeight = Math.min(minHeight, finalRight); + } + return minHeight; + } + +} diff --git a/体系学习班/class31/Code01_SegmentTree.java b/体系学习班/class31/Code01_SegmentTree.java new file mode 100644 index 0000000..282c20a --- /dev/null +++ b/体系学习班/class31/Code01_SegmentTree.java @@ -0,0 +1,245 @@ +package class31; + +public class Code01_SegmentTree { + + public static class SegmentTree { + // arr[]为原序列的信息从0开始，但在arr里是从1开始的 + // sum[]模拟线段树维护区间和 + // lazy[]为累加和懒惰标记 + // change[]为更新的值 + // update[]为更新慵懒标记 + private int MAXN; + private int[] arr; + private int[] sum; + private int[] lazy; + private int[] change; + private boolean[] update; + + public SegmentTree(int[] origin) { + MAXN = origin.length + 1; + arr = new int[MAXN]; // arr[0] 不用 从1开始使用 + for (int i = 1; i < MAXN; i++) { + arr[i] = origin[i - 1]; + } + sum = new int[MAXN << 2]; // 用来支持脑补概念中，某一个范围的累加和信息 + lazy = new int[MAXN << 2]; // 用来支持脑补概念中，某一个范围沒有往下傳遞的纍加任務 + change = new int[MAXN << 2]; // 用来支持脑补概念中，某一个范围有没有更新操作的任务 + update = new boolean[MAXN << 2]; // 用来支持脑补概念中，某一个范围更新任务，更新成了什么 + } + + private void pushUp(int rt) { + sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; + } + + // 之前的，所有懒增加，和懒更新，从父范围，发给左右两个子范围 + // 分发策略是什么 + // ln表示左子树元素结点个数，rn表示右子树结点个数 + private void pushDown(int rt, int ln, int rn) { + if (update[rt]) { + update[rt << 1] = true; + update[rt << 1 | 1] = true; + change[rt << 1] = change[rt]; + change[rt << 1 | 1] = change[rt]; + lazy[rt << 1] = 0; + lazy[rt << 1 | 1] = 0; + sum[rt << 1] = change[rt] * ln; + sum[rt << 1 | 1] = change[rt] * rn; + update[rt] = false; + } + if (lazy[rt] != 0) { + lazy[rt << 1] += lazy[rt]; + sum[rt << 1] += lazy[rt] * ln; + lazy[rt << 1 | 1] += lazy[rt]; + sum[rt << 1 | 1] += lazy[rt] * rn; + lazy[rt] = 0; + } + } + + // 在初始化阶段，先把sum数组，填好 + // 在arr[l~r]范围上，去build，1~N， + // rt : 这个范围在sum中的下标 + public void build(int l, int r, int rt) { + if (l == r) { + sum[rt] = arr[l]; + return; + } + int mid = (l + r) >> 1; + build(l, mid, rt << 1); + build(mid + 1, r, rt << 1 | 1); + pushUp(rt); + } + + + // L~R 所有的值变成C + // l~r rt + public void update(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + update[rt] = true; + change[rt] = C; + sum[rt] = C * (r - l + 1); + lazy[rt] = 0; + return; + } + // 当前任务躲不掉，无法懒更新，要往下发 + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + update(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + update(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + // L~R, C 任务！ + // rt，l~r + public void add(int L, int R, int C, int l, int r, int rt) { + // 任务如果把此时的范围全包了！ + if (L <= l && r <= R) { + sum[rt] += C * (r - l + 1); + lazy[rt] += C; + return; + } + // 任务没有把你全包！ + // l r mid = (l+r)/2 + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + // L~R + if (L <= mid) { + add(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + add(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + // 1~6 累加和是多少？ 1~8 rt + public long query(int L, int R, int l, int r, int rt) { + if (L <= l && r <= R) { + return sum[rt]; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + long ans = 0; + if (L <= mid) { + ans += query(L, R, l, mid, rt << 1); + } + if (R > mid) { + ans += query(L, R, mid + 1, r, rt << 1 | 1); + } + return ans; + } + + } + + public static class Right { + public int[] arr; + + public Right(int[] origin) { + arr = new int[origin.length + 1]; + for (int i = 0; i < origin.length; i++) { + arr[i + 1] = origin[i]; + } + } + + public void update(int L, int R, int C) { + for (int i = L; i <= R; i++) { + arr[i] = C; + } + } + + public void add(int L, int R, int C) { + for (int i = L; i <= R; i++) { + arr[i] += C; + } + } + + public long query(int L, int R) { + long ans = 0; + for (int i = L; i <= R; i++) { + ans += arr[i]; + } + return ans; + } + + } + + public static int[] genarateRandomArray(int len, int max) { + int size = (int) (Math.random() * len) + 1; + int[] origin = new int[size]; + for (int i = 0; i < size; i++) { + origin[i] = (int) (Math.random() * max) - (int) (Math.random() * max); + } + return origin; + } + + public static boolean test() { + int len = 100; + int max = 1000; + int testTimes = 5000; + int addOrUpdateTimes = 1000; + int queryTimes = 500; + for (int i = 0; i < testTimes; i++) { + int[] origin = genarateRandomArray(len, max); + SegmentTree seg = new SegmentTree(origin); + int S = 1; + int N = origin.length; + int root = 1; + seg.build(S, N, root); + Right rig = new Right(origin); + for (int j = 0; j < addOrUpdateTimes; j++) { + int num1 = (int) (Math.random() * N) + 1; + int num2 = (int) (Math.random() * N) + 1; + int L = Math.min(num1, num2); + int R = Math.max(num1, num2); + int C = (int) (Math.random() * max) - (int) (Math.random() * max); + if (Math.random() < 0.5) { + seg.add(L, R, C, S, N, root); + rig.add(L, R, C); + } else { + seg.update(L, R, C, S, N, root); + rig.update(L, R, C); + } + } + for (int k = 0; k < queryTimes; k++) { + int num1 = (int) (Math.random() * N) + 1; + int num2 = (int) (Math.random() * N) + 1; + int L = Math.min(num1, num2); + int R = Math.max(num1, num2); + long ans1 = seg.query(L, R, S, N, root); + long ans2 = rig.query(L, R); + if (ans1 != ans2) { + return false; + } + } + } + return true; + } + + public static void main(String[] args) { + int[] origin = { 2, 1, 1, 2, 3, 4, 5 }; + SegmentTree seg = new SegmentTree(origin); + int S = 1; // 整个区间的开始位置，规定从1开始，不从0开始 -> 固定 + int N = origin.length; // 整个区间的结束位置，规定能到N，不是N-1 -> 固定 + int root = 1; // 整棵树的头节点位置，规定是1，不是0 -> 固定 + int L = 2; // 操作区间的开始位置 -> 可变 + int R = 5; // 操作区间的结束位置 -> 可变 + int C = 4; // 要加的数字或者要更新的数字 -> 可变 + // 区间生成，必须在[S,N]整个范围上build + seg.build(S, N, root); + // 区间修改，可以改变L、R和C的值，其他值不可改变 + seg.add(L, R, C, S, N, root); + // 区间更新，可以改变L、R和C的值，其他值不可改变 + seg.update(L, R, C, S, N, root); + // 区间查询，可以改变L和R的值，其他值不可改变 + long sum = seg.query(L, R, S, N, root); + System.out.println(sum); + + System.out.println("对数器测试开始..."); + System.out.println("测试结果 : " + (test() ? "通过" : "未通过")); + + } + +} diff --git a/体系学习班/class31/Code02_FallingSquares.java b/体系学习班/class31/Code02_FallingSquares.java new file mode 100644 index 0000000..2f3b2a2 --- /dev/null +++ b/体系学习班/class31/Code02_FallingSquares.java @@ -0,0 +1,109 @@ +package class31; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.List; +import java.util.TreeSet; + +public class Code02_FallingSquares { + + public static class SegmentTree { + private int[] max; + private int[] change; + private boolean[] update; + + public SegmentTree(int size) { + int N = size + 1; + max = new int[N << 2]; + + change = new int[N << 2]; + update = new boolean[N << 2]; + } + + private void pushUp(int rt) { + max[rt] = Math.max(max[rt << 1], max[rt << 1 | 1]); + } + + // ln表示左子树元素结点个数，rn表示右子树结点个数 + private void pushDown(int rt, int ln, int rn) { + if (update[rt]) { + update[rt << 1] = true; + update[rt << 1 | 1] = true; + change[rt << 1] = change[rt]; + change[rt << 1 | 1] = change[rt]; + max[rt << 1] = change[rt]; + max[rt << 1 | 1] = change[rt]; + update[rt] = false; + } + } + + public void update(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + update[rt] = true; + change[rt] = C; + max[rt] = C; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + update(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + update(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + public int query(int L, int R, int l, int r, int rt) { + if (L <= l && r <= R) { + return max[rt]; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + int left = 0; + int right = 0; + if (L <= mid) { + left = query(L, R, l, mid, rt << 1); + } + if (R > mid) { + right = query(L, R, mid + 1, r, rt << 1 | 1); + } + return Math.max(left, right); + } + + } + + public HashMap index(int[][] positions) { + TreeSet pos = new TreeSet<>(); + for (int[] arr : positions) { + pos.add(arr[0]); + pos.add(arr[0] + arr[1] - 1); + } + HashMap map = new HashMap<>(); + int count = 0; + for (Integer index : pos) { + map.put(index, ++count); + } + return map; + } + + public List fallingSquares(int[][] positions) { + HashMap map = index(positions); + int N = map.size(); + SegmentTree segmentTree = new SegmentTree(N); + int max = 0; + List res = new ArrayList<>(); + // 每落一个正方形，收集一下，所有东西组成的图像，最高高度是什么 + for (int[] arr : positions) { + int L = map.get(arr[0]); + int R = map.get(arr[0] + arr[1] - 1); + int height = segmentTree.query(L, R, 1, N, 1) + arr[1]; + max = Math.max(max, height); + res.add(max); + segmentTree.update(L, R, height, 1, N, 1); + } + return res; + } + +} diff --git a/体系学习班/class32/Code01_IndexTree.java b/体系学习班/class32/Code01_IndexTree.java new file mode 100644 index 0000000..4c4bdab --- /dev/null +++ b/体系学习班/class32/Code01_IndexTree.java @@ -0,0 +1,83 @@ +package class32; + +public class Code01_IndexTree { + + // 下标从1开始！ + public static class IndexTree { + + private int[] tree; + private int N; + + // 0位置弃而不用！ + public IndexTree(int size) { + N = size; + tree = new int[N + 1]; + } + + // 1~index 累加和是多少？ + public int sum(int index) { + int ret = 0; + while (index > 0) { + ret += tree[index]; + index -= index & -index; + } + return ret; + } + + // index & -index : 提取出index最右侧的1出来 + // index : 0011001000 + // index & -index : 0000001000 + public void add(int index, int d) { + while (index <= N) { + tree[index] += d; + index += index & -index; + } + } + } + + public static class Right { + private int[] nums; + private int N; + + public Right(int size) { + N = size + 1; + nums = new int[N + 1]; + } + + public int sum(int index) { + int ret = 0; + for (int i = 1; i <= index; i++) { + ret += nums[i]; + } + return ret; + } + + public void add(int index, int d) { + nums[index] += d; + } + + } + + public static void main(String[] args) { + int N = 100; + int V = 100; + int testTime = 2000000; + IndexTree tree = new IndexTree(N); + Right test = new Right(N); + System.out.println("test begin"); + for (int i = 0; i < testTime; i++) { + int index = (int) (Math.random() * N) + 1; + if (Math.random() <= 0.5) { + int add = (int) (Math.random() * V); + tree.add(index, add); + test.add(index, add); + } else { + if (tree.sum(index) != test.sum(index)) { + System.out.println("Oops!"); + } + } + } + System.out.println("test finish"); + } + +} diff --git a/体系学习班/class32/Code02_IndexTree2D.java b/体系学习班/class32/Code02_IndexTree2D.java new file mode 100644 index 0000000..12bb6e2 --- /dev/null +++ b/体系学习班/class32/Code02_IndexTree2D.java @@ -0,0 +1,57 @@ +package class32; + +// 测试链接：https://leetcode.com/problems/range-sum-query-2d-mutable +// 但这个题是付费题目 +// 提交时把类名、构造函数名从Code02_IndexTree2D改成NumMatrix +public class Code02_IndexTree2D { + private int[][] tree; + private int[][] nums; + private int N; + private int M; + + public Code02_IndexTree2D(int[][] matrix) { + if (matrix.length == 0 || matrix[0].length == 0) { + return; + } + N = matrix.length; + M = matrix[0].length; + tree = new int[N + 1][M + 1]; + nums = new int[N][M]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + update(i, j, matrix[i][j]); + } + } + } + + private int sum(int row, int col) { + int sum = 0; + for (int i = row + 1; i > 0; i -= i & (-i)) { + for (int j = col + 1; j > 0; j -= j & (-j)) { + sum += tree[i][j]; + } + } + return sum; + } + + public void update(int row, int col, int val) { + if (N == 0 || M == 0) { + return; + } + int add = val - nums[row][col]; + nums[row][col] = val; + for (int i = row + 1; i <= N; i += i & (-i)) { + for (int j = col + 1; j <= M; j += j & (-j)) { + tree[i][j] += add; + } + } + } + + public int sumRegion(int row1, int col1, int row2, int col2) { + if (N == 0 || M == 0) { + return 0; + } + return sum(row2, col2) + sum(row1 - 1, col1 - 1) - sum(row1 - 1, col2) - sum(row2, col1 - 1); + } + +} diff --git a/体系学习班/class32/Code03_AC1.java b/体系学习班/class32/Code03_AC1.java new file mode 100644 index 0000000..a6874e5 --- /dev/null +++ b/体系学习班/class32/Code03_AC1.java @@ -0,0 +1,105 @@ +package class32; + +import java.util.LinkedList; +import java.util.Queue; + +public class Code03_AC1 { + + public static class Node { + public int end; // 有多少个字符串以该节点结尾 + public Node fail; + public Node[] nexts; + + public Node() { + end = 0; + fail = null; + nexts = new Node[26]; + } + } + + public static class ACAutomation { + private Node root; + + public ACAutomation() { + root = new Node(); + } + + // 你有多少个匹配串，就调用多少次insert + public void insert(String s) { + char[] str = s.toCharArray(); + Node cur = root; + int index = 0; + for (int i = 0; i < str.length; i++) { + index = str[i] - 'a'; + if (cur.nexts[index] == null) { + Node next = new Node(); + cur.nexts[index] = next; + } + cur = cur.nexts[index]; + } + cur.end++; + } + + public void build() { + Queue queue = new LinkedList<>(); + queue.add(root); + Node cur = null; + Node cfail = null; + while (!queue.isEmpty()) { + cur = queue.poll(); // 父 + for (int i = 0; i < 26; i++) { // 下级所有的路 + if (cur.nexts[i] != null) { // 该路下有子节点 + cur.nexts[i].fail = root; // 初始时先设置一个值 + cfail = cur.fail; + while (cfail != null) { // cur不是头节点 + if (cfail.nexts[i] != null) { + cur.nexts[i].fail = cfail.nexts[i]; + break; + } + cfail = cfail.fail; + } + queue.add(cur.nexts[i]); + } + } + } + } + + public int containNum(String content) { + char[] str = content.toCharArray(); + Node cur = root; + Node follow = null; + int index = 0; + int ans = 0; + for (int i = 0; i < str.length; i++) { + index = str[i] - 'a'; + while (cur.nexts[index] == null && cur != root) { + cur = cur.fail; + } + cur = cur.nexts[index] != null ? cur.nexts[index] : root; + follow = cur; + while (follow != root) { + if (follow.end == -1) { + break; + } + { // 不同的需求，在这一段{ }之间修改 + ans += follow.end; + follow.end = -1; + } // 不同的需求，在这一段{ }之间修改 + follow = follow.fail; + } + } + return ans; + } + + } + + public static void main(String[] args) { + ACAutomation ac = new ACAutomation(); + ac.insert("dhe"); + ac.insert("he"); + ac.insert("c"); + ac.build(); + System.out.println(ac.containNum("cdhe")); + } + +} diff --git a/体系学习班/class32/Code04_AC2.java b/体系学习班/class32/Code04_AC2.java new file mode 100644 index 0000000..e684de3 --- /dev/null +++ b/体系学习班/class32/Code04_AC2.java @@ -0,0 +1,125 @@ +package class32; + +import java.util.ArrayList; +import java.util.LinkedList; +import java.util.List; +import java.util.Queue; + +public class Code04_AC2 { + + // 前缀树的节点 + public static class Node { + // 如果一个node，end为空，不是结尾 + // 如果end不为空，表示这个点是某个字符串的结尾，end的值就是这个字符串 + public String end; + // 只有在上面的end变量不为空的时候，endUse才有意义 + // 表示，这个字符串之前有没有加入过答案 + public boolean endUse; + public Node fail; + public Node[] nexts; + + public Node() { + endUse = false; + end = null; + fail = null; + nexts = new Node[26]; + } + } + + public static class ACAutomation { + private Node root; + + public ACAutomation() { + root = new Node(); + } + + public void insert(String s) { + char[] str = s.toCharArray(); + Node cur = root; + int index = 0; + for (int i = 0; i < str.length; i++) { + index = str[i] - 'a'; + if (cur.nexts[index] == null) { + cur.nexts[index] = new Node(); + } + cur = cur.nexts[index]; + } + cur.end = s; + } + + public void build() { + Queue queue = new LinkedList<>(); + queue.add(root); + Node cur = null; + Node cfail = null; + while (!queue.isEmpty()) { + // 某个父亲，cur + cur = queue.poll(); + for (int i = 0; i < 26; i++) { // 所有的路 + // cur -> 父亲 i号儿子，必须把i号儿子的fail指针设置好！ + if (cur.nexts[i] != null) { // 如果真的有i号儿子 + cur.nexts[i].fail = root; + cfail = cur.fail; + while (cfail != null) { + if (cfail.nexts[i] != null) { + cur.nexts[i].fail = cfail.nexts[i]; + break; + } + cfail = cfail.fail; + } + queue.add(cur.nexts[i]); + } + } + } + } + + // 大文章：content + public List containWords(String content) { + char[] str = content.toCharArray(); + Node cur = root; + Node follow = null; + int index = 0; + List ans = new ArrayList<>(); + for (int i = 0; i < str.length; i++) { + index = str[i] - 'a'; // 路 + // 如果当前字符在这条路上没配出来，就随着fail方向走向下条路径 + while (cur.nexts[index] == null && cur != root) { + cur = cur.fail; + } + // 1) 现在来到的路径，是可以继续匹配的 + // 2) 现在来到的节点，就是前缀树的根节点 + cur = cur.nexts[index] != null ? cur.nexts[index] : root; + follow = cur; + while (follow != root) { + if (follow.endUse) { + break; + } + // 不同的需求，在这一段之间修改 + if (follow.end != null) { + ans.add(follow.end); + follow.endUse = true; + } + // 不同的需求，在这一段之间修改 + follow = follow.fail; + } + } + return ans; + } + + } + + public static void main(String[] args) { + ACAutomation ac = new ACAutomation(); + ac.insert("dhe"); + ac.insert("he"); + ac.insert("abcdheks"); + // 设置fail指针 + ac.build(); + + List contains = ac.containWords("abcdhekskdjfafhasldkflskdjhwqaeruv"); + for (String word : contains) { + System.out.println(word); + } + } + +} diff --git a/体系学习班/class33/Hash.java b/体系学习班/class33/Hash.java new file mode 100644 index 0000000..c9c01f3 --- /dev/null +++ b/体系学习班/class33/Hash.java @@ -0,0 +1,49 @@ +package class33; + +import java.security.MessageDigest; +import java.security.NoSuchAlgorithmException; +import java.security.Security; + +// 需要自己找一下javax.xml.bind的jar，然后导入到项目 +import javax.xml.bind.DatatypeConverter; + +public class Hash { + + private MessageDigest hash; + + public Hash(String algorithm) { + try { + hash = MessageDigest.getInstance(algorithm); + } catch (NoSuchAlgorithmException e) { + e.printStackTrace(); + } + } + + public String hashCode(String input) { + return DatatypeConverter.printHexBinary(hash.digest(input.getBytes())).toUpperCase(); + } + + public static void main(String[] args) { + System.out.println("支持的算法 : "); + for (String str : Security.getAlgorithms("MessageDigest")) { + System.out.println(str); + } + System.out.println("======="); + + String algorithm = "MD5"; + Hash hash = new Hash(algorithm); + + String input1 = "zuochengyunzuochengyun1"; + String input2 = "zuochengyunzuochengyun2"; + String input3 = "zuochengyunzuochengyun3"; + String input4 = "zuochengyunzuochengyun4"; + String input5 = "zuochengyunzuochengyun5"; + System.out.println(hash.hashCode(input1)); + System.out.println(hash.hashCode(input2)); + System.out.println(hash.hashCode(input3)); + System.out.println(hash.hashCode(input4)); + System.out.println(hash.hashCode(input5)); + + } + +} diff --git a/体系学习班/class34/ReadMe.java b/体系学习班/class34/ReadMe.java new file mode 100644 index 0000000..16e42b9 --- /dev/null +++ b/体系学习班/class34/ReadMe.java @@ -0,0 +1,2 @@ +// 本章并无code，因为资源限制类题目输入需要的条件较多并且真的实现代码量巨大 +// 面试中这类题目出现也就和面试官聊解法，不会有代码实现的要求 \ No newline at end of file diff --git a/体系学习班/class35/Code01_AVLTreeMap.java b/体系学习班/class35/Code01_AVLTreeMap.java new file mode 100644 index 0000000..caafd46 --- /dev/null +++ b/体系学习班/class35/Code01_AVLTreeMap.java @@ -0,0 +1,257 @@ +package class35; + +public class Code01_AVLTreeMap { + + public static class AVLNode, V> { + public K k; + public V v; + public AVLNode l; + public AVLNode r; + public int h; + + public AVLNode(K key, V value) { + k = key; + v = value; + h = 1; + } + } + + public static class AVLTreeMap, V> { + private AVLNode root; + private int size; + + public AVLTreeMap() { + root = null; + size = 0; + } + + private AVLNode rightRotate(AVLNode cur) { + AVLNode left = cur.l; + cur.l = left.r; + left.r = cur; + cur.h = Math.max((cur.l != null ? cur.l.h : 0), (cur.r != null ? cur.r.h : 0)) + 1; + left.h = Math.max((left.l != null ? left.l.h : 0), (left.r != null ? left.r.h : 0)) + 1; + return left; + } + + private AVLNode leftRotate(AVLNode cur) { + AVLNode right = cur.r; + cur.r = right.l; + right.l = cur; + cur.h = Math.max((cur.l != null ? cur.l.h : 0), (cur.r != null ? cur.r.h : 0)) + 1; + right.h = Math.max((right.l != null ? right.l.h : 0), (right.r != null ? right.r.h : 0)) + 1; + return right; + } + + private AVLNode maintain(AVLNode cur) { + if (cur == null) { + return null; + } + int leftHeight = cur.l != null ? cur.l.h : 0; + int rightHeight = cur.r != null ? cur.r.h : 0; + if (Math.abs(leftHeight - rightHeight) > 1) { + if (leftHeight > rightHeight) { + int leftLeftHeight = cur.l != null && cur.l.l != null ? cur.l.l.h : 0; + int leftRightHeight = cur.l != null && cur.l.r != null ? cur.l.r.h : 0; + if (leftLeftHeight >= leftRightHeight) { + cur = rightRotate(cur); + } else { + cur.l = leftRotate(cur.l); + cur = rightRotate(cur); + } + } else { + int rightLeftHeight = cur.r != null && cur.r.l != null ? cur.r.l.h : 0; + int rightRightHeight = cur.r != null && cur.r.r != null ? cur.r.r.h : 0; + if (rightRightHeight >= rightLeftHeight) { + cur = leftRotate(cur); + } else { + cur.r = rightRotate(cur.r); + cur = leftRotate(cur); + } + } + } + return cur; + } + + private AVLNode findLastIndex(K key) { + AVLNode pre = root; + AVLNode cur = root; + while (cur != null) { + pre = cur; + if (key.compareTo(cur.k) == 0) { + break; + } else if (key.compareTo(cur.k) < 0) { + cur = cur.l; + } else { + cur = cur.r; + } + } + return pre; + } + + private AVLNode findLastNoSmallIndex(K key) { + AVLNode ans = null; + AVLNode cur = root; + while (cur != null) { + if (key.compareTo(cur.k) == 0) { + ans = cur; + break; + } else if (key.compareTo(cur.k) < 0) { + ans = cur; + cur = cur.l; + } else { + cur = cur.r; + } + } + return ans; + } + + private AVLNode findLastNoBigIndex(K key) { + AVLNode ans = null; + AVLNode cur = root; + while (cur != null) { + if (key.compareTo(cur.k) == 0) { + ans = cur; + break; + } else if (key.compareTo(cur.k) < 0) { + cur = cur.l; + } else { + ans = cur; + cur = cur.r; + } + } + return ans; + } + + private AVLNode add(AVLNode cur, K key, V value) { + if (cur == null) { + return new AVLNode(key, value); + } else { + if (key.compareTo(cur.k) < 0) { + cur.l = add(cur.l, key, value); + } else { + cur.r = add(cur.r, key, value); + } + cur.h = Math.max(cur.l != null ? cur.l.h : 0, cur.r != null ? cur.r.h : 0) + 1; + return maintain(cur); + } + } + + // 在cur这棵树上，删掉key所代表的节点 + // 返回cur这棵树的新头部 + private AVLNode delete(AVLNode cur, K key) { + if (key.compareTo(cur.k) > 0) { + cur.r = delete(cur.r, key); + } else if (key.compareTo(cur.k) < 0) { + cur.l = delete(cur.l, key); + } else { + if (cur.l == null && cur.r == null) { + cur = null; + } else if (cur.l == null && cur.r != null) { + cur = cur.r; + } else if (cur.l != null && cur.r == null) { + cur = cur.l; + } else { + AVLNode des = cur.r; + while (des.l != null) { + des = des.l; + } + cur.r = delete(cur.r, des.k); + des.l = cur.l; + des.r = cur.r; + cur = des; + } + } + if (cur != null) { + cur.h = Math.max(cur.l != null ? cur.l.h : 0, cur.r != null ? cur.r.h : 0) + 1; + } + return maintain(cur); + } + + public int size() { + return size; + } + + public boolean containsKey(K key) { + if (key == null) { + return false; + } + AVLNode lastNode = findLastIndex(key); + return lastNode != null && key.compareTo(lastNode.k) == 0 ? true : false; + } + + public void put(K key, V value) { + if (key == null) { + return; + } + AVLNode lastNode = findLastIndex(key); + if (lastNode != null && key.compareTo(lastNode.k) == 0) { + lastNode.v = value; + } else { + size++; + root = add(root, key, value); + } + } + + public void remove(K key) { + if (key == null) { + return; + } + if (containsKey(key)) { + size--; + root = delete(root, key); + } + } + + public V get(K key) { + if (key == null) { + return null; + } + AVLNode lastNode = findLastIndex(key); + if (lastNode != null && key.compareTo(lastNode.k) == 0) { + return lastNode.v; + } + return null; + } + + public K firstKey() { + if (root == null) { + return null; + } + AVLNode cur = root; + while (cur.l != null) { + cur = cur.l; + } + return cur.k; + } + + public K lastKey() { + if (root == null) { + return null; + } + AVLNode cur = root; + while (cur.r != null) { + cur = cur.r; + } + return cur.k; + } + + public K floorKey(K key) { + if (key == null) { + return null; + } + AVLNode lastNoBigNode = findLastNoBigIndex(key); + return lastNoBigNode == null ? null : lastNoBigNode.k; + } + + public K ceilingKey(K key) { + if (key == null) { + return null; + } + AVLNode lastNoSmallNode = findLastNoSmallIndex(key); + return lastNoSmallNode == null ? null : lastNoSmallNode.k; + } + + } + +} diff --git a/体系学习班/class36/Code01_SizeBalancedTreeMap.java b/体系学习班/class36/Code01_SizeBalancedTreeMap.java new file mode 100644 index 0000000..5077b7a --- /dev/null +++ b/体系学习班/class36/Code01_SizeBalancedTreeMap.java @@ -0,0 +1,360 @@ +package class36; + +public class Code01_SizeBalancedTreeMap { + + public static class SBTNode, V> { + public K key; + public V value; + public SBTNode l; + public SBTNode r; + public int size; // 不同的key的数量 + + public SBTNode(K key, V value) { + this.key = key; + this.value = value; + size = 1; + } + } + + public static class SizeBalancedTreeMap, V> { + private SBTNode root; + + private SBTNode rightRotate(SBTNode cur) { + SBTNode leftNode = cur.l; + cur.l = leftNode.r; + leftNode.r = cur; + leftNode.size = cur.size; + cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1; + return leftNode; + } + + private SBTNode leftRotate(SBTNode cur) { + SBTNode rightNode = cur.r; + cur.r = rightNode.l; + rightNode.l = cur; + rightNode.size = cur.size; + cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1; + return rightNode; + } + + private SBTNode maintain(SBTNode cur) { + if (cur == null) { + return null; + } + int leftSize = cur.l != null ? cur.l.size : 0; + int leftLeftSize = cur.l != null && cur.l.l != null ? cur.l.l.size : 0; + int leftRightSize = cur.l != null && cur.l.r != null ? cur.l.r.size : 0; + int rightSize = cur.r != null ? cur.r.size : 0; + int rightLeftSize = cur.r != null && cur.r.l != null ? cur.r.l.size : 0; + int rightRightSize = cur.r != null && cur.r.r != null ? cur.r.r.size : 0; + if (leftLeftSize > rightSize) { + cur = rightRotate(cur); + cur.r = maintain(cur.r); + cur = maintain(cur); + } else if (leftRightSize > rightSize) { + cur.l = leftRotate(cur.l); + cur = rightRotate(cur); + cur.l = maintain(cur.l); + cur.r = maintain(cur.r); + cur = maintain(cur); + } else if (rightRightSize > leftSize) { + cur = leftRotate(cur); + cur.l = maintain(cur.l); + cur = maintain(cur); + } else if (rightLeftSize > leftSize) { + cur.r = rightRotate(cur.r); + cur = leftRotate(cur); + cur.l = maintain(cur.l); + cur.r = maintain(cur.r); + cur = maintain(cur); + } + return cur; + } + + private SBTNode findLastIndex(K key) { + SBTNode pre = root; + SBTNode cur = root; + while (cur != null) { + pre = cur; + if (key.compareTo(cur.key) == 0) { + break; + } else if (key.compareTo(cur.key) < 0) { + cur = cur.l; + } else { + cur = cur.r; + } + } + return pre; + } + + private SBTNode findLastNoSmallIndex(K key) { + SBTNode ans = null; + SBTNode cur = root; + while (cur != null) { + if (key.compareTo(cur.key) == 0) { + ans = cur; + break; + } else if (key.compareTo(cur.key) < 0) { + ans = cur; + cur = cur.l; + } else { + cur = cur.r; + } + } + return ans; + } + + private SBTNode findLastNoBigIndex(K key) { + SBTNode ans = null; + SBTNode cur = root; + while (cur != null) { + if (key.compareTo(cur.key) == 0) { + ans = cur; + break; + } else if (key.compareTo(cur.key) < 0) { + cur = cur.l; + } else { + ans = cur; + cur = cur.r; + } + } + return ans; + } + + // 现在，以cur为头的树上，新增，加(key, value)这样的记录 + // 加完之后，会对cur做检查，该调整调整 + // 返回，调整完之后，整棵树的新头部 + private SBTNode add(SBTNode cur, K key, V value) { + if (cur == null) { + return new SBTNode(key, value); + } else { + cur.size++; + if (key.compareTo(cur.key) < 0) { + cur.l = add(cur.l, key, value); + } else { + cur.r = add(cur.r, key, value); + } + return maintain(cur); + } + } + + // 在cur这棵树上，删掉key所代表的节点 + // 返回cur这棵树的新头部 + private SBTNode delete(SBTNode cur, K key) { + cur.size--; + if (key.compareTo(cur.key) > 0) { + cur.r = delete(cur.r, key); + } else if (key.compareTo(cur.key) < 0) { + cur.l = delete(cur.l, key); + } else { // 当前要删掉cur + if (cur.l == null && cur.r == null) { + // free cur memory -> C++ + cur = null; + } else if (cur.l == null && cur.r != null) { + // free cur memory -> C++ + cur = cur.r; + } else if (cur.l != null && cur.r == null) { + // free cur memory -> C++ + cur = cur.l; + } else { // 有左有右 + SBTNode pre = null; + SBTNode des = cur.r; + des.size--; + while (des.l != null) { + pre = des; + des = des.l; + des.size--; + } + if (pre != null) { + pre.l = des.r; + des.r = cur.r; + } + des.l = cur.l; + des.size = des.l.size + (des.r == null ? 0 : des.r.size) + 1; + // free cur memory -> C++ + cur = des; + } + } + // cur = maintain(cur); + return cur; + } + + private SBTNode getIndex(SBTNode cur, int kth) { + if (kth == (cur.l != null ? cur.l.size : 0) + 1) { + return cur; + } else if (kth <= (cur.l != null ? cur.l.size : 0)) { + return getIndex(cur.l, kth); + } else { + return getIndex(cur.r, kth - (cur.l != null ? cur.l.size : 0) - 1); + } + } + + public int size() { + return root == null ? 0 : root.size; + } + + public boolean containsKey(K key) { + if (key == null) { + throw new RuntimeException("invalid parameter."); + } + SBTNode lastNode = findLastIndex(key); + return lastNode != null && key.compareTo(lastNode.key) == 0 ? true : false; + } + + // （key，value） put -> 有序表 新增、改value + public void put(K key, V value) { + if (key == null) { + throw new RuntimeException("invalid parameter."); + } + SBTNode lastNode = findLastIndex(key); + if (lastNode != null && key.compareTo(lastNode.key) == 0) { + lastNode.value = value; + } else { + root = add(root, key, value); + } + } + + public void remove(K key) { + if (key == null) { + throw new RuntimeException("invalid parameter."); + } + if (containsKey(key)) { + root = delete(root, key); + } + } + + public K getIndexKey(int index) { + if (index < 0 || index >= this.size()) { + throw new RuntimeException("invalid parameter."); + } + return getIndex(root, index + 1).key; + } + + public V getIndexValue(int index) { + if (index < 0 || index >= this.size()) { + throw new RuntimeException("invalid parameter."); + } + return getIndex(root, index + 1).value; + } + + public V get(K key) { + if (key == null) { + throw new RuntimeException("invalid parameter."); + } + SBTNode lastNode = findLastIndex(key); + if (lastNode != null && key.compareTo(lastNode.key) == 0) { + return lastNode.value; + } else { + return null; + } + } + + public K firstKey() { + if (root == null) { + return null; + } + SBTNode cur = root; + while (cur.l != null) { + cur = cur.l; + } + return cur.key; + } + + public K lastKey() { + if (root == null) { + return null; + } + SBTNode cur = root; + while (cur.r != null) { + cur = cur.r; + } + return cur.key; + } + + public K floorKey(K key) { + if (key == null) { + throw new RuntimeException("invalid parameter."); + } + SBTNode lastNoBigNode = findLastNoBigIndex(key); + return lastNoBigNode == null ? null : lastNoBigNode.key; + } + + public K ceilingKey(K key) { + if (key == null) { + throw new RuntimeException("invalid parameter."); + } + SBTNode lastNoSmallNode = findLastNoSmallIndex(key); + return lastNoSmallNode == null ? null : lastNoSmallNode.key; + } + + } + + // for test + public static void printAll(SBTNode head) { + System.out.println("Binary Tree:"); + printInOrder(head, 0, "H", 17); + System.out.println(); + } + + // for test + public static void printInOrder(SBTNode head, int height, String to, int len) { + if (head == null) { + return; + } + printInOrder(head.r, height + 1, "v", len); + String val = to + "(" + head.key + "," + head.value + ")" + to; + int lenM = val.length(); + int lenL = (len - lenM) / 2; + int lenR = len - lenM - lenL; + val = getSpace(lenL) + val + getSpace(lenR); + System.out.println(getSpace(height * len) + val); + printInOrder(head.l, height + 1, "^", len); + } + + // for test + public static String getSpace(int num) { + String space = " "; + StringBuffer buf = new StringBuffer(""); + for (int i = 0; i < num; i++) { + buf.append(space); + } + return buf.toString(); + } + + public static void main(String[] args) { + SizeBalancedTreeMap sbt = new SizeBalancedTreeMap(); + sbt.put("d", 4); + sbt.put("c", 3); + sbt.put("a", 1); + sbt.put("b", 2); + // sbt.put("e", 5); + sbt.put("g", 7); + sbt.put("f", 6); + sbt.put("h", 8); + sbt.put("i", 9); + sbt.put("a", 111); + System.out.println(sbt.get("a")); + sbt.put("a", 1); + System.out.println(sbt.get("a")); + for (int i = 0; i < sbt.size(); i++) { + System.out.println(sbt.getIndexKey(i) + " , " + sbt.getIndexValue(i)); + } + printAll(sbt.root); + System.out.println(sbt.firstKey()); + System.out.println(sbt.lastKey()); + System.out.println(sbt.floorKey("g")); + System.out.println(sbt.ceilingKey("g")); + System.out.println(sbt.floorKey("e")); + System.out.println(sbt.ceilingKey("e")); + System.out.println(sbt.floorKey("")); + System.out.println(sbt.ceilingKey("")); + System.out.println(sbt.floorKey("j")); + System.out.println(sbt.ceilingKey("j")); + sbt.remove("d"); + printAll(sbt.root); + sbt.remove("f"); + printAll(sbt.root); + + } + +} diff --git a/体系学习班/class36/Code02_SkipListMap.java b/体系学习班/class36/Code02_SkipListMap.java new file mode 100644 index 0000000..a450e78 --- /dev/null +++ b/体系学习班/class36/Code02_SkipListMap.java @@ -0,0 +1,248 @@ +package class36; + +import java.util.ArrayList; + +public class Code02_SkipListMap { + + // 跳表的节点定义 + public static class SkipListNode, V> { + public K key; + public V val; + public ArrayList> nextNodes; + + public SkipListNode(K k, V v) { + key = k; + val = v; + nextNodes = new ArrayList>(); + } + + // 遍历的时候，如果是往右遍历到的null(next == null), 遍历结束 + // 头(null), 头节点的null，认为最小 + // node -> 头，node(null, "") node.isKeyLess(!null) true + // node里面的key是否比otherKey小，true，不是false + public boolean isKeyLess(K otherKey) { + // otherKey == null -> false + return otherKey != null && (key == null || key.compareTo(otherKey) < 0); + } + + public boolean isKeyEqual(K otherKey) { + return (key == null && otherKey == null) + || (key != null && otherKey != null && key.compareTo(otherKey) == 0); + } + + } + + public static class SkipListMap, V> { + private static final double PROBABILITY = 0.5; // < 0.5 继续做，>=0.5 停 + private SkipListNode head; + private int size; + private int maxLevel; + + public SkipListMap() { + head = new SkipListNode(null, null); + head.nextNodes.add(null); // 0 + size = 0; + maxLevel = 0; + } + + // 从最高层开始，一路找下去， + // 最终，找到第0层的 mostRightLessNodeInTree(K key) { + if (key == null) { + return null; + } + int level = maxLevel; + SkipListNode cur = head; + while (level >= 0) { // 从上层跳下层 + // cur level -> level-1 + cur = mostRightLessNodeInLevel(key, cur, level--); + } + return cur; + } + + // 在level层里，如何往右移动 + // 现在来到的节点是cur，来到了cur的level层，在level层上，找到 mostRightLessNodeInLevel(K key, + SkipListNode cur, + int level) { + SkipListNode next = cur.nextNodes.get(level); + while (next != null && next.isKeyLess(key)) { + cur = next; + next = cur.nextNodes.get(level); + } + return cur; + } + + public boolean containsKey(K key) { + if (key == null) { + return false; + } + SkipListNode less = mostRightLessNodeInTree(key); + SkipListNode next = less.nextNodes.get(0); + return next != null && next.isKeyEqual(key); + } + + // 新增、改value + public void put(K key, V value) { + if (key == null) { + return; + } + // 0层上，最右一个，< key 的Node -> >key + SkipListNode less = mostRightLessNodeInTree(key); + SkipListNode find = less.nextNodes.get(0); + if (find != null && find.isKeyEqual(key)) { + find.val = value; + } else { // find == null 8 7 9 + size++; + int newNodeLevel = 0; + while (Math.random() < PROBABILITY) { + newNodeLevel++; + } + // newNodeLevel + while (newNodeLevel > maxLevel) { + head.nextNodes.add(null); + maxLevel++; + } + SkipListNode newNode = new SkipListNode(key, value); + for (int i = 0; i <= newNodeLevel; i++) { + newNode.nextNodes.add(null); + } + int level = maxLevel; + SkipListNode pre = head; + while (level >= 0) { + // level 层中，找到最右的 < key 的节点 + pre = mostRightLessNodeInLevel(key, pre, level); + if (level <= newNodeLevel) { + newNode.nextNodes.set(level, pre.nextNodes.get(level)); + pre.nextNodes.set(level, newNode); + } + level--; + } + } + } + + public V get(K key) { + if (key == null) { + return null; + } + SkipListNode less = mostRightLessNodeInTree(key); + SkipListNode next = less.nextNodes.get(0); + return next != null && next.isKeyEqual(key) ? next.val : null; + } + + public void remove(K key) { + if (containsKey(key)) { + size--; + int level = maxLevel; + SkipListNode pre = head; + while (level >= 0) { + pre = mostRightLessNodeInLevel(key, pre, level); + SkipListNode next = pre.nextNodes.get(level); + // 1）在这一层中，pre下一个就是key + // 2）在这一层中，pre的下一个key是>要删除key + if (next != null && next.isKeyEqual(key)) { + // free delete node memory -> C++ + // level : pre -> next(key) -> ... + pre.nextNodes.set(level, next.nextNodes.get(level)); + } + // 在level层只有一个节点了，就是默认节点head + if (level != 0 && pre == head && pre.nextNodes.get(level) == null) { + head.nextNodes.remove(level); + maxLevel--; + } + level--; + } + } + } + + public K firstKey() { + return head.nextNodes.get(0) != null ? head.nextNodes.get(0).key : null; + } + + public K lastKey() { + int level = maxLevel; + SkipListNode cur = head; + while (level >= 0) { + SkipListNode next = cur.nextNodes.get(level); + while (next != null) { + cur = next; + next = cur.nextNodes.get(level); + } + level--; + } + return cur.key; + } + + public K ceilingKey(K key) { + if (key == null) { + return null; + } + SkipListNode less = mostRightLessNodeInTree(key); + SkipListNode next = less.nextNodes.get(0); + return next != null ? next.key : null; + } + + public K floorKey(K key) { + if (key == null) { + return null; + } + SkipListNode less = mostRightLessNodeInTree(key); + SkipListNode next = less.nextNodes.get(0); + return next != null && next.isKeyEqual(key) ? next.key : less.key; + } + + public int size() { + return size; + } + + } + + // for test + public static void printAll(SkipListMap obj) { + for (int i = obj.maxLevel; i >= 0; i--) { + System.out.print("Level " + i + " : "); + SkipListNode cur = obj.head; + while (cur.nextNodes.get(i) != null) { + SkipListNode next = cur.nextNodes.get(i); + System.out.print("(" + next.key + " , " + next.val + ") "); + cur = next; + } + System.out.println(); + } + } + + public static void main(String[] args) { + SkipListMap test = new SkipListMap<>(); + printAll(test); + System.out.println("======================"); + test.put("A", "10"); + printAll(test); + System.out.println("======================"); + test.remove("A"); + printAll(test); + System.out.println("======================"); + test.put("E", "E"); + test.put("B", "B"); + test.put("A", "A"); + test.put("F", "F"); + test.put("C", "C"); + test.put("D", "D"); + printAll(test); + System.out.println("======================"); + System.out.println(test.containsKey("B")); + System.out.println(test.containsKey("Z")); + System.out.println(test.firstKey()); + System.out.println(test.lastKey()); + System.out.println(test.floorKey("D")); + System.out.println(test.ceilingKey("D")); + System.out.println("======================"); + test.remove("D"); + printAll(test); + System.out.println("======================"); + System.out.println(test.floorKey("D")); + System.out.println(test.ceilingKey("D")); + + + } + +} diff --git a/体系学习班/class37/Code01_CountofRangeSum.java b/体系学习班/class37/Code01_CountofRangeSum.java new file mode 100644 index 0000000..d79fa02 --- /dev/null +++ b/体系学习班/class37/Code01_CountofRangeSum.java @@ -0,0 +1,223 @@ +package class37; + +import java.util.HashSet; + +public class Code01_CountofRangeSum { + + public static int countRangeSum1(int[] nums, int lower, int upper) { + int n = nums.length; + long[] sums = new long[n + 1]; + for (int i = 0; i < n; ++i) + sums[i + 1] = sums[i] + nums[i]; + return countWhileMergeSort(sums, 0, n + 1, lower, upper); + } + + private static int countWhileMergeSort(long[] sums, int start, int end, int lower, int upper) { + if (end - start <= 1) + return 0; + int mid = (start + end) / 2; + int count = countWhileMergeSort(sums, start, mid, lower, upper) + + countWhileMergeSort(sums, mid, end, lower, upper); + int j = mid, k = mid, t = mid; + long[] cache = new long[end - start]; + for (int i = start, r = 0; i < mid; ++i, ++r) { + while (k < end && sums[k] - sums[i] < lower) + k++; + while (j < end && sums[j] - sums[i] <= upper) + j++; + while (t < end && sums[t] < sums[i]) + cache[r++] = sums[t++]; + cache[r] = sums[i]; + count += j - k; + } + System.arraycopy(cache, 0, sums, start, t - start); + return count; + } + + public static class SBTNode { + public long key; + public SBTNode l; + public SBTNode r; + public long size; // 不同key的size + public long all; // 总的size + + public SBTNode(long k) { + key = k; + size = 1; + all = 1; + } + } + + public static class SizeBalancedTreeSet { + private SBTNode root; + private HashSet set = new HashSet<>(); + + private SBTNode rightRotate(SBTNode cur) { + long same = cur.all - (cur.l != null ? cur.l.all : 0) - (cur.r != null ? cur.r.all : 0); + SBTNode leftNode = cur.l; + cur.l = leftNode.r; + leftNode.r = cur; + leftNode.size = cur.size; + cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1; + // all modify + leftNode.all = cur.all; + cur.all = (cur.l != null ? cur.l.all : 0) + (cur.r != null ? cur.r.all : 0) + same; + return leftNode; + } + + private SBTNode leftRotate(SBTNode cur) { + long same = cur.all - (cur.l != null ? cur.l.all : 0) - (cur.r != null ? cur.r.all : 0); + SBTNode rightNode = cur.r; + cur.r = rightNode.l; + rightNode.l = cur; + rightNode.size = cur.size; + cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1; + // all modify + rightNode.all = cur.all; + cur.all = (cur.l != null ? cur.l.all : 0) + (cur.r != null ? cur.r.all : 0) + same; + return rightNode; + } + + private SBTNode maintain(SBTNode cur) { + if (cur == null) { + return null; + } + long leftSize = cur.l != null ? cur.l.size : 0; + long leftLeftSize = cur.l != null && cur.l.l != null ? cur.l.l.size : 0; + long leftRightSize = cur.l != null && cur.l.r != null ? cur.l.r.size : 0; + long rightSize = cur.r != null ? cur.r.size : 0; + long rightLeftSize = cur.r != null && cur.r.l != null ? cur.r.l.size : 0; + long rightRightSize = cur.r != null && cur.r.r != null ? cur.r.r.size : 0; + if (leftLeftSize > rightSize) { + cur = rightRotate(cur); + cur.r = maintain(cur.r); + cur = maintain(cur); + } else if (leftRightSize > rightSize) { + cur.l = leftRotate(cur.l); + cur = rightRotate(cur); + cur.l = maintain(cur.l); + cur.r = maintain(cur.r); + cur = maintain(cur); + } else if (rightRightSize > leftSize) { + cur = leftRotate(cur); + cur.l = maintain(cur.l); + cur = maintain(cur); + } else if (rightLeftSize > leftSize) { + cur.r = rightRotate(cur.r); + cur = leftRotate(cur); + cur.l = maintain(cur.l); + cur.r = maintain(cur.r); + cur = maintain(cur); + } + return cur; + } + + private SBTNode add(SBTNode cur, long key, boolean contains) { + if (cur == null) { + return new SBTNode(key); + } else { + cur.all++; + if (key == cur.key) { + return cur; + } else { // 还在左滑或者右滑 + if (!contains) { + cur.size++; + } + if (key < cur.key) { + cur.l = add(cur.l, key, contains); + } else { + cur.r = add(cur.r, key, contains); + } + return maintain(cur); + } + } + } + + public void add(long sum) { + boolean contains = set.contains(sum); + root = add(root, sum, contains); + set.add(sum); + } + + public long lessKeySize(long key) { + SBTNode cur = root; + long ans = 0; + while (cur != null) { + if (key == cur.key) { + return ans + (cur.l != null ? cur.l.all : 0); + } else if (key < cur.key) { + cur = cur.l; + } else { + ans += cur.all - (cur.r != null ? cur.r.all : 0); + cur = cur.r; + } + } + return ans; + } + + // > 7 8... + // <8 ...<=7 + public long moreKeySize(long key) { + return root != null ? (root.all - lessKeySize(key + 1)) : 0; + } + + } + + public static int countRangeSum2(int[] nums, int lower, int upper) { + // 黑盒，加入数字（前缀和），不去重，可以接受重复数字 + // < num , 有几个数？ + SizeBalancedTreeSet treeSet = new SizeBalancedTreeSet(); + long sum = 0; + int ans = 0; + treeSet.add(0);// 一个数都没有的时候，就已经有一个前缀和累加和为0， + for (int i = 0; i < nums.length; i++) { + sum += nums[i]; + // [sum - upper, sum - lower] + // [10, 20] ? + // < 10 ? < 21 ? + long a = treeSet.lessKeySize(sum - lower + 1); + long b = treeSet.lessKeySize(sum - upper); + ans += a - b; + treeSet.add(sum); + } + return ans; + } + + // for test + public static void printArray(int[] arr) { + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static int[] generateArray(int len, int varible) { + int[] arr = new int[len]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) (Math.random() * varible); + } + return arr; + } + + public static void main(String[] args) { + int len = 200; + int varible = 50; + for (int i = 0; i < 10000; i++) { + int[] test = generateArray(len, varible); + int lower = (int) (Math.random() * varible) - (int) (Math.random() * varible); + int upper = lower + (int) (Math.random() * varible); + int ans1 = countRangeSum1(test, lower, upper); + int ans2 = countRangeSum2(test, lower, upper); + if (ans1 != ans2) { + printArray(test); + System.out.println(lower); + System.out.println(upper); + System.out.println(ans1); + System.out.println(ans2); + } + } + + } + +} diff --git a/体系学习班/class37/Code02_SlidingWindowMedian.java b/体系学习班/class37/Code02_SlidingWindowMedian.java new file mode 100644 index 0000000..2f3fa63 --- /dev/null +++ b/体系学习班/class37/Code02_SlidingWindowMedian.java @@ -0,0 +1,228 @@ +package class37; + +public class Code02_SlidingWindowMedian { + + public static class SBTNode> { + public K key; + public SBTNode l; + public SBTNode r; + public int size; + + public SBTNode(K k) { + key = k; + size = 1; + } + } + + public static class SizeBalancedTreeMap> { + private SBTNode root; + + private SBTNode rightRotate(SBTNode cur) { + SBTNode leftNode = cur.l; + cur.l = leftNode.r; + leftNode.r = cur; + leftNode.size = cur.size; + cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1; + return leftNode; + } + + private SBTNode leftRotate(SBTNode cur) { + SBTNode rightNode = cur.r; + cur.r = rightNode.l; + rightNode.l = cur; + rightNode.size = cur.size; + cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1; + return rightNode; + } + + private SBTNode maintain(SBTNode cur) { + if (cur == null) { + return null; + } + int leftSize = cur.l != null ? cur.l.size : 0; + int leftLeftSize = cur.l != null && cur.l.l != null ? cur.l.l.size : 0; + int leftRightSize = cur.l != null && cur.l.r != null ? cur.l.r.size : 0; + int rightSize = cur.r != null ? cur.r.size : 0; + int rightLeftSize = cur.r != null && cur.r.l != null ? cur.r.l.size : 0; + int rightRightSize = cur.r != null && cur.r.r != null ? cur.r.r.size : 0; + if (leftLeftSize > rightSize) { + cur = rightRotate(cur); + cur.r = maintain(cur.r); + cur = maintain(cur); + } else if (leftRightSize > rightSize) { + cur.l = leftRotate(cur.l); + cur = rightRotate(cur); + cur.l = maintain(cur.l); + cur.r = maintain(cur.r); + cur = maintain(cur); + } else if (rightRightSize > leftSize) { + cur = leftRotate(cur); + cur.l = maintain(cur.l); + cur = maintain(cur); + } else if (rightLeftSize > leftSize) { + cur.r = rightRotate(cur.r); + cur = leftRotate(cur); + cur.l = maintain(cur.l); + cur.r = maintain(cur.r); + cur = maintain(cur); + } + return cur; + } + + private SBTNode findLastIndex(K key) { + SBTNode pre = root; + SBTNode cur = root; + while (cur != null) { + pre = cur; + if (key.compareTo(cur.key) == 0) { + break; + } else if (key.compareTo(cur.key) < 0) { + cur = cur.l; + } else { + cur = cur.r; + } + } + return pre; + } + + private SBTNode add(SBTNode cur, K key) { + if (cur == null) { + return new SBTNode(key); + } else { + cur.size++; + if (key.compareTo(cur.key) < 0) { + cur.l = add(cur.l, key); + } else { + cur.r = add(cur.r, key); + } + return maintain(cur); + } + } + + private SBTNode delete(SBTNode cur, K key) { + cur.size--; + if (key.compareTo(cur.key) > 0) { + cur.r = delete(cur.r, key); + } else if (key.compareTo(cur.key) < 0) { + cur.l = delete(cur.l, key); + } else { + if (cur.l == null && cur.r == null) { + // free cur memory -> C++ + cur = null; + } else if (cur.l == null && cur.r != null) { + // free cur memory -> C++ + cur = cur.r; + } else if (cur.l != null && cur.r == null) { + // free cur memory -> C++ + cur = cur.l; + } else { + SBTNode pre = null; + SBTNode des = cur.r; + des.size--; + while (des.l != null) { + pre = des; + des = des.l; + des.size--; + } + if (pre != null) { + pre.l = des.r; + des.r = cur.r; + } + des.l = cur.l; + des.size = des.l.size + (des.r == null ? 0 : des.r.size) + 1; + // free cur memory -> C++ + cur = des; + } + } + return cur; + } + + private SBTNode getIndex(SBTNode cur, int kth) { + if (kth == (cur.l != null ? cur.l.size : 0) + 1) { + return cur; + } else if (kth <= (cur.l != null ? cur.l.size : 0)) { + return getIndex(cur.l, kth); + } else { + return getIndex(cur.r, kth - (cur.l != null ? cur.l.size : 0) - 1); + } + } + + public int size() { + return root == null ? 0 : root.size; + } + + public boolean containsKey(K key) { + if (key == null) { + throw new RuntimeException("invalid parameter."); + } + SBTNode lastNode = findLastIndex(key); + return lastNode != null && key.compareTo(lastNode.key) == 0 ? true : false; + } + + public void add(K key) { + if (key == null) { + throw new RuntimeException("invalid parameter."); + } + SBTNode lastNode = findLastIndex(key); + if (lastNode == null || key.compareTo(lastNode.key) != 0) { + root = add(root, key); + } + } + + public void remove(K key) { + if (key == null) { + throw new RuntimeException("invalid parameter."); + } + if (containsKey(key)) { + root = delete(root, key); + } + } + + public K getIndexKey(int index) { + if (index < 0 || index >= this.size()) { + throw new RuntimeException("invalid parameter."); + } + return getIndex(root, index + 1).key; + } + + } + + public static class Node implements Comparable { + public int index; + public int value; + + public Node(int i, int v) { + index = i; + value = v; + } + + @Override + public int compareTo(Node o) { + return value != o.value ? Integer.valueOf(value).compareTo(o.value) + : Integer.valueOf(index).compareTo(o.index); + } + } + + public static double[] medianSlidingWindow(int[] nums, int k) { + SizeBalancedTreeMap map = new SizeBalancedTreeMap<>(); + for (int i = 0; i < k - 1; i++) { + map.add(new Node(i, nums[i])); + } + double[] ans = new double[nums.length - k + 1]; + int index = 0; + for (int i = k - 1; i < nums.length; i++) { + map.add(new Node(i, nums[i])); + if (map.size() % 2 == 0) { + Node upmid = map.getIndexKey(map.size() / 2 - 1); + Node downmid = map.getIndexKey(map.size() / 2); + ans[index++] = ((double) upmid.value + (double) downmid.value) / 2; + } else { + Node mid = map.getIndexKey(map.size() / 2); + ans[index++] = (double) mid.value; + } + map.remove(new Node(i - k + 1, nums[i - k + 1])); + } + return ans; + } + +} diff --git a/体系学习班/class37/Code03_AddRemoveGetIndexGreat.java b/体系学习班/class37/Code03_AddRemoveGetIndexGreat.java new file mode 100644 index 0000000..2c2436a --- /dev/null +++ b/体系学习班/class37/Code03_AddRemoveGetIndexGreat.java @@ -0,0 +1,259 @@ +package class37; + +import java.util.ArrayList; + +public class Code03_AddRemoveGetIndexGreat { + + public static class SBTNode { + public V value; + public SBTNode l; + public SBTNode r; + public int size; + + public SBTNode(V v) { + value = v; + size = 1; + } + } + + public static class SbtList { + private SBTNode root; + + private SBTNode rightRotate(SBTNode cur) { + SBTNode leftNode = cur.l; + cur.l = leftNode.r; + leftNode.r = cur; + leftNode.size = cur.size; + cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1; + return leftNode; + } + + private SBTNode leftRotate(SBTNode cur) { + SBTNode rightNode = cur.r; + cur.r = rightNode.l; + rightNode.l = cur; + rightNode.size = cur.size; + cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1; + return rightNode; + } + + private SBTNode maintain(SBTNode cur) { + if (cur == null) { + return null; + } + int leftSize = cur.l != null ? cur.l.size : 0; + int leftLeftSize = cur.l != null && cur.l.l != null ? cur.l.l.size : 0; + int leftRightSize = cur.l != null && cur.l.r != null ? cur.l.r.size : 0; + int rightSize = cur.r != null ? cur.r.size : 0; + int rightLeftSize = cur.r != null && cur.r.l != null ? cur.r.l.size : 0; + int rightRightSize = cur.r != null && cur.r.r != null ? cur.r.r.size : 0; + if (leftLeftSize > rightSize) { + cur = rightRotate(cur); + cur.r = maintain(cur.r); + cur = maintain(cur); + } else if (leftRightSize > rightSize) { + cur.l = leftRotate(cur.l); + cur = rightRotate(cur); + cur.l = maintain(cur.l); + cur.r = maintain(cur.r); + cur = maintain(cur); + } else if (rightRightSize > leftSize) { + cur = leftRotate(cur); + cur.l = maintain(cur.l); + cur = maintain(cur); + } else if (rightLeftSize > leftSize) { + cur.r = rightRotate(cur.r); + cur = leftRotate(cur); + cur.l = maintain(cur.l); + cur.r = maintain(cur.r); + cur = maintain(cur); + } + return cur; + } + + private SBTNode add(SBTNode root, int index, SBTNode cur) { + if (root == null) { + return cur; + } + root.size++; + int leftAndHeadSize = (root.l != null ? root.l.size : 0) + 1; + if (index < leftAndHeadSize) { + root.l = add(root.l, index, cur); + } else { + root.r = add(root.r, index - leftAndHeadSize, cur); + } + root = maintain(root); + return root; + } + + private SBTNode remove(SBTNode root, int index) { + root.size--; + int rootIndex = root.l != null ? root.l.size : 0; + if (index != rootIndex) { + if (index < rootIndex) { + root.l = remove(root.l, index); + } else { + root.r = remove(root.r, index - rootIndex - 1); + } + return root; + } + if (root.l == null && root.r == null) { + return null; + } + if (root.l == null) { + return root.r; + } + if (root.r == null) { + return root.l; + } + SBTNode pre = null; + SBTNode suc = root.r; + suc.size--; + while (suc.l != null) { + pre = suc; + suc = suc.l; + suc.size--; + } + if (pre != null) { + pre.l = suc.r; + suc.r = root.r; + } + suc.l = root.l; + suc.size = suc.l.size + (suc.r == null ? 0 : suc.r.size) + 1; + return suc; + } + + private SBTNode get(SBTNode root, int index) { + int leftSize = root.l != null ? root.l.size : 0; + if (index < leftSize) { + return get(root.l, index); + } else if (index == leftSize) { + return root; + } else { + return get(root.r, index - leftSize - 1); + } + } + + public void add(int index, V num) { + SBTNode cur = new SBTNode(num); + if (root == null) { + root = cur; + } else { + if (index <= root.size) { + root = add(root, index, cur); + } + } + } + + public V get(int index) { + SBTNode ans = get(root, index); + return ans.value; + } + + public void remove(int index) { + if (index >= 0 && size() > index) { + root = remove(root, index); + } + } + + public int size() { + return root == null ? 0 : root.size; + } + + } + + // 通过以下这个测试， + // 可以很明显的看到LinkedList的插入、删除、get效率不如SbtList + // LinkedList需要找到index所在的位置之后才能插入或者读取，时间复杂度O(N) + // SbtList是平衡搜索二叉树，所以插入或者读取时间复杂度都是O(logN) + public static void main(String[] args) { + // 功能测试 + int test = 50000; + int max = 1000000; + boolean pass = true; + ArrayList list = new ArrayList<>(); + SbtList sbtList = new SbtList<>(); + for (int i = 0; i < test; i++) { + if (list.size() != sbtList.size()) { + pass = false; + break; + } + if (list.size() > 1 && Math.random() < 0.5) { + int removeIndex = (int) (Math.random() * list.size()); + list.remove(removeIndex); + sbtList.remove(removeIndex); + } else { + int randomIndex = (int) (Math.random() * (list.size() + 1)); + int randomValue = (int) (Math.random() * (max + 1)); + list.add(randomIndex, randomValue); + sbtList.add(randomIndex, randomValue); + } + } + for (int i = 0; i < list.size(); i++) { + if (!list.get(i).equals(sbtList.get(i))) { + pass = false; + break; + } + } + System.out.println("功能测试是否通过 : " + pass); + + // 性能测试 + test = 500000; + list = new ArrayList<>(); + sbtList = new SbtList<>(); + long start = 0; + long end = 0; + + start = System.currentTimeMillis(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * (list.size() + 1)); + int randomValue = (int) (Math.random() * (max + 1)); + list.add(randomIndex, randomValue); + } + end = System.currentTimeMillis(); + System.out.println("ArrayList插入总时长(毫秒) ： " + (end - start)); + + start = System.currentTimeMillis(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * (i + 1)); + list.get(randomIndex); + } + end = System.currentTimeMillis(); + System.out.println("ArrayList读取总时长(毫秒) : " + (end - start)); + + start = System.currentTimeMillis(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * list.size()); + list.remove(randomIndex); + } + end = System.currentTimeMillis(); + System.out.println("ArrayList删除总时长(毫秒) : " + (end - start)); + + start = System.currentTimeMillis(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * (sbtList.size() + 1)); + int randomValue = (int) (Math.random() * (max + 1)); + sbtList.add(randomIndex, randomValue); + } + end = System.currentTimeMillis(); + System.out.println("SbtList插入总时长(毫秒) : " + (end - start)); + + start = System.currentTimeMillis(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * (i + 1)); + sbtList.get(randomIndex); + } + end = System.currentTimeMillis(); + System.out.println("SbtList读取总时长(毫秒) : " + (end - start)); + + start = System.currentTimeMillis(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * sbtList.size()); + sbtList.remove(randomIndex); + } + end = System.currentTimeMillis(); + System.out.println("SbtList删除总时长(毫秒) : " + (end - start)); + + } + +} diff --git a/体系学习班/class37/Code04_QueueReconstructionByHeight.java b/体系学习班/class37/Code04_QueueReconstructionByHeight.java new file mode 100644 index 0000000..2513402 --- /dev/null +++ b/体系学习班/class37/Code04_QueueReconstructionByHeight.java @@ -0,0 +1,265 @@ +package class37; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.Comparator; +import java.util.LinkedList; + +// 本题测试链接：https://leetcode.com/problems/queue-reconstruction-by-height/ +public class Code04_QueueReconstructionByHeight { + + public static int[][] reconstructQueue1(int[][] people) { + int N = people.length; + Unit[] units = new Unit[N]; + for (int i = 0; i < N; i++) { + units[i] = new Unit(people[i][0], people[i][1]); + } + Arrays.sort(units, new UnitComparator()); + ArrayList arrList = new ArrayList<>(); + for (Unit unit : units) { + arrList.add(unit.k, unit); + } + int[][] ans = new int[N][2]; + int index = 0; + for (Unit unit : arrList) { + ans[index][0] = unit.h; + ans[index++][1] = unit.k; + } + return ans; + } + + public static int[][] reconstructQueue2(int[][] people) { + int N = people.length; + Unit[] units = new Unit[N]; + for (int i = 0; i < N; i++) { + units[i] = new Unit(people[i][0], people[i][1]); + } + Arrays.sort(units, new UnitComparator()); + SBTree tree = new SBTree(); + for (int i = 0; i < N; i++) { + tree.insert(units[i].k, i); + } + LinkedList allIndexes = tree.allIndexes(); + int[][] ans = new int[N][2]; + int index = 0; + for (Integer arri : allIndexes) { + ans[index][0] = units[arri].h; + ans[index++][1] = units[arri].k; + } + return ans; + } + + public static class Unit { + public int h; + public int k; + + public Unit(int height, int greater) { + h = height; + k = greater; + } + } + + public static class UnitComparator implements Comparator { + + @Override + public int compare(Unit o1, Unit o2) { + return o1.h != o2.h ? (o2.h - o1.h) : (o1.k - o2.k); + } + + } + + public static class SBTNode { + public int value; + public SBTNode l; + public SBTNode r; + public int size; + + public SBTNode(int arrIndex) { + value = arrIndex; + size = 1; + } + } + + public static class SBTree { + private SBTNode root; + + private SBTNode rightRotate(SBTNode cur) { + SBTNode leftNode = cur.l; + cur.l = leftNode.r; + leftNode.r = cur; + leftNode.size = cur.size; + cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1; + return leftNode; + } + + private SBTNode leftRotate(SBTNode cur) { + SBTNode rightNode = cur.r; + cur.r = rightNode.l; + rightNode.l = cur; + rightNode.size = cur.size; + cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1; + return rightNode; + } + + private SBTNode maintain(SBTNode cur) { + if (cur == null) { + return null; + } + int leftSize = cur.l != null ? cur.l.size : 0; + int leftLeftSize = cur.l != null && cur.l.l != null ? cur.l.l.size : 0; + int leftRightSize = cur.l != null && cur.l.r != null ? cur.l.r.size : 0; + int rightSize = cur.r != null ? cur.r.size : 0; + int rightLeftSize = cur.r != null && cur.r.l != null ? cur.r.l.size : 0; + int rightRightSize = cur.r != null && cur.r.r != null ? cur.r.r.size : 0; + if (leftLeftSize > rightSize) { + cur = rightRotate(cur); + cur.r = maintain(cur.r); + cur = maintain(cur); + } else if (leftRightSize > rightSize) { + cur.l = leftRotate(cur.l); + cur = rightRotate(cur); + cur.l = maintain(cur.l); + cur.r = maintain(cur.r); + cur = maintain(cur); + } else if (rightRightSize > leftSize) { + cur = leftRotate(cur); + cur.l = maintain(cur.l); + cur = maintain(cur); + } else if (rightLeftSize > leftSize) { + cur.r = rightRotate(cur.r); + cur = leftRotate(cur); + cur.l = maintain(cur.l); + cur.r = maintain(cur.r); + cur = maintain(cur); + } + return cur; + } + + private SBTNode insert(SBTNode root, int index, SBTNode cur) { + if (root == null) { + return cur; + } + root.size++; + int leftAndHeadSize = (root.l != null ? root.l.size : 0) + 1; + if (index < leftAndHeadSize) { + root.l = insert(root.l, index, cur); + } else { + root.r = insert(root.r, index - leftAndHeadSize, cur); + } + root = maintain(root); + return root; + } + + private SBTNode get(SBTNode root, int index) { + int leftSize = root.l != null ? root.l.size : 0; + if (index < leftSize) { + return get(root.l, index); + } else if (index == leftSize) { + return root; + } else { + return get(root.r, index - leftSize - 1); + } + } + + private void process(SBTNode head, LinkedList indexes) { + if (head == null) { + return; + } + process(head.l, indexes); + indexes.addLast(head.value); + process(head.r, indexes); + } + + public void insert(int index, int value) { + SBTNode cur = new SBTNode(value); + if (root == null) { + root = cur; + } else { + if (index <= root.size) { + root = insert(root, index, cur); + } + } + } + + public int get(int index) { + SBTNode ans = get(root, index); + return ans.value; + } + + public LinkedList allIndexes() { + LinkedList indexes = new LinkedList<>(); + process(root, indexes); + return indexes; + } + + } + + // 通过以下这个测试， + // 可以很明显的看到LinkedList的插入和get效率不如SBTree + // LinkedList需要找到index所在的位置之后才能插入或者读取，时间复杂度O(N) + // SBTree是平衡搜索二叉树，所以插入或者读取时间复杂度都是O(logN) + public static void main(String[] args) { + // 功能测试 + int test = 10000; + int max = 1000000; + boolean pass = true; + LinkedList list = new LinkedList<>(); + SBTree sbtree = new SBTree(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * (i + 1)); + int randomValue = (int) (Math.random() * (max + 1)); + list.add(randomIndex, randomValue); + sbtree.insert(randomIndex, randomValue); + } + for (int i = 0; i < test; i++) { + if (list.get(i) != sbtree.get(i)) { + pass = false; + break; + } + } + System.out.println("功能测试是否通过 : " + pass); + + // 性能测试 + test = 50000; + list = new LinkedList<>(); + sbtree = new SBTree(); + long start = 0; + long end = 0; + + start = System.currentTimeMillis(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * (i + 1)); + int randomValue = (int) (Math.random() * (max + 1)); + list.add(randomIndex, randomValue); + } + end = System.currentTimeMillis(); + System.out.println("LinkedList插入总时长(毫秒) ： " + (end - start)); + + start = System.currentTimeMillis(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * (i + 1)); + list.get(randomIndex); + } + end = System.currentTimeMillis(); + System.out.println("LinkedList读取总时长(毫秒) : " + (end - start)); + + start = System.currentTimeMillis(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * (i + 1)); + int randomValue = (int) (Math.random() * (max + 1)); + sbtree.insert(randomIndex, randomValue); + } + end = System.currentTimeMillis(); + System.out.println("SBTree插入总时长(毫秒) : " + (end - start)); + + start = System.currentTimeMillis(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * (i + 1)); + sbtree.get(randomIndex); + } + end = System.currentTimeMillis(); + System.out.println("SBTree读取总时长(毫秒) : " + (end - start)); + + } + +} diff --git a/体系学习班/class37/Compare.java b/体系学习班/class37/Compare.java new file mode 100644 index 0000000..5bbfd0d --- /dev/null +++ b/体系学习班/class37/Compare.java @@ -0,0 +1,408 @@ +package class37; + +import java.util.TreeMap; + +import class35.Code01_AVLTreeMap.AVLTreeMap; +import class36.Code01_SizeBalancedTreeMap.SizeBalancedTreeMap; +import class36.Code02_SkipListMap.SkipListMap; + +// 本文件为avl、sbt、skiplist三种结构的测试文件 +public class Compare { + + public static void functionTest() { + System.out.println("功能测试开始"); + TreeMap treeMap = new TreeMap<>(); + AVLTreeMap avl = new AVLTreeMap<>(); + SizeBalancedTreeMap sbt = new SizeBalancedTreeMap<>(); + SkipListMap skip = new SkipListMap<>(); + int maxK = 500; + int maxV = 50000; + int testTime = 1000000; + for (int i = 0; i < testTime; i++) { + int addK = (int) (Math.random() * maxK); + int addV = (int) (Math.random() * maxV); + treeMap.put(addK, addV); + avl.put(addK, addV); + sbt.put(addK, addV); + skip.put(addK, addV); + + int removeK = (int) (Math.random() * maxK); + treeMap.remove(removeK); + avl.remove(removeK); + sbt.remove(removeK); + skip.remove(removeK); + + int querryK = (int) (Math.random() * maxK); + if (treeMap.containsKey(querryK) != avl.containsKey(querryK) + || sbt.containsKey(querryK) != skip.containsKey(querryK) + || treeMap.containsKey(querryK) != sbt.containsKey(querryK)) { + System.out.println("containsKey Oops"); + System.out.println(treeMap.containsKey(querryK)); + System.out.println(avl.containsKey(querryK)); + System.out.println(sbt.containsKey(querryK)); + System.out.println(skip.containsKey(querryK)); + break; + } + + if (treeMap.containsKey(querryK)) { + int v1 = treeMap.get(querryK); + int v2 = avl.get(querryK); + int v3 = sbt.get(querryK); + int v4 = skip.get(querryK); + if (v1 != v2 || v3 != v4 || v1 != v3) { + System.out.println("get Oops"); + System.out.println(treeMap.get(querryK)); + System.out.println(avl.get(querryK)); + System.out.println(sbt.get(querryK)); + System.out.println(skip.get(querryK)); + break; + } + Integer f1 = treeMap.floorKey(querryK); + Integer f2 = avl.floorKey(querryK); + Integer f3 = sbt.floorKey(querryK); + Integer f4 = skip.floorKey(querryK); + if (f1 == null && (f2 != null || f3 != null || f4 != null)) { + System.out.println("floorKey Oops"); + System.out.println(treeMap.floorKey(querryK)); + System.out.println(avl.floorKey(querryK)); + System.out.println(sbt.floorKey(querryK)); + System.out.println(skip.floorKey(querryK)); + break; + } + if (f1 != null && (f2 == null || f3 == null || f4 == null)) { + System.out.println("floorKey Oops"); + System.out.println(treeMap.floorKey(querryK)); + System.out.println(avl.floorKey(querryK)); + System.out.println(sbt.floorKey(querryK)); + System.out.println(skip.floorKey(querryK)); + break; + } + if (f1 != null) { + int ans1 = f1; + int ans2 = f2; + int ans3 = f3; + int ans4 = f4; + if (ans1 != ans2 || ans3 != ans4 || ans1 != ans3) { + System.out.println("floorKey Oops"); + System.out.println(treeMap.floorKey(querryK)); + System.out.println(avl.floorKey(querryK)); + System.out.println(sbt.floorKey(querryK)); + System.out.println(skip.floorKey(querryK)); + break; + } + } + f1 = treeMap.ceilingKey(querryK); + f2 = avl.ceilingKey(querryK); + f3 = sbt.ceilingKey(querryK); + f4 = skip.ceilingKey(querryK); + if (f1 == null && (f2 != null || f3 != null || f4 != null)) { + System.out.println("ceilingKey Oops"); + System.out.println(treeMap.ceilingKey(querryK)); + System.out.println(avl.ceilingKey(querryK)); + System.out.println(sbt.ceilingKey(querryK)); + System.out.println(skip.ceilingKey(querryK)); + break; + } + if (f1 != null && (f2 == null || f3 == null || f4 == null)) { + System.out.println("ceilingKey Oops"); + System.out.println(treeMap.ceilingKey(querryK)); + System.out.println(avl.ceilingKey(querryK)); + System.out.println(sbt.ceilingKey(querryK)); + System.out.println(skip.ceilingKey(querryK)); + break; + } + if (f1 != null) { + int ans1 = f1; + int ans2 = f2; + int ans3 = f3; + int ans4 = f4; + if (ans1 != ans2 || ans3 != ans4 || ans1 != ans3) { + System.out.println("ceilingKey Oops"); + System.out.println(treeMap.ceilingKey(querryK)); + System.out.println(avl.ceilingKey(querryK)); + System.out.println(sbt.ceilingKey(querryK)); + System.out.println(skip.ceilingKey(querryK)); + break; + } + } + + } + + Integer f1 = treeMap.firstKey(); + Integer f2 = avl.firstKey(); + Integer f3 = sbt.firstKey(); + Integer f4 = skip.firstKey(); + if (f1 == null && (f2 != null || f3 != null || f4 != null)) { + System.out.println("firstKey Oops"); + System.out.println(treeMap.firstKey()); + System.out.println(avl.firstKey()); + System.out.println(sbt.firstKey()); + System.out.println(skip.firstKey()); + break; + } + if (f1 != null && (f2 == null || f3 == null || f4 == null)) { + System.out.println("firstKey Oops"); + System.out.println(treeMap.firstKey()); + System.out.println(avl.firstKey()); + System.out.println(sbt.firstKey()); + System.out.println(skip.firstKey()); + break; + } + if (f1 != null) { + int ans1 = f1; + int ans2 = f2; + int ans3 = f3; + int ans4 = f4; + if (ans1 != ans2 || ans3 != ans4 || ans1 != ans3) { + System.out.println("firstKey Oops"); + System.out.println(treeMap.firstKey()); + System.out.println(avl.firstKey()); + System.out.println(sbt.firstKey()); + System.out.println(skip.firstKey()); + break; + } + } + + f1 = treeMap.lastKey(); + f2 = avl.lastKey(); + f3 = sbt.lastKey(); + f4 = skip.lastKey(); + if (f1 == null && (f2 != null || f3 != null || f4 != null)) { + System.out.println("lastKey Oops"); + System.out.println(treeMap.lastKey()); + System.out.println(avl.lastKey()); + System.out.println(sbt.lastKey()); + System.out.println(skip.lastKey()); + break; + } + if (f1 != null && (f2 == null || f3 == null || f4 == null)) { + System.out.println("firstKey Oops"); + System.out.println(treeMap.lastKey()); + System.out.println(avl.lastKey()); + System.out.println(sbt.lastKey()); + System.out.println(skip.lastKey()); + break; + } + if (f1 != null) { + int ans1 = f1; + int ans2 = f2; + int ans3 = f3; + int ans4 = f4; + if (ans1 != ans2 || ans3 != ans4 || ans1 != ans3) { + System.out.println("lastKey Oops"); + System.out.println(treeMap.lastKey()); + System.out.println(avl.lastKey()); + System.out.println(sbt.lastKey()); + System.out.println(skip.lastKey()); + break; + } + } + if (treeMap.size() != avl.size() || sbt.size() != skip.size() || treeMap.size() != sbt.size()) { + System.out.println("size Oops"); + System.out.println(treeMap.size()); + System.out.println(avl.size()); + System.out.println(sbt.size()); + System.out.println(skip.size()); + break; + } + } + System.out.println("功能测试结束"); + } + + public static void performanceTest() { + System.out.println("性能测试开始"); + TreeMap treeMap; + AVLTreeMap avl; + SizeBalancedTreeMap sbt; + SkipListMap skip; + long start; + long end; + int max = 1000000; + treeMap = new TreeMap<>(); + avl = new AVLTreeMap<>(); + sbt = new SizeBalancedTreeMap<>(); + skip = new SkipListMap<>(); + System.out.println("顺序递增加入测试，数据规模 : " + max); + start = System.currentTimeMillis(); + for (int i = 0; i < max; i++) { + treeMap.put(i, i); + } + end = System.currentTimeMillis(); + System.out.println("treeMap 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = 0; i < max; i++) { + avl.put(i, i); + } + end = System.currentTimeMillis(); + System.out.println("avl 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = 0; i < max; i++) { + sbt.put(i, i); + } + end = System.currentTimeMillis(); + System.out.println("sbt 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = 0; i < max; i++) { + skip.put(i, i); + } + end = System.currentTimeMillis(); + System.out.println("skip 运行时间 : " + (end - start) + "ms"); + + System.out.println("顺序递增删除测试，数据规模 : " + max); + start = System.currentTimeMillis(); + for (int i = 0; i < max; i++) { + treeMap.remove(i); + } + end = System.currentTimeMillis(); + System.out.println("treeMap 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = 0; i < max; i++) { + avl.remove(i); + } + end = System.currentTimeMillis(); + System.out.println("avl 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = 0; i < max; i++) { + sbt.remove(i); + } + end = System.currentTimeMillis(); + System.out.println("sbt 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = 0; i < max; i++) { + skip.remove(i); + } + end = System.currentTimeMillis(); + System.out.println("skip 运行时间 : " + (end - start) + "ms"); + + System.out.println("顺序递减加入测试，数据规模 : " + max); + start = System.currentTimeMillis(); + for (int i = max; i >= 0; i--) { + treeMap.put(i, i); + } + end = System.currentTimeMillis(); + System.out.println("treeMap 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = max; i >= 0; i--) { + avl.put(i, i); + } + end = System.currentTimeMillis(); + System.out.println("avl 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = max; i >= 0; i--) { + sbt.put(i, i); + } + end = System.currentTimeMillis(); + System.out.println("sbt 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = max; i >= 0; i--) { + skip.put(i, i); + } + end = System.currentTimeMillis(); + System.out.println("skip 运行时间 : " + (end - start) + "ms"); + + System.out.println("顺序递减删除测试，数据规模 : " + max); + start = System.currentTimeMillis(); + for (int i = max; i >= 0; i--) { + treeMap.remove(i); + } + end = System.currentTimeMillis(); + System.out.println("treeMap 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = max; i >= 0; i--) { + avl.remove(i); + } + end = System.currentTimeMillis(); + System.out.println("avl 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = max; i >= 0; i--) { + sbt.remove(i); + } + end = System.currentTimeMillis(); + System.out.println("sbt 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = max; i >= 0; i--) { + skip.remove(i); + } + end = System.currentTimeMillis(); + System.out.println("skip 运行时间 : " + (end - start) + "ms"); + + System.out.println("随机加入测试，数据规模 : " + max); + start = System.currentTimeMillis(); + for (int i = 0; i < max; i++) { + treeMap.put((int) (Math.random() * i), i); + } + end = System.currentTimeMillis(); + System.out.println("treeMap 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = max; i >= 0; i--) { + avl.put((int) (Math.random() * i), i); + } + end = System.currentTimeMillis(); + System.out.println("avl 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = max; i >= 0; i--) { + sbt.put((int) (Math.random() * i), i); + } + end = System.currentTimeMillis(); + System.out.println("sbt 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = max; i >= 0; i--) { + skip.put((int) (Math.random() * i), i); + } + end = System.currentTimeMillis(); + System.out.println("skip 运行时间 : " + (end - start) + "ms"); + + System.out.println("随机删除测试，数据规模 : " + max); + start = System.currentTimeMillis(); + for (int i = 0; i < max; i++) { + treeMap.remove((int) (Math.random() * i)); + } + end = System.currentTimeMillis(); + System.out.println("treeMap 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = max; i >= 0; i--) { + avl.remove((int) (Math.random() * i)); + } + end = System.currentTimeMillis(); + System.out.println("avl 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = max; i >= 0; i--) { + sbt.remove((int) (Math.random() * i)); + } + end = System.currentTimeMillis(); + System.out.println("sbt 运行时间 : " + (end - start) + "ms"); + + start = System.currentTimeMillis(); + for (int i = max; i >= 0; i--) { + skip.remove((int) (Math.random() * i)); + } + end = System.currentTimeMillis(); + System.out.println("skip 运行时间 : " + (end - start) + "ms"); + + System.out.println("性能测试结束"); + } + + public static void main(String[] args) { + functionTest(); + System.out.println("======"); + performanceTest(); + } + +} diff --git a/体系学习班/class38/Code01_AppleMinBags.java b/体系学习班/class38/Code01_AppleMinBags.java new file mode 100644 index 0000000..e1b5f21 --- /dev/null +++ b/体系学习班/class38/Code01_AppleMinBags.java @@ -0,0 +1,41 @@ +package class38; + +public class Code01_AppleMinBags { + + public static int minBags(int apple) { + if (apple < 0) { + return -1; + } + int bag8 = (apple >> 3); + int rest = apple - (bag8 << 3); + while(bag8 >= 0) { + // rest 个 + if(rest % 6 ==0) { + return bag8 + (rest / 6); + } else { + bag8--; + rest += 8; + } + } + return -1; + } + + public static int minBagAwesome(int apple) { + if ((apple & 1) != 0) { // 如果是奇数，返回-1 + return -1; + } + if (apple < 18) { + return apple == 0 ? 0 : (apple == 6 || apple == 8) ? 1 + : (apple == 12 || apple == 14 || apple == 16) ? 2 : -1; + } + return (apple - 18) / 8 + 3; + } + + public static void main(String[] args) { + for(int apple = 1; apple < 200;apple++) { + System.out.println(apple + " : "+ minBags(apple)); + } + + } + +} diff --git a/体系学习班/class38/Code02_EatGrass.java b/体系学习班/class38/Code02_EatGrass.java new file mode 100644 index 0000000..8b400fb --- /dev/null +++ b/体系学习班/class38/Code02_EatGrass.java @@ -0,0 +1,57 @@ +package class38; + +public class Code02_EatGrass { + + // 如果n份草，最终先手赢，返回"先手" + // 如果n份草，最终后手赢，返回"后手" + public static String whoWin(int n) { + if (n < 5) { + return n == 0 || n == 2 ? "后手" : "先手"; + } + // 进到这个过程里来，当前的先手，先选 + int want = 1; + while (want <= n) { + if (whoWin(n - want).equals("后手")) { + return "先手"; + } + if (want <= (n / 4)) { + want *= 4; + } else { + break; + } + } + return "后手"; + } + + public static String winner1(int n) { + if (n < 5) { + return (n == 0 || n == 2) ? "后手" : "先手"; + } + int base = 1; + while (base <= n) { + if (winner1(n - base).equals("后手")) { + return "先手"; + } + if (base > n / 4) { // 防止base*4之后溢出 + break; + } + base *= 4; + } + return "后手"; + } + + public static String winner2(int n) { + if (n % 5 == 0 || n % 5 == 2) { + return "后手"; + } else { + return "先手"; + } + } + + public static void main(String[] args) { + for (int i = 0; i <= 50; i++) { + System.out.println(i + " : " + whoWin(i)); + } + } + +} diff --git a/体系学习班/class38/Code03_MSumToN.java b/体系学习班/class38/Code03_MSumToN.java new file mode 100644 index 0000000..50c72be --- /dev/null +++ b/体系学习班/class38/Code03_MSumToN.java @@ -0,0 +1,44 @@ +package class38; + +public class Code03_MSumToN { + + public static boolean isMSum1(int num) { + for (int start = 1; start <= num; start++) { + int sum = start; + for (int j = start + 1; j <= num; j++) { + if (sum + j > num) { + break; + } + if (sum + j == num) { + return true; + } + sum += j; + } + } + return false; + } + + public static boolean isMSum2(int num) { +// +// return num == (num & (~num + 1)); +// +// return num == (num & (-num)); +// +// + return (num & (num - 1)) != 0; + } + + public static void main(String[] args) { + for (int num = 1; num < 200; num++) { + System.out.println(num + " : " + isMSum1(num)); + } + System.out.println("test begin"); + for (int num = 1; num < 5000; num++) { + if (isMSum1(num) != isMSum2(num)) { + System.out.println("Oops!"); + } + } + System.out.println("test end"); + + } +} diff --git a/体系学习班/class38/Code04_MoneyProblem.java b/体系学习班/class38/Code04_MoneyProblem.java new file mode 100644 index 0000000..722c5a6 --- /dev/null +++ b/体系学习班/class38/Code04_MoneyProblem.java @@ -0,0 +1,169 @@ +package class38; + +public class Code04_MoneyProblem { + + // int[] d d[i]：i号怪兽的武力 + // int[] p p[i]：i号怪兽要求的钱 + // ability 当前你所具有的能力 + // index 来到了第index个怪兽的面前 + + // 目前，你的能力是ability，你来到了index号怪兽的面前，如果要通过后续所有的怪兽， + // 请返回需要花的最少钱数 + public static long process1(int[] d, int[] p, int ability, int index) { + if (index == d.length) { + return 0; + } + if (ability < d[index]) { + return p[index] + process1(d, p, ability + d[index], index + 1); + } else { // ability >= d[index] 可以贿赂，也可以不贿赂 + return Math.min( + + p[index] + process1(d, p, ability + d[index], index + 1), + + 0 + process1(d, p, ability, index + 1)); + } + } + + public static long func1(int[] d, int[] p) { + return process1(d, p, 0, 0); + } + + // 从0....index号怪兽，花的钱，必须严格==money + // 如果通过不了，返回-1 + // 如果可以通过，返回能通过情况下的最大能力值 + public static long process2(int[] d, int[] p, int index, int money) { + if (index == -1) { // 一个怪兽也没遇到呢 + return money == 0 ? 0 : -1; + } + // index >= 0 + // 1) 不贿赂当前index号怪兽 + long preMaxAbility = process2(d, p, index - 1, money); + long p1 = -1; + if (preMaxAbility != -1 && preMaxAbility >= d[index]) { + p1 = preMaxAbility; + } + // 2) 贿赂当前的怪兽 当前的钱 p[index] + long preMaxAbility2 = process2(d, p, index - 1, money - p[index]); + long p2 = -1; + if (preMaxAbility2 != -1) { + p2 = d[index] + preMaxAbility2; + } + return Math.max(p1, p2); + } + + public static int minMoney2(int[] d, int[] p) { + int allMoney = 0; + for (int i = 0; i < p.length; i++) { + allMoney += p[i]; + } + int N = d.length; + for (int money = 0; money < allMoney; money++) { + if (process2(d, p, N - 1, money) != -1) { + return money; + } + } + return allMoney; + } + + public static long func2(int[] d, int[] p) { + int sum = 0; + for (int num : d) { + sum += num; + } + long[][] dp = new long[d.length + 1][sum + 1]; + for (int i = 0; i <= sum; i++) { + dp[0][i] = 0; + } + for (int cur = d.length - 1; cur >= 0; cur--) { + for (int hp = 0; hp <= sum; hp++) { + // 如果这种情况发生，那么这个hp必然是递归过程中不会出现的状态 + // 既然动态规划是尝试过程的优化，尝试过程碰不到的状态，不必计算 + if (hp + d[cur] > sum) { + continue; + } + if (hp < d[cur]) { + dp[cur][hp] = p[cur] + dp[cur + 1][hp + d[cur]]; + } else { + dp[cur][hp] = Math.min(p[cur] + dp[cur + 1][hp + d[cur]], dp[cur + 1][hp]); + } + } + } + return dp[0][0]; + } + + public static long func3(int[] d, int[] p) { + int sum = 0; + for (int num : p) { + sum += num; + } + // dp[i][j]含义： + // 能经过0～i的怪兽，且花钱为j（花钱的严格等于j）时的武力值最大是多少？ + // 如果dp[i][j]==-1，表示经过0～i的怪兽，花钱为j是无法通过的，或者之前的钱怎么组合也得不到正好为j的钱数 + int[][] dp = new int[d.length][sum + 1]; + for (int i = 0; i < dp.length; i++) { + for (int j = 0; j <= sum; j++) { + dp[i][j] = -1; + } + } + // 经过0～i的怪兽，花钱数一定为p[0]，达到武力值d[0]的地步。其他第0行的状态一律是无效的 + dp[0][p[0]] = d[0]; + for (int i = 1; i < d.length; i++) { + for (int j = 0; j <= sum; j++) { + // 可能性一，为当前怪兽花钱 + // 存在条件： + // j - p[i]要不越界，并且在钱数为j - p[i]时，要能通过0～i-1的怪兽，并且钱数组合是有效的。 + if (j >= p[i] && dp[i - 1][j - p[i]] != -1) { + dp[i][j] = dp[i - 1][j - p[i]] + d[i]; + } + // 可能性二，不为当前怪兽花钱 + // 存在条件： + // 0~i-1怪兽在花钱为j的情况下，能保证通过当前i位置的怪兽 + if (dp[i - 1][j] >= d[i]) { + // 两种可能性中，选武力值最大的 + dp[i][j] = Math.max(dp[i][j], dp[i - 1][j]); + } + } + } + int ans = 0; + // dp表最后一行上，dp[N-1][j]代表： + // 能经过0～N-1的怪兽，且花钱为j（花钱的严格等于j）时的武力值最大是多少？ + // 那么最后一行上，最左侧的不为-1的列数(j)，就是答案 + for (int j = 0; j <= sum; j++) { + if (dp[d.length - 1][j] != -1) { + ans = j; + break; + } + } + return ans; + } + + public static int[][] generateTwoRandomArray(int len, int value) { + int size = (int) (Math.random() * len) + 1; + int[][] arrs = new int[2][size]; + for (int i = 0; i < size; i++) { + arrs[0][i] = (int) (Math.random() * value) + 1; + arrs[1][i] = (int) (Math.random() * value) + 1; + } + return arrs; + } + + public static void main(String[] args) { + int len = 10; + int value = 20; + int testTimes = 10000; + for (int i = 0; i < testTimes; i++) { + int[][] arrs = generateTwoRandomArray(len, value); + int[] d = arrs[0]; + int[] p = arrs[1]; + long ans1 = func1(d, p); + long ans2 = func2(d, p); + long ans3 = func3(d, p); + long ans4 = minMoney2(d,p); + if (ans1 != ans2 || ans2 != ans3 || ans1 != ans4) { + System.out.println("oops!"); + } + } + + } + +} diff --git a/体系学习班/class39/Code01_SubsquenceMaxModM.java b/体系学习班/class39/Code01_SubsquenceMaxModM.java new file mode 100644 index 0000000..9890bf4 --- /dev/null +++ b/体系学习班/class39/Code01_SubsquenceMaxModM.java @@ -0,0 +1,141 @@ +package class39; + +import java.util.HashSet; +import java.util.TreeSet; + +// 给定一个非负数组arr，和一个正数m。 返回arr的所有子序列中累加和%m之后的最大值。 +public class Code01_SubsquenceMaxModM { + + public static int max1(int[] arr, int m) { + HashSet set = new HashSet<>(); + process(arr, 0, 0, set); + int max = 0; + for (Integer sum : set) { + max = Math.max(max, sum % m); + } + return max; + } + + public static void process(int[] arr, int index, int sum, HashSet set) { + if (index == arr.length) { + set.add(sum); + } else { + process(arr, index + 1, sum, set); + process(arr, index + 1, sum + arr[index], set); + } + } + + public static int max2(int[] arr, int m) { + int sum = 0; + int N = arr.length; + for (int i = 0; i < N; i++) { + sum += arr[i]; + } + boolean[][] dp = new boolean[N][sum + 1]; + for (int i = 0; i < N; i++) { + dp[i][0] = true; + } + dp[0][arr[0]] = true; + for (int i = 1; i < N; i++) { + for (int j = 1; j <= sum; j++) { + dp[i][j] = dp[i - 1][j]; + if (j - arr[i] >= 0) { + dp[i][j] |= dp[i - 1][j - arr[i]]; + } + } + } + int ans = 0; + for (int j = 0; j <= sum; j++) { + if (dp[N - 1][j]) { + ans = Math.max(ans, j % m); + } + } + return ans; + } + + public static int max3(int[] arr, int m) { + int N = arr.length; + // 0...m-1 + boolean[][] dp = new boolean[N][m]; + for (int i = 0; i < N; i++) { + dp[i][0] = true; + } + dp[0][arr[0] % m] = true; + for (int i = 1; i < N; i++) { + for (int j = 1; j < m; j++) { + // dp[i][j] T or F + dp[i][j] = dp[i - 1][j]; + int cur = arr[i] % m; + if (cur <= j) { + dp[i][j] |= dp[i - 1][j - cur]; + } else { + dp[i][j] |= dp[i - 1][m + j - cur]; + } + } + } + int ans = 0; + for (int i = 0; i < m; i++) { + if (dp[N - 1][i]) { + ans = i; + } + } + return ans; + } + + // 如果arr的累加和很大，m也很大 + // 但是arr的长度相对不大 + public static int max4(int[] arr, int m) { + if (arr.length == 1) { + return arr[0] % m; + } + int mid = (arr.length - 1) / 2; + TreeSet sortSet1 = new TreeSet<>(); + process4(arr, 0, 0, mid, m, sortSet1); + TreeSet sortSet2 = new TreeSet<>(); + process4(arr, mid + 1, 0, arr.length - 1, m, sortSet2); + int ans = 0; + for (Integer leftMod : sortSet1) { + ans = Math.max(ans, leftMod + sortSet2.floor(m - 1 - leftMod)); + } + return ans; + } + + // 从index出发，最后有边界是end+1，arr[index...end] + public static void process4(int[] arr, int index, int sum, int end, int m, TreeSet sortSet) { + if (index == end + 1) { + sortSet.add(sum % m); + } else { + process4(arr, index + 1, sum, end, m, sortSet); + process4(arr, index + 1, sum + arr[index], end, m, sortSet); + } + } + + public static int[] generateRandomArray(int len, int value) { + int[] ans = new int[(int) (Math.random() * len) + 1]; + for (int i = 0; i < ans.length; i++) { + ans[i] = (int) (Math.random() * value); + } + return ans; + } + + public static void main(String[] args) { + int len = 10; + int value = 100; + int m = 76; + int testTime = 500000; + System.out.println("test begin"); + for (int i = 0; i < testTime; i++) { + int[] arr = generateRandomArray(len, value); + int ans1 = max1(arr, m); + int ans2 = max2(arr, m); + int ans3 = max3(arr, m); + int ans4 = max4(arr, m); + if (ans1 != ans2 || ans2 != ans3 || ans3 != ans4) { + System.out.println("Oops!"); + } + } + System.out.println("test finish!"); + + } + +} diff --git a/体系学习班/class39/Code02_SnacksWays.java b/体系学习班/class39/Code02_SnacksWays.java new file mode 100644 index 0000000..bb2c538 --- /dev/null +++ b/体系学习班/class39/Code02_SnacksWays.java @@ -0,0 +1,79 @@ +package class39; + +public class Code02_SnacksWays { + + public static int ways1(int[] arr, int w) { + // arr[0...] + return process(arr, 0, w); + } + + // 从左往右的经典模型 + // 还剩的容量是rest，arr[index...]自由选择， + // 返回选择方案 + // index ： 0～N + // rest : 0~w + public static int process(int[] arr, int index, int rest) { + if (rest < 0) { // 没有容量了 + // -1 无方案的意思 + return -1; + } + // rest>=0, + if (index == arr.length) { // 无零食可选 + return 1; + } + // rest >=0 + // 有零食index + // index号零食，要 or 不要 + // index, rest + // (index+1, rest) + // (index+1, rest-arr[i]) + int next1 = process(arr, index + 1, rest); // 不要 + int next2 = process(arr, index + 1, rest - arr[index]); // 要 + return next1 + (next2 == -1 ? 0 : next2); + } + + public static int ways2(int[] arr, int w) { + int N = arr.length; + int[][] dp = new int[N + 1][w + 1]; + for (int j = 0; j <= w; j++) { + dp[N][j] = 1; + } + for (int i = N - 1; i >= 0; i--) { + for (int j = 0; j <= w; j++) { + dp[i][j] = dp[i + 1][j] + ((j - arr[i] >= 0) ? dp[i + 1][j - arr[i]] : 0); + } + } + return dp[0][w]; + } + + public static int ways3(int[] arr, int w) { + int N = arr.length; + int[][] dp = new int[N][w + 1]; + for (int i = 0; i < N; i++) { + dp[i][0] = 1; + } + if (arr[0] <= w) { + dp[0][arr[0]] = 1; + } + for (int i = 1; i < N; i++) { + for (int j = 1; j <= w; j++) { + dp[i][j] = dp[i - 1][j] + ((j - arr[i]) >= 0 ? dp[i - 1][j - arr[i]] : 0); + } + } + int ans = 0; + for (int j = 0; j <= w; j++) { + ans += dp[N - 1][j]; + } + return ans; + } + + public static void main(String[] args) { + int[] arr = { 4, 3, 2, 9 }; + int w = 8; + System.out.println(ways1(arr, w)); + System.out.println(ways2(arr, w)); + System.out.println(ways3(arr, w)); + + } + +} diff --git a/体系学习班/class39/Code02_SnacksWaysMain1.java b/体系学习班/class39/Code02_SnacksWaysMain1.java new file mode 100644 index 0000000..f282098 --- /dev/null +++ b/体系学习班/class39/Code02_SnacksWaysMain1.java @@ -0,0 +1,126 @@ +// 不要拷贝包信息的内容 +package class39; + +// 课堂版本 +// 本文件是Code02_SnacksWays问题的牛客题目解答 +// 但是用的分治的方法 +// 这是牛客的测试链接： +// https://www.nowcoder.com/questionTerminal/d94bb2fa461d42bcb4c0f2b94f5d4281 +// 把如下的全部代码拷贝进编辑器（java） +// 可以直接通过 +import java.util.Map.Entry; +import java.util.Scanner; +import java.util.TreeMap; + +public class Code02_SnacksWaysMain1 { + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + int N = sc.nextInt(); + int bag = sc.nextInt(); + int[] arr = new int[N]; + for (int i = 0; i < arr.length; i++) { + arr[i] = sc.nextInt(); + } + long ways = ways(arr, bag); + System.out.println(ways); + sc.close(); + } + + public static long ways(int[] arr, int bag) { + if (arr == null || arr.length == 0) { + return 0; + } + if (arr.length == 1) { + return arr[0] <= bag ? 2 : 1; + } + int mid = (arr.length - 1) >> 1; + TreeMap lmap = new TreeMap<>(); + long ways = process(arr, 0, 0, mid, bag, lmap); + TreeMap rmap = new TreeMap<>(); + ways += process(arr, mid + 1, 0, arr.length - 1, bag, rmap); + TreeMap rpre = new TreeMap<>(); + long pre = 0; + for (Entry entry : rmap.entrySet()) { + pre += entry.getValue(); + rpre.put(entry.getKey(), pre); + } + for (Entry entry : lmap.entrySet()) { + long lweight = entry.getKey(); + long lways = entry.getValue(); + Long floor = rpre.floorKey(bag - lweight); + if (floor != null) { + long rways = rpre.get(floor); + ways += lways * rways; + } + } + return ways + 1; + } + + + + + // arr 30 + // func(arr, 0, 14, 0, bag, map) + + // func(arr, 15, 29, 0, bag, map) + + // 从index出发，到end结束 + // 之前的选择，已经形成的累加和sum + // 零食[index....end]自由选择，出来的所有累加和，不能超过bag，每一种累加和对应的方法数，填在map里 + // 最后不能什么货都没选 + // [3,3,3,3] bag = 6 + // 0 1 2 3 + // - - - - 0 -> （0 : 1） + // - - - $ 3 -> （0 : 1）(3, 1) + // - - $ - 3 -> （0 : 1）(3, 2) + public static long func(int[] arr, int index, int end, long sum, long bag, TreeMap map) { + if(sum > bag) { + return 0; + } + // sum <= bag + if(index > end) { // 所有商品自由选择完了！ + // sum + if(sum != 0) { + if (!map.containsKey(sum)) { + map.put(sum, 1L); + } else { + map.put(sum, map.get(sum) + 1); + } + return 1; + } else { + return 0; + } + } + // sum <= bag 并且 index <= end(还有货) + // 1) 不要当前index位置的货 + long ways = func(arr, index + 1, end, sum, bag, map); + + // 2) 要当前index位置的货 + ways += func(arr, index + 1, end, sum + arr[index], bag, map); + return ways; + } + + public static long process(int[] arr, int index, long w, int end, int bag, TreeMap map) { + if (w > bag) { + return 0; + } + if (index > end) { + if (w != 0) { + if (!map.containsKey(w)) { + map.put(w, 1L); + } else { + map.put(w, map.get(w) + 1); + } + return 1; + } else { + return 0; + } + } else { + long ways = process(arr, index + 1, w, end, bag, map); + ways += process(arr, index + 1, w + arr[index], end, bag, map); + return ways; + } + } + +} \ No newline at end of file diff --git a/体系学习班/class39/Code02_SnacksWaysMain2.java b/体系学习班/class39/Code02_SnacksWaysMain2.java new file mode 100644 index 0000000..0d4bf87 --- /dev/null +++ b/体系学习班/class39/Code02_SnacksWaysMain2.java @@ -0,0 +1,135 @@ +// 不要拷贝包信息的内容 +package class39; + +//优化版本 +import java.util.Arrays; +import java.util.Scanner; + +public class Code02_SnacksWaysMain2 { + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + while (sc.hasNext()) { + size = sc.nextInt(); + long w = (long) sc.nextInt(); + for (int i = 0; i < size; i++) { + arr[i] = (long) sc.nextInt(); + } + long ways = ways(w); + System.out.println(ways); + } + sc.close(); + } + + // 用来收集所有输入的数字 + public static long[] arr = new long[31]; + public static int size = 0; + // 用来生成左部分可能的所有累加和 + public static long[] leftSum = new long[1 << 16]; + // 准备的数组可能用不完，左部分生成了多少累加和，用leftSize表示 + public static int leftSize = 0; + // 用来生成右部分可能的所有累加和 + public static long[] rightSum = new long[1 << 16]; + // 准备的数组可能用不完，左部分生成了多少累加和，用leftSize表示 + public static int rightSize = 0; + + public static long ways(long w) { + if (size == 0) { + return 0; + } + if (size == 1) { + return arr[0] <= w ? 2 : 1; + } + // 求中点 + int mid = size >> 1; + // 生成左侧的累加和 + leftSize = 0; + dfsLeft(0, mid + 1, 0L); + // 生成右侧的累加和 + rightSize = 0; + dfsRight(mid + 1, size, 0L); + // 把左侧累加和排序 + Arrays.sort(leftSum, 0, leftSize); + // 把右侧累加和排序 + Arrays.sort(rightSum, 0, rightSize); + // 解释一下，接下来的流程。 + // 举个例子，比如： + // 左侧累加和是:{0, 1, 1, 1, 2, 2, 3, 4, 4} + // 右侧累加和是:{0, 1, 2, 3, 3, 3, 4, 4, 5} + // w = 5 + // 左侧严格得到0的方法数：1 + // 右侧得到<=5的方法数（二分求出）：9 + // 1 * 9 + // 左侧严格得到1的方法数：3 + // 右侧得到<=4的方法数（二分求出）：8 + // 3 * 8 + // 左侧严格得到2的方法数：2 + // 右侧得到<=3的方法数（二分求出）：6 + // 2 * 6 + // 左侧严格得到3的方法数：1 + // 右侧得到<=2的方法数（二分求出）：3 + // 1 * 3 + // 左侧严格得到4的方法数：2 + // 右侧得到<=1的方法数（二分求出）：2 + // 2 * 2 + // 都累加起来 + // 其实和课上讲的一样！多看一下例子 + long ans = 0; + long count = 1; + for (int i = 1; i < leftSize; i++) { + if (leftSum[i] != leftSum[i - 1]) { + ans += count * (long) find(w - leftSum[i - 1]); + count = 1; + } else { + count++; + } + } + ans += count * (long) find(w - leftSum[leftSize - 1]); + return ans; + } + + // 生成左部分的累加和，每一个累加和出来，都记录 + public static void dfsLeft(int cur, int end, long sum) { + if (cur == end) { // 已经终止位置了 + // 记录累加和 + leftSum[leftSize++] = sum; + } else { + // 可能性1，不要当前数 + dfsLeft(cur + 1, end, sum); + // 可能性2，要当前数 + dfsLeft(cur + 1, end, sum + arr[cur]); + } + } + + // 生成右部分的累加和，每一个累加和出来，都记录 + public static void dfsRight(int cur, int end, long sum) { + if (cur == end) { // 已经终止位置了 + // 记录累加和 + rightSum[rightSize++] = sum; + } else { + // 可能性1，不要当前数 + dfsRight(cur + 1, end, sum); + // 可能性2，要当前数 + dfsRight(cur + 1, end, sum + arr[cur]); + } + } + + // <= num的数的个数，返回 + public static int find(long num) { + int ans = -1; + int l = 0; + int r = rightSize - 1; + int m = 0; + while (l <= r) { + m = (l + r) / 2; + if (rightSum[m] <= num) { + ans = m; + l = m + 1; + } else { + r = m - 1; + } + } + return ans + 1; + } + +} \ No newline at end of file diff --git a/体系学习班/class39/Code03_10Ways.java b/体系学习班/class39/Code03_10Ways.java new file mode 100644 index 0000000..1afffb7 --- /dev/null +++ b/体系学习班/class39/Code03_10Ways.java @@ -0,0 +1,92 @@ +package class39; + +import java.util.LinkedList; + +public class Code03_10Ways { + + public static long ways1(int N) { + int zero = N; + int one = N; + LinkedList path = new LinkedList<>(); + LinkedList> ans = new LinkedList<>(); + process(zero, one, path, ans); + long count = 0; + for (LinkedList cur : ans) { + int status = 0; + for (Integer num : cur) { + if (num == 0) { + status++; + } else { + status--; + } + if (status < 0) { + break; + } + } + if (status == 0) { + count++; + } + } + return count; + } + + public static void process(int zero, int one, LinkedList path, LinkedList> ans) { + if (zero == 0 && one == 0) { + LinkedList cur = new LinkedList<>(); + for (Integer num : path) { + cur.add(num); + } + ans.add(cur); + } else { + if (zero == 0) { + path.addLast(1); + process(zero, one - 1, path, ans); + path.removeLast(); + } else if (one == 0) { + path.addLast(0); + process(zero - 1, one, path, ans); + path.removeLast(); + } else { + path.addLast(1); + process(zero, one - 1, path, ans); + path.removeLast(); + path.addLast(0); + process(zero - 1, one, path, ans); + path.removeLast(); + } + } + } + + public static long ways2(int N) { + if (N < 0) { + return 0; + } + if (N < 2) { + return 1; + } + long a = 1; + long b = 1; + long limit = N << 1; + for (long i = 1; i <= limit; i++) { + if (i <= N) { + a *= i; + } else { + b *= i; + } + } + return (b / a) / (N + 1); + } + + public static void main(String[] args) { + System.out.println("test begin"); + for (int i = 0; i < 10; i++) { + long ans1 = ways1(i); + long ans2 = ways2(i); + if (ans1 != ans2) { + System.out.println("Oops!"); + } + } + System.out.println("test finish"); + } + +} diff --git a/体系学习班/class39/Code04_DifferentBTNum.java b/体系学习班/class39/Code04_DifferentBTNum.java new file mode 100644 index 0000000..3c48911 --- /dev/null +++ b/体系学习班/class39/Code04_DifferentBTNum.java @@ -0,0 +1,66 @@ +package class39; + +public class Code04_DifferentBTNum { + +// k(0) = 1, k(1) = 1 +// +// k(n) = k(0) * k(n - 1) + k(1) * k(n - 2) + ... + k(n - 2) * k(1) + k(n - 1) * k(0) +// 或者 +// k(n) = c(2n, n) / (n + 1) +// 或者 +// k(n) = c(2n, n) - c(2n, n-1) + + public static long num1(int N) { + if (N < 0) { + return 0; + } + if (N < 2) { + return 1; + } + long[] dp = new long[N + 1]; + dp[0] = 1; + dp[1] = 1; + for (int i = 2; i <= N; i++) { + for (int leftSize = 0; leftSize < i; leftSize++) { + dp[i] += dp[leftSize] * dp[i - 1 - leftSize]; + } + } + return dp[N]; + } + + public static long num2(int N) { + if (N < 0) { + return 0; + } + if (N < 2) { + return 1; + } + long a = 1; + long b = 1; + for (int i = 1, j = N + 1; i <= N; i++, j++) { + a *= i; + b *= j; + long gcd = gcd(a, b); + a /= gcd; + b /= gcd; + } + return (b / a) / (N + 1); + } + + public static long gcd(long m, long n) { + return n == 0 ? m : gcd(n, m % n); + } + + public static void main(String[] args) { + System.out.println("test begin"); + for (int i = 0; i < 15; i++) { + long ans1 = num1(i); + long ans2 = num2(i); + if (ans1 != ans2) { + System.out.println("Oops!"); + } + } + System.out.println("test finish"); + } + +} diff --git a/体系学习班/class39/IsSum.java b/体系学习班/class39/IsSum.java new file mode 100644 index 0000000..9ff7a3c --- /dev/null +++ b/体系学习班/class39/IsSum.java @@ -0,0 +1,180 @@ +package class39; + +import java.util.HashMap; +import java.util.HashSet; + +// 这道题是一个小小的补充，课上没有讲 +// 但是如果你听过体系学习班动态规划专题和本节课的话 +// 这道题就是一道水题 +public class IsSum { + + // arr中的值可能为正，可能为负，可能为0 + // 自由选择arr中的数字，能不能累加得到sum + // 暴力递归方法 + public static boolean isSum1(int[] arr, int sum) { + if (sum == 0) { + return true; + } + if (arr == null || arr.length == 0) { + return false; + } + return process1(arr, arr.length - 1, sum); + } + + // 可以自由使用arr[0...i]上的数字，能不能累加得到sum + public static boolean process1(int[] arr, int i, int sum) { + if (sum == 0) { + return true; + } + if (i == -1) { + return false; + } + return process1(arr, i - 1, sum) || process1(arr, i - 1, sum - arr[i]); + } + + // arr中的值可能为正，可能为负，可能为0 + // 自由选择arr中的数字，能不能累加得到sum + // 记忆化搜索方法 + // 从暴力递归方法来，加了记忆化缓存，就是动态规划了 + public static boolean isSum2(int[] arr, int sum) { + if (sum == 0) { + return true; + } + if (arr == null || arr.length == 0) { + return false; + } + return process2(arr, arr.length - 1, sum, new HashMap<>()); + } + + public static boolean process2(int[] arr, int i, int sum, HashMap> dp) { + if (dp.containsKey(i) && dp.get(i).containsKey(sum)) { + return dp.get(i).get(sum); + } + boolean ans = false; + if (sum == 0) { + ans = true; + } else if (i != -1) { + ans = process2(arr, i - 1, sum, dp) || process2(arr, i - 1, sum - arr[i], dp); + } + if (!dp.containsKey(i)) { + dp.put(i, new HashMap<>()); + } + dp.get(i).put(sum, ans); + return ans; + } + + // arr中的值可能为正，可能为负，可能为0 + // 自由选择arr中的数字，能不能累加得到sum + // 经典动态规划 + public static boolean isSum3(int[] arr, int sum) { + if (sum == 0) { + return true; + } + if (arr == null || arr.length == 0) { + return false; + } + int min = 0; + int max = 0; + for (int num : arr) { + min += num < 0 ? num : 0; + max += num > 0 ? num : 0; + } + if (sum < min || sum > max) { + return false; + } + int N = arr.length; + boolean[][] dp = new boolean[N][max - min + 1]; + dp[0][-min] = true; + dp[0][arr[0] - min] = true; + for (int i = 1; i < N; i++) { + for (int j = min; j <= max; j++) { + dp[i][j - min] = dp[i - 1][j - min]; + int next = j - min - arr[i]; + dp[i][j - min] |= (next >= 0 && next <= max - min && dp[i - 1][next]); + } + } + return dp[N - 1][sum - min]; + } + + // arr中的值可能为正，可能为负，可能为0 + // 自由选择arr中的数字，能不能累加得到sum + // 分治的方法 + // 如果arr中的数值特别大，动态规划方法依然会很慢 + // 此时如果arr的数字个数不算多(40以内)，哪怕其中的数值很大，分治的方法也将是最优解 + public static boolean isSum4(int[] arr, int sum) { + if (sum == 0) { + return true; + } + if (arr == null || arr.length == 0) { + return false; + } + if (arr.length == 1) { + return arr[0] == sum; + } + int N = arr.length; + int mid = N >> 1; + HashSet leftSum = new HashSet<>(); + HashSet rightSum = new HashSet<>(); + process4(arr, 0, mid, 0, leftSum); + process4(arr, mid, N, 0, rightSum); + for (int l : leftSum) { + if (rightSum.contains(sum - l)) { + return true; + } + } + return false; + } + + public static void process4(int[] arr, int i, int end, int pre, HashSet ans) { + if (i == end) { + ans.add(pre); + } else { + process4(arr, i + 1, end, pre, ans); + process4(arr, i + 1, end, pre + arr[i], ans); + } + } + + // 为了测试 + // 生成长度为len的随机数组 + // 值在[-max, max]上随机 + public static int[] randomArray(int len, int max) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * ((max << 1) + 1)) - max; + } + return arr; + } + + // 对数器验证所有方法 + public static void main(String[] args) { + int N = 20; + int M = 100; + int testTime = 100000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int size = (int) (Math.random() * (N + 1)); + int[] arr = randomArray(size, M); + int sum = (int) (Math.random() * ((M << 1) + 1)) - M; + boolean ans1 = isSum1(arr, sum); + boolean ans2 = isSum2(arr, sum); + boolean ans3 = isSum3(arr, sum); + boolean ans4 = isSum4(arr, sum); + if (ans1 ^ ans2 || ans3 ^ ans4 || ans1 ^ ans3) { + System.out.println("出错了！"); + System.out.print("arr : "); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println("sum : " + sum); + System.out.println("方法一答案 : " + ans1); + System.out.println("方法二答案 : " + ans2); + System.out.println("方法三答案 : " + ans3); + System.out.println("方法四答案 : " + ans4); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class40/Code01_LongestSumSubArrayLengthInPositiveArray.java b/体系学习班/class40/Code01_LongestSumSubArrayLengthInPositiveArray.java new file mode 100644 index 0000000..a2389fa --- /dev/null +++ b/体系学习班/class40/Code01_LongestSumSubArrayLengthInPositiveArray.java @@ -0,0 +1,91 @@ +package class40; + +public class Code01_LongestSumSubArrayLengthInPositiveArray { + + public static int getMaxLength(int[] arr, int K) { + if (arr == null || arr.length == 0 || K <= 0) { + return 0; + } + int left = 0; + int right = 0; + int sum = arr[0]; + int len = 0; + while (right < arr.length) { + if (sum == K) { + len = Math.max(len, right - left + 1); + sum -= arr[left++]; + } else if (sum < K) { + right++; + if (right == arr.length) { + break; + } + sum += arr[right]; + } else { + sum -= arr[left++]; + } + } + return len; + } + + // for test + public static int right(int[] arr, int K) { + int max = 0; + for (int i = 0; i < arr.length; i++) { + for (int j = i; j < arr.length; j++) { + if (valid(arr, i, j, K)) { + max = Math.max(max, j - i + 1); + } + } + } + return max; + } + + // for test + public static boolean valid(int[] arr, int L, int R, int K) { + int sum = 0; + for (int i = L; i <= R; i++) { + sum += arr[i]; + } + return sum == K; + } + + // for test + public static int[] generatePositiveArray(int size, int value) { + int[] ans = new int[size]; + for (int i = 0; i != size; i++) { + ans[i] = (int) (Math.random() * value) + 1; + } + return ans; + } + + // for test + public static void printArray(int[] arr) { + for (int i = 0; i != arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + public static void main(String[] args) { + int len = 50; + int value = 100; + int testTime = 500000; + System.out.println("test begin"); + for (int i = 0; i < testTime; i++) { + int[] arr = generatePositiveArray(len, value); + int K = (int) (Math.random() * value) + 1; + int ans1 = getMaxLength(arr, K); + int ans2 = right(arr, K); + if (ans1 != ans2) { + System.out.println("Oops!"); + printArray(arr); + System.out.println("K : " + K); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("test end"); + } + +} diff --git a/体系学习班/class40/Code02_LongestSumSubArrayLength.java b/体系学习班/class40/Code02_LongestSumSubArrayLength.java new file mode 100644 index 0000000..a89fb14 --- /dev/null +++ b/体系学习班/class40/Code02_LongestSumSubArrayLength.java @@ -0,0 +1,92 @@ +package class40; + +import java.util.HashMap; + +public class Code02_LongestSumSubArrayLength { + + public static int maxLength(int[] arr, int k) { + if (arr == null || arr.length == 0) { + return 0; + } + // key:前缀和 + // value : 0~value这个前缀和是最早出现key这个值的 + HashMap map = new HashMap(); + map.put(0, -1); // important + int len = 0; + int sum = 0; + for (int i = 0; i < arr.length; i++) { + sum += arr[i]; + if (map.containsKey(sum - k)) { + len = Math.max(i - map.get(sum - k), len); + } + if (!map.containsKey(sum)) { + map.put(sum, i); + } + } + return len; + } + + // for test + public static int right(int[] arr, int K) { + int max = 0; + for (int i = 0; i < arr.length; i++) { + for (int j = i; j < arr.length; j++) { + if (valid(arr, i, j, K)) { + max = Math.max(max, j - i + 1); + } + } + } + return max; + } + + // for test + public static boolean valid(int[] arr, int L, int R, int K) { + int sum = 0; + for (int i = L; i <= R; i++) { + sum += arr[i]; + } + return sum == K; + } + + // for test + public static int[] generateRandomArray(int size, int value) { + int[] ans = new int[(int) (Math.random() * size) + 1]; + for (int i = 0; i < ans.length; i++) { + ans[i] = (int) (Math.random() * value) - (int) (Math.random() * value); + } + return ans; + } + + // for test + public static void printArray(int[] arr) { + for (int i = 0; i != arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + public static void main(String[] args) { + int len = 50; + int value = 100; + int testTime = 500000; + + System.out.println("test begin"); + for (int i = 0; i < testTime; i++) { + int[] arr = generateRandomArray(len, value); + int K = (int) (Math.random() * value) - (int) (Math.random() * value); + int ans1 = maxLength(arr, K); + int ans2 = right(arr, K); + if (ans1 != ans2) { + System.out.println("Oops!"); + printArray(arr); + System.out.println("K : " + K); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("test end"); + + } + +} diff --git a/体系学习班/class40/Code03_LongestLessSumSubArrayLength.java b/体系学习班/class40/Code03_LongestLessSumSubArrayLength.java new file mode 100644 index 0000000..c4ebc56 --- /dev/null +++ b/体系学习班/class40/Code03_LongestLessSumSubArrayLength.java @@ -0,0 +1,103 @@ +package class40; + +public class Code03_LongestLessSumSubArrayLength { + + public static int maxLengthAwesome(int[] arr, int k) { + if (arr == null || arr.length == 0) { + return 0; + } + int[] minSums = new int[arr.length]; + int[] minSumEnds = new int[arr.length]; + minSums[arr.length - 1] = arr[arr.length - 1]; + minSumEnds[arr.length - 1] = arr.length - 1; + for (int i = arr.length - 2; i >= 0; i--) { + if (minSums[i + 1] < 0) { + minSums[i] = arr[i] + minSums[i + 1]; + minSumEnds[i] = minSumEnds[i + 1]; + } else { + minSums[i] = arr[i]; + minSumEnds[i] = i; + } + } + // 迟迟扩不进来那一块儿的开头位置 + int end = 0; + int sum = 0; + int ans = 0; + for (int i = 0; i < arr.length; i++) { + // while循环结束之后： + // 1) 如果以i开头的情况下，累加和<=k的最长子数组是arr[i..end-1]，看看这个子数组长度能不能更新res； + // 2) 如果以i开头的情况下，累加和<=k的最长子数组比arr[i..end-1]短，更新还是不更新res都不会影响最终结果； + while (end < arr.length && sum + minSums[end] <= k) { + sum += minSums[end]; + end = minSumEnds[end] + 1; + } + ans = Math.max(ans, end - i); + if (end > i) { // 还有窗口，哪怕窗口没有数字 [i~end) [4,4) + sum -= arr[i]; + } else { // i == end, 即将 i++, i > end, 此时窗口概念维持不住了，所以end跟着i一起走 + end = i + 1; + } + } + return ans; + } + + public static int maxLength(int[] arr, int k) { + int[] h = new int[arr.length + 1]; + int sum = 0; + h[0] = sum; + for (int i = 0; i != arr.length; i++) { + sum += arr[i]; + h[i + 1] = Math.max(sum, h[i]); + } + sum = 0; + int res = 0; + int pre = 0; + int len = 0; + for (int i = 0; i != arr.length; i++) { + sum += arr[i]; + pre = getLessIndex(h, sum - k); + len = pre == -1 ? 0 : i - pre + 1; + res = Math.max(res, len); + } + return res; + } + + public static int getLessIndex(int[] arr, int num) { + int low = 0; + int high = arr.length - 1; + int mid = 0; + int res = -1; + while (low <= high) { + mid = (low + high) / 2; + if (arr[mid] >= num) { + res = mid; + high = mid - 1; + } else { + low = mid + 1; + } + } + return res; + } + + // for test + public static int[] generateRandomArray(int len, int maxValue) { + int[] res = new int[len]; + for (int i = 0; i != res.length; i++) { + res[i] = (int) (Math.random() * maxValue) - (maxValue / 3); + } + return res; + } + + public static void main(String[] args) { + System.out.println("test begin"); + for (int i = 0; i < 10000000; i++) { + int[] arr = generateRandomArray(10, 20); + int k = (int) (Math.random() * 20) - 5; + if (maxLengthAwesome(arr, k) != maxLength(arr, k)) { + System.out.println("Oops!"); + } + } + System.out.println("test finish"); + } + +} diff --git a/体系学习班/class40/Code04_AvgLessEqualValueLongestSubarray.java b/体系学习班/class40/Code04_AvgLessEqualValueLongestSubarray.java new file mode 100644 index 0000000..c56eef9 --- /dev/null +++ b/体系学习班/class40/Code04_AvgLessEqualValueLongestSubarray.java @@ -0,0 +1,152 @@ +package class40; + +import java.util.TreeMap; + +public class Code04_AvgLessEqualValueLongestSubarray { + + // 暴力解，时间复杂度O(N^3)，用于做对数器 + public static int ways1(int[] arr, int v) { + int ans = 0; + for (int L = 0; L < arr.length; L++) { + for (int R = L; R < arr.length; R++) { + int sum = 0; + int k = R - L + 1; + for (int i = L; i <= R; i++) { + sum += arr[i]; + } + double avg = (double) sum / (double) k; + if (avg <= v) { + ans = Math.max(ans, k); + } + } + } + return ans; + } + + // 想实现的解法2，时间复杂度O(N*logN) + public static int ways2(int[] arr, int v) { + if (arr == null || arr.length == 0) { + return 0; + } + TreeMap origins = new TreeMap<>(); + int ans = 0; + int modify = 0; + for (int i = 0; i < arr.length; i++) { + int p1 = arr[i] <= v ? 1 : 0; + int p2 = 0; + int querry = -arr[i] - modify; + if (origins.floorKey(querry) != null) { + p2 = i - origins.get(origins.floorKey(querry)) + 1; + } + ans = Math.max(ans, Math.max(p1, p2)); + int curOrigin = -modify - v; + if (origins.floorKey(curOrigin) == null) { + origins.put(curOrigin, i); + } + modify += arr[i] - v; + } + return ans; + } + + // 想实现的解法3，时间复杂度O(N) + public static int ways3(int[] arr, int v) { + if (arr == null || arr.length == 0) { + return 0; + } + for (int i = 0; i < arr.length; i++) { + arr[i] -= v; + } + return maxLengthAwesome(arr, 0); + } + + // 找到数组中累加和<=k的最长子数组 + public static int maxLengthAwesome(int[] arr, int k) { + int N = arr.length; + int[] sums = new int[N]; + int[] ends = new int[N]; + sums[N - 1] = arr[N - 1]; + ends[N - 1] = N - 1; + for (int i = N - 2; i >= 0; i--) { + if (sums[i + 1] < 0) { + sums[i] = arr[i] + sums[i + 1]; + ends[i] = ends[i + 1]; + } else { + sums[i] = arr[i]; + ends[i] = i; + } + } + int end = 0; + int sum = 0; + int res = 0; + for (int i = 0; i < N; i++) { + while (end < N && sum + sums[end] <= k) { + sum += sums[end]; + end = ends[end] + 1; + } + res = Math.max(res, end - i); + if (end > i) { + sum -= arr[i]; + } else { + end = i + 1; + } + } + return res; + } + + // 用于测试 + public static int[] randomArray(int maxLen, int maxValue) { + int len = (int) (Math.random() * maxLen) + 1; + int[] ans = new int[len]; + for (int i = 0; i < len; i++) { + ans[i] = (int) (Math.random() * maxValue); + } + return ans; + } + + // 用于测试 + public static int[] copyArray(int[] arr) { + int[] ans = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + ans[i] = arr[i]; + } + return ans; + } + + // 用于测试 + public static void printArray(int[] arr) { + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // 用于测试 + public static void main(String[] args) { + System.out.println("测试开始"); + int maxLen = 20; + int maxValue = 100; + int testTime = 500000; + for (int i = 0; i < testTime; i++) { + int[] arr = randomArray(maxLen, maxValue); + int value = (int) (Math.random() * maxValue); + int[] arr1 = copyArray(arr); + int[] arr2 = copyArray(arr); + int[] arr3 = copyArray(arr); + int ans1 = ways1(arr1, value); + int ans2 = ways2(arr2, value); + int ans3 = ways3(arr3, value); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("测试出错！"); + System.out.print("测试数组："); + printArray(arr); + System.out.println("子数组平均值不小于 ：" + value); + System.out.println("方法1得到的最大长度：" + ans1); + System.out.println("方法2得到的最大长度：" + ans2); + System.out.println("方法3得到的最大长度：" + ans3); + System.out.println("========================="); + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class40/Code05_PrintMatrixSpiralOrder.java b/体系学习班/class40/Code05_PrintMatrixSpiralOrder.java new file mode 100644 index 0000000..cb3f80c --- /dev/null +++ b/体系学习班/class40/Code05_PrintMatrixSpiralOrder.java @@ -0,0 +1,53 @@ +package class40; + +public class Code05_PrintMatrixSpiralOrder { + + public static void spiralOrderPrint(int[][] matrix) { + int tR = 0; + int tC = 0; + int dR = matrix.length - 1; + int dC = matrix[0].length - 1; + while (tR <= dR && tC <= dC) { + printEdge(matrix, tR++, tC++, dR--, dC--); + } + } + + public static void printEdge(int[][] m, int tR, int tC, int dR, int dC) { + if (tR == dR) { + for (int i = tC; i <= dC; i++) { + System.out.print(m[tR][i] + " "); + } + } else if (tC == dC) { + for (int i = tR; i <= dR; i++) { + System.out.print(m[i][tC] + " "); + } + } else { + int curC = tC; + int curR = tR; + while (curC != dC) { + System.out.print(m[tR][curC] + " "); + curC++; + } + while (curR != dR) { + System.out.print(m[curR][dC] + " "); + curR++; + } + while (curC != tC) { + System.out.print(m[dR][curC] + " "); + curC--; + } + while (curR != tR) { + System.out.print(m[curR][tC] + " "); + curR--; + } + } + } + + public static void main(String[] args) { + int[][] matrix = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, + { 13, 14, 15, 16 } }; + spiralOrderPrint(matrix); + + } + +} diff --git a/体系学习班/class40/Code06_RotateMatrix.java b/体系学习班/class40/Code06_RotateMatrix.java new file mode 100644 index 0000000..4e9534f --- /dev/null +++ b/体系学习班/class40/Code06_RotateMatrix.java @@ -0,0 +1,44 @@ +package class40; + +public class Code06_RotateMatrix { + + public static void rotate(int[][] matrix) { + int a = 0; + int b = 0; + int c = matrix.length - 1; + int d = matrix[0].length - 1; + while (a < c) { + rotateEdge(matrix, a++, b++, c--, d--); + } + } + + public static void rotateEdge(int[][] m, int a, int b, int c, int d) { + int tmp = 0; + for (int i = 0; i < d - b; i++) { + tmp = m[a][b + i]; + m[a][b + i] = m[c - i][b]; + m[c - i][b] = m[c][d - i]; + m[c][d - i] = m[a + i][d]; + m[a + i][d] = tmp; + } + } + + public static void printMatrix(int[][] matrix) { + for (int i = 0; i != matrix.length; i++) { + for (int j = 0; j != matrix[0].length; j++) { + System.out.print(matrix[i][j] + " "); + } + System.out.println(); + } + } + + public static void main(String[] args) { + int[][] matrix = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; + printMatrix(matrix); + rotate(matrix); + System.out.println("========="); + printMatrix(matrix); + + } + +} diff --git a/体系学习班/class40/Code07_ZigZagPrintMatrix.java b/体系学习班/class40/Code07_ZigZagPrintMatrix.java new file mode 100644 index 0000000..01b9921 --- /dev/null +++ b/体系学习班/class40/Code07_ZigZagPrintMatrix.java @@ -0,0 +1,42 @@ +package class40; + +public class Code07_ZigZagPrintMatrix { + + public static void printMatrixZigZag(int[][] matrix) { + int tR = 0; + int tC = 0; + int dR = 0; + int dC = 0; + int endR = matrix.length - 1; + int endC = matrix[0].length - 1; + boolean fromUp = false; + while (tR != endR + 1) { + printLevel(matrix, tR, tC, dR, dC, fromUp); + tR = tC == endC ? tR + 1 : tR; + tC = tC == endC ? tC : tC + 1; + dC = dR == endR ? dC + 1 : dC; + dR = dR == endR ? dR : dR + 1; + fromUp = !fromUp; + } + System.out.println(); + } + + public static void printLevel(int[][] m, int tR, int tC, int dR, int dC, boolean f) { + if (f) { + while (tR != dR + 1) { + System.out.print(m[tR++][tC--] + " "); + } + } else { + while (dR != tR - 1) { + System.out.print(m[dR--][dC++] + " "); + } + } + } + + public static void main(String[] args) { + int[][] matrix = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 } }; + printMatrixZigZag(matrix); + + } + +} diff --git a/体系学习班/class40/Code08_PrintStar.java b/体系学习班/class40/Code08_PrintStar.java new file mode 100644 index 0000000..77d41ef --- /dev/null +++ b/体系学习班/class40/Code08_PrintStar.java @@ -0,0 +1,46 @@ +package class40; + +public class Code08_PrintStar { + + public static void printStar(int N) { + int leftUp = 0; + int rightDown = N - 1; + char[][] m = new char[N][N]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < N; j++) { + m[i][j] = ' '; + } + } + while (leftUp <= rightDown) { + set(m, leftUp, rightDown); + leftUp += 2; + rightDown -= 2; + } + for (int i = 0; i < N; i++) { + for (int j = 0; j < N; j++) { + System.out.print(m[i][j] + " "); + } + System.out.println(); + } + } + + public static void set(char[][] m, int leftUp, int rightDown) { + for (int col = leftUp; col <= rightDown; col++) { + m[leftUp][col] = '*'; + } + for (int row = leftUp + 1; row <= rightDown; row++) { + m[row][rightDown] = '*'; + } + for (int col = rightDown - 1; col > leftUp; col--) { + m[rightDown][col] = '*'; + } + for (int row = rightDown - 1; row > leftUp + 1; row--) { + m[row][leftUp + 1] = '*'; + } + } + + public static void main(String[] args) { + printStar(5); + } + +} diff --git a/体系学习班/class41/Code01_BestSplitForAll.java b/体系学习班/class41/Code01_BestSplitForAll.java new file mode 100644 index 0000000..a4eddb9 --- /dev/null +++ b/体系学习班/class41/Code01_BestSplitForAll.java @@ -0,0 +1,72 @@ +package class41; + +public class Code01_BestSplitForAll { + + public static int bestSplit1(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int N = arr.length; + int ans = 0; + for (int s = 0; s < N - 1; s++) { + int sumL = 0; + for (int L = 0; L <= s; L++) { + sumL += arr[L]; + } + int sumR = 0; + for (int R = s + 1; R < N; R++) { + sumR += arr[R]; + } + ans = Math.max(ans, Math.min(sumL, sumR)); + } + return ans; + } + + public static int bestSplit2(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int N = arr.length; + int sumAll = 0; + for (int num : arr) { + sumAll += num; + } + int ans = 0; + int sumL = 0; + // [0...s] [s+1...N-1] + for (int s = 0; s < N - 1; s++) { + sumL += arr[s]; + int sumR = sumAll - sumL; + ans = Math.max(ans, Math.min(sumL, sumR)); + } + return ans; + } + + public static int[] randomArray(int len, int max) { + int[] ans = new int[len]; + for (int i = 0; i < len; i++) { + ans[i] = (int) (Math.random() * max); + } + return ans; + } + + public static void main(String[] args) { + int N = 20; + int max = 30; + int testTime = 1000000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * N); + int[] arr = randomArray(len, max); + int ans1 = bestSplit1(arr); + int ans2 = bestSplit2(arr); + if (ans1 != ans2) { + System.out.println(ans1); + System.out.println(ans2); + System.out.println("Oops!"); + } + } + System.out.println("测试结束"); + } + +} \ No newline at end of file diff --git a/体系学习班/class41/Code02_BestSplitForEveryPosition.java b/体系学习班/class41/Code02_BestSplitForEveryPosition.java new file mode 100644 index 0000000..d98b389 --- /dev/null +++ b/体系学习班/class41/Code02_BestSplitForEveryPosition.java @@ -0,0 +1,130 @@ +package class41; + +public class Code02_BestSplitForEveryPosition { + + public static int[] bestSplit1(int[] arr) { + if (arr == null || arr.length == 0) { + return new int[0]; + } + int N = arr.length; + int[] ans = new int[N]; + ans[0] = 0; + for (int range = 1; range < N; range++) { + for (int s = 0; s < range; s++) { + int sumL = 0; + for (int L = 0; L <= s; L++) { + sumL += arr[L]; + } + int sumR = 0; + for (int R = s + 1; R <= range; R++) { + sumR += arr[R]; + } + ans[range] = Math.max(ans[range], Math.min(sumL, sumR)); + } + } + return ans; + } + + // 求原来的数组arr中，arr[L...R]的累加和 + public static int sum(int[] sum, int L, int R) { + return sum[R + 1] - sum[L]; + } + + public static int[] bestSplit2(int[] arr) { + if (arr == null || arr.length == 0) { + return new int[0]; + } + int N = arr.length; + int[] ans = new int[N]; + ans[0] = 0; + int[] sum = new int[N + 1]; + for (int i = 0; i < N; i++) { + sum[i + 1] = sum[i] + arr[i]; + } + for (int range = 1; range < N; range++) { + for (int s = 0; s < range; s++) { + int sumL = sum(sum, 0, s); + int sumR = sum(sum, s + 1, range); + ans[range] = Math.max(ans[range], Math.min(sumL, sumR)); + } + } + return ans; + } + + public static int[] bestSplit3(int[] arr) { + if (arr == null || arr.length == 0) { + return new int[0]; + } + int N = arr.length; + int[] ans = new int[N]; + ans[0] = 0; + // arr = {5, 3, 1, 3} + // 0 1 2 3 + // sum ={0, 5, 8, 9, 12} + // 0 1 2 3 4 + // 0~2 -> sum[3] - sum[0] + // 1~3 -> sum[4] - sum[1] + int[] sum = new int[N + 1]; + for (int i = 0; i < N; i++) { + sum[i + 1] = sum[i] + arr[i]; + } + // 最优划分 + // 0~range-1上，最优划分是左部分[0~best] 右部分[best+1~range-1] + int best = 0; + for (int range = 1; range < N; range++) { + while (best + 1 < range) { + int before = Math.min(sum(sum, 0, best), sum(sum, best + 1, range)); + int after = Math.min(sum(sum, 0, best + 1), sum(sum, best + 2, range)); + // 注意，一定要是>=，只是>会出错 + // 课上会讲解 + if (after >= before) { + best++; + } else { + break; + } + } + ans[range] = Math.min(sum(sum, 0, best), sum(sum, best + 1, range)); + } + return ans; + } + + public static int[] randomArray(int len, int max) { + int[] ans = new int[len]; + for (int i = 0; i < len; i++) { + ans[i] = (int) (Math.random() * max); + } + return ans; + } + + public static boolean isSameArray(int[] arr1, int[] arr2) { + if (arr1.length != arr2.length) { + return false; + } + int N = arr1.length; + for (int i = 0; i < N; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + public static void main(String[] args) { + int N = 20; + int max = 30; + int testTime = 1000000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * N); + int[] arr = randomArray(len, max); + int[] ans1 = bestSplit1(arr); + int[] ans2 = bestSplit2(arr); + int[] ans3 = bestSplit3(arr); + if (!isSameArray(ans1, ans2) || !isSameArray(ans1, ans3)) { + System.out.println("Oops!"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class41/Code03_StoneMerge.java b/体系学习班/class41/Code03_StoneMerge.java new file mode 100644 index 0000000..793d7d2 --- /dev/null +++ b/体系学习班/class41/Code03_StoneMerge.java @@ -0,0 +1,117 @@ +package class41; + +// 四边形不等式：合并石子问题 +public class Code03_StoneMerge { + + public static int[] sum(int[] arr) { + int N = arr.length; + int[] s = new int[N + 1]; + s[0] = 0; + for (int i = 0; i < N; i++) { + s[i + 1] = s[i] + arr[i]; + } + return s; + } + + public static int w(int[] s, int l, int r) { + return s[r + 1] - s[l]; + } + + public static int min1(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int N = arr.length; + int[] s = sum(arr); + return process1(0, N - 1, s); + } + + public static int process1(int L, int R, int[] s) { + if (L == R) { + return 0; + } + int next = Integer.MAX_VALUE; + for (int leftEnd = L; leftEnd < R; leftEnd++) { + next = Math.min(next, process1(L, leftEnd, s) + process1(leftEnd + 1, R, s)); + } + return next + w(s, L, R); + } + + public static int min2(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int N = arr.length; + int[] s = sum(arr); + int[][] dp = new int[N][N]; + // dp[i][i] = 0 + for (int L = N - 2; L >= 0; L--) { + for (int R = L + 1; R < N; R++) { + int next = Integer.MAX_VALUE; + // dp(L..leftEnd) + dp[leftEnd+1...R] + 累加和[L...R] + for (int leftEnd = L; leftEnd < R; leftEnd++) { + next = Math.min(next, dp[L][leftEnd] + dp[leftEnd + 1][R]); + } + dp[L][R] = next + w(s, L, R); + } + } + return dp[0][N - 1]; + } + + public static int min3(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int N = arr.length; + int[] s = sum(arr); + int[][] dp = new int[N][N]; + int[][] best = new int[N][N]; + for (int i = 0; i < N - 1; i++) { + best[i][i + 1] = i; + dp[i][i + 1] = w(s, i, i + 1); + } + for (int L = N - 3; L >= 0; L--) { + for (int R = L + 2; R < N; R++) { + int next = Integer.MAX_VALUE; + int choose = -1; + for (int leftEnd = best[L][R - 1]; leftEnd <= best[L + 1][R]; leftEnd++) { + int cur = dp[L][leftEnd] + dp[leftEnd + 1][R]; + if (cur <= next) { + next = cur; + choose = leftEnd; + } + } + best[L][R] = choose; + dp[L][R] = next + w(s, L, R); + } + } + return dp[0][N - 1]; + } + + public static int[] randomArray(int len, int maxValue) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * maxValue); + } + return arr; + } + + public static void main(String[] args) { + int N = 15; + int maxValue = 100; + int testTime = 1000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * N); + int[] arr = randomArray(len, maxValue); + int ans1 = min1(arr); + int ans2 = min2(arr); + int ans3 = min3(arr); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("Oops!"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class41/Code04_SplitArrayLargestSum.java b/体系学习班/class41/Code04_SplitArrayLargestSum.java new file mode 100644 index 0000000..8d644e6 --- /dev/null +++ b/体系学习班/class41/Code04_SplitArrayLargestSum.java @@ -0,0 +1,193 @@ +package class41; + +// leetcode原题 +// 测试链接：https://leetcode.com/problems/split-array-largest-sum/ +public class Code04_SplitArrayLargestSum { + + // 求原数组arr[L...R]的累加和 + public static int sum(int[] sum, int L, int R) { + return sum[R + 1] - sum[L]; + } + + // 不优化枚举的动态规划方法，O(N^2 * K) + public static int splitArray1(int[] nums, int K) { + int N = nums.length; + int[] sum = new int[N + 1]; + for (int i = 0; i < N; i++) { + sum[i + 1] = sum[i] + nums[i]; + } + int[][] dp = new int[N][K + 1]; + for (int j = 1; j <= K; j++) { + dp[0][j] = nums[0]; + } + for (int i = 1; i < N; i++) { + dp[i][1] = sum(sum, 0, i); + } + // 每一行从上往下 + // 每一列从左往右 + // 根本不去凑优化位置对儿！ + for (int i = 1; i < N; i++) { + for (int j = 2; j <= K; j++) { + int ans = Integer.MAX_VALUE; + // 枚举是完全不优化的！ + for (int leftEnd = 0; leftEnd <= i; leftEnd++) { + int leftCost = leftEnd == -1 ? 0 : dp[leftEnd][j - 1]; + int rightCost = leftEnd == i ? 0 : sum(sum, leftEnd + 1, i); + int cur = Math.max(leftCost, rightCost); + if (cur < ans) { + ans = cur; + } + } + dp[i][j] = ans; + } + } + return dp[N - 1][K]; + } + + // 课上现场写的方法，用了枚举优化，O(N * K) + public static int splitArray2(int[] nums, int K) { + int N = nums.length; + int[] sum = new int[N + 1]; + for (int i = 0; i < N; i++) { + sum[i + 1] = sum[i] + nums[i]; + } + int[][] dp = new int[N][K + 1]; + int[][] best = new int[N][K + 1]; + for (int j = 1; j <= K; j++) { + dp[0][j] = nums[0]; + best[0][j] = -1; + } + for (int i = 1; i < N; i++) { + dp[i][1] = sum(sum, 0, i); + best[i][1] = -1; + } + // 从第2列开始，从左往右 + // 每一列，从下往上 + // 为什么这样的顺序？因为要去凑（左，下）优化位置对儿！ + for (int j = 2; j <= K; j++) { + for (int i = N - 1; i >= 1; i--) { + int down = best[i][j - 1]; + // 如果i==N-1，则不优化上限 + int up = i == N - 1 ? N - 1 : best[i + 1][j]; + int ans = Integer.MAX_VALUE; + int bestChoose = -1; + for (int leftEnd = down; leftEnd <= up; leftEnd++) { + int leftCost = leftEnd == -1 ? 0 : dp[leftEnd][j - 1]; + int rightCost = leftEnd == i ? 0 : sum(sum, leftEnd + 1, i); + int cur = Math.max(leftCost, rightCost); + // 注意下面的if一定是 < 课上的错误就是此处！当时写的 <= ！ + // 也就是说，只有取得明显的好处才移动！ + // 举个例子来说明，比如[2,6,4,4]，3个画匠时候，如下两种方案都是最优: + // (2,6) (4) 两个画匠负责 | (4) 最后一个画匠负责 + // (2,6) (4,4)两个画匠负责 | 最后一个画匠什么也不负责 + // 第一种方案划分为，[0~2] [3~3] + // 第二种方案划分为，[0~3] [无] + // 两种方案的答案都是8，但是划分点位置一定不要移动! + // 只有明显取得好处时(<)，划分点位置才移动! + // 也就是说后面的方案如果==前面的最优，不要移动！只有优于前面的最优，才移动 + // 比如上面的两个方案，如果你移动到了方案二，你会得到: + // [2,6,4,4] 三个画匠时，最优为[0~3](前两个画家) [无](最后一个画家)， + // 最优划分点为3位置(best[3][3]) + // 那么当4个画匠时，也就是求解dp[3][4]时 + // 因为best[3][3] = 3，这个值提供了dp[3][4]的下限 + // 而事实上dp[3][4]的最优划分为: + // [0~2]（三个画家处理） [3~3] (一个画家处理)，此时最优解为6 + // 所以，你就得不到dp[3][4]的最优解了，因为划分点已经越过2了 + // 提供了对数器验证，你可以改成<=，对数器和leetcode都过不了 + // 这里是<，对数器和leetcode都能通过 + // 这里面会让同学们感到困惑的点： + // 为啥==的时候，不移动，只有<的时候，才移动呢？例子懂了，但是道理何在？ + // 哈哈哈哈哈，看了邮局选址问题，你更懵，请看42节！ + if (cur < ans) { + ans = cur; + bestChoose = leftEnd; + } + } + dp[i][j] = ans; + best[i][j] = bestChoose; + } + } + return dp[N - 1][K]; + } + + public static int splitArray3(int[] nums, int M) { + long sum = 0; + for (int i = 0; i < nums.length; i++) { + sum += nums[i]; + } + long l = 0; + long r = sum; + long ans = 0; + while (l <= r) { + long mid = (l + r) / 2; + long cur = getNeedParts(nums, mid); + if (cur <= M) { + ans = mid; + r = mid - 1; + } else { + l = mid + 1; + } + } + return (int) ans; + } + + public static int getNeedParts(int[] arr, long aim) { + for (int i = 0; i < arr.length; i++) { + if (arr[i] > aim) { + return Integer.MAX_VALUE; + } + } + int parts = 1; + int all = arr[0]; + for (int i = 1; i < arr.length; i++) { + if (all + arr[i] > aim) { + parts++; + all = arr[i]; + } else { + all += arr[i]; + } + } + return parts; + } + + public static int[] randomArray(int len, int maxValue) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * maxValue); + } + return arr; + } + + public static void printArray(int[] arr) { + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + public static void main(String[] args) { + int N = 100; + int maxValue = 100; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * N) + 1; + int M = (int) (Math.random() * N) + 1; + int[] arr = randomArray(len, maxValue); + int ans1 = splitArray1(arr, M); + int ans2 = splitArray2(arr, M); + int ans3 = splitArray3(arr, M); + if (ans1 != ans2 || ans1 != ans3) { + System.out.print("arr : "); + printArray(arr); + System.out.println("M : " + M); + System.out.println("ans1 : " + ans1); + System.out.println("ans2 : " + ans2); + System.out.println("ans3 : " + ans3); + System.out.println("Oops!"); + break; + } + } + System.out.println("测试结束"); + } +} diff --git a/体系学习班/class42/Code01_PostOfficeProblem.java b/体系学习班/class42/Code01_PostOfficeProblem.java new file mode 100644 index 0000000..c6fa893 --- /dev/null +++ b/体系学习班/class42/Code01_PostOfficeProblem.java @@ -0,0 +1,115 @@ +package class42; + +import java.util.Arrays; + +public class Code01_PostOfficeProblem { + + public static int min1(int[] arr, int num) { + if (arr == null || num < 1 || arr.length < num) { + return 0; + } + int N = arr.length; + int[][] w = new int[N + 1][N + 1]; + for (int L = 0; L < N; L++) { + for (int R = L + 1; R < N; R++) { + w[L][R] = w[L][R - 1] + arr[R] - arr[(L + R) >> 1]; + } + } + int[][] dp = new int[N][num + 1]; + for (int i = 0; i < N; i++) { + dp[i][1] = w[0][i]; + } + for (int i = 1; i < N; i++) { + for (int j = 2; j <= Math.min(i, num); j++) { + int ans = Integer.MAX_VALUE; + for (int k = 0; k <= i; k++) { + ans = Math.min(ans, dp[k][j - 1] + w[k + 1][i]); + } + dp[i][j] = ans; + } + } + return dp[N - 1][num]; + } + + public static int min2(int[] arr, int num) { + if (arr == null || num < 1 || arr.length < num) { + return 0; + } + int N = arr.length; + int[][] w = new int[N + 1][N + 1]; + for (int L = 0; L < N; L++) { + for (int R = L + 1; R < N; R++) { + w[L][R] = w[L][R - 1] + arr[R] - arr[(L + R) >> 1]; + } + } + int[][] dp = new int[N][num + 1]; + int[][] best = new int[N][num + 1]; + for (int i = 0; i < N; i++) { + dp[i][1] = w[0][i]; + best[i][1] = -1; + } + for (int j = 2; j <= num; j++) { + for (int i = N - 1; i >= j; i--) { + int down = best[i][j - 1]; + int up = i == N - 1 ? N - 1 : best[i + 1][j]; + int ans = Integer.MAX_VALUE; + int bestChoose = -1; + for (int leftEnd = down; leftEnd <= up; leftEnd++) { + int leftCost = leftEnd == -1 ? 0 : dp[leftEnd][j - 1]; + int rightCost = leftEnd == i ? 0 : w[leftEnd + 1][i]; + int cur = leftCost + rightCost; + if (cur <= ans) { + ans = cur; + bestChoose = leftEnd; + } + } + dp[i][j] = ans; + best[i][j] = bestChoose; + } + } + return dp[N - 1][num]; + } + + // for test + public static int[] randomSortedArray(int len, int range) { + int[] arr = new int[len]; + for (int i = 0; i != len; i++) { + arr[i] = (int) (Math.random() * range); + } + Arrays.sort(arr); + return arr; + } + + // for test + public static void printArray(int[] arr) { + for (int i = 0; i != arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + int N = 30; + int maxValue = 100; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * N) + 1; + int[] arr = randomSortedArray(len, maxValue); + int num = (int) (Math.random() * N) + 1; + int ans1 = min1(arr, num); + int ans2 = min2(arr, num); + if (ans1 != ans2) { + printArray(arr); + System.out.println(num); + System.out.println(ans1); + System.out.println(ans2); + System.out.println("Oops!"); + } + } + System.out.println("测试结束"); + + } + +} diff --git a/体系学习班/class42/Code02_ThrowChessPiecesProblem.java b/体系学习班/class42/Code02_ThrowChessPiecesProblem.java new file mode 100644 index 0000000..d996c70 --- /dev/null +++ b/体系学习班/class42/Code02_ThrowChessPiecesProblem.java @@ -0,0 +1,166 @@ +package class42; + +// leetcode测试链接：https://leetcode.com/problems/super-egg-drop +// 方法1和方法2会超时 +// 方法3勉强通过 +// 方法4打败100% +// 方法5打败100%，方法5是在方法4的基础上做了进一步的常数优化 +public class Code02_ThrowChessPiecesProblem { + + public static int superEggDrop1(int kChess, int nLevel) { + if (nLevel < 1 || kChess < 1) { + return 0; + } + return Process1(nLevel, kChess); + } + + // rest还剩多少层楼需要去验证 + // k还有多少颗棋子能够使用 + // 一定要验证出最高的不会碎的楼层！但是每次都是坏运气。 + // 返回至少需要扔几次？ + public static int Process1(int rest, int k) { + if (rest == 0) { + return 0; + } + if (k == 1) { + return rest; + } + int min = Integer.MAX_VALUE; + for (int i = 1; i != rest + 1; i++) { // 第一次扔的时候，仍在了i层 + min = Math.min(min, Math.max(Process1(i - 1, k - 1), Process1(rest - i, k))); + } + return min + 1; + } + + public static int superEggDrop2(int kChess, int nLevel) { + if (nLevel < 1 || kChess < 1) { + return 0; + } + if (kChess == 1) { + return nLevel; + } + int[][] dp = new int[nLevel + 1][kChess + 1]; + for (int i = 1; i != dp.length; i++) { + dp[i][1] = i; + } + for (int i = 1; i != dp.length; i++) { + for (int j = 2; j != dp[0].length; j++) { + int min = Integer.MAX_VALUE; + for (int k = 1; k != i + 1; k++) { + min = Math.min(min, Math.max(dp[k - 1][j - 1], dp[i - k][j])); + } + dp[i][j] = min + 1; + } + } + return dp[nLevel][kChess]; + } + + public static int superEggDrop3(int kChess, int nLevel) { + if (nLevel < 1 || kChess < 1) { + return 0; + } + if (kChess == 1) { + return nLevel; + } + int[][] dp = new int[nLevel + 1][kChess + 1]; + for (int i = 1; i != dp.length; i++) { + dp[i][1] = i; + } + int[][] best = new int[nLevel + 1][kChess + 1]; + for (int i = 1; i != dp[0].length; i++) { + dp[1][i] = 1; + best[1][i] = 1; + } + for (int i = 2; i < nLevel + 1; i++) { + for (int j = kChess; j > 1; j--) { + int ans = Integer.MAX_VALUE; + int bestChoose = -1; + int down = best[i - 1][j]; + int up = j == kChess ? i : best[i][j + 1]; + for (int first = down; first <= up; first++) { + int cur = Math.max(dp[first - 1][j - 1], dp[i - first][j]); + if (cur <= ans) { + ans = cur; + bestChoose = first; + } + } + dp[i][j] = ans + 1; + best[i][j] = bestChoose; + } + } + return dp[nLevel][kChess]; + } + + public static int superEggDrop4(int kChess, int nLevel) { + if (nLevel < 1 || kChess < 1) { + return 0; + } + int[] dp = new int[kChess]; + int res = 0; + while (true) { + res++; + int previous = 0; + for (int i = 0; i < dp.length; i++) { + int tmp = dp[i]; + dp[i] = dp[i] + previous + 1; + previous = tmp; + if (dp[i] >= nLevel) { + return res; + } + } + } + } + + public static int superEggDrop5(int kChess, int nLevel) { + if (nLevel < 1 || kChess < 1) { + return 0; + } + int bsTimes = log2N(nLevel) + 1; + if (kChess >= bsTimes) { + return bsTimes; + } + int[] dp = new int[kChess]; + int res = 0; + while (true) { + res++; + int previous = 0; + for (int i = 0; i < dp.length; i++) { + int tmp = dp[i]; + dp[i] = dp[i] + previous + 1; + previous = tmp; + if (dp[i] >= nLevel) { + return res; + } + } + } + } + + public static int log2N(int n) { + int res = -1; + while (n != 0) { + res++; + n >>>= 1; + } + return res; + } + + public static void main(String[] args) { + int maxN = 500; + int maxK = 30; + int testTime = 1000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int N = (int) (Math.random() * maxN) + 1; + int K = (int) (Math.random() * maxK) + 1; + int ans2 = superEggDrop2(K, N); + int ans3 = superEggDrop3(K, N); + int ans4 = superEggDrop4(K, N); + int ans5 = superEggDrop5(K, N); + if (ans2 != ans3 || ans4 != ans5 || ans2 != ans4) { + System.out.println("出错了!"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class43/Code01_CanIWin.java b/体系学习班/class43/Code01_CanIWin.java new file mode 100644 index 0000000..608b1f1 --- /dev/null +++ b/体系学习班/class43/Code01_CanIWin.java @@ -0,0 +1,116 @@ +package class43; + +// leetcode 464题 +public class Code01_CanIWin { + + // 1~choose 拥有的数字 + // total 一开始的剩余 + // 返回先手会不会赢 + public static boolean canIWin0(int choose, int total) { + if (total == 0) { + return true; + } + if ((choose * (choose + 1) >> 1) < total) { + return false; + } + int[] arr = new int[choose]; + for (int i = 0; i < choose; i++) { + arr[i] = i + 1; + } + // arr[i] != -1 表示arr[i]这个数字还没被拿走 + // arr[i] == -1 表示arr[i]这个数字已经被拿走 + // 集合，arr，1~choose + return process(arr, total); + } + + // 当前轮到先手拿， + // 先手只能选择在arr中还存在的数字， + // 还剩rest这么值， + // 返回先手会不会赢 + public static boolean process(int[] arr, int rest) { + if (rest <= 0) { + return false; + } + // 先手去尝试所有的情况 + for (int i = 0; i < arr.length; i++) { + if (arr[i] != -1) { + int cur = arr[i]; + arr[i] = -1; + boolean next = process(arr, rest - cur); + arr[i] = cur; + if (!next) { + return true; + } + } + } + return false; + } + + // 这个是暴力尝试，思路是正确的，超时而已 + public static boolean canIWin1(int choose, int total) { + if (total == 0) { + return true; + } + if ((choose * (choose + 1) >> 1) < total) { + return false; + } + return process1(choose, 0, total); + } + + // 当前轮到先手拿， + // 先手可以拿1~choose中的任何一个数字 + // status i位如果为0，代表没拿，当前可以拿 + // i位为1，代表已经拿过了，当前不能拿 + // 还剩rest这么值， + // 返回先手会不会赢 + public static boolean process1(int choose, int status, int rest) { + if (rest <= 0) { + return false; + } + for (int i = 1; i <= choose; i++) { + if (((1 << i) & status) == 0) { // i 这个数字，是此时先手的决定！ + if (!process1(choose, (status | (1 << i)), rest - i)) { + return true; + } + } + } + return false; + } + + // 暴力尝试改动态规划而已 + public static boolean canIWin2(int choose, int total) { + if (total == 0) { + return true; + } + if ((choose * (choose + 1) >> 1) < total) { + return false; + } + int[] dp = new int[1 << (choose + 1)]; + // dp[status] == 1 true + // dp[status] == -1 false + // dp[status] == 0 process(status) 没算过！去算！ + return process2(choose, 0, total, dp); + } + + // 为什么明明status和rest是两个可变参数，却只用status来代表状态(也就是dp) + // 因为选了一批数字之后，得到的和一定是一样的，所以rest是由status决定的，所以rest不需要参与记忆化搜索 + public static boolean process2(int choose, int status, int rest, int[] dp) { + if (dp[status] != 0) { + return dp[status] == 1 ? true : false; + } + boolean ans = false; + if (rest > 0) { + for (int i = 1; i <= choose; i++) { + if (((1 << i) & status) == 0) { + if (!process2(choose, (status | (1 << i)), rest - i, dp)) { + ans = true; + break; + } + } + } + } + dp[status] = ans ? 1 : -1; + return ans; + } + +} diff --git a/体系学习班/class43/Code02_TSP.java b/体系学习班/class43/Code02_TSP.java new file mode 100644 index 0000000..bf984a6 --- /dev/null +++ b/体系学习班/class43/Code02_TSP.java @@ -0,0 +1,295 @@ +package class43; + +import java.util.ArrayList; +import java.util.List; + +public class Code02_TSP { + + public static int t1(int[][] matrix) { + int N = matrix.length; // 0...N-1 + // set + // set.get(i) != null i这座城市在集合里 + // set.get(i) == null i这座城市不在集合里 + List set = new ArrayList<>(); + for (int i = 0; i < N; i++) { + set.add(1); + } + return func1(matrix, set, 0); + } + + // 任何两座城市之间的距离，可以在matrix里面拿到 + // set中表示着哪些城市的集合， + // start这座城一定在set里， + // 从start出发，要把set中所有的城市过一遍，最终回到0这座城市，最小距离是多少 + public static int func1(int[][] matrix, List set, int start) { + int cityNum = 0; + for (int i = 0; i < set.size(); i++) { + if (set.get(i) != null) { + cityNum++; + } + } + if (cityNum == 1) { + return matrix[start][0]; + } + // cityNum > 1 不只start这一座城 + set.set(start, null); + int min = Integer.MAX_VALUE; + for (int i = 0; i < set.size(); i++) { + if (set.get(i) != null) { + // start -> i i... -> 0 + int cur = matrix[start][i] + func1(matrix, set, i); + min = Math.min(min, cur); + } + } + set.set(start, 1); + return min; + } + + public static int t2(int[][] matrix) { + int N = matrix.length; // 0...N-1 + // 7座城 1111111 + int allCity = (1 << N) - 1; + return f2(matrix, allCity, 0); + } + + // 任何两座城市之间的距离，可以在matrix里面拿到 + // set中表示着哪些城市的集合， + // start这座城一定在set里， + // 从start出发，要把set中所有的城市过一遍，最终回到0这座城市，最小距离是多少 + public static int f2(int[][] matrix, int cityStatus, int start) { + // cityStatus == cityStatux & (~cityStaus + 1) + + if (cityStatus == (cityStatus & (~cityStatus + 1))) { + return matrix[start][0]; + } + + // 把start位的1去掉， + cityStatus &= (~(1 << start)); + int min = Integer.MAX_VALUE; + // 枚举所有的城市 + for (int move = 0; move < matrix.length; move++) { + if ((cityStatus & (1 << move)) != 0) { + int cur = matrix[start][move] + f2(matrix, cityStatus, move); + min = Math.min(min, cur); + } + } + cityStatus |= (1 << start); + return min; + } + + public static int t3(int[][] matrix) { + int N = matrix.length; // 0...N-1 + // 7座城 1111111 + int allCity = (1 << N) - 1; + int[][] dp = new int[1 << N][N]; + for (int i = 0; i < (1 << N); i++) { + for (int j = 0; j < N; j++) { + dp[i][j] = -1; + } + } + return f3(matrix, allCity, 0, dp); + } + + // 任何两座城市之间的距离，可以在matrix里面拿到 + // set中表示着哪些城市的集合， + // start这座城一定在set里， + // 从start出发，要把set中所有的城市过一遍，最终回到0这座城市，最小距离是多少 + public static int f3(int[][] matrix, int cityStatus, int start, int[][] dp) { + if (dp[cityStatus][start] != -1) { + return dp[cityStatus][start]; + } + if (cityStatus == (cityStatus & (~cityStatus + 1))) { + dp[cityStatus][start] = matrix[start][0]; + } else { + // 把start位的1去掉， + cityStatus &= (~(1 << start)); + int min = Integer.MAX_VALUE; + // 枚举所有的城市 + for (int move = 0; move < matrix.length; move++) { + if (move != start && (cityStatus & (1 << move)) != 0) { + int cur = matrix[start][move] + f3(matrix, cityStatus, move, dp); + min = Math.min(min, cur); + } + } + cityStatus |= (1 << start); + dp[cityStatus][start] = min; + } + return dp[cityStatus][start]; + } + + public static int t4(int[][] matrix) { + int N = matrix.length; // 0...N-1 + int statusNums = 1 << N; + int[][] dp = new int[statusNums][N]; + + for (int status = 0; status < statusNums; status++) { + for (int start = 0; start < N; start++) { + if ((status & (1 << start)) != 0) { + if (status == (status & (~status + 1))) { + dp[status][start] = matrix[start][0]; + } else { + int min = Integer.MAX_VALUE; + // start 城市在status里去掉之后，的状态 + int preStatus = status & (~(1 << start)); + // start -> i + for (int i = 0; i < N; i++) { + if ((preStatus & (1 << i)) != 0) { + int cur = matrix[start][i] + dp[preStatus][i]; + min = Math.min(min, cur); + } + } + dp[status][start] = min; + } + } + } + } + return dp[statusNums - 1][0]; + } + + // matrix[i][j] -> i城市到j城市的距离 + public static int tsp1(int[][] matrix, int origin) { + if (matrix == null || matrix.length < 2 || origin < 0 || origin >= matrix.length) { + return 0; + } + // 要考虑的集合 + ArrayList cities = new ArrayList<>(); + // cities[0] != null 表示0城在集合里 + // cities[i] != null 表示i城在集合里 + for (int i = 0; i < matrix.length; i++) { + cities.add(1); + } + // null,1,1,1,1,1,1 + // origin城不参与集合 + cities.set(origin, null); + return process(matrix, origin, cities, origin); + } + + // matrix 所有距离，存在其中 + // origin 固定参数，唯一的目标 + // cities 要考虑的集合，一定不含有origin + // 当前来到的城市是谁，cur + public static int process(int[][] matrix, int aim, ArrayList cities, int cur) { + boolean hasCity = false; // 集团中还是否有城市 + int ans = Integer.MAX_VALUE; + for (int i = 0; i < cities.size(); i++) { + if (cities.get(i) != null) { + hasCity = true; + cities.set(i, null); + // matrix[cur][i] + f(i, 集团(去掉i) ) + ans = Math.min(ans, matrix[cur][i] + process(matrix, aim, cities, i)); + cities.set(i, 1); + } + } + return hasCity ? ans : matrix[cur][aim]; + } + + // cities 里，一定含有cur这座城 + // 解决的是，集合从cur出发，通过集合里所有的城市，最终来到aim，最短距离 + public static int process2(int[][] matrix, int aim, ArrayList cities, int cur) { + if (cities.size() == 1) { + return matrix[cur][aim]; + } + cities.set(cur, null); + int ans = Integer.MAX_VALUE; + for (int i = 0; i < cities.size(); i++) { + if (cities.get(i) != null) { + int dis = matrix[cur][i] + process2(matrix, aim, cities, i); + ans = Math.min(ans, dis); + } + } + cities.set(cur, 1); + return ans; + } + + public static int tsp2(int[][] matrix, int origin) { + if (matrix == null || matrix.length < 2 || origin < 0 || origin >= matrix.length) { + return 0; + } + int N = matrix.length - 1; // 除去origin之后是n-1个点 + int S = 1 << N; // 状态数量 + int[][] dp = new int[S][N]; + int icity = 0; + int kcity = 0; + for (int i = 0; i < N; i++) { + icity = i < origin ? i : i + 1; + // 00000000 i + dp[0][i] = matrix[icity][origin]; + } + for (int status = 1; status < S; status++) { + // 尝试每一种状态 status = 0 0 1 0 0 0 0 0 0 + // 下标 8 7 6 5 4 3 2 1 0 + for (int i = 0; i < N; i++) { + // i 枚举的出发城市 + dp[status][i] = Integer.MAX_VALUE; + if ((1 << i & status) != 0) { + // 如果i这座城是可以枚举的，i = 6 ， i对应的原始城的编号，icity + icity = i < origin ? i : i + 1; + for (int k = 0; k < N; k++) { // i 这一步连到的点，k + if ((1 << k & status) != 0) { // i 这一步可以连到k + kcity = k < origin ? k : k + 1; // k对应的原始城的编号，kcity + dp[status][i] = Math.min(dp[status][i], dp[status ^ (1 << i)][k] + matrix[icity][kcity]); + } + } + } + } + } + int ans = Integer.MAX_VALUE; + for (int i = 0; i < N; i++) { + icity = i < origin ? i : i + 1; + ans = Math.min(ans, dp[S - 1][i] + matrix[origin][icity]); + } + return ans; + } + + public static int[][] generateGraph(int maxSize, int maxValue) { + int len = (int) (Math.random() * maxSize) + 1; + int[][] matrix = new int[len][len]; + for (int i = 0; i < len; i++) { + for (int j = 0; j < len; j++) { + matrix[i][j] = (int) (Math.random() * maxValue) + 1; + } + } + for (int i = 0; i < len; i++) { + matrix[i][i] = 0; + } + return matrix; + } + + public static void main(String[] args) { + int len = 10; + int value = 100; + System.out.println("功能测试开始"); + for (int i = 0; i < 20000; i++) { + int[][] matrix = generateGraph(len, value); + int origin = (int) (Math.random() * matrix.length); + int ans1 = t3(matrix); + int ans2 = t4(matrix); + int ans3 = tsp2(matrix, origin); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("fuck"); + } + } + System.out.println("功能测试结束"); + + len = 22; + System.out.println("性能测试开始，数据规模 : " + len); + int[][] matrix = new int[len][len]; + for (int i = 0; i < len; i++) { + for (int j = 0; j < len; j++) { + matrix[i][j] = (int) (Math.random() * value) + 1; + } + } + for (int i = 0; i < len; i++) { + matrix[i][i] = 0; + } + long start; + long end; + start = System.currentTimeMillis(); + t4(matrix); + end = System.currentTimeMillis(); + System.out.println("运行时间 : " + (end - start) + " 毫秒"); + System.out.println("性能测试结束"); + + } + +} diff --git a/体系学习班/class43/Code03_PavingTile.java b/体系学习班/class43/Code03_PavingTile.java new file mode 100644 index 0000000..c4bf363 --- /dev/null +++ b/体系学习班/class43/Code03_PavingTile.java @@ -0,0 +1,215 @@ +package class43; + +public class Code03_PavingTile { + + /* + * 2*M铺地的问题非常简单，这个是解决N*M铺地的问题 + */ + + public static int ways1(int N, int M) { + if (N < 1 || M < 1 || ((N * M) & 1) != 0) { + return 0; + } + if (N == 1 || M == 1) { + return 1; + } + int[] pre = new int[M]; // pre代表-1行的状况 + for (int i = 0; i < pre.length; i++) { + pre[i] = 1; + } + return process(pre, 0, N); + } + + // pre 表示level-1行的状态 + // level表示，正在level行做决定 + // N 表示一共有多少行 固定的 + // level-2行及其之上所有行，都摆满砖了 + // level做决定，让所有区域都满，方法数返回 + public static int process(int[] pre, int level, int N) { + if (level == N) { // base case + for (int i = 0; i < pre.length; i++) { + if (pre[i] == 0) { + return 0; + } + } + return 1; + } + + // 没到终止行，可以选择在当前的level行摆瓷砖 + int[] op = getOp(pre); + return dfs(op, 0, level, N); + } + + // op[i] == 0 可以考虑摆砖 + // op[i] == 1 只能竖着向上 + public static int dfs(int[] op, int col, int level, int N) { + // 在列上自由发挥，玩深度优先遍历，当col来到终止列，i行的决定做完了 + // 轮到i+1行，做决定 + if (col == op.length) { + return process(op, level + 1, N); + } + int ans = 0; + // col位置不横摆 + ans += dfs(op, col + 1, level, N); // col位置上不摆横转 + // col位置横摆, 向右 + if (col + 1 < op.length && op[col] == 0 && op[col + 1] == 0) { + op[col] = 1; + op[col + 1] = 1; + ans += dfs(op, col + 2, level, N); + op[col] = 0; + op[col + 1] = 0; + } + return ans; + } + + public static int[] getOp(int[] pre) { + int[] cur = new int[pre.length]; + for (int i = 0; i < pre.length; i++) { + cur[i] = pre[i] ^ 1; + } + return cur; + } + + // Min (N,M) 不超过 32 + public static int ways2(int N, int M) { + if (N < 1 || M < 1 || ((N * M) & 1) != 0) { + return 0; + } + if (N == 1 || M == 1) { + return 1; + } + int max = Math.max(N, M); + int min = Math.min(N, M); + int pre = (1 << min) - 1; + return process2(pre, 0, max, min); + } + + // 上一行的状态，是pre，limit是用来对齐的，固定参数不用管 + // 当前来到i行，一共N行，返回填满的方法数 + public static int process2(int pre, int i, int N, int M) { + if (i == N) { // base case + return pre == ((1 << M) - 1) ? 1 : 0; + } + int op = ((~pre) & ((1 << M) - 1)); + return dfs2(op, M - 1, i, N, M); + } + + public static int dfs2(int op, int col, int level, int N, int M) { + if (col == -1) { + return process2(op, level + 1, N, M); + } + int ans = 0; + ans += dfs2(op, col - 1, level, N, M); + if ((op & (1 << col)) == 0 && col - 1 >= 0 && (op & (1 << (col - 1))) == 0) { + ans += dfs2((op | (3 << (col - 1))), col - 2, level, N, M); + } + return ans; + } + + // 记忆化搜索的解 + // Min(N,M) 不超过 32 + public static int ways3(int N, int M) { + if (N < 1 || M < 1 || ((N * M) & 1) != 0) { + return 0; + } + if (N == 1 || M == 1) { + return 1; + } + int max = Math.max(N, M); + int min = Math.min(N, M); + int pre = (1 << min) - 1; + int[][] dp = new int[1 << min][max + 1]; + for (int i = 0; i < dp.length; i++) { + for (int j = 0; j < dp[0].length; j++) { + dp[i][j] = -1; + } + } + return process3(pre, 0, max, min, dp); + } + + public static int process3(int pre, int i, int N, int M, int[][] dp) { + if (dp[pre][i] != -1) { + return dp[pre][i]; + } + int ans = 0; + if (i == N) { + ans = pre == ((1 << M) - 1) ? 1 : 0; + } else { + int op = ((~pre) & ((1 << M) - 1)); + ans = dfs3(op, M - 1, i, N, M, dp); + } + dp[pre][i] = ans; + return ans; + } + + public static int dfs3(int op, int col, int level, int N, int M, int[][] dp) { + if (col == -1) { + return process3(op, level + 1, N, M, dp); + } + int ans = 0; + ans += dfs3(op, col - 1, level, N, M, dp); + if (col > 0 && (op & (3 << (col - 1))) == 0) { + ans += dfs3((op | (3 << (col - 1))), col - 2, level, N, M, dp); + } + return ans; + } + + // 严格位置依赖的动态规划解 + public static int ways4(int N, int M) { + if (N < 1 || M < 1 || ((N * M) & 1) != 0) { + return 0; + } + if (N == 1 || M == 1) { + return 1; + } + int big = N > M ? N : M; + int small = big == N ? M : N; + int sn = 1 << small; + int limit = sn - 1; + int[] dp = new int[sn]; + dp[limit] = 1; + int[] cur = new int[sn]; + for (int level = 0; level < big; level++) { + for (int status = 0; status < sn; status++) { + if (dp[status] != 0) { + int op = (~status) & limit; + dfs4(dp[status], op, 0, small - 1, cur); + } + } + for (int i = 0; i < sn; i++) { + dp[i] = 0; + } + int[] tmp = dp; + dp = cur; + cur = tmp; + } + return dp[limit]; + } + + public static void dfs4(int way, int op, int index, int end, int[] cur) { + if (index == end) { + cur[op] += way; + } else { + dfs4(way, op, index + 1, end, cur); + if (((3 << index) & op) == 0) { // 11 << index 可以放砖 + dfs4(way, op | (3 << index), index + 1, end, cur); + } + } + } + + public static void main(String[] args) { + int N = 8; + int M = 6; + System.out.println(ways1(N, M)); + System.out.println(ways2(N, M)); + System.out.println(ways3(N, M)); + System.out.println(ways4(N, M)); + + N = 10; + M = 10; + System.out.println("========="); + System.out.println(ways3(N, M)); + System.out.println(ways4(N, M)); + } + +} diff --git a/体系学习班/class44/Code01_LastSubstringInLexicographicalOrder.java b/体系学习班/class44/Code01_LastSubstringInLexicographicalOrder.java new file mode 100644 index 0000000..27ac147 --- /dev/null +++ b/体系学习班/class44/Code01_LastSubstringInLexicographicalOrder.java @@ -0,0 +1,136 @@ +package class44; + +// 测试链接: https://leetcode.com/problems/last-substring-in-lexicographical-order/ +public class Code01_LastSubstringInLexicographicalOrder { + + public static String lastSubstring(String s) { + if (s == null || s.length() == 0) { + return s; + } + int N = s.length(); + char[] str = s.toCharArray(); + int min = Integer.MAX_VALUE; + int max = Integer.MIN_VALUE; + for (char cha : str) { + min = Math.min(min, cha); + max = Math.max(max, cha); + } + int[] arr = new int[N]; + for (int i = 0; i < N; i++) { + arr[i] = str[i] - min + 1; + } + DC3 dc3 = new DC3(arr, max - min + 1); + return s.substring(dc3.sa[N - 1]); + } + + public static class DC3 { + + public int[] sa; + + public DC3(int[] nums, int max) { + sa = sa(nums, max); + } + + private int[] sa(int[] nums, int max) { + int n = nums.length; + int[] arr = new int[n + 3]; + for (int i = 0; i < n; i++) { + arr[i] = nums[i]; + } + return skew(arr, n, max); + } + + private int[] skew(int[] nums, int n, int K) { + int n0 = (n + 2) / 3, n1 = (n + 1) / 3, n2 = n / 3, n02 = n0 + n2; + int[] s12 = new int[n02 + 3], sa12 = new int[n02 + 3]; + for (int i = 0, j = 0; i < n + (n0 - n1); ++i) { + if (0 != i % 3) { + s12[j++] = i; + } + } + radixPass(nums, s12, sa12, 2, n02, K); + radixPass(nums, sa12, s12, 1, n02, K); + radixPass(nums, s12, sa12, 0, n02, K); + int name = 0, c0 = -1, c1 = -1, c2 = -1; + for (int i = 0; i < n02; ++i) { + if (c0 != nums[sa12[i]] || c1 != nums[sa12[i] + 1] || c2 != nums[sa12[i] + 2]) { + name++; + c0 = nums[sa12[i]]; + c1 = nums[sa12[i] + 1]; + c2 = nums[sa12[i] + 2]; + } + if (1 == sa12[i] % 3) { + s12[sa12[i] / 3] = name; + } else { + s12[sa12[i] / 3 + n0] = name; + } + } + if (name < n02) { + sa12 = skew(s12, n02, name); + for (int i = 0; i < n02; i++) { + s12[sa12[i]] = i + 1; + } + } else { + for (int i = 0; i < n02; i++) { + sa12[s12[i] - 1] = i; + } + } + int[] s0 = new int[n0], sa0 = new int[n0]; + for (int i = 0, j = 0; i < n02; i++) { + if (sa12[i] < n0) { + s0[j++] = 3 * sa12[i]; + } + } + radixPass(nums, s0, sa0, 0, n0, K); + int[] sa = new int[n]; + for (int p = 0, t = n0 - n1, k = 0; k < n; k++) { + int i = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2; + int j = sa0[p]; + if (sa12[t] < n0 ? leq(nums[i], s12[sa12[t] + n0], nums[j], s12[j / 3]) + : leq(nums[i], nums[i + 1], s12[sa12[t] - n0 + 1], nums[j], nums[j + 1], s12[j / 3 + n0])) { + sa[k] = i; + t++; + if (t == n02) { + for (k++; p < n0; p++, k++) { + sa[k] = sa0[p]; + } + } + } else { + sa[k] = j; + p++; + if (p == n0) { + for (k++; t < n02; t++, k++) { + sa[k] = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2; + } + } + } + } + return sa; + } + + private void radixPass(int[] nums, int[] input, int[] output, int offset, int n, int k) { + int[] cnt = new int[k + 1]; + for (int i = 0; i < n; ++i) { + cnt[nums[input[i] + offset]]++; + } + for (int i = 0, sum = 0; i < cnt.length; ++i) { + int t = cnt[i]; + cnt[i] = sum; + sum += t; + } + for (int i = 0; i < n; ++i) { + output[cnt[nums[input[i] + offset]]++] = input[i]; + } + } + + private boolean leq(int a1, int a2, int b1, int b2) { + return a1 < b1 || (a1 == b1 && a2 <= b2); + } + + private boolean leq(int a1, int a2, int a3, int b1, int b2, int b3) { + return a1 < b1 || (a1 == b1 && leq(a2, a3, b2, b3)); + } + + } + +} diff --git a/体系学习班/class44/DC3.java b/体系学习班/class44/DC3.java new file mode 100644 index 0000000..47e7601 --- /dev/null +++ b/体系学习班/class44/DC3.java @@ -0,0 +1,168 @@ +package class44; + +public class DC3 { + + public int[] sa; + + public int[] rank; + + public int[] height; + + // 构造方法的约定: + // 数组叫nums，如果你是字符串，请转成整型数组nums + // 数组中，最小值>=1 + // 如果不满足，处理成满足的，也不会影响使用 + // max, nums里面最大值是多少 + public DC3(int[] nums, int max) { + sa = sa(nums, max); + rank = rank(); + height = height(nums); + } + + private int[] sa(int[] nums, int max) { + int n = nums.length; + int[] arr = new int[n + 3]; + for (int i = 0; i < n; i++) { + arr[i] = nums[i]; + } + return skew(arr, n, max); + } + + private int[] skew(int[] nums, int n, int K) { + int n0 = (n + 2) / 3, n1 = (n + 1) / 3, n2 = n / 3, n02 = n0 + n2; + int[] s12 = new int[n02 + 3], sa12 = new int[n02 + 3]; + for (int i = 0, j = 0; i < n + (n0 - n1); ++i) { + if (0 != i % 3) { + s12[j++] = i; + } + } + radixPass(nums, s12, sa12, 2, n02, K); + radixPass(nums, sa12, s12, 1, n02, K); + radixPass(nums, s12, sa12, 0, n02, K); + int name = 0, c0 = -1, c1 = -1, c2 = -1; + for (int i = 0; i < n02; ++i) { + if (c0 != nums[sa12[i]] || c1 != nums[sa12[i] + 1] || c2 != nums[sa12[i] + 2]) { + name++; + c0 = nums[sa12[i]]; + c1 = nums[sa12[i] + 1]; + c2 = nums[sa12[i] + 2]; + } + if (1 == sa12[i] % 3) { + s12[sa12[i] / 3] = name; + } else { + s12[sa12[i] / 3 + n0] = name; + } + } + if (name < n02) { + sa12 = skew(s12, n02, name); + for (int i = 0; i < n02; i++) { + s12[sa12[i]] = i + 1; + } + } else { + for (int i = 0; i < n02; i++) { + sa12[s12[i] - 1] = i; + } + } + int[] s0 = new int[n0], sa0 = new int[n0]; + for (int i = 0, j = 0; i < n02; i++) { + if (sa12[i] < n0) { + s0[j++] = 3 * sa12[i]; + } + } + radixPass(nums, s0, sa0, 0, n0, K); + int[] sa = new int[n]; + for (int p = 0, t = n0 - n1, k = 0; k < n; k++) { + int i = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2; + int j = sa0[p]; + if (sa12[t] < n0 ? leq(nums[i], s12[sa12[t] + n0], nums[j], s12[j / 3]) + : leq(nums[i], nums[i + 1], s12[sa12[t] - n0 + 1], nums[j], nums[j + 1], s12[j / 3 + n0])) { + sa[k] = i; + t++; + if (t == n02) { + for (k++; p < n0; p++, k++) { + sa[k] = sa0[p]; + } + } + } else { + sa[k] = j; + p++; + if (p == n0) { + for (k++; t < n02; t++, k++) { + sa[k] = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2; + } + } + } + } + return sa; + } + + private void radixPass(int[] nums, int[] input, int[] output, int offset, int n, int k) { + int[] cnt = new int[k + 1]; + for (int i = 0; i < n; ++i) { + cnt[nums[input[i] + offset]]++; + } + for (int i = 0, sum = 0; i < cnt.length; ++i) { + int t = cnt[i]; + cnt[i] = sum; + sum += t; + } + for (int i = 0; i < n; ++i) { + output[cnt[nums[input[i] + offset]]++] = input[i]; + } + } + + private boolean leq(int a1, int a2, int b1, int b2) { + return a1 < b1 || (a1 == b1 && a2 <= b2); + } + + private boolean leq(int a1, int a2, int a3, int b1, int b2, int b3) { + return a1 < b1 || (a1 == b1 && leq(a2, a3, b2, b3)); + } + + private int[] rank() { + int n = sa.length; + int[] ans = new int[n]; + for (int i = 0; i < n; i++) { + ans[sa[i]] = i; + } + return ans; + } + + private int[] height(int[] s) { + int n = s.length; + int[] ans = new int[n]; + for (int i = 0, k = 0; i < n; ++i) { + if (rank[i] != 0) { + if (k > 0) { + --k; + } + int j = sa[rank[i] - 1]; + while (i + k < n && j + k < n && s[i + k] == s[j + k]) { + ++k; + } + ans[rank[i]] = k; + } + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int len, int maxValue) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * maxValue) + 1; + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int len = 100000; + int maxValue = 100; + long start = System.currentTimeMillis(); + new DC3(randomArray(len, maxValue), maxValue); + long end = System.currentTimeMillis(); + System.out.println("数据量 " + len + ", 运行时间 " + (end - start) + " ms"); + } + +} \ No newline at end of file diff --git a/体系学习班/class44/DC3_Algorithm.pdf b/体系学习班/class44/DC3_Algorithm.pdf new file mode 100644 index 0000000000000000000000000000000000000000..f053a6749606ad2ae8abbdffb6bc2a959c2f1617 GIT binary patch literal 192716 zcmbSz1zeTO*0&(tQW6TA28o@pK|#7px{=y+cdLYiNT<>%NP{4PQX-84(jX~_A|Q%1 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100644 index 0000000..0962781 --- /dev/null +++ b/体系学习班/class45/Code01_InsertS2MakeMostAlphabeticalOrder.java @@ -0,0 +1,261 @@ +package class45; + +public class Code01_InsertS2MakeMostAlphabeticalOrder { + + // 暴力方法 + public static String right(String s1, String s2) { + if (s1 == null || s1.length() == 0) { + return s2; + } + if (s2 == null || s2.length() == 0) { + return s1; + } + String p1 = s1 + s2; + String p2 = s2 + s1; + String ans = p1.compareTo(p2) > 0 ? p1 : p2; + for (int end = 1; end < s1.length(); end++) { + String cur = s1.substring(0, end) + s2 + s1.substring(end); + if (cur.compareTo(ans) > 0) { + ans = cur; + } + } + return ans; + } + + // 正式方法 O(N+M) + O(M^2) + // N : s1长度 + // M : s2长度 + public static String maxCombine(String s1, String s2) { + if (s1 == null || s1.length() == 0) { + return s2; + } + if (s2 == null || s2.length() == 0) { + return s1; + } + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + int N = str1.length; + int M = str2.length; + int min = str1[0]; + int max = str1[0]; + for (int i = 1; i < N; i++) { + min = Math.min(min, str1[i]); + max = Math.max(max, str1[i]); + } + for (int i = 0; i < M; i++) { + min = Math.min(min, str2[i]); + max = Math.max(max, str2[i]); + } + int[] all = new int[N + M + 1]; + int index = 0; + for (int i = 0; i < N; i++) { + all[index++] = str1[i] - min + 2; + } + all[index++] = 1; + for (int i = 0; i < M; i++) { + all[index++] = str2[i] - min + 2; + } + DC3 dc3 = new DC3(all, max - min + 2); + int[] rank = dc3.rank; + int comp = N + 1; + for (int i = 0; i < N; i++) { + if (rank[i] < rank[comp]) { + int best = bestSplit(s1, s2, i); + return s1.substring(0, best) + s2 + s1.substring(best); + } + } + return s1 + s2; + } + + public static int bestSplit(String s1, String s2, int first) { + int N = s1.length(); + int M = s2.length(); + int end = N; + for (int i = first, j = 0; i < N && j < M; i++, j++) { + if (s1.charAt(i) < s2.charAt(j)) { + end = i; + break; + } + } + String bestPrefix = s2; + int bestSplit = first; + for (int i = first + 1, j = M - 1; i <= end; i++, j--) { + String curPrefix = s1.substring(first, i) + s2.substring(0, j); + if (curPrefix.compareTo(bestPrefix) >= 0) { + bestPrefix = curPrefix; + bestSplit = i; + } + } + return bestSplit; + } + + public static class DC3 { + + public int[] sa; + + public int[] rank; + + public DC3(int[] nums, int max) { + sa = sa(nums, max); + rank = rank(); + } + + private int[] sa(int[] nums, int max) { + int n = nums.length; + int[] arr = new int[n + 3]; + for (int i = 0; i < n; i++) { + arr[i] = nums[i]; + } + return skew(arr, n, max); + } + + private int[] skew(int[] nums, int n, int K) { + int n0 = (n + 2) / 3, n1 = (n + 1) / 3, n2 = n / 3, n02 = n0 + n2; + int[] s12 = new int[n02 + 3], sa12 = new int[n02 + 3]; + for (int i = 0, j = 0; i < n + (n0 - n1); ++i) { + if (0 != i % 3) { + s12[j++] = i; + } + } + radixPass(nums, s12, sa12, 2, n02, K); + radixPass(nums, sa12, s12, 1, n02, K); + radixPass(nums, s12, sa12, 0, n02, K); + int name = 0, c0 = -1, c1 = -1, c2 = -1; + for (int i = 0; i < n02; ++i) { + if (c0 != nums[sa12[i]] || c1 != nums[sa12[i] + 1] || c2 != nums[sa12[i] + 2]) { + name++; + c0 = nums[sa12[i]]; + c1 = nums[sa12[i] + 1]; + c2 = nums[sa12[i] + 2]; + } + if (1 == sa12[i] % 3) { + s12[sa12[i] / 3] = name; + } else { + s12[sa12[i] / 3 + n0] = name; + } + } + if (name < n02) { + sa12 = skew(s12, n02, name); + for (int i = 0; i < n02; i++) { + s12[sa12[i]] = i + 1; + } + } else { + for (int i = 0; i < n02; i++) { + sa12[s12[i] - 1] = i; + } + } + int[] s0 = new int[n0], sa0 = new int[n0]; + for (int i = 0, j = 0; i < n02; i++) { + if (sa12[i] < n0) { + s0[j++] = 3 * sa12[i]; + } + } + radixPass(nums, s0, sa0, 0, n0, K); + int[] sa = new int[n]; + for (int p = 0, t = n0 - n1, k = 0; k < n; k++) { + int i = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2; + int j = sa0[p]; + if (sa12[t] < n0 ? leq(nums[i], s12[sa12[t] + n0], nums[j], s12[j / 3]) + : leq(nums[i], nums[i + 1], s12[sa12[t] - n0 + 1], nums[j], nums[j + 1], s12[j / 3 + n0])) { + sa[k] = i; + t++; + if (t == n02) { + for (k++; p < n0; p++, k++) { + sa[k] = sa0[p]; + } + } + } else { + sa[k] = j; + p++; + if (p == n0) { + for (k++; t < n02; t++, k++) { + sa[k] = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2; + } + } + } + } + return sa; + } + + private void radixPass(int[] nums, int[] input, int[] output, int offset, int n, int k) { + int[] cnt = new int[k + 1]; + for (int i = 0; i < n; ++i) { + cnt[nums[input[i] + offset]]++; + } + for (int i = 0, sum = 0; i < cnt.length; ++i) { + int t = cnt[i]; + cnt[i] = sum; + sum += t; + } + for (int i = 0; i < n; ++i) { + output[cnt[nums[input[i] + offset]]++] = input[i]; + } + } + + private boolean leq(int a1, int a2, int b1, int b2) { + return a1 < b1 || (a1 == b1 && a2 <= b2); + } + + private boolean leq(int a1, int a2, int a3, int b1, int b2, int b3) { + return a1 < b1 || (a1 == b1 && leq(a2, a3, b2, b3)); + } + + private int[] rank() { + int n = sa.length; + int[] ans = new int[n]; + for (int i = 0; i < n; i++) { + ans[sa[i]] = i; + } + return ans; + } + + } + + // for test + public static String randomNumberString(int len, int range) { + char[] str = new char[len]; + for (int i = 0; i < len; i++) { + str[i] = (char) ((int) (Math.random() * range) + '0'); + } + return String.valueOf(str); + } + + // for test + public static void main(String[] args) { + int range = 10; + int len = 50; + int testTime = 100000; + System.out.println("功能测试开始"); + for (int i = 0; i < testTime; i++) { + int s1Len = (int) (Math.random() * len); + int s2Len = (int) (Math.random() * len); + String s1 = randomNumberString(s1Len, range); + String s2 = randomNumberString(s2Len, range); + String ans1 = right(s1, s2); + String ans2 = maxCombine(s1, s2); + if (!ans1.equals(ans2)) { + System.out.println("Oops!"); + System.out.println(s1); + System.out.println(s2); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("功能测试结束"); + + System.out.println("=========="); + + System.out.println("性能测试开始"); + int s1Len = 1000000; + int s2Len = 500; + String s1 = randomNumberString(s1Len, range); + String s2 = randomNumberString(s2Len, range); + long start = System.currentTimeMillis(); + maxCombine(s1, s2); + long end = System.currentTimeMillis(); + System.out.println("运行时间 : " + (end - start) + " ms"); + System.out.println("性能测试结束"); + } + +} diff --git a/体系学习班/class45/Code02_CreateMaximumNumber.java b/体系学习班/class45/Code02_CreateMaximumNumber.java new file mode 100644 index 0000000..1e9cc28 --- /dev/null +++ b/体系学习班/class45/Code02_CreateMaximumNumber.java @@ -0,0 +1,250 @@ +package class45; + +// 测试链接: https://leetcode.com/problems/create-maximum-number/ +public class Code02_CreateMaximumNumber { + + public static int[] maxNumber1(int[] nums1, int[] nums2, int k) { + int len1 = nums1.length; + int len2 = nums2.length; + if (k < 0 || k > len1 + len2) { + return null; + } + int[] res = new int[k]; + int[][] dp1 = getdp(nums1); // 生成dp1这个表，以后从nums1中，只要固定拿N个数， + int[][] dp2 = getdp(nums2); + // get1 从arr1里拿的数量 + // K - get1 从arr2里拿的数量 + for (int get1 = Math.max(0, k - len2); get1 <= Math.min(k, len1); get1++) { + // arr1 挑 get1个，怎么得到一个最优结果 + int[] pick1 = maxPick(nums1, dp1, get1); + int[] pick2 = maxPick(nums2, dp2, k - get1); + int[] merge = merge(pick1, pick2); + res = preMoreThanLast(res, 0, merge, 0) ? res : merge; + } + return res; + } + + public static int[] merge(int[] nums1, int[] nums2) { + int k = nums1.length + nums2.length; + int[] ans = new int[k]; + for (int i = 0, j = 0, r = 0; r < k; ++r) { + ans[r] = preMoreThanLast(nums1, i, nums2, j) ? nums1[i++] : nums2[j++]; + } + return ans; + } + + public static boolean preMoreThanLast(int[] nums1, int i, int[] nums2, int j) { + while (i < nums1.length && j < nums2.length && nums1[i] == nums2[j]) { + i++; + j++; + } + return j == nums2.length || (i < nums1.length && nums1[i] > nums2[j]); + } + + public static int[] maxNumber2(int[] nums1, int[] nums2, int k) { + int len1 = nums1.length; + int len2 = nums2.length; + if (k < 0 || k > len1 + len2) { + return null; + } + int[] res = new int[k]; + int[][] dp1 = getdp(nums1); + int[][] dp2 = getdp(nums2); + for (int get1 = Math.max(0, k - len2); get1 <= Math.min(k, len1); get1++) { + int[] pick1 = maxPick(nums1, dp1, get1); + int[] pick2 = maxPick(nums2, dp2, k - get1); + int[] merge = mergeBySuffixArray(pick1, pick2); + res = moreThan(res, merge) ? res : merge; + } + return res; + } + + public static boolean moreThan(int[] pre, int[] last) { + int i = 0; + int j = 0; + while (i < pre.length && j < last.length && pre[i] == last[j]) { + i++; + j++; + } + return j == last.length || (i < pre.length && pre[i] > last[j]); + } + + public static int[] mergeBySuffixArray(int[] nums1, int[] nums2) { + int size1 = nums1.length; + int size2 = nums2.length; + int[] nums = new int[size1 + 1 + size2]; + for (int i = 0; i < size1; i++) { + nums[i] = nums1[i] + 2; + } + nums[size1] = 1; + for (int j = 0; j < size2; j++) { + nums[j + size1 + 1] = nums2[j] + 2; + } + DC3 dc3 = new DC3(nums, 11); + int[] rank = dc3.rank; + int[] ans = new int[size1 + size2]; + int i = 0; + int j = 0; + int r = 0; + while (i < size1 && j < size2) { + ans[r++] = rank[i] > rank[j + size1 + 1] ? nums1[i++] : nums2[j++]; + } + while (i < size1) { + ans[r++] = nums1[i++]; + } + while (j < size2) { + ans[r++] = nums2[j++]; + } + return ans; + } + + public static class DC3 { + + public int[] sa; + + public int[] rank; + + public DC3(int[] nums, int max) { + sa = sa(nums, max); + rank = rank(); + } + + private int[] sa(int[] nums, int max) { + int n = nums.length; + int[] arr = new int[n + 3]; + for (int i = 0; i < n; i++) { + arr[i] = nums[i]; + } + return skew(arr, n, max); + } + + private int[] skew(int[] nums, int n, int K) { + int n0 = (n + 2) / 3, n1 = (n + 1) / 3, n2 = n / 3, n02 = n0 + n2; + int[] s12 = new int[n02 + 3], sa12 = new int[n02 + 3]; + for (int i = 0, j = 0; i < n + (n0 - n1); ++i) { + if (0 != i % 3) { + s12[j++] = i; + } + } + radixPass(nums, s12, sa12, 2, n02, K); + radixPass(nums, sa12, s12, 1, n02, K); + radixPass(nums, s12, sa12, 0, n02, K); + int name = 0, c0 = -1, c1 = -1, c2 = -1; + for (int i = 0; i < n02; ++i) { + if (c0 != nums[sa12[i]] || c1 != nums[sa12[i] + 1] || c2 != nums[sa12[i] + 2]) { + name++; + c0 = nums[sa12[i]]; + c1 = nums[sa12[i] + 1]; + c2 = nums[sa12[i] + 2]; + } + if (1 == sa12[i] % 3) { + s12[sa12[i] / 3] = name; + } else { + s12[sa12[i] / 3 + n0] = name; + } + } + if (name < n02) { + sa12 = skew(s12, n02, name); + for (int i = 0; i < n02; i++) { + s12[sa12[i]] = i + 1; + } + } else { + for (int i = 0; i < n02; i++) { + sa12[s12[i] - 1] = i; + } + } + int[] s0 = new int[n0], sa0 = new int[n0]; + for (int i = 0, j = 0; i < n02; i++) { + if (sa12[i] < n0) { + s0[j++] = 3 * sa12[i]; + } + } + radixPass(nums, s0, sa0, 0, n0, K); + int[] sa = new int[n]; + for (int p = 0, t = n0 - n1, k = 0; k < n; k++) { + int i = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2; + int j = sa0[p]; + if (sa12[t] < n0 ? leq(nums[i], s12[sa12[t] + n0], nums[j], s12[j / 3]) + : leq(nums[i], nums[i + 1], s12[sa12[t] - n0 + 1], nums[j], nums[j + 1], s12[j / 3 + n0])) { + sa[k] = i; + t++; + if (t == n02) { + for (k++; p < n0; p++, k++) { + sa[k] = sa0[p]; + } + } + } else { + sa[k] = j; + p++; + if (p == n0) { + for (k++; t < n02; t++, k++) { + sa[k] = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2; + } + } + } + } + return sa; + } + + private void radixPass(int[] nums, int[] input, int[] output, int offset, int n, int k) { + int[] cnt = new int[k + 1]; + for (int i = 0; i < n; ++i) { + cnt[nums[input[i] + offset]]++; + } + for (int i = 0, sum = 0; i < cnt.length; ++i) { + int t = cnt[i]; + cnt[i] = sum; + sum += t; + } + for (int i = 0; i < n; ++i) { + output[cnt[nums[input[i] + offset]]++] = input[i]; + } + } + + private boolean leq(int a1, int a2, int b1, int b2) { + return a1 < b1 || (a1 == b1 && a2 <= b2); + } + + private boolean leq(int a1, int a2, int a3, int b1, int b2, int b3) { + return a1 < b1 || (a1 == b1 && leq(a2, a3, b2, b3)); + } + + private int[] rank() { + int n = sa.length; + int[] ans = new int[n]; + for (int i = 0; i < n; i++) { + ans[sa[i]] = i; + } + return ans; + } + + } + + public static int[][] getdp(int[] arr) { + int size = arr.length; // 0~N-1 + int pick = arr.length + 1; // 1 ~ N + int[][] dp = new int[size][pick]; + // get 不从0开始，因为拿0个无意义 + for (int get = 1; get < pick; get++) { // 1 ~ N + int maxIndex = size - get; + // i~N-1 + for (int i = size - get; i >= 0; i--) { + if (arr[i] >= arr[maxIndex]) { + maxIndex = i; + } + dp[i][get] = maxIndex; + } + } + return dp; + } + + public static int[] maxPick(int[] arr, int[][] dp, int pick) { + int[] res = new int[pick]; + for (int resIndex = 0, dpRow = 0; pick > 0; pick--, resIndex++) { + res[resIndex] = arr[dp[dpRow][pick]]; + dpRow = dp[dpRow][pick] + 1; + } + return res; + } + +} diff --git a/体系学习班/class45/Code03_LongestCommonSubstringConquerByHeight.java b/体系学习班/class45/Code03_LongestCommonSubstringConquerByHeight.java new file mode 100644 index 0000000..2b92fff --- /dev/null +++ b/体系学习班/class45/Code03_LongestCommonSubstringConquerByHeight.java @@ -0,0 +1,286 @@ +package class45; + +// 最长公共子串问题是面试常见题目之一 +// 假设str1长度N，str2长度M +// 因为最优解的难度所限，一般在面试场上回答出O(N*M)的解法已经是比较优秀了 +// 因为得到O(N*M)的解法，就已经需要用到动态规划了 +// 但其实这个问题的最优解是O(N+M)，为了达到这个复杂度可是不容易 +// 首先需要用到DC3算法得到后缀数组(sa) +// 进而用sa数组去生成height数组 +// 而且在生成的时候，还有一个不回退的优化，都非常不容易理解 +// 这就是后缀数组在面试算法中的地位 : 德高望重的噩梦 +public class Code03_LongestCommonSubstringConquerByHeight { + + public static int lcs1(String s1, String s2) { + if (s1 == null || s2 == null || s1.length() == 0 || s2.length() == 0) { + return 0; + } + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + int row = 0; + int col = str2.length - 1; + int max = 0; + while (row < str1.length) { + int i = row; + int j = col; + int len = 0; + while (i < str1.length && j < str2.length) { + if (str1[i] != str2[j]) { + len = 0; + } else { + len++; + } + if (len > max) { + max = len; + } + i++; + j++; + } + if (col > 0) { + col--; + } else { + row++; + } + } + return max; + } + + public static int lcs2(String s1, String s2) { + if (s1 == null || s2 == null || s1.length() == 0 || s2.length() == 0) { + return 0; + } + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + int N = str1.length; + int M = str2.length; + int min = str1[0]; + int max = str1[0]; + for (int i = 1; i < N; i++) { + min = Math.min(min, str1[i]); + max = Math.max(max, str1[i]); + } + for (int i = 0; i < M; i++) { + min = Math.min(min, str2[i]); + max = Math.max(max, str2[i]); + } + int[] all = new int[N + M + 1]; + int index = 0; + for (int i = 0; i < N; i++) { + all[index++] = str1[i] - min + 2; + } + all[index++] = 1; + for (int i = 0; i < M; i++) { + all[index++] = str2[i] - min + 2; + } + DC3 dc3 = new DC3(all, max - min + 2); + int n = all.length; + int[] sa = dc3.sa; + int[] height = dc3.height; + int ans = 0; + for (int i = 1; i < n; i++) { + int Y = sa[i - 1]; + int X = sa[i]; + if (Math.min(X, Y) < N && Math.max(X, Y) > N) { + ans = Math.max(ans, height[i]); + } + } + return ans; + } + + public static class DC3 { + + public int[] sa; + + public int[] rank; + + public int[] height; + + public DC3(int[] nums, int max) { + sa = sa(nums, max); + rank = rank(); + height = height(nums); + } + + private int[] sa(int[] nums, int max) { + int n = nums.length; + int[] arr = new int[n + 3]; + for (int i = 0; i < n; i++) { + arr[i] = nums[i]; + } + return skew(arr, n, max); + } + + private int[] skew(int[] nums, int n, int K) { + int n0 = (n + 2) / 3, n1 = (n + 1) / 3, n2 = n / 3, n02 = n0 + n2; + int[] s12 = new int[n02 + 3], sa12 = new int[n02 + 3]; + for (int i = 0, j = 0; i < n + (n0 - n1); ++i) { + if (0 != i % 3) { + s12[j++] = i; + } + } + radixPass(nums, s12, sa12, 2, n02, K); + radixPass(nums, sa12, s12, 1, n02, K); + radixPass(nums, s12, sa12, 0, n02, K); + int name = 0, c0 = -1, c1 = -1, c2 = -1; + for (int i = 0; i < n02; ++i) { + if (c0 != nums[sa12[i]] || c1 != nums[sa12[i] + 1] || c2 != nums[sa12[i] + 2]) { + name++; + c0 = nums[sa12[i]]; + c1 = nums[sa12[i] + 1]; + c2 = nums[sa12[i] + 2]; + } + if (1 == sa12[i] % 3) { + s12[sa12[i] / 3] = name; + } else { + s12[sa12[i] / 3 + n0] = name; + } + } + if (name < n02) { + sa12 = skew(s12, n02, name); + for (int i = 0; i < n02; i++) { + s12[sa12[i]] = i + 1; + } + } else { + for (int i = 0; i < n02; i++) { + sa12[s12[i] - 1] = i; + } + } + int[] s0 = new int[n0], sa0 = new int[n0]; + for (int i = 0, j = 0; i < n02; i++) { + if (sa12[i] < n0) { + s0[j++] = 3 * sa12[i]; + } + } + radixPass(nums, s0, sa0, 0, n0, K); + int[] sa = new int[n]; + for (int p = 0, t = n0 - n1, k = 0; k < n; k++) { + int i = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2; + int j = sa0[p]; + if (sa12[t] < n0 ? leq(nums[i], s12[sa12[t] + n0], nums[j], s12[j / 3]) + : leq(nums[i], nums[i + 1], s12[sa12[t] - n0 + 1], nums[j], nums[j + 1], s12[j / 3 + n0])) { + sa[k] = i; + t++; + if (t == n02) { + for (k++; p < n0; p++, k++) { + sa[k] = sa0[p]; + } + } + } else { + sa[k] = j; + p++; + if (p == n0) { + for (k++; t < n02; t++, k++) { + sa[k] = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2; + } + } + } + } + return sa; + } + + private void radixPass(int[] nums, int[] input, int[] output, int offset, int n, int k) { + int[] cnt = new int[k + 1]; + for (int i = 0; i < n; ++i) { + cnt[nums[input[i] + offset]]++; + } + for (int i = 0, sum = 0; i < cnt.length; ++i) { + int t = cnt[i]; + cnt[i] = sum; + sum += t; + } + for (int i = 0; i < n; ++i) { + output[cnt[nums[input[i] + offset]]++] = input[i]; + } + } + + private boolean leq(int a1, int a2, int b1, int b2) { + return a1 < b1 || (a1 == b1 && a2 <= b2); + } + + private boolean leq(int a1, int a2, int a3, int b1, int b2, int b3) { + return a1 < b1 || (a1 == b1 && leq(a2, a3, b2, b3)); + } + + private int[] rank() { + int n = sa.length; + int[] ans = new int[n]; + for (int i = 0; i < n; i++) { + ans[sa[i]] = i; + } + return ans; + } + + private int[] height(int[] s) { + int n = s.length; + int[] ans = new int[n]; + // 依次求h[i] , k = 0 + for (int i = 0, k = 0; i < n; ++i) { + if (rank[i] != 0) { + if (k > 0) { + --k; + } + int j = sa[rank[i] - 1]; + while (i + k < n && j + k < n && s[i + k] == s[j + k]) { + ++k; + } + // h[i] = k + ans[rank[i]] = k; + } + } + return ans; + } + + } + + // for test + public static String randomNumberString(int len, int range) { + char[] str = new char[len]; + for (int i = 0; i < len; i++) { + str[i] = (char) ((int) (Math.random() * range) + 'a'); + } + return String.valueOf(str); + } + + public static void main(String[] args) { + int len = 30; + int range = 5; + int testTime = 100000; + System.out.println("功能测试开始"); + for (int i = 0; i < testTime; i++) { + int N1 = (int) (Math.random() * len); + int N2 = (int) (Math.random() * len); + String str1 = randomNumberString(N1, range); + String str2 = randomNumberString(N2, range); + int ans1 = lcs1(str1, str2); + int ans2 = lcs2(str1, str2); + if (ans1 != ans2) { + System.out.println("Oops!"); + } + } + System.out.println("功能测试结束"); + System.out.println("=========="); + + System.out.println("性能测试开始"); + len = 80000; + range = 26; + long start; + long end; + + String str1 = randomNumberString(len, range); + String str2 = randomNumberString(len, range); + + start = System.currentTimeMillis(); + int ans1 = lcs1(str1, str2); + end = System.currentTimeMillis(); + System.out.println("方法1结果 : " + ans1 + " , 运行时间 : " + (end - start) + " ms"); + + start = System.currentTimeMillis(); + int ans2 = lcs2(str1, str2); + end = System.currentTimeMillis(); + System.out.println("方法2结果 : " + ans2 + " , 运行时间 : " + (end - start) + " ms"); + + System.out.println("性能测试结束"); + + } + +} diff --git a/体系学习班/class45/Code04_LongestRepeatingSubstring.java b/体系学习班/class45/Code04_LongestRepeatingSubstring.java new file mode 100644 index 0000000..a6090e3 --- /dev/null +++ b/体系学习班/class45/Code04_LongestRepeatingSubstring.java @@ -0,0 +1,196 @@ +package class45; + +// 一个非常经典的题 +// 这道题课上没有讲 +// 后缀数组的模版题 +// 需要学会DC3算法生成后缀数组 +// 需要学会课上讲的如何生成高度数组 +// 时间复杂度O(N)，连官方题解都没有做到的时间复杂度，但这才是最优解 +// 测试链接 : https://leetcode.cn/problems/longest-repeating-substring/ +public class Code04_LongestRepeatingSubstring { + + public static int longestRepeatingSubstring(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + int n = str.length; + int min = str[0]; + int max = str[0]; + for (int i = 1; i < n; i++) { + min = Math.min(min, str[i]); + max = Math.max(max, str[i]); + } + int[] all = new int[n]; + for (int i = 0; i < n; i++) { + all[i] = str[i] - min + 1; + } + DC3 dc3 = new DC3(all, max - min + 1); + int ans = 0; + for (int i = 1; i < n; i++) { + ans = Math.max(ans, dc3.height[i]); + } + return ans; + } + + public static class DC3 { + public int[] sa; + public int[] rank; + public int[] height; + + public DC3(int[] nums, int max) { + sa = sa(nums, max); + rank = rank(); + height = height(nums); + } + + private int[] sa(int[] nums, int max) { + int n = nums.length; + int[] arr = new int[n + 3]; + for (int i = 0; i < n; i++) { + arr[i] = nums[i]; + } + return skew(arr, n, max); + } + + private int[] skew(int[] nums, int n, int K) { + int n0 = (n + 2) / 3, n1 = (n + 1) / 3, n2 = n / 3, n02 = n0 + n2; + int[] s12 = new int[n02 + 3], sa12 = new int[n02 + 3]; + for (int i = 0, j = 0; i < n + (n0 - n1); ++i) { + if (0 != i % 3) { + s12[j++] = i; + } + } + radixPass(nums, s12, sa12, 2, n02, K); + radixPass(nums, sa12, s12, 1, n02, K); + radixPass(nums, s12, sa12, 0, n02, K); + int name = 0, c0 = -1, c1 = -1, c2 = -1; + for (int i = 0; i < n02; ++i) { + if (c0 != nums[sa12[i]] || c1 != nums[sa12[i] + 1] || c2 != nums[sa12[i] + 2]) { + name++; + c0 = nums[sa12[i]]; + c1 = nums[sa12[i] + 1]; + c2 = nums[sa12[i] + 2]; + } + if (1 == sa12[i] % 3) { + s12[sa12[i] / 3] = name; + } else { + s12[sa12[i] / 3 + n0] = name; + } + } + if (name < n02) { + sa12 = skew(s12, n02, name); + for (int i = 0; i < n02; i++) { + s12[sa12[i]] = i + 1; + } + } else { + for (int i = 0; i < n02; i++) { + sa12[s12[i] - 1] = i; + } + } + int[] s0 = new int[n0], sa0 = new int[n0]; + for (int i = 0, j = 0; i < n02; i++) { + if (sa12[i] < n0) { + s0[j++] = 3 * sa12[i]; + } + } + radixPass(nums, s0, sa0, 0, n0, K); + int[] sa = new int[n]; + for (int p = 0, t = n0 - n1, k = 0; k < n; k++) { + int i = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2; + int j = sa0[p]; + if (sa12[t] < n0 ? leq(nums[i], s12[sa12[t] + n0], nums[j], s12[j / 3]) + : leq(nums[i], nums[i + 1], s12[sa12[t] - n0 + 1], nums[j], nums[j + 1], s12[j / 3 + n0])) { + sa[k] = i; + t++; + if (t == n02) { + for (k++; p < n0; p++, k++) { + sa[k] = sa0[p]; + } + } + } else { + sa[k] = j; + p++; + if (p == n0) { + for (k++; t < n02; t++, k++) { + sa[k] = sa12[t] < n0 ? sa12[t] * 3 + 1 : (sa12[t] - n0) * 3 + 2; + } + } + } + } + return sa; + } + + private void radixPass(int[] nums, int[] input, int[] output, int offset, int n, int k) { + int[] cnt = new int[k + 1]; + for (int i = 0; i < n; ++i) { + cnt[nums[input[i] + offset]]++; + } + for (int i = 0, sum = 0; i < cnt.length; ++i) { + int t = cnt[i]; + cnt[i] = sum; + sum += t; + } + for (int i = 0; i < n; ++i) { + output[cnt[nums[input[i] + offset]]++] = input[i]; + } + } + + private boolean leq(int a1, int a2, int b1, int b2) { + return a1 < b1 || (a1 == b1 && a2 <= b2); + } + + private boolean leq(int a1, int a2, int a3, int b1, int b2, int b3) { + return a1 < b1 || (a1 == b1 && leq(a2, a3, b2, b3)); + } + + private int[] rank() { + int n = sa.length; + int[] ans = new int[n]; + for (int i = 0; i < n; i++) { + ans[sa[i]] = i; + } + return ans; + } + + private int[] height(int[] s) { + int n = s.length; + int[] ans = new int[n]; + for (int i = 0, k = 0; i < n; ++i) { + if (rank[i] != 0) { + if (k > 0) { + --k; + } + int j = sa[rank[i] - 1]; + while (i + k < n && j + k < n && s[i + k] == s[j + k]) { + ++k; + } + ans[rank[i]] = k; + } + } + return ans; + } + + } + + // 为了测试, 不用提交 + public static String randomString(int n, int r) { + char[] str = new char[n]; + for (int i = 0; i < n; i++) { + str[i] = (char) ((int) (Math.random() * r) + 'a'); + } + return String.valueOf(str); + } + + // 为了测试, 不用提交 + public static void main(String[] args) { + int n = 500000; + int r = 3; + long start = System.currentTimeMillis(); + longestRepeatingSubstring(randomString(n, r)); + long end = System.currentTimeMillis(); + System.out.println("字符长度为 " + n + ", 字符种类数为 " + r + " 时"); + System.out.println("求最长重复子串的运行时间 : " + (end - start) + " 毫秒"); + } + +} diff --git a/体系学习班/class46/Code01_BurstBalloons.java b/体系学习班/class46/Code01_BurstBalloons.java new file mode 100644 index 0000000..71d2602 --- /dev/null +++ b/体系学习班/class46/Code01_BurstBalloons.java @@ -0,0 +1,107 @@ +package class46; + +// 本题测试链接 : https://leetcode.com/problems/burst-balloons/ +public class Code01_BurstBalloons { + + public static int maxCoins0(int[] arr) { + // [3,2,1,3] + // [1,3,2,1,3,1] + int N = arr.length; + int[] help = new int[N + 2]; + for (int i = 0; i < N; i++) { + help[i + 1] = arr[i]; + } + help[0] = 1; + help[N + 1] = 1; + return func(help, 1, N); + } + + // L-1位置，和R+1位置，永远不越界，并且，[L-1] 和 [R+1] 一定没爆呢！ + // 返回，arr[L...R]打爆所有气球，最大得分是什么 + public static int func(int[] arr, int L, int R) { + if (L == R) { + return arr[L - 1] * arr[L] * arr[R + 1]; + } + // 尝试每一种情况，最后打爆的气球，是什么位置 + // L...R + // L位置的气球，最后打爆 + int max = func(arr, L + 1, R) + arr[L - 1] * arr[L] * arr[R + 1]; + // R位置的气球，最后打爆 + max = Math.max(max, func(arr, L, R - 1) + arr[L - 1] * arr[R] * arr[R + 1]); + // 尝试所有L...R，中间的位置，(L,R) + for (int i = L + 1; i < R; i++) { + // i位置的气球，最后打爆 + int left = func(arr, L, i - 1); + int right = func(arr, i + 1, R); + int last = arr[L - 1] * arr[i] * arr[R + 1]; + int cur = left + right + last; + max = Math.max(max, cur); + } + return max; + } + + public static int maxCoins1(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + if (arr.length == 1) { + return arr[0]; + } + int N = arr.length; + int[] help = new int[N + 2]; + help[0] = 1; + help[N + 1] = 1; + for (int i = 0; i < N; i++) { + help[i + 1] = arr[i]; + } + return process(help, 1, N); + } + + // 打爆arr[L..R]范围上的所有气球，返回最大的分数 + // 假设arr[L-1]和arr[R+1]一定没有被打爆 + public static int process(int[] arr, int L, int R) { + if (L == R) {// 如果arr[L..R]范围上只有一个气球，直接打爆即可 + return arr[L - 1] * arr[L] * arr[R + 1]; + } + // 最后打爆arr[L]的方案，和最后打爆arr[R]的方案，先比较一下 + int max = Math.max(arr[L - 1] * arr[L] * arr[R + 1] + process(arr, L + 1, R), + arr[L - 1] * arr[R] * arr[R + 1] + process(arr, L, R - 1)); + // 尝试中间位置的气球最后被打爆的每一种方案 + for (int i = L + 1; i < R; i++) { + max = Math.max(max, arr[L - 1] * arr[i] * arr[R + 1] + process(arr, L, i - 1) + process(arr, i + 1, R)); + } + return max; + } + + public static int maxCoins2(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + if (arr.length == 1) { + return arr[0]; + } + int N = arr.length; + int[] help = new int[N + 2]; + help[0] = 1; + help[N + 1] = 1; + for (int i = 0; i < N; i++) { + help[i + 1] = arr[i]; + } + int[][] dp = new int[N + 2][N + 2]; + for (int i = 1; i <= N; i++) { + dp[i][i] = help[i - 1] * help[i] * help[i + 1]; + } + for (int L = N; L >= 1; L--) { + for (int R = L + 1; R <= N; R++) { + int ans = help[L - 1] * help[L] * help[R + 1] + dp[L + 1][R]; + ans = Math.max(ans, help[L - 1] * help[R] * help[R + 1] + dp[L][R - 1]); + for (int i = L + 1; i < R; i++) { + ans = Math.max(ans, help[L - 1] * help[i] * help[R + 1] + dp[L][i - 1] + dp[i + 1][R]); + } + dp[L][R] = ans; + } + } + return dp[1][N]; + } + +} diff --git a/体系学习班/class46/Code02_RemoveBoxes.java b/体系学习班/class46/Code02_RemoveBoxes.java new file mode 100644 index 0000000..8db4d8f --- /dev/null +++ b/体系学习班/class46/Code02_RemoveBoxes.java @@ -0,0 +1,81 @@ +package class46; + +// 本题测试链接 : https://leetcode.com/problems/remove-boxes/ +public class Code02_RemoveBoxes { + + // arr[L...R]消除，而且前面跟着K个arr[L]这个数 + // 返回：所有东西都消掉，最大得分 + public static int func1(int[] arr, int L, int R, int K) { + if (L > R) { + return 0; + } + int ans = func1(arr, L + 1, R, 0) + (K + 1) * (K + 1); + + // 前面的K个X，和arr[L]数，合在一起了，现在有K+1个arr[L]位置的数 + for (int i = L + 1; i <= R; i++) { + if (arr[i] == arr[L]) { + ans = Math.max(ans, func1(arr, L + 1, i - 1, 0) + func1(arr, i, R, K + 1)); + } + } + return ans; + } + + public static int removeBoxes1(int[] boxes) { + int N = boxes.length; + int[][][] dp = new int[N][N][N]; + int ans = process1(boxes, 0, N - 1, 0, dp); + return ans; + } + + public static int process1(int[] boxes, int L, int R, int K, int[][][] dp) { + if (L > R) { + return 0; + } + if (dp[L][R][K] > 0) { + return dp[L][R][K]; + } + int ans = process1(boxes, L + 1, R, 0, dp) + (K + 1) * (K + 1); + for (int i = L + 1; i <= R; i++) { + if (boxes[i] == boxes[L]) { + ans = Math.max(ans, process1(boxes, L + 1, i - 1, 0, dp) + process1(boxes, i, R, K + 1, dp)); + } + } + dp[L][R][K] = ans; + return ans; + } + + public static int removeBoxes2(int[] boxes) { + int N = boxes.length; + int[][][] dp = new int[N][N][N]; + int ans = process2(boxes, 0, N - 1, 0, dp); + return ans; + } + + public static int process2(int[] boxes, int L, int R, int K, int[][][] dp) { + if (L > R) { + return 0; + } + if (dp[L][R][K] > 0) { + return dp[L][R][K]; + } + // 找到开头， + // 1,1,1,1,1,5 + // 3 4 5 6 7 8 + // ! + int last = L; + while (last + 1 <= R && boxes[last + 1] == boxes[L]) { + last++; + } + // K个1 (K + last - L) last + int pre = K + last - L; + int ans = (pre + 1) * (pre + 1) + process2(boxes, last + 1, R, 0, dp); + for (int i = last + 2; i <= R; i++) { + if (boxes[i] == boxes[L] && boxes[i - 1] != boxes[L]) { + ans = Math.max(ans, process2(boxes, last + 1, i - 1, 0, dp) + process2(boxes, i, R, pre + 1, dp)); + } + } + dp[L][R][K] = ans; + return ans; + } + +} diff --git a/体系学习班/class46/Code03_DeleteAdjacentSameCharacter.java b/体系学习班/class46/Code03_DeleteAdjacentSameCharacter.java new file mode 100644 index 0000000..0b1e7f3 --- /dev/null +++ b/体系学习班/class46/Code03_DeleteAdjacentSameCharacter.java @@ -0,0 +1,176 @@ +package class46; + +// 如果一个字符相邻的位置没有相同字符，那么这个位置的字符出现不能被消掉 +// 比如:"ab"，其中a和b都不能被消掉 +// 如果一个字符相邻的位置有相同字符，就可以一起消掉 +// 比如:"abbbc"，中间一串的b是可以被消掉的，消除之后剩下"ac" +// 某些字符如果消掉了，剩下的字符认为重新靠在一起 +// 给定一个字符串，你可以决定每一步消除的顺序，目标是请尽可能多的消掉字符，返回最少的剩余字符数量 +// 比如："aacca", 如果先消掉最左侧的"aa"，那么将剩下"cca"，然后把"cc"消掉，剩下的"a"将无法再消除，返回1 +// 但是如果先消掉中间的"cc"，那么将剩下"aaa"，最后都消掉就一个字符也不剩了，返回0，这才是最优解。 +// 再比如："baaccabb"， +// 如果先消除最左侧的两个a，剩下"bccabb"， +// 如果再消除最左侧的两个c，剩下"babb"， +// 最后消除最右侧的两个b，剩下"ba"无法再消除，返回2 +// 而最优策略是： +// 如果先消除中间的两个c，剩下"baaabb"， +// 如果再消除中间的三个a，剩下"bbb"， +// 最后消除三个b，不留下任何字符，返回0，这才是最优解 +public class Code03_DeleteAdjacentSameCharacter { + + // 暴力解 + public static int restMin1(String s) { + if (s == null) { + return 0; + } + if (s.length() < 2) { + return s.length(); + } + int minLen = s.length(); + for (int L = 0; L < s.length(); L++) { + for (int R = L + 1; R < s.length(); R++) { + if (canDelete(s.substring(L, R + 1))) { + minLen = Math.min(minLen, restMin1(s.substring(0, L) + s.substring(R + 1, s.length()))); + } + } + } + return minLen; + } + + public static boolean canDelete(String s) { + char[] str = s.toCharArray(); + for (int i = 1; i < str.length; i++) { + if (str[i - 1] != str[i]) { + return false; + } + } + return true; + } + + // 优良尝试的暴力递归版本 + public static int restMin2(String s) { + if (s == null) { + return 0; + } + if (s.length() < 2) { + return s.length(); + } + char[] str = s.toCharArray(); + return process(str, 0, str.length - 1, false); + } + + // str[L...R] 前面有没有跟着[L]字符，has T 有 F 无 + // L,R,has + // 最少能剩多少字符，消不了 + public static int process(char[] str, int L, int R, boolean has) { + if (L > R) { + return 0; + } + if (L == R) { + return has ? 0 : 1; + } + int index = L; + int K = has ? 1 : 0; + while (index <= R && str[index] == str[L]) { + K++; + index++; + } + // index表示，第一个不是[L]字符的位置 + int way1 = (K > 1 ? 0 : 1) + process(str, index, R, false); + int way2 = Integer.MAX_VALUE; + for (int split = index; split <= R; split++) { + if (str[split] == str[L] && str[split] != str[split - 1]) { + if (process(str, index, split - 1, false) == 0) { + way2 = Math.min(way2, process(str, split, R, K != 0)); + } + } + } + return Math.min(way1, way2); + } + + // 优良尝试的动态规划版本 + public static int restMin3(String s) { + if (s == null) { + return 0; + } + if (s.length() < 2) { + return s.length(); + } + char[] str = s.toCharArray(); + int N = str.length; + int[][][] dp = new int[N][N][2]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < N; j++) { + for (int k = 0; k < 2; k++) { + dp[i][j][k] = -1; + } + } + } + return dpProcess(str, 0, N - 1, false, dp); + } + + public static int dpProcess(char[] str, int L, int R, boolean has, int[][][] dp) { + if (L > R) { + return 0; + } + int K = has ? 1 : 0; + if (dp[L][R][K] != -1) { + return dp[L][R][K]; + } + int ans = 0; + if (L == R) { + ans = (K == 0 ? 1 : 0); + } else { + int index = L; + int all = K; + while (index <= R && str[index] == str[L]) { + all++; + index++; + } + int way1 = (all > 1 ? 0 : 1) + dpProcess(str, index, R, false, dp); + int way2 = Integer.MAX_VALUE; + for (int split = index; split <= R; split++) { + if (str[split] == str[L] && str[split] != str[split - 1]) { + if (dpProcess(str, index, split - 1, false, dp) == 0) { + way2 = Math.min(way2, dpProcess(str, split, R, all > 0, dp)); + } + } + } + ans = Math.min(way1, way2); + } + dp[L][R][K] = ans; + return ans; + } + + public static String randomString(int len, int variety) { + char[] str = new char[len]; + for (int i = 0; i < len; i++) { + str[i] = (char) ((int) (Math.random() * variety) + 'a'); + } + return String.valueOf(str); + } + + public static void main(String[] args) { + int maxLen = 16; + int variety = 3; + int testTime = 100000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * maxLen); + String str = randomString(len, variety); + int ans1 = restMin1(str); + int ans2 = restMin2(str); + int ans3 = restMin3(str); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println(str); + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + System.out.println("出错了！"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class46/Code04_MaxSumLengthNoMore.java b/体系学习班/class46/Code04_MaxSumLengthNoMore.java new file mode 100644 index 0000000..be50615 --- /dev/null +++ b/体系学习班/class46/Code04_MaxSumLengthNoMore.java @@ -0,0 +1,100 @@ +package class46; + +import java.util.LinkedList; + +// 给定一个数组arr，和一个正数M +// 返回在子数组长度不大于M的情况下，最大的子数组累加和 +public class Code04_MaxSumLengthNoMore { + + // O(N^2)的解法，暴力解，用作对数器 + public static int test(int[] arr, int M) { + if (arr == null || arr.length == 0 || M < 1) { + return 0; + } + int N = arr.length; + int max = Integer.MIN_VALUE; + for (int L = 0; L < N; L++) { + int sum = 0; + for (int R = L; R < N; R++) { + if (R - L + 1 > M) { + break; + } + sum += arr[R]; + max = Math.max(max, sum); + } + } + return max; + } + + // O(N)的解法，最优解 + public static int maxSum(int[] arr, int M) { + if (arr == null || arr.length == 0 || M < 1) { + return 0; + } + int N = arr.length; + int[] sum = new int[N]; + sum[0] = arr[0]; + for (int i = 1; i < N; i++) { + sum[i] = sum[i - 1] + arr[i]; + } + LinkedList qmax = new LinkedList<>(); + int i = 0; + int end = Math.min(N, M); + for (; i < end; i++) { + while (!qmax.isEmpty() && sum[qmax.peekLast()] <= sum[i]) { + qmax.pollLast(); + } + qmax.add(i); + } + int max = sum[qmax.peekFirst()]; + int L = 0; + for (; i < N; L++, i++) { + if (qmax.peekFirst() == L) { + qmax.pollFirst(); + } + while (!qmax.isEmpty() && sum[qmax.peekLast()] <= sum[i]) { + qmax.pollLast(); + } + qmax.add(i); + max = Math.max(max, sum[qmax.peekFirst()] - sum[L]); + } + for (; L < N - 1; L++) { + if (qmax.peekFirst() == L) { + qmax.pollFirst(); + } + max = Math.max(max, sum[qmax.peekFirst()] - sum[L]); + } + return max; + } + + // 用作测试 + public static int[] randomArray(int len, int max) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * max) - (int) (Math.random() * max); + } + return arr; + } + + // 用作测试 + public static void main(String[] args) { + int maxN = 50; + int maxValue = 100; + int testTime = 1000000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int N = (int) (Math.random() * maxN); + int M = (int) (Math.random() * maxN); + int[] arr = randomArray(N, maxValue); + int ans1 = test(arr, M); + int ans2 = maxSum(arr, M); + if (ans1 != ans2) { + System.out.println(ans1); + System.out.println(ans2); + System.out.println("Oops!"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/体系学习班/class46/Code05_HuffmanTree.java b/体系学习班/class46/Code05_HuffmanTree.java new file mode 100644 index 0000000..65b6be8 --- /dev/null +++ b/体系学习班/class46/Code05_HuffmanTree.java @@ -0,0 +1,214 @@ +package class46; + +import java.util.Comparator; +import java.util.HashMap; +import java.util.Map.Entry; +import java.util.PriorityQueue; + +// 本文件不牵扯任何byte类型的转化 +// 怎么转byte自己来，我只负责huffman算法本身的正确实现 +// 字符串为空的时候，自己处理边界吧 +// 实现的代码通过了大样本随机测试的对数器 +// 可以从main函数的内容开始看起 +public class Code05_HuffmanTree { + + // 根据文章str, 生成词频统计表 + public static HashMap countMap(String str) { + HashMap ans = new HashMap<>(); + char[] s = str.toCharArray(); + for (char cha : s) { + if (!ans.containsKey(cha)) { + ans.put(cha, 1); + } else { + ans.put(cha, ans.get(cha) + 1); + } + } + return ans; + } + + public static class Node { + public int count; + public Node left; + public Node right; + + public Node(int c) { + count = c; + } + } + + public static class NodeComp implements Comparator { + + @Override + public int compare(Node o1, Node o2) { + return o1.count - o2.count; + } + + } + + // 根据由文章生成词频表countMap，生成哈夫曼编码表 + // key : 字符 + // value: 该字符编码后的二进制形式 + // 比如，频率表 A：60, B:45, C:13 D:69 E:14 F:5 G:3 + // A 10 + // B 01 + // C 0011 + // D 11 + // E 000 + // F 00101 + // G 00100 + public static HashMap huffmanForm(HashMap countMap) { + HashMap ans = new HashMap<>(); + if (countMap.size() == 1) { + for (char key : countMap.keySet()) { + ans.put(key, "0"); + } + return ans; + } + HashMap nodes = new HashMap<>(); + PriorityQueue heap = new PriorityQueue<>(new NodeComp()); + for (Entry entry : countMap.entrySet()) { + Node cur = new Node(entry.getValue()); + char cha = entry.getKey(); + nodes.put(cur, cha); + heap.add(cur); + } + while (heap.size() != 1) { + Node a = heap.poll(); + Node b = heap.poll(); + Node h = new Node(a.count + b.count); + h.left = a; + h.right = b; + heap.add(h); + } + Node head = heap.poll(); + fillForm(head, "", nodes, ans); + return ans; + } + + public static void fillForm(Node head, String pre, HashMap nodes, HashMap ans) { + if (nodes.containsKey(head)) { + ans.put(nodes.get(head), pre); + } else { + fillForm(head.left, pre + "0", nodes, ans); + fillForm(head.right, pre + "1", nodes, ans); + } + } + + // 原始字符串str，根据哈夫曼编码表，转译成哈夫曼编码返回 + public static String huffmanEncode(String str, HashMap huffmanForm) { + char[] s = str.toCharArray(); + StringBuilder builder = new StringBuilder(); + for (char cha : s) { + builder.append(huffmanForm.get(cha)); + } + return builder.toString(); + } + + // 原始字符串的哈夫曼编码huffmanEncode，根据哈夫曼编码表，还原成原始字符串 + public static String huffmanDecode(String huffmanEncode, HashMap huffmanForm) { + TrieNode root = createTrie(huffmanForm); + TrieNode cur = root; + char[] encode = huffmanEncode.toCharArray(); + StringBuilder builder = new StringBuilder(); + for (int i = 0; i < encode.length; i++) { + int index = encode[i] == '0' ? 0 : 1; + cur = cur.nexts[index]; + if (cur.nexts[0] == null && cur.nexts[1] == null) { + builder.append(cur.value); + cur = root; + } + } + return builder.toString(); + } + + public static TrieNode createTrie(HashMap huffmanForm) { + TrieNode root = new TrieNode(); + for (char key : huffmanForm.keySet()) { + char[] path = huffmanForm.get(key).toCharArray(); + TrieNode cur = root; + for (int i = 0; i < path.length; i++) { + int index = path[i] == '0' ? 0 : 1; + if (cur.nexts[index] == null) { + cur.nexts[index] = new TrieNode(); + } + cur = cur.nexts[index]; + } + cur.value = key; + } + return root; + } + + public static class TrieNode { + public char value; + public TrieNode[] nexts; + + public TrieNode() { + value = 0; + nexts = new TrieNode[2]; + } + } + + // 为了测试 + public static String randomNumberString(int len, int range) { + char[] str = new char[len]; + for (int i = 0; i < len; i++) { + str[i] = (char) ((int) (Math.random() * range) + 'a'); + } + return String.valueOf(str); + } + + // 为了测试 + public static void main(String[] args) { + // 根据词频表生成哈夫曼编码表 + HashMap map = new HashMap<>(); + map.put('A', 60); + map.put('B', 45); + map.put('C', 13); + map.put('D', 69); + map.put('E', 14); + map.put('F', 5); + map.put('G', 3); + HashMap huffmanForm = huffmanForm(map); + for (Entry entry : huffmanForm.entrySet()) { + System.out.println(entry.getKey() + " : " + entry.getValue()); + } + System.out.println("===================="); + // str是原始字符串 + String str = "CBBBAABBACAABDDEFBA"; + System.out.println(str); + // countMap是根据str建立的词频表 + HashMap countMap = countMap(str); + // hf是根据countMap生成的哈夫曼编码表 + HashMap hf = huffmanForm(countMap); + // huffmanEncode是原始字符串转译后的哈夫曼编码 + String huffmanEncode = huffmanEncode(str, hf); + System.out.println(huffmanEncode); + // huffmanDecode是哈夫曼编码还原成的原始字符串 + String huffmanDecode = huffmanDecode(huffmanEncode, hf); + System.out.println(huffmanDecode); + System.out.println("===================="); + System.out.println("大样本随机测试开始"); + // 字符串最大长度 + int len = 500; + // 所含字符种类 + int range = 26; + // 随机测试进行的次数 + int testTime = 100000; + for (int i = 0; i < testTime; i++) { + int N = (int) (Math.random() * len) + 1; + String test = randomNumberString(N, range); + HashMap counts = countMap(test); + HashMap form = huffmanForm(counts); + String encode = huffmanEncode(test, form); + String decode = huffmanDecode(encode, form); + if (!test.equals(decode)) { + System.out.println(test); + System.out.println(encode); + System.out.println(decode); + System.out.println("出错了!"); + } + } + System.out.println("大样本随机测试结束"); + } + +} diff --git a/体系学习班/class47/Code01_StrangePrinter.java b/体系学习班/class47/Code01_StrangePrinter.java new file mode 100644 index 0000000..b7172db --- /dev/null +++ b/体系学习班/class47/Code01_StrangePrinter.java @@ -0,0 +1,78 @@ +package class47; + +// 本题测试链接 : https://leetcode.com/problems/strange-printer/ +public class Code01_StrangePrinter { + + public static int strangePrinter1(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + return process1(str, 0, str.length - 1); + } + + // 要想刷出str[L...R]的样子！ + // 返回最少的转数 + public static int process1(char[] str, int L, int R) { + if (L == R) { + return 1; + } + // L...R + int ans = R - L + 1; + for (int k = L + 1; k <= R; k++) { + // L...k-1 k....R + ans = Math.min(ans, process1(str, L, k - 1) + process1(str, k, R) - (str[L] == str[k] ? 1 : 0)); + } + return ans; + } + + public static int strangePrinter2(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + int N = str.length; + int[][] dp = new int[N][N]; + return process2(str, 0, N - 1, dp); + } + + public static int process2(char[] str, int L, int R, int[][] dp) { + if (dp[L][R] != 0) { + return dp[L][R]; + } + int ans = R - L + 1; + if (L == R) { + ans = 1; + } else { + for (int k = L + 1; k <= R; k++) { + ans = Math.min(ans, process2(str, L, k - 1, dp) + process2(str, k, R, dp) - (str[L] == str[k] ? 1 : 0)); + } + } + dp[L][R] = ans; + return ans; + } + + public static int strangePrinter3(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + int N = str.length; + int[][] dp = new int[N][N]; + dp[N - 1][N - 1] = 1; + for (int i = 0; i < N - 1; i++) { + dp[i][i] = 1; + dp[i][i + 1] = str[i] == str[i + 1] ? 1 : 2; + } + for (int L = N - 3; L >= 0; L--) { + for (int R = L + 2; R < N; R++) { + dp[L][R] = R - L + 1; + for (int k = L + 1; k <= R; k++) { + dp[L][R] = Math.min(dp[L][R], dp[L][k - 1] + dp[k][R] - (str[L] == str[k] ? 1 : 0)); + } + } + } + return dp[0][N - 1]; + } + +} diff --git a/体系学习班/class47/Code02_RestoreWays.java b/体系学习班/class47/Code02_RestoreWays.java new file mode 100644 index 0000000..c9c583a --- /dev/null +++ b/体系学习班/class47/Code02_RestoreWays.java @@ -0,0 +1,246 @@ +package class47; + +// 整型数组arr长度为n(3 <= n <= 10^4)，最初每个数字是<=200的正数且满足如下条件： +// 1. 0位置的要求：arr[0]<=arr[1] +// 2. n-1位置的要求：arr[n-1]<=arr[n-2] +// 3. 中间i位置的要求：arr[i]<=max(arr[i-1],arr[i+1]) +// 但是在arr有些数字丢失了，比如k位置的数字之前是正数，丢失之后k位置的数字为0 +// 请你根据上述条件，计算可能有多少种不同的arr可以满足以上条件 +// 比如 [6,0,9] 只有还原成 [6,9,9]满足全部三个条件，所以返回1种，即[6,9,9]达标 +public class Code02_RestoreWays { + + public static int ways0(int[] arr) { + return process0(arr, 0); + } + + public static int process0(int[] arr, int index) { + if (index == arr.length) { + return isValid(arr) ? 1 : 0; + } else { + if (arr[index] != 0) { + return process0(arr, index + 1); + } else { + int ways = 0; + for (int v = 1; v < 201; v++) { + arr[index] = v; + ways += process0(arr, index + 1); + } + arr[index] = 0; + return ways; + } + } + } + + public static boolean isValid(int[] arr) { + if (arr[0] > arr[1]) { + return false; + } + if (arr[arr.length - 1] > arr[arr.length - 2]) { + return false; + } + for (int i = 1; i < arr.length - 1; i++) { + if (arr[i] > Math.max(arr[i - 1], arr[i + 1])) { + return false; + } + } + return true; + } + + public static int ways1(int[] arr) { + int N = arr.length; + if (arr[N - 1] != 0) { + return process1(arr, N - 1, arr[N - 1], 2); + } else { + int ways = 0; + for (int v = 1; v < 201; v++) { + ways += process1(arr, N - 1, v, 2); + } + return ways; + } + } + + // 如果i位置的数字变成了v, + // 并且arr[i]和arr[i+1]的关系为s， + // s==0，代表arr[i] < arr[i+1] 右大 + // s==1，代表arr[i] == arr[i+1] 右=当前 + // s==2，代表arr[i] > arr[i+1] 右小 + // 返回0...i范围上有多少种有效的转化方式？ + public static int process1(int[] arr, int i, int v, int s) { + if (i == 0) { // 0...i 只剩一个数了，0...0 + return ((s == 0 || s == 1) && (arr[0] == 0 || v == arr[0])) ? 1 : 0; + } + // i > 0 + if (arr[i] != 0 && v != arr[i]) { + return 0; + } + // i>0 ，并且， i位置的数真的可以变成V， + int ways = 0; + if (s == 0 || s == 1) { // [i] -> V <= [i+1] + for (int pre = 1; pre < 201; pre++) { + ways += process1(arr, i - 1, pre, pre < v ? 0 : (pre == v ? 1 : 2)); + } + } else { // ? 当前 > 右 当前 <= max{左，右} + for (int pre = v; pre < 201; pre++) { + ways += process1(arr, i - 1, pre, pre == v ? 1 : 2); + } + } + return ways; + } + + public static int zuo(int[] arr, int i, int v, int s) { + if (i == 0) { // 0...i 只剩一个数了，0...0 + return ((s == 0 || s == 1) && (arr[0] == 0 || v == arr[0])) ? 1 : 0; + } + // i > 0 + if (arr[i] != 0 && v != arr[i]) { + return 0; + } + // i>0 ，并且， i位置的数真的可以变成V， + int ways = 0; + if (s == 0 || s == 1) { // [i] -> V <= [i+1] + for (int pre = 1; pre < v; pre++) { + ways += zuo(arr, i - 1, pre, 0); + } + } + ways += zuo(arr, i - 1, v, 1); + for (int pre = v + 1; pre < 201; pre++) { + ways += zuo(arr, i - 1, pre, 2); + } + return ways; + } + + public static int ways2(int[] arr) { + int N = arr.length; + int[][][] dp = new int[N][201][3]; + if (arr[0] != 0) { + dp[0][arr[0]][0] = 1; + dp[0][arr[0]][1] = 1; + } else { + for (int v = 1; v < 201; v++) { + dp[0][v][0] = 1; + dp[0][v][1] = 1; + } + } + for (int i = 1; i < N; i++) { + for (int v = 1; v < 201; v++) { + for (int s = 0; s < 3; s++) { + if (arr[i] == 0 || v == arr[i]) { + if (s == 0 || s == 1) { + for (int pre = 1; pre < v; pre++) { + dp[i][v][s] += dp[i - 1][pre][0]; + } + } + dp[i][v][s] += dp[i - 1][v][1]; + for (int pre = v + 1; pre < 201; pre++) { + dp[i][v][s] += dp[i - 1][pre][2]; + } + } + } + } + } + if (arr[N - 1] != 0) { + return dp[N - 1][arr[N - 1]][2]; + } else { + int ways = 0; + for (int v = 1; v < 201; v++) { + ways += dp[N - 1][v][2]; + } + return ways; + } + } + + public static int ways3(int[] arr) { + int N = arr.length; + int[][][] dp = new int[N][201][3]; + if (arr[0] != 0) { + dp[0][arr[0]][0] = 1; + dp[0][arr[0]][1] = 1; + } else { + for (int v = 1; v < 201; v++) { + dp[0][v][0] = 1; + dp[0][v][1] = 1; + } + } + int[][] presum = new int[201][3]; + for (int v = 1; v < 201; v++) { + for (int s = 0; s < 3; s++) { + presum[v][s] = presum[v - 1][s] + dp[0][v][s]; + } + } + for (int i = 1; i < N; i++) { + for (int v = 1; v < 201; v++) { + for (int s = 0; s < 3; s++) { + if (arr[i] == 0 || v == arr[i]) { + if (s == 0 || s == 1) { + dp[i][v][s] += sum(1, v - 1, 0, presum); + } + dp[i][v][s] += dp[i - 1][v][1]; + dp[i][v][s] += sum(v + 1, 200, 2, presum); + } + } + } + for (int v = 1; v < 201; v++) { + for (int s = 0; s < 3; s++) { + presum[v][s] = presum[v - 1][s] + dp[i][v][s]; + } + } + } + if (arr[N - 1] != 0) { + return dp[N - 1][arr[N - 1]][2]; + } else { + return sum(1, 200, 2, presum); + } + } + + public static int sum(int begin, int end, int relation, int[][] presum) { + return presum[end][relation] - presum[begin - 1][relation]; + } + + // for test + public static int[] generateRandomArray(int len) { + int[] ans = new int[len]; + for (int i = 0; i < ans.length; i++) { + if (Math.random() < 0.5) { + ans[i] = 0; + } else { + ans[i] = (int) (Math.random() * 200) + 1; + } + } + return ans; + } + + // for test + public static void printArray(int[] arr) { + System.out.println("arr size : " + arr.length); + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + public static void main(String[] args) { + int len = 4; + int testTime = 15; + System.out.println("功能测试开始"); + for (int i = 0; i < testTime; i++) { + int N = (int) (Math.random() * len) + 2; + int[] arr = generateRandomArray(N); + int ans0 = ways0(arr); + int ans1 = ways1(arr); + int ans2 = ways2(arr); + int ans3 = ways3(arr); + if (ans0 != ans1 || ans2 != ans3 || ans0 != ans2) { + System.out.println("Oops!"); + } + } + System.out.println("功能测试结束"); + System.out.println("==========="); + int N = 100000; + int[] arr = generateRandomArray(N); + long begin = System.currentTimeMillis(); + ways3(arr); + long end = System.currentTimeMillis(); + System.out.println("run time : " + (end - begin) + " ms"); + } + +} diff --git a/体系学习班/class47/Code03_DinicAlgorithm.java b/体系学习班/class47/Code03_DinicAlgorithm.java new file mode 100644 index 0000000..efed284 --- /dev/null +++ b/体系学习班/class47/Code03_DinicAlgorithm.java @@ -0,0 +1,139 @@ +// 本题测试链接: +// https://lightoj.com/problem/internet-bandwidth +// 这是一道DinicAlgorithm算法的题 +// 把如下代码粘贴进网页所提供的java编译器环境中 +// 不需要修改任何内容可以直接通过 +// 请看网页上的题目描述并结合main函数的写法去了解这个模板的用法 + +package class47; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.LinkedList; +import java.util.Scanner; + +public class Code03_DinicAlgorithm { + + public static class Edge { + public int from; + public int to; + public int available; + + public Edge(int a, int b, int c) { + from = a; + to = b; + available = c; + } + } + + public static class Dinic { + private int N; + private ArrayList> nexts; + private ArrayList edges; + private int[] depth; + private int[] cur; + + public Dinic(int nums) { + N = nums + 1; + nexts = new ArrayList<>(); + for (int i = 0; i <= N; i++) { + nexts.add(new ArrayList<>()); + } + edges = new ArrayList<>(); + depth = new int[N]; + cur = new int[N]; + } + + public void addEdge(int u, int v, int r) { + int m = edges.size(); + edges.add(new Edge(u, v, r)); + nexts.get(u).add(m); + edges.add(new Edge(v, u, 0)); + nexts.get(v).add(m + 1); + } + + public int maxFlow(int s, int t) { + int flow = 0; + while (bfs(s, t)) { + Arrays.fill(cur, 0); + flow += dfs(s, t, Integer.MAX_VALUE); + Arrays.fill(depth, 0); + } + return flow; + } + + private boolean bfs(int s, int t) { + LinkedList queue = new LinkedList<>(); + queue.addFirst(s); + boolean[] visited = new boolean[N]; + visited[s] = true; + while (!queue.isEmpty()) { + int u = queue.pollLast(); + for (int i = 0; i < nexts.get(u).size(); i++) { + Edge e = edges.get(nexts.get(u).get(i)); + int v = e.to; + if (!visited[v] && e.available > 0) { + visited[v] = true; + depth[v] = depth[u] + 1; + if (v == t) { + break; + } + queue.addFirst(v); + } + } + } + return visited[t]; + } + + // 当前来到了s点，s可变 + // 最终目标是t，t固定参数 + // r，收到的任务 + // 收集到的流，作为结果返回，ans <= r + private int dfs(int s, int t, int r) { + if (s == t || r == 0) { + return r; + } + int f = 0; + int flow = 0; + // s点从哪条边开始试 -> cur[s] + for (; cur[s] < nexts.get(s).size(); cur[s]++) { + int ei = nexts.get(s).get(cur[s]); + Edge e = edges.get(ei); + Edge o = edges.get(ei ^ 1); + if (depth[e.to] == depth[s] + 1 && (f = dfs(e.to, t, Math.min(e.available, r))) != 0) { + e.available -= f; + o.available += f; + flow += f; + r -= f; + if (r <= 0) { + break; + } + } + } + return flow; + } + } + + public static void main(String[] args) { + Scanner cin = new Scanner(System.in); + int cases = cin.nextInt(); + for (int i = 1; i <= cases; i++) { + int n = cin.nextInt(); + int s = cin.nextInt(); + int t = cin.nextInt(); + int m = cin.nextInt(); + Dinic dinic = new Dinic(n); + for (int j = 0; j < m; j++) { + int from = cin.nextInt(); + int to = cin.nextInt(); + int weight = cin.nextInt(); + dinic.addEdge(from, to, weight); + dinic.addEdge(to, from, weight); + } + int ans = dinic.maxFlow(s, t); + System.out.println("Case " + i + ": " + ans); + } + cin.close(); + } + +} \ No newline at end of file diff --git a/大厂刷题班/class01/Code01_CordCoverMaxPoint.java b/大厂刷题班/class01/Code01_CordCoverMaxPoint.java new file mode 100644 index 0000000..a95eaf4 --- /dev/null +++ b/大厂刷题班/class01/Code01_CordCoverMaxPoint.java @@ -0,0 +1,87 @@ +package class01; + +import java.util.Arrays; + +public class Code01_CordCoverMaxPoint { + + public static int maxPoint1(int[] arr, int L) { + int res = 1; + for (int i = 0; i < arr.length; i++) { + int nearest = nearestIndex(arr, i, arr[i] - L); + res = Math.max(res, i - nearest + 1); + } + return res; + } + + public static int nearestIndex(int[] arr, int R, int value) { + int L = 0; + int index = R; + while (L <= R) { + int mid = L + ((R - L) >> 1); + if (arr[mid] >= value) { + index = mid; + R = mid - 1; + } else { + L = mid + 1; + } + } + return index; + } + + public static int maxPoint2(int[] arr, int L) { + int left = 0; + int right = 0; + int N = arr.length; + int max = 0; + while (left < N) { + while (right < N && arr[right] - arr[left] <= L) { + right++; + } + max = Math.max(max, right - (left++)); + } + return max; + } + + // for test + public static int test(int[] arr, int L) { + int max = 0; + for (int i = 0; i < arr.length; i++) { + int pre = i - 1; + while (pre >= 0 && arr[i] - arr[pre] <= L) { + pre--; + } + max = Math.max(max, i - pre); + } + return max; + } + + // for test + public static int[] generateArray(int len, int max) { + int[] ans = new int[(int) (Math.random() * len) + 1]; + for (int i = 0; i < ans.length; i++) { + ans[i] = (int) (Math.random() * max); + } + Arrays.sort(ans); + return ans; + } + + public static void main(String[] args) { + int len = 100; + int max = 1000; + int testTime = 100000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int L = (int) (Math.random() * max); + int[] arr = generateArray(len, max); + int ans1 = maxPoint1(arr, L); + int ans2 = maxPoint2(arr, L); + int ans3 = test(arr, L); + if (ans1 != ans2 || ans2 != ans3) { + System.out.println("oops!"); + break; + } + } + + } + +} diff --git a/大厂刷题班/class01/Code02_CountFiles.java b/大厂刷题班/class01/Code02_CountFiles.java new file mode 100644 index 0000000..f09d132 --- /dev/null +++ b/大厂刷题班/class01/Code02_CountFiles.java @@ -0,0 +1,40 @@ +package class01; + +import java.io.File; +import java.util.Stack; + +public class Code02_CountFiles { + + // 注意这个函数也会统计隐藏文件 + public static int getFileNumber(String folderPath) { + File root = new File(folderPath); + if (!root.isDirectory() && !root.isFile()) { + return 0; + } + if (root.isFile()) { + return 1; + } + Stack stack = new Stack<>(); + stack.add(root); + int files = 0; + while (!stack.isEmpty()) { + File folder = stack.pop(); + for (File next : folder.listFiles()) { + if (next.isFile()) { + files++; + } + if (next.isDirectory()) { + stack.push(next); + } + } + } + return files; + } + + public static void main(String[] args) { + // 你可以自己更改目录 + String path = "/Users/zuochengyun/Desktop/"; + System.out.println(getFileNumber(path)); + } + +} diff --git a/大厂刷题班/class01/Code03_Near2Power.java b/大厂刷题班/class01/Code03_Near2Power.java new file mode 100644 index 0000000..43aca3b --- /dev/null +++ b/大厂刷题班/class01/Code03_Near2Power.java @@ -0,0 +1,22 @@ +package class01; + +public class Code03_Near2Power { + + // 已知n是正数 + // 返回大于等于，且最接近n的，2的某次方的值 + public static final int tableSizeFor(int n) { + n--; + n |= n >>> 1; + n |= n >>> 2; + n |= n >>> 4; + n |= n >>> 8; + n |= n >>> 16; + return (n < 0) ? 1 : n + 1; + } + + public static void main(String[] args) { + int cap = 120; + System.out.println(tableSizeFor(cap)); + } + +} diff --git a/大厂刷题班/class01/Code04_MinSwapStep.java b/大厂刷题班/class01/Code04_MinSwapStep.java new file mode 100644 index 0000000..d659451 --- /dev/null +++ b/大厂刷题班/class01/Code04_MinSwapStep.java @@ -0,0 +1,74 @@ +package class01; + +public class Code04_MinSwapStep { + + // 一个数组中只有两种字符'G'和'B'， + // 可以让所有的G都放在左侧，所有的B都放在右侧 + // 或者可以让所有的G都放在右侧，所有的B都放在左侧 + // 但是只能在相邻字符之间进行交换操作，请问请问至少需要交换几次， + public static int minSteps1(String s) { + if (s == null || s.equals("")) { + return 0; + } + char[] str = s.toCharArray(); + int step1 = 0; + int gi = 0; + for (int i = 0; i < str.length; i++) { + if (str[i] == 'G') { + step1 += i - (gi++); + } + } + int step2 = 0; + int bi = 0; + for (int i = 0; i < str.length; i++) { + if (str[i] == 'B') { + step2 += i - (bi++); + } + } + return Math.min(step1, step2); + } + + // 可以让G在左，或者在右 + public static int minSteps2(String s) { + if (s == null || s.equals("")) { + return 0; + } + char[] str = s.toCharArray(); + int step1 = 0; + int step2 = 0; + int gi = 0; + int bi = 0; + for (int i = 0; i < str.length; i++) { + if (str[i] == 'G') { // 当前的G，去左边 方案1 + step1 += i - (gi++); + } else {// 当前的B，去左边 方案2 + step2 += i - (bi++); + } + } + return Math.min(step1, step2); + } + + // 为了测试 + public static String randomString(int maxLen) { + char[] str = new char[(int) (Math.random() * maxLen)]; + for (int i = 0; i < str.length; i++) { + str[i] = Math.random() < 0.5 ? 'G' : 'B'; + } + return String.valueOf(str); + } + + public static void main(String[] args) { + int maxLen = 100; + int testTime = 1000000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + String str = randomString(maxLen); + int ans1 = minSteps1(str); + int ans2 = minSteps2(str); + if (ans1 != ans2) { + System.out.println("Oops!"); + } + } + System.out.println("测试结束"); + } +} \ No newline at end of file diff --git a/大厂刷题班/class01/Code05_LongestIncreasingPath.java b/大厂刷题班/class01/Code05_LongestIncreasingPath.java new file mode 100644 index 0000000..682f6b5 --- /dev/null +++ b/大厂刷题班/class01/Code05_LongestIncreasingPath.java @@ -0,0 +1,54 @@ +package class01; + +public class Code05_LongestIncreasingPath { + + public static int longestIncreasingPath1(int[][] matrix) { + int ans = 0; + int N = matrix.length; + int M = matrix[0].length; + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + ans = Math.max(ans, process1(matrix, i, j)); + } + } + return ans; + } + + // 从m[i][j]开始走，走出来的最长递增链，返回！ + public static int process1(int[][] m, int i, int j) { + int up = i > 0 && m[i][j] < m[i - 1][j] ? process1(m, i - 1, j) : 0; + int down = i < (m.length - 1) && m[i][j] < m[i + 1][j] ? process1(m, i + 1, j) : 0; + int left = j > 0 && m[i][j] < m[i][j - 1] ? process1(m, i, j - 1) : 0; + int right = j < (m[0].length - 1) && m[i][j] < m[i][j + 1] ? process1(m, i, j + 1) : 0; + return Math.max(Math.max(up, down), Math.max(left, right)) + 1; + } + + public static int longestIncreasingPath2(int[][] matrix) { + int ans = 0; + int N = matrix.length; + int M = matrix[0].length; + int[][] dp = new int[N][M]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + ans = Math.max(ans, process2(matrix, i, j, dp)); + } + } + return ans; + } + + // 从m[i][j]开始走，走出来的最长递增链，返回！ + public static int process2(int[][] m, int i, int j, int[][] dp) { + if (dp[i][j] != 0) { + return dp[i][j]; + } + // (i,j)不越界 + int up = i > 0 && m[i][j] < m[i - 1][j] ? process2(m, i - 1, j, dp) : 0; + int down = i < (m.length - 1) && m[i][j] < m[i + 1][j] ? process2(m, i + 1, j, dp) : 0; + int left = j > 0 && m[i][j] < m[i][j - 1] ? process2(m, i, j - 1, dp) : 0; + int right = j < (m[0].length - 1) && m[i][j] < m[i][j + 1] ? process2(m, i, j + 1, dp) : 0; + int ans = Math.max(Math.max(up, down), Math.max(left, right)) + 1; + dp[i][j] = ans; + return ans; + } + +} diff --git a/大厂刷题班/class01/Code06_AOE.java b/大厂刷题班/class01/Code06_AOE.java new file mode 100644 index 0000000..6b8e72c --- /dev/null +++ b/大厂刷题班/class01/Code06_AOE.java @@ -0,0 +1,304 @@ +package class01; + +import java.util.Arrays; + +/* + * 给定两个数组x和hp，长度都是N。 + * x数组一定是有序的，x[i]表示i号怪兽在x轴上的位置 + * hp数组不要求有序，hp[i]表示i号怪兽的血量 + * 为了方便起见，可以认为x数组和hp数组中没有负数。 + * 再给定一个正数range，表示如果法师释放技能的范围长度(直径！) + * 被打到的每只怪兽损失1点血量。 + * 返回要把所有怪兽血量清空，至少需要释放多少次aoe技能？ + * 三个参数：int[] x, int[] hp, int range + * 返回：int 次数 + * */ +public class Code06_AOE { + + // 纯暴力解法 + // 太容易超时 + // 只能小样本量使用 + public static int minAoe1(int[] x, int[] hp, int range) { + boolean allClear = true; + for (int i = 0; i < hp.length; i++) { + if (hp[i] > 0) { + allClear = false; + break; + } + } + if (allClear) { + return 0; + } else { + int ans = Integer.MAX_VALUE; + for (int left = 0; left < x.length; left++) { + if (hasHp(x, hp, left, range)) { + minusOneHp(x, hp, left, range); + ans = Math.min(ans, 1 + minAoe1(x, hp, range)); + addOneHp(x, hp, left, range); + } + } + return ans; + } + } + + public static boolean hasHp(int[] x, int[] hp, int left, int range) { + for (int index = left; index < x.length && x[index] - x[left] <= range; index++) { + if (hp[index] > 0) { + return true; + } + } + return false; + } + + public static void minusOneHp(int[] x, int[] hp, int left, int range) { + for (int index = left; index < x.length && x[index] - x[left] <= range; index++) { + hp[index]--; + } + } + + public static void addOneHp(int[] x, int[] hp, int left, int range) { + for (int index = left; index < x.length && x[index] - x[left] <= range; index++) { + hp[index]++; + } + } + + // 为了验证 + // 不用线段树，但是贪心的思路，和课上一样 + // 1) 总是用技能的最左边缘刮死当前最左侧的没死的怪物 + // 2) 然后向右找下一个没死的怪物，重复步骤1) + public static int minAoe2(int[] x, int[] hp, int range) { + // 举个例子： + // 如果怪兽情况如下， + // 怪兽所在，x数组 : 2 3 5 6 7 9 + // 怪兽血量，hp数组 : 2 4 1 2 3 1 + // 怪兽编号 : 0 1 2 3 4 5 + // 技能直径，range = 2 + int n = x.length; + int[] cover = new int[n]; + // 首先求cover数组， + // 如果技能左边界就在0号怪兽，那么技能到2号怪兽就覆盖不到了 + // 所以cover[0] = 2; + // 如果技能左边界就在1号怪兽，那么技能到3号怪兽就覆盖不到了 + // 所以cover[1] = 3; + // 如果技能左边界就在2号怪兽，那么技能到5号怪兽就覆盖不到了 + // 所以cover[2] = 5; + // 如果技能左边界就在3号怪兽，那么技能到5号怪兽就覆盖不到了 + // 所以cover[3] = 5; + // 如果技能左边界就在4号怪兽，那么技能到6号怪兽（越界位置）就覆盖不到了 + // 所以cover[4] = 6(越界位置); + // 如果技能左边界就在5号怪兽，那么技能到6号怪兽（越界位置）就覆盖不到了 + // 所以cover[5] = 6(越界位置); + // 综上： + // 如果怪兽情况如下， + // 怪兽所在，x数组 : 2 3 5 6 7 9 + // 怪兽血量，hp数组 : 2 4 1 2 3 1 + // 怪兽编号 : 0 1 2 3 4 5 + // cover数组情况 : 2 3 5 5 6 6 + // 技能直径，range = 2 + // cover[i] = j，表示如果技能左边界在i怪兽，那么技能会影响i...j-1号所有的怪兽 + // 就是如下的for循环，在求cover数组 + int r = 0; + for (int i = 0; i < n; i++) { + while (r < n && x[r] - x[i] <= range) { + r++; + } + cover[i] = r; + } + int ans = 0; + for (int i = 0; i < n; i++) { + // 假设来到i号怪兽了 + // 如果i号怪兽的血量>0，说明i号怪兽没死 + // 根据我们课上讲的贪心： + // 我们要让技能的左边界，刮死当前的i号怪兽 + // 这样能够让技能尽可能的往右释放，去尽可能的打掉右侧的怪兽 + // 此时cover[i]，正好的告诉我们，技能影响多大范围。 + // 比如当前来到100号怪兽，血量30 + // 假设cover[100] == 200 + // 说明，技能左边界在100位置，可以影响100号到199号怪兽的血量。 + // 为了打死100号怪兽，我们释放技能30次， + // 释放的时候，100号到199号怪兽都掉血，30点 + // 然后继续向右寻找没死的怪兽，像课上讲的一样 + if (hp[i] > 0) { + int minus = hp[i]; + for (int index = i; index < cover[i]; index++) { + hp[index] -= minus; + } + ans += minus; + } + } + return ans; + } + + // 正式方法 + // 关键点就是： + // 1) 线段树 + // 2) 总是用技能的最左边缘刮死当前最左侧的没死的怪物 + // 3) 然后向右找下一个没死的怪物，重复步骤2) + public static int minAoe3(int[] x, int[] hp, int range) { + int n = x.length; + int[] cover = new int[n]; + int r = 0; + // cover[i] : 如果i位置是技能的最左侧，技能往右的range范围内，最右影响到哪 + for (int i = 0; i < n; i++) { + while (r < n && x[r] - x[i] <= range) { + r++; + } + cover[i] = r - 1; + } + SegmentTree st = new SegmentTree(hp); + st.build(1, n, 1); + int ans = 0; + for (int i = 1; i <= n; i++) { + int leftHP = st.query(i, i, 1, n, 1); + if (leftHP > 0) { + ans += leftHP; + st.add(i, cover[i - 1] + 1, -leftHP, 1, n, 1); + } + } + return ans; + } + + public static class SegmentTree { + private int MAXN; + private int[] arr; + private int[] sum; + private int[] lazy; + + public SegmentTree(int[] origin) { + MAXN = origin.length + 1; + arr = new int[MAXN]; + for (int i = 1; i < MAXN; i++) { + arr[i] = origin[i - 1]; + } + sum = new int[MAXN << 2]; + lazy = new int[MAXN << 2]; + } + + private void pushUp(int rt) { + sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; + } + + private void pushDown(int rt, int ln, int rn) { + if (lazy[rt] != 0) { + lazy[rt << 1] += lazy[rt]; + sum[rt << 1] += lazy[rt] * ln; + lazy[rt << 1 | 1] += lazy[rt]; + sum[rt << 1 | 1] += lazy[rt] * rn; + lazy[rt] = 0; + } + } + + public void build(int l, int r, int rt) { + if (l == r) { + sum[rt] = arr[l]; + return; + } + int mid = (l + r) >> 1; + build(l, mid, rt << 1); + build(mid + 1, r, rt << 1 | 1); + pushUp(rt); + } + + public void add(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + sum[rt] += C * (r - l + 1); + lazy[rt] += C; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + add(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + add(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + public int query(int L, int R, int l, int r, int rt) { + if (L <= l && r <= R) { + return sum[rt]; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + int ans = 0; + if (L <= mid) { + ans += query(L, R, l, mid, rt << 1); + } + if (R > mid) { + ans += query(L, R, mid + 1, r, rt << 1 | 1); + } + return ans; + } + + } + + // 为了测试 + public static int[] randomArray(int n, int valueMax) { + int[] ans = new int[n]; + for (int i = 0; i < n; i++) { + ans[i] = (int) (Math.random() * valueMax) + 1; + } + return ans; + } + + // 为了测试 + public static int[] copyArray(int[] arr) { + int N = arr.length; + int[] ans = new int[N]; + for (int i = 0; i < N; i++) { + ans[i] = arr[i]; + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int N = 50; + int X = 500; + int H = 60; + int R = 10; + int testTime = 50000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * N) + 1; + int[] x2 = randomArray(len, X); + Arrays.sort(x2); + int[] hp2 = randomArray(len, H); + int[] x3 = copyArray(x2); + int[] hp3 = copyArray(hp2); + int range = (int) (Math.random() * R) + 1; + int ans2 = minAoe2(x2, hp2, range); + int ans3 = minAoe3(x3, hp3, range); + if (ans2 != ans3) { + System.out.println("出错了！"); + } + } + System.out.println("测试结束"); + + N = 500000; + long start; + long end; + int[] x2 = randomArray(N, N); + Arrays.sort(x2); + int[] hp2 = new int[N]; + for (int i = 0; i < N; i++) { + hp2[i] = i * 5 + 10; + } + int[] x3 = copyArray(x2); + int[] hp3 = copyArray(hp2); + int range = 10000; + + start = System.currentTimeMillis(); + System.out.println(minAoe2(x2, hp2, range)); + end = System.currentTimeMillis(); + System.out.println("运行时间 : " + (end - start) + " 毫秒"); + + start = System.currentTimeMillis(); + System.out.println(minAoe3(x3, hp3, range)); + end = System.currentTimeMillis(); + System.out.println("运行时间 : " + (end - start) + " 毫秒"); + } + +} diff --git a/大厂刷题班/class01/Code07_TargetSum.java b/大厂刷题班/class01/Code07_TargetSum.java new file mode 100644 index 0000000..a91f560 --- /dev/null +++ b/大厂刷题班/class01/Code07_TargetSum.java @@ -0,0 +1,127 @@ +package class01; + +import java.util.HashMap; + +// leetcode 494题 +public class Code07_TargetSum { + + public static int findTargetSumWays1(int[] arr, int s) { + return process1(arr, 0, s); + } + + // 可以自由使用arr[index....]所有的数字！ + // 搞出rest这个数，方法数是多少？返回 + // index == 7 rest = 13 + // map "7_13" 256 + public static int process1(int[] arr, int index, int rest) { + if (index == arr.length) { // 没数了！ + return rest == 0 ? 1 : 0; + } + // 还有数！arr[index] arr[index+1 ... ] + return process1(arr, index + 1, rest - arr[index]) + process1(arr, index + 1, rest + arr[index]); + } + + public static int findTargetSumWays2(int[] arr, int s) { + return process2(arr, 0, s, new HashMap<>()); + } + + public static int process2(int[] arr, int index, int rest, HashMap> dp) { + if (dp.containsKey(index) && dp.get(index).containsKey(rest)) { + return dp.get(index).get(rest); + } + // 否则，没命中！ + int ans = 0; + if (index == arr.length) { + ans = rest == 0 ? 1 : 0; + } else { + ans = process2(arr, index + 1, rest - arr[index], dp) + process2(arr, index + 1, rest + arr[index], dp); + } + if (!dp.containsKey(index)) { + dp.put(index, new HashMap<>()); + } + dp.get(index).put(rest, ans); + return ans; + } + + // 优化点一 : + // 你可以认为arr中都是非负数 + // 因为即便是arr中有负数，比如[3,-4,2] + // 因为你能在每个数前面用+或者-号 + // 所以[3,-4,2]其实和[3,4,2]达成一样的效果 + // 那么我们就全把arr变成非负数，不会影响结果的 + // 优化点二 : + // 如果arr都是非负数，并且所有数的累加和是sum + // 那么如果target> 1); + } + + // 求非负数组nums有多少个子集，累加和是s + // 二维动态规划 + // 不用空间压缩 + public static int subset1(int[] nums, int s) { + if (s < 0) { + return 0; + } + int n = nums.length; + // dp[i][j] : nums前缀长度为i的所有子集，有多少累加和是j？ + int[][] dp = new int[n + 1][s + 1]; + // nums前缀长度为0的所有子集，有多少累加和是0？一个：空集 + dp[0][0] = 1; + for (int i = 1; i <= n; i++) { + for (int j = 0; j <= s; j++) { + dp[i][j] = dp[i - 1][j]; + if (j - nums[i - 1] >= 0) { + dp[i][j] += dp[i - 1][j - nums[i - 1]]; + } + } + } + return dp[n][s]; + } + + // 求非负数组nums有多少个子集，累加和是s + // 二维动态规划 + // 用空间压缩: + // 核心就是for循环里面的：for (int i = s; i >= n; i--) { + // 为啥不枚举所有可能的累加和？只枚举 n...s 这些累加和？ + // 因为如果 i - n < 0，dp[i]怎么更新？和上一步的dp[i]一样！所以不用更新 + // 如果 i - n >= 0，dp[i]怎么更新？上一步的dp[i] + 上一步dp[i - n]的值，这才需要更新 + public static int subset2(int[] nums, int s) { + if (s < 0) { + return 0; + } + int[] dp = new int[s + 1]; + dp[0] = 1; + for (int n : nums) { + for (int i = s; i >= n; i--) { + dp[i] += dp[i - n]; + } + } + return dp[s]; + } + +} diff --git a/大厂刷题班/class02/Code01_ChooseWork.java b/大厂刷题班/class02/Code01_ChooseWork.java new file mode 100644 index 0000000..4630a33 --- /dev/null +++ b/大厂刷题班/class02/Code01_ChooseWork.java @@ -0,0 +1,48 @@ +package class02; + +import java.util.Arrays; +import java.util.Comparator; +import java.util.TreeMap; + +public class Code01_ChooseWork { + + public static class Job { + public int money; + public int hard; + + public Job(int m, int h) { + money = m; + hard = h; + } + } + + public static class JobComparator implements Comparator { + @Override + public int compare(Job o1, Job o2) { + return o1.hard != o2.hard ? (o1.hard - o2.hard) : (o2.money - o1.money); + } + } + + public static int[] getMoneys(Job[] job, int[] ability) { + Arrays.sort(job, new JobComparator()); + // key : 难度 value：报酬 + TreeMap map = new TreeMap<>(); + map.put(job[0].hard, job[0].money); + // pre : 上一份进入map的工作 + Job pre = job[0]; + for (int i = 1; i < job.length; i++) { + if (job[i].hard != pre.hard && job[i].money > pre.money) { + pre = job[i]; + map.put(pre.hard, pre.money); + } + } + int[] ans = new int[ability.length]; + for (int i = 0; i < ability.length; i++) { + // ability[i] 当前人的能力 <= ability[i] 且离它最近的 + Integer key = map.floorKey(ability[i]); + ans[i] = key != null ? map.get(key) : 0; + } + return ans; + } + +} diff --git a/大厂刷题班/class02/Code02_Cola.java b/大厂刷题班/class02/Code02_Cola.java new file mode 100644 index 0000000..c41d0fb --- /dev/null +++ b/大厂刷题班/class02/Code02_Cola.java @@ -0,0 +1,151 @@ +package class02; + +public class Code02_Cola { + /* + * 买饮料 时间限制： 3000MS 内存限制： 589824KB 题目描述： + * 游游今年就要毕业了，和同学们在携程上定制了日本毕业旅行。愉快的一天行程结束后大家回到了酒店房间，这时候同学们都很口渴， + * 石头剪刀布选出游游去楼下的自动贩卖机给大家买可乐。 贩卖机只支持硬币支付，且收退都只支持10 ，50，100 + * 三种面额。一次购买行为只能出一瓶可乐，且每次购买后总是找零最小枚数的硬币。（例如投入100圆，可乐30圆，则找零50圆一枚，10圆两枚） + * 游游需要购买的可乐数量是 m，其中手头拥有的 10,50,100 面额硬币的枚数分别是 a,b,c，可乐的价格是x(x是10的倍数)。 + * 如果游游优先使用大面额购买且钱是够的情况下,请计算出需要投入硬币次数？ 输入描述 依次输入， 需要可乐的数量为 m 10元的张数为 a 50元的张数为 b + * 100元的张树为 c 1瓶可乐的价格为 x 输出描述 输出当前金额下需要投入硬币的次数 + * 例如需要购买2瓶可乐，每瓶可乐250圆，手里有100圆3枚，50圆4枚，10圆1枚。 购买第1瓶投递100圆3枚，找50圆 购买第2瓶投递50圆5枚 + * 所以是总共需要操作8次金额投递操作 样例输入 2 1 4 3 250 样例输出 8 + */ + + // 暴力尝试，为了验证正式方法而已 + public static int right(int m, int a, int b, int c, int x) { + int[] qian = { 100, 50, 10 }; + int[] zhang = { c, b, a }; + int puts = 0; + while (m != 0) { + int cur = buy(qian, zhang, x); + if (cur == -1) { + return -1; + } + puts += cur; + m--; + } + return puts; + } + + public static int buy(int[] qian, int[] zhang, int rest) { + int first = -1; + for (int i = 0; i < 3; i++) { + if (zhang[i] != 0) { + first = i; + break; + } + } + if (first == -1) { + return -1; + } + if (qian[first] >= rest) { + zhang[first]--; + giveRest(qian, zhang, first + 1, qian[first] - rest, 1); + return 1; + } else { + zhang[first]--; + int next = buy(qian, zhang, rest - qian[first]); + if (next == -1) { + return -1; + } + return 1 + next; + } + } + + // 正式的方法 + // 要买的可乐数量，m + // 100元有a张 + // 50元有b张 + // 10元有c张 + // 可乐单价x + public static int putTimes(int m, int a, int b, int c, int x) { + // 0 1 2 + int[] qian = { 100, 50, 10 }; + int[] zhang = { c, b, a }; + // 总共需要多少次投币 + int puts = 0; + // 之前面值的钱还剩下多少总钱数 + int preQianRest = 0; + // 之前面值的钱还剩下多少总张数 + int preQianZhang = 0; + for (int i = 0; i < 3 && m != 0; i++) { + // 要用之前剩下的钱、当前面值的钱，共同买第一瓶可乐 + // 之前的面值剩下多少钱，是preQianRest + // 之前的面值剩下多少张，是preQianZhang + // 之所以之前的面值会剩下来，一定是剩下的钱，一直攒不出一瓶可乐的单价 + // 当前的面值付出一些钱+之前剩下的钱，此时有可能凑出一瓶可乐来 + // 那么当前面值参与搞定第一瓶可乐，需要掏出多少张呢？就是curQianFirstBuyZhang + int curQianFirstBuyZhang = (x - preQianRest + qian[i] - 1) / qian[i]; + if (zhang[i] >= curQianFirstBuyZhang) { // 如果之前的钱和当前面值的钱，能凑出第一瓶可乐 + // 凑出来了一瓶可乐也可能存在找钱的情况， + giveRest(qian, zhang, i + 1, (preQianRest + qian[i] * curQianFirstBuyZhang) - x, 1); + puts += curQianFirstBuyZhang + preQianZhang; + zhang[i] -= curQianFirstBuyZhang; + m--; + } else { // 如果之前的钱和当前面值的钱，不能凑出第一瓶可乐 + preQianRest += qian[i] * zhang[i]; + preQianZhang += zhang[i]; + continue; + } + // 凑出第一瓶可乐之后，当前的面值有可能能继续买更多的可乐 + // 以下过程就是后续的可乐怎么用当前面值的钱来买 + // 用当前面值的钱，买一瓶可乐需要几张 + int curQianBuyOneColaZhang = (x + qian[i] - 1) / qian[i]; + // 用当前面值的钱，一共可以搞定几瓶可乐 + int curQianBuyColas = Math.min(zhang[i] / curQianBuyOneColaZhang, m); + // 用当前面值的钱，每搞定一瓶可乐，收货机会吐出多少零钱 + int oneTimeRest = qian[i] * curQianBuyOneColaZhang - x; + // 每次买一瓶可乐，吐出的找零总钱数是oneTimeRest + // 一共买的可乐数是curQianBuyColas，所以把零钱去提升后面几种面值的硬币数， + // 就是giveRest的含义 + giveRest(qian, zhang, i + 1, oneTimeRest, curQianBuyColas); + // 当前面值去搞定可乐这件事，一共投了几次币 + puts += curQianBuyOneColaZhang * curQianBuyColas; + // 还剩下多少瓶可乐需要去搞定，继续用后面的面值搞定去吧 + m -= curQianBuyColas; + // 当前面值可能剩下若干张，要参与到后续买可乐的过程中去， + // 所以要更新preQianRest和preQianZhang + zhang[i] -= curQianBuyOneColaZhang * curQianBuyColas; + preQianRest = qian[i] * zhang[i]; + preQianZhang = zhang[i]; + } + return m == 0 ? puts : -1; + } + + public static void giveRest(int[] qian, int[] zhang, int i, int oneTimeRest, int times) { + for (; i < 3; i++) { + zhang[i] += (oneTimeRest / qian[i]) * times; + oneTimeRest %= qian[i]; + } + } + + public static void main(String[] args) { + int testTime = 1000; + int zhangMax = 10; + int colaMax = 10; + int priceMax = 20; + System.out.println("如果错误会打印错误数据，否则就是正确"); + System.out.println("test begin"); + for (int i = 0; i < testTime; i++) { + int m = (int) (Math.random() * colaMax); + int a = (int) (Math.random() * zhangMax); + int b = (int) (Math.random() * zhangMax); + int c = (int) (Math.random() * zhangMax); + int x = ((int) (Math.random() * priceMax) + 1) * 10; + int ans1 = putTimes(m, a, b, c, x); + int ans2 = right(m, a, b, c, x); + if (ans1 != ans2) { + System.out.println("int m = " + m + ";"); + System.out.println("int a = " + a + ";"); + System.out.println("int b = " + b + ";"); + System.out.println("int c = " + c + ";"); + System.out.println("int x = " + x + ";"); + break; + } + } + System.out.println("test end"); + } + +} diff --git a/大厂刷题班/class02/Code03_ReceiveAndPrintOrderLine.java b/大厂刷题班/class02/Code03_ReceiveAndPrintOrderLine.java new file mode 100644 index 0000000..7db9ba3 --- /dev/null +++ b/大厂刷题班/class02/Code03_ReceiveAndPrintOrderLine.java @@ -0,0 +1,89 @@ +package class02; + +import java.util.HashMap; + +public class Code03_ReceiveAndPrintOrderLine { + + public static class Node { + public String info; + public Node next; + + public Node(String str) { + info = str; + } + } + + public static class MessageBox { + private HashMap headMap; + private HashMap tailMap; + private int waitPoint; + + public MessageBox() { + headMap = new HashMap(); + tailMap = new HashMap(); + waitPoint = 1; + } + + // 消息的编号，info消息的内容, 消息一定从1开始 + public void receive(int num, String info) { + if (num < 1) { + return; + } + Node cur = new Node(info); + // num~num + headMap.put(num, cur); + tailMap.put(num, cur); + // 建立了num~num这个连续区间的头和尾 + // 查询有没有某个连续区间以num-1结尾 + if (tailMap.containsKey(num - 1)) { + tailMap.get(num - 1).next = cur; + tailMap.remove(num - 1); + headMap.remove(num); + } + // 查询有没有某个连续区间以num+1开头的 + if (headMap.containsKey(num + 1)) { + cur.next = headMap.get(num + 1); + tailMap.remove(num); + headMap.remove(num + 1); + } + if (num == waitPoint) { + print(); + } + } + + private void print() { + Node node = headMap.get(waitPoint); + headMap.remove(waitPoint); + while (node != null) { + System.out.print(node.info + " "); + node = node.next; + waitPoint++; + } + tailMap.remove(waitPoint-1); + System.out.println(); + } + + } + + public static void main(String[] args) { + // MessageBox only receive 1~N + MessageBox box = new MessageBox(); + // 1.... + System.out.println("这是2来到的时候"); + box.receive(2,"B"); // - 2" + System.out.println("这是1来到的时候"); + box.receive(1,"A"); // 1 2 -> print, trigger is 1 + box.receive(4,"D"); // - 4 + box.receive(5,"E"); // - 4 5 + box.receive(7,"G"); // - 4 5 - 7 + box.receive(8,"H"); // - 4 5 - 7 8 + box.receive(6,"F"); // - 4 5 6 7 8 + box.receive(3,"C"); // 3 4 5 6 7 8 -> print, trigger is 3 + box.receive(9,"I"); // 9 -> print, trigger is 9 + box.receive(10,"J"); // 10 -> print, trigger is 10 + box.receive(12,"L"); // - 12 + box.receive(13,"M"); // - 12 13 + box.receive(11,"K"); // 11 12 13 -> print, trigger is 11 + + } +} diff --git a/大厂刷题班/class02/Code04_Drive.java b/大厂刷题班/class02/Code04_Drive.java new file mode 100644 index 0000000..a5a3a07 --- /dev/null +++ b/大厂刷题班/class02/Code04_Drive.java @@ -0,0 +1,116 @@ +package class02; + +import java.util.Arrays; + +public class Code04_Drive { + + // 课上的现场版本 + // income -> N * 2 的矩阵 N是偶数！ + // 0 [9, 13] + // 1 [45,60] + public static int maxMoney1(int[][] income) { + if (income == null || income.length < 2 || (income.length & 1) != 0) { + return 0; + } + int N = income.length; // 司机数量一定是偶数，所以才能平分，A N /2 B N/2 + int M = N >> 1; // M = N / 2 要去A区域的人 + return process1(income, 0, M); + } + + // index.....所有的司机，往A和B区域分配！ + // A区域还有rest个名额! + // 返回把index...司机，分配完，并且最终A和B区域同样多的情况下，index...这些司机，整体收入最大是多少！ + public static int process1(int[][] income, int index, int rest) { + if (index == income.length) { + return 0; + } + // 还剩下司机！ + if (income.length - index == rest) { + return income[index][0] + process1(income, index + 1, rest - 1); + } + if (rest == 0) { + return income[index][1] + process1(income, index + 1, rest); + } + // 当前司机，可以去A，或者去B + int p1 = income[index][0] + process1(income, index + 1, rest - 1); + int p2 = income[index][1] + process1(income, index + 1, rest); + return Math.max(p1, p2); + } + + // 严格位置依赖的动态规划版本 + public static int maxMoney2(int[][] income) { + int N = income.length; + int M = N >> 1; + int[][] dp = new int[N + 1][M + 1]; + for (int i = N - 1; i >= 0; i--) { + for (int j = 0; j <= M; j++) { + if (N - i == j) { + dp[i][j] = income[i][0] + dp[i + 1][j - 1]; + } else if (j == 0) { + dp[i][j] = income[i][1] + dp[i + 1][j]; + } else { + int p1 = income[i][0] + dp[i + 1][j - 1]; + int p2 = income[i][1] + dp[i + 1][j]; + dp[i][j] = Math.max(p1, p2); + } + } + } + return dp[0][M]; + } + + // 这题有贪心策略 : + // 假设一共有10个司机，思路是先让所有司机去A，得到一个总收益 + // 然后看看哪5个司机改换门庭(去B)，可以获得最大的额外收益 + // 这道题有贪心策略，打了我的脸 + // 但是我课上提到的技巧请大家重视 + // 根据数据量猜解法可以省去大量的多余分析，节省时间 + // 这里感谢卢圣文同学 + public static int maxMoney3(int[][] income) { + int N = income.length; + int[] arr = new int[N]; + int sum = 0; + for (int i = 0; i < N; i++) { + arr[i] = income[i][1] - income[i][0]; + sum += income[i][0]; + } + Arrays.sort(arr); + int M = N >> 1; + for (int i = N - 1; i >= M; i--) { + sum += arr[i]; + } + return sum; + } + + // 返回随机len*2大小的正数矩阵 + // 值在0~value-1之间 + public static int[][] randomMatrix(int len, int value) { + int[][] ans = new int[len << 1][2]; + for (int i = 0; i < ans.length; i++) { + ans[i][0] = (int) (Math.random() * value); + ans[i][1] = (int) (Math.random() * value); + } + return ans; + } + + public static void main(String[] args) { + int N = 10; + int value = 100; + int testTime = 500; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * N) + 1; + int[][] matrix = randomMatrix(len, value); + int ans1 = maxMoney1(matrix); + int ans2 = maxMoney2(matrix); + int ans3 = maxMoney3(matrix); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + System.out.println("Oops!"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class02/Code05_SetAll.java b/大厂刷题班/class02/Code05_SetAll.java new file mode 100644 index 0000000..329964a --- /dev/null +++ b/大厂刷题班/class02/Code05_SetAll.java @@ -0,0 +1,48 @@ +package class02; + +import java.util.HashMap; + +public class Code05_SetAll { + + public static class MyValue { + public V value; + public long time; + + public MyValue(V v, long t) { + value = v; + time = t; + } + } + + public static class MyHashMap { + private HashMap> map; + private long time; + private MyValue setAll; + + public MyHashMap() { + map = new HashMap<>(); + time = 0; + setAll = new MyValue(null, -1); + } + + public void put(K key, V value) { + map.put(key, new MyValue(value, time++)); + } + + public void setAll(V value) { + setAll = new MyValue(value, time++); + } + + public V get(K key) { + if (!map.containsKey(key)) { + return null; + } + if (map.get(key).time > setAll.time) { + return map.get(key).value; + } else { + return setAll.value; + } + } + } + +} diff --git a/大厂刷题班/class02/Code06_MinLengthForSort.java b/大厂刷题班/class02/Code06_MinLengthForSort.java new file mode 100644 index 0000000..5c3f74b --- /dev/null +++ b/大厂刷题班/class02/Code06_MinLengthForSort.java @@ -0,0 +1,30 @@ +package class02; + +// 本题测试链接 : https://leetcode.com/problems/shortest-unsorted-continuous-subarray/ +public class Code06_MinLengthForSort { + + public static int findUnsortedSubarray(int[] nums) { + if (nums == null || nums.length < 2) { + return 0; + } + int N = nums.length; + int right = -1; + int max = Integer.MIN_VALUE; + for (int i = 0; i < N; i++) { + if (max > nums[i]) { + right = i; + } + max = Math.max(max, nums[i]); + } + int min = Integer.MAX_VALUE; + int left = N; + for (int i = N - 1; i >= 0; i--) { + if (min < nums[i]) { + left = i; + } + min = Math.min(min, nums[i]); + } + return Math.max(0, right - left + 1); + } + +} diff --git a/大厂刷题班/class03/Code01_LongestSubstringWithoutRepeatingCharacters.java b/大厂刷题班/class03/Code01_LongestSubstringWithoutRepeatingCharacters.java new file mode 100644 index 0000000..120a441 --- /dev/null +++ b/大厂刷题班/class03/Code01_LongestSubstringWithoutRepeatingCharacters.java @@ -0,0 +1,27 @@ +package class03; + +// 本题测试链接 : https://leetcode.com/problems/longest-substring-without-repeating-characters/ +public class Code01_LongestSubstringWithoutRepeatingCharacters { + + public static int lengthOfLongestSubstring(String s) { + if (s == null || s.equals("")) { + return 0; + } + char[] str = s.toCharArray(); + int[] map = new int[256]; + for (int i = 0; i < 256; i++) { + map[i] = -1; + } + map[str[0]] = 0; + int N = str.length; + int ans = 1; + int pre = 1; + for (int i = 1; i < N; i++) { + pre = Math.min(i - map[str[i]], pre + 1); + ans = Math.max(ans, pre); + map[str[i]] = i; + } + return ans; + } + +} diff --git a/大厂刷题班/class03/Code02_HowManyTypes.java b/大厂刷题班/class03/Code02_HowManyTypes.java new file mode 100644 index 0000000..9d40789 --- /dev/null +++ b/大厂刷题班/class03/Code02_HowManyTypes.java @@ -0,0 +1,82 @@ +package class03; + +import java.util.HashSet; + +public class Code02_HowManyTypes { + + /* + * 只由小写字母（a~z）组成的一批字符串，都放在字符类型的数组String[] arr中， + * 如果其中某两个字符串，所含有的字符种类完全一样，就将两个字符串算作一类 比如：baacba和bac就算作一类 + * 虽然长度不一样，但是所含字符的种类完全一样（a、b、c） 返回arr中有多少类？ + * + */ + + public static int types1(String[] arr) { + HashSet types = new HashSet<>(); + for (String str : arr) { + char[] chs = str.toCharArray(); + boolean[] map = new boolean[26]; + for (int i = 0; i < chs.length; i++) { + map[chs[i] - 'a'] = true; + } + String key = ""; + for (int i = 0; i < 26; i++) { + if (map[i]) { + key += String.valueOf((char) (i + 'a')); + } + } + types.add(key); + } + return types.size(); + } + + public static int types2(String[] arr) { + HashSet types = new HashSet<>(); + for (String str : arr) { + char[] chs = str.toCharArray(); + int key = 0; + for(int i = 0 ; i < chs.length;i++) { + key |= (1 << (chs[i] - 'a')); + } + types.add(key); + } + return types.size(); + } + + // for test + public static String[] getRandomStringArray(int possibilities, int strMaxSize, int arrMaxSize) { + String[] ans = new String[(int) (Math.random() * arrMaxSize) + 1]; + for (int i = 0; i < ans.length; i++) { + ans[i] = getRandomString(possibilities, strMaxSize); + } + return ans; + } + + // for test + public static String getRandomString(int possibilities, int strMaxSize) { + char[] ans = new char[(int) (Math.random() * strMaxSize) + 1]; + for (int i = 0; i < ans.length; i++) { + ans[i] = (char) ((int) (Math.random() * possibilities) + 'a'); + } + return String.valueOf(ans); + } + + public static void main(String[] args) { + int possibilities = 5; + int strMaxSize = 10; + int arrMaxSize = 100; + int testTimes = 500000; + System.out.println("test begin, test time : " + testTimes); + for (int i = 0; i < testTimes; i++) { + String[] arr = getRandomStringArray(possibilities, strMaxSize, arrMaxSize); + int ans1 = types1(arr); + int ans2 = types2(arr); + if (ans1 != ans2) { + System.out.println("Oops!"); + } + } + System.out.println("test finish"); + + } + +} diff --git a/大厂刷题班/class03/Code03_Largest1BorderedSquare.java b/大厂刷题班/class03/Code03_Largest1BorderedSquare.java new file mode 100644 index 0000000..9d1f053 --- /dev/null +++ b/大厂刷题班/class03/Code03_Largest1BorderedSquare.java @@ -0,0 +1,59 @@ +package class03; + +// 本题测试链接 : https://leetcode.com/problems/largest-1-bordered-square/ +public class Code03_Largest1BorderedSquare { + + public static int largest1BorderedSquare(int[][] m) { + int[][] right = new int[m.length][m[0].length]; + int[][] down = new int[m.length][m[0].length]; + setBorderMap(m, right, down); + for (int size = Math.min(m.length, m[0].length); size != 0; size--) { + if (hasSizeOfBorder(size, right, down)) { + return size * size; + } + } + return 0; + } + + public static void setBorderMap(int[][] m, int[][] right, int[][] down) { + int r = m.length; + int c = m[0].length; + if (m[r - 1][c - 1] == 1) { + right[r - 1][c - 1] = 1; + down[r - 1][c - 1] = 1; + } + for (int i = r - 2; i != -1; i--) { + if (m[i][c - 1] == 1) { + right[i][c - 1] = 1; + down[i][c - 1] = down[i + 1][c - 1] + 1; + } + } + for (int i = c - 2; i != -1; i--) { + if (m[r - 1][i] == 1) { + right[r - 1][i] = right[r - 1][i + 1] + 1; + down[r - 1][i] = 1; + } + } + for (int i = r - 2; i != -1; i--) { + for (int j = c - 2; j != -1; j--) { + if (m[i][j] == 1) { + right[i][j] = right[i][j + 1] + 1; + down[i][j] = down[i + 1][j] + 1; + } + } + } + } + + public static boolean hasSizeOfBorder(int size, int[][] right, int[][] down) { + for (int i = 0; i != right.length - size + 1; i++) { + for (int j = 0; j != right[0].length - size + 1; j++) { + if (right[i][j] >= size && down[i][j] >= size && right[i + size - 1][j] >= size + && down[i][j + size - 1] >= size) { + return true; + } + } + } + return false; + } + +} diff --git a/大厂刷题班/class03/Code04_MaxPairNumber.java b/大厂刷题班/class03/Code04_MaxPairNumber.java new file mode 100644 index 0000000..a77909c --- /dev/null +++ b/大厂刷题班/class03/Code04_MaxPairNumber.java @@ -0,0 +1,126 @@ +package class03; + +import java.util.Arrays; + +// 给定一个数组arr，代表每个人的能力值。再给定一个非负数k。 +// 如果两个人能力差值正好为k，那么可以凑在一起比赛，一局比赛只有两个人 +// 返回最多可以同时有多少场比赛 +public class Code04_MaxPairNumber { + + // 暴力解 + public static int maxPairNum1(int[] arr, int k) { + if (k < 0) { + return -1; + } + return process1(arr, 0, k); + } + + public static int process1(int[] arr, int index, int k) { + int ans = 0; + if (index == arr.length) { + for (int i = 1; i < arr.length; i += 2) { + if (arr[i] - arr[i - 1] == k) { + ans++; + } + } + } else { + for (int r = index; r < arr.length; r++) { + swap(arr, index, r); + ans = Math.max(ans, process1(arr, index + 1, k)); + swap(arr, index, r); + } + } + return ans; + } + + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + + // 时间复杂度O(N*logN) + public static int maxPairNum2(int[] arr, int k) { + if (k < 0 || arr == null || arr.length < 2) { + return 0; + } + Arrays.sort(arr); + int ans = 0; + int N = arr.length; + int L = 0; + int R = 0; + boolean[] usedR = new boolean[N]; + while (L < N && R < N) { + if (usedR[L]) { + L++; + } else if (L >= R) { + R++; + } else { // 不止一个数，而且都没用过！ + int distance = arr[R] - arr[L]; + if (distance == k) { + ans++; + usedR[R++] = true; + L++; + } else if (distance < k) { + R++; + } else { + L++; + } + } + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int len, int value) { + int[] arr = new int[len]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) (Math.random() * value); + } + return arr; + } + + // 为了测试 + public static void printArray(int[] arr) { + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // 为了测试 + public static int[] copyArray(int[] arr) { + int[] ans = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + ans[i] = arr[i]; + } + return ans; + } + + public static void main(String[] args) { + int maxLen = 10; + int maxValue = 20; + int maxK = 5; + int testTime = 1000; + System.out.println("功能测试开始"); + for (int i = 0; i < testTime; i++) { + int N = (int) (Math.random() * (maxLen + 1)); + int[] arr = randomArray(N, maxValue); + int[] arr1 = copyArray(arr); + int[] arr2 = copyArray(arr); + int k = (int) (Math.random() * (maxK + 1)); + int ans1 = maxPairNum1(arr1, k); + int ans2 = maxPairNum2(arr2, k); + if (ans1 != ans2) { + System.out.println("Oops!"); + printArray(arr); + System.out.println(k); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("功能测试结束"); + } + +} diff --git a/大厂刷题班/class03/Code05_BoatsToSavePeople.java b/大厂刷题班/class03/Code05_BoatsToSavePeople.java new file mode 100644 index 0000000..d5cc68f --- /dev/null +++ b/大厂刷题班/class03/Code05_BoatsToSavePeople.java @@ -0,0 +1,74 @@ +package class03; + +import java.util.Arrays; + +// 给定一个正数数组arr，代表若干人的体重 +// 再给定一个正数limit，表示所有船共同拥有的载重量 +// 每艘船最多坐两人，且不能超过载重 +// 想让所有的人同时过河，并且用最好的分配方法让船尽量少 +// 返回最少的船数 +// 测试链接 : https://leetcode.com/problems/boats-to-save-people/ +public class Code05_BoatsToSavePeople { + + public static int numRescueBoats1(int[] arr, int limit) { + if (arr == null || arr.length == 0) { + return 0; + } + int N = arr.length; + Arrays.sort(arr); + if (arr[N - 1] > limit) { + return -1; + } + int lessR = -1; + for (int i = N - 1; i >= 0; i--) { + if (arr[i] <= (limit / 2)) { + lessR = i; + break; + } + } + if (lessR == -1) { + return N; + } + int L = lessR; + int R = lessR + 1; + int noUsed = 0; + while (L >= 0) { + int solved = 0; + while (R < N && arr[L] + arr[R] <= limit) { + R++; + solved++; + } + if (solved == 0) { + noUsed++; + L--; + } else { + L = Math.max(-1, L - solved); + } + } + int all = lessR + 1; + int used = all - noUsed; + int moreUnsolved = (N - all) - used; + return used + ((noUsed + 1) >> 1) + moreUnsolved; + } + + // 首尾双指针的解法 + public static int numRescueBoats2(int[] people, int limit) { + Arrays.sort(people); + int ans = 0; + int l = 0; + int r = people.length - 1; + int sum = 0; + while (l <= r) { + sum = l == r ? people[l] : people[l] + people[r]; + if (sum > limit) { + r--; + } else { + l++; + r--; + } + ans++; + } + return ans; + } + +} diff --git a/大厂刷题班/class03/Code06_ClosestSubsequenceSum.java b/大厂刷题班/class03/Code06_ClosestSubsequenceSum.java new file mode 100644 index 0000000..1cb2c0b --- /dev/null +++ b/大厂刷题班/class03/Code06_ClosestSubsequenceSum.java @@ -0,0 +1,46 @@ +package class03; + +import java.util.Arrays; + +// 本题测试链接 : https://leetcode.com/problems/closest-subsequence-sum/ +// 本题数据量描述: +// 1 <= nums.length <= 40 +// -10^7 <= nums[i] <= 10^7 +// -10^9 <= goal <= 10^9 +// 通过这个数据量描述可知，需要用到分治，因为数组长度不大 +// 而值很大，用动态规划的话，表会爆 +public class Code06_ClosestSubsequenceSum { + + public static int[] l = new int[1 << 20]; + public static int[] r = new int[1 << 20]; + + public static int minAbsDifference(int[] nums, int goal) { + if (nums == null || nums.length == 0) { + return goal; + } + int le = process(nums, 0, nums.length >> 1, 0, 0, l); + int re = process(nums, nums.length >> 1, nums.length, 0, 0, r); + Arrays.sort(l, 0, le); + Arrays.sort(r, 0, re--); + int ans = Math.abs(goal); + for (int i = 0; i < le; i++) { + int rest = goal - l[i]; + while (re > 0 && Math.abs(rest - r[re - 1]) <= Math.abs(rest - r[re])) { + re--; + } + ans = Math.min(ans, Math.abs(rest - r[re])); + } + return ans; + } + + public static int process(int[] nums, int index, int end, int sum, int fill, int[] arr) { + if (index == end) { + arr[fill++] = sum; + } else { + fill = process(nums, index + 1, end, sum, fill, arr); + fill = process(nums, index + 1, end, sum + nums[index], fill, arr); + } + return fill; + } + +} diff --git a/大厂刷题班/class03/Code07_FreedomTrail.java b/大厂刷题班/class03/Code07_FreedomTrail.java new file mode 100644 index 0000000..a3ef36d --- /dev/null +++ b/大厂刷题班/class03/Code07_FreedomTrail.java @@ -0,0 +1,62 @@ +package class03; + +import java.util.ArrayList; +import java.util.HashMap; + +// 本题测试链接 : https://leetcode.com/problems/freedom-trail/ +public class Code07_FreedomTrail { + + public static int findRotateSteps(String r, String k) { + char[] ring = r.toCharArray(); + int N = ring.length; + HashMap> map = new HashMap<>(); + for (int i = 0; i < N; i++) { + if (!map.containsKey(ring[i])) { + map.put(ring[i], new ArrayList<>()); + } + map.get(ring[i]).add(i); + } + char[] str = k.toCharArray(); + int M = str.length; + int[][] dp = new int[N][M + 1]; + // hashmap + // dp[][] == -1 : 表示没算过！ + for (int i = 0; i < N; i++) { + for (int j = 0; j <= M; j++) { + dp[i][j] = -1; + } + } + return process(0, 0, str, map, N, dp); + } + + // 电话里：指针指着的上一个按键preButton + // 目标里：此时要搞定哪个字符？keyIndex + // map : key 一种字符 value: 哪些位置拥有这个字符 + // N: 电话大小 + // f(0, 0, aim, map, N) + public static int process(int preButton, int index, char[] str, HashMap> map, int N, + int[][] dp) { + if (dp[preButton][index] != -1) { + return dp[preButton][index]; + } + int ans = Integer.MAX_VALUE; + if (index == str.length) { + ans = 0; + } else { + // 还有字符需要搞定呢！ + char cur = str[index]; + ArrayList nextPositions = map.get(cur); + for (int next : nextPositions) { + int cost = dial(preButton, next, N) + 1 + process(next, index + 1, str, map, N, dp); + ans = Math.min(ans, cost); + } + } + dp[preButton][index] = ans; + return ans; + } + + public static int dial(int i1, int i2, int size) { + return Math.min(Math.abs(i1 - i2), Math.min(i1, i2) + size - Math.max(i1, i2)); + } + +} diff --git a/大厂刷题班/class03/Code08_DistanceKNodes.java b/大厂刷题班/class03/Code08_DistanceKNodes.java new file mode 100644 index 0000000..30de2e8 --- /dev/null +++ b/大厂刷题班/class03/Code08_DistanceKNodes.java @@ -0,0 +1,105 @@ +package class03; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.HashSet; +import java.util.LinkedList; +import java.util.List; +import java.util.Queue; + +public class Code08_DistanceKNodes { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int v) { + value = v; + } + } + + public static List distanceKNodes(Node root, Node target, int K) { + HashMap parents = new HashMap<>(); + parents.put(root, null); + createParentMap(root, parents); + Queue queue = new LinkedList<>(); + HashSet visited = new HashSet<>(); + queue.offer(target); + visited.add(target); + int curLevel = 0; + List ans = new ArrayList<>(); + while (!queue.isEmpty()) { + int size = queue.size(); + while (size-- > 0) { + Node cur = queue.poll(); + if (curLevel == K) { + ans.add(cur); + } + if (cur.left != null && !visited.contains(cur.left)) { + visited.add(cur.left); + queue.offer(cur.left); + } + if (cur.right != null && !visited.contains(cur.right)) { + visited.add(cur.right); + queue.offer(cur.right); + } + if (parents.get(cur) != null && !visited.contains(parents.get(cur))) { + visited.add(parents.get(cur)); + queue.offer(parents.get(cur)); + } + } + curLevel++; + if (curLevel > K) { + break; + } + } + return ans; + } + + public static void createParentMap(Node cur, HashMap parents) { + if (cur == null) { + return; + } + if (cur.left != null) { + parents.put(cur.left, cur); + createParentMap(cur.left, parents); + } + if (cur.right != null) { + parents.put(cur.right, cur); + createParentMap(cur.right, parents); + } + } + + public static void main(String[] args) { + Node n0 = new Node(0); + Node n1 = new Node(1); + Node n2 = new Node(2); + Node n3 = new Node(3); + Node n4 = new Node(4); + Node n5 = new Node(5); + Node n6 = new Node(6); + Node n7 = new Node(7); + Node n8 = new Node(8); + + n3.left = n5; + n3.right = n1; + n5.left = n6; + n5.right = n2; + n1.left = n0; + n1.right = n8; + n2.left = n7; + n2.right = n4; + + Node root = n3; + Node target = n5; + int K = 2; + + List ans = distanceKNodes(root, target, K); + for (Node o1 : ans) { + System.out.println(o1.value); + } + + } + +} diff --git a/大厂刷题班/class04/Code01_QueryHobby.java b/大厂刷题班/class04/Code01_QueryHobby.java new file mode 100644 index 0000000..20babf4 --- /dev/null +++ b/大厂刷题班/class04/Code01_QueryHobby.java @@ -0,0 +1,115 @@ +package class04; + +import java.util.ArrayList; +import java.util.HashMap; + +public class Code01_QueryHobby { + + /* + * 今日头条原题 + * + * 数组为{3, 2, 2, 3, 1}，查询为(0, 3, 2)。意思是在数组里下标0~3这个范围上，有几个2？返回2。 + * 假设给你一个数组arr，对这个数组的查询非常频繁，请返回所有查询的结果 + * + */ + + public static class QueryBox1 { + private int[] arr; + + public QueryBox1(int[] array) { + arr = new int[array.length]; + for (int i = 0; i < arr.length; i++) { + arr[i] = array[i]; + } + } + + public int query(int L, int R, int v) { + int ans = 0; + for (; L <= R; L++) { + if (arr[L] == v) { + ans++; + } + } + return ans; + } + } + + public static class QueryBox2 { + private HashMap> map; + + public QueryBox2(int[] arr) { + map = new HashMap<>(); + for (int i = 0; i < arr.length; i++) { + if (!map.containsKey(arr[i])) { + map.put(arr[i], new ArrayList<>()); + } + map.get(arr[i]).add(i); + } + } + + public int query(int L, int R, int value) { + if (!map.containsKey(value)) { + return 0; + } + ArrayList indexArr = map.get(value); + // 查询 < L 的下标有几个 + int a = countLess(indexArr, L); + // 查询 < R+1 的下标有几个 + int b = countLess(indexArr, R + 1); + return b - a; + } + + // 在有序数组arr中，用二分的方法数出 arr, int limit) { + int L = 0; + int R = arr.size() - 1; + int mostRight = -1; + while (L <= R) { + int mid = L + ((R - L) >> 1); + if (arr.get(mid) < limit) { + mostRight = mid; + L = mid + 1; + } else { + R = mid - 1; + } + } + return mostRight + 1; + } + + } + + public static int[] generateRandomArray(int len, int value) { + int[] ans = new int[(int) (Math.random() * len) + 1]; + for (int i = 0; i < ans.length; i++) { + ans[i] = (int) (Math.random() * value) + 1; + } + return ans; + } + + public static void main(String[] args) { + int len = 300; + int value = 20; + int testTimes = 1000; + int queryTimes = 1000; + System.out.println("test begin"); + for (int i = 0; i < testTimes; i++) { + int[] arr = generateRandomArray(len, value); + int N = arr.length; + QueryBox1 box1 = new QueryBox1(arr); + QueryBox2 box2 = new QueryBox2(arr); + for (int j = 0; j < queryTimes; j++) { + int a = (int) (Math.random() * N); + int b = (int) (Math.random() * N); + int L = Math.min(a, b); + int R = Math.max(a, b); + int v = (int) (Math.random() * value) + 1; + if (box1.query(L, R, v) != box2.query(L, R, v)) { + System.out.println("Oops!"); + } + } + } + System.out.println("test end"); + } + +} diff --git a/大厂刷题班/class04/Code02_SubArrayMaxSum.java b/大厂刷题班/class04/Code02_SubArrayMaxSum.java new file mode 100644 index 0000000..25477fa --- /dev/null +++ b/大厂刷题班/class04/Code02_SubArrayMaxSum.java @@ -0,0 +1,35 @@ +package class04; + +// 本题测试链接 : https://leetcode.com/problems/maximum-subarray/ +public class Code02_SubArrayMaxSum { + + public static int maxSubArray(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int max = Integer.MIN_VALUE; + int cur = 0; + for (int i = 0; i < arr.length; i++) { + cur += arr[i]; + max = Math.max(max, cur); + cur = cur < 0 ? 0 : cur; + } + return max; + } + + public static int maxSubArray2(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + // 上一步，dp的值 + // dp[0] + int pre = arr[0]; + int max = arr[0]; + for (int i = 1; i < arr.length; i++) { + pre = Math.max(arr[i], arr[i] + pre); + max = Math.max(max, pre); + } + return max; + } + +} diff --git a/大厂刷题班/class04/Code03_SubMatrixMaxSum.java b/大厂刷题班/class04/Code03_SubMatrixMaxSum.java new file mode 100644 index 0000000..2940774 --- /dev/null +++ b/大厂刷题班/class04/Code03_SubMatrixMaxSum.java @@ -0,0 +1,75 @@ +package class04; + +public class Code03_SubMatrixMaxSum { + + public static int maxSum(int[][] m) { + if (m == null || m.length == 0 || m[0].length == 0) { + return 0; + } + // O(N^2 * M) + int N = m.length; + int M = m[0].length; + int max = Integer.MIN_VALUE; + for (int i = 0; i < N; i++) { + // i~j + int[] s = new int[M]; + for (int j = i; j < N; j++) { + for (int k = 0; k < M; k++) { + s[k] += m[j][k]; + } + max = Math.max(max, maxSubArray(s)); + } + } + return max; + } + + public static int maxSubArray(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int max = Integer.MIN_VALUE; + int cur = 0; + for (int i = 0; i < arr.length; i++) { + cur += arr[i]; + max = Math.max(max, cur); + cur = cur < 0 ? 0 : cur; + } + return max; + } + + // 本题测试链接 : https://leetcode-cn.com/problems/max-submatrix-lcci/ + public static int[] getMaxMatrix(int[][] m) { + int N = m.length; + int M = m[0].length; + int max = Integer.MIN_VALUE; + int cur = 0; + int a = 0; + int b = 0; + int c = 0; + int d = 0; + for (int i = 0; i < N; i++) { + int[] s = new int[M]; + for (int j = i; j < N; j++) { + cur = 0; + int begin = 0; + for (int k = 0; k < M; k++) { + s[k] += m[j][k]; + cur += s[k]; + if (max < cur) { + max = cur; + a = i; + b = begin; + c = j; + d = k; + } + if (cur < 0) { + cur = 0; + begin = k + 1; + } + } + } + } + return new int[] { a, b, c, d }; + } + +} diff --git a/大厂刷题班/class04/Code04_SubArrayMaxSumFollowUp.java b/大厂刷题班/class04/Code04_SubArrayMaxSumFollowUp.java new file mode 100644 index 0000000..0d5cccd --- /dev/null +++ b/大厂刷题班/class04/Code04_SubArrayMaxSumFollowUp.java @@ -0,0 +1,59 @@ +package class04; + +// 在线测试链接 : https://leetcode.com/problems/house-robber/ +public class Code04_SubArrayMaxSumFollowUp { + + public static int rob1(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + if (arr.length == 1) { + return arr[0]; + } + int[] dp = new int[arr.length]; + // dp[i] : arr[0..i]挑选，满足不相邻设定的情况下，随意挑选，最大的累加和 + dp[0] = arr[0]; + dp[1] = Math.max(arr[0], arr[1]); + for (int i = 2; i < arr.length; i++) { + int p1 = dp[i - 1]; + int p2 = arr[i] + Math.max(dp[i - 2], 0); + dp[i] = Math.max(p1, p2); + } + return dp[arr.length - 1]; + } + + // 给定一个数组arr，在不能取相邻数的情况下，返回所有组合中的最大累加和 + // 思路： + // 定义dp[i] : 表示arr[0...i]范围上，在不能取相邻数的情况下，返回所有组合中的最大累加和 + // 在arr[0...i]范围上，在不能取相邻数的情况下，得到的最大累加和，可能性分类： + // 可能性 1) 选出的组合，不包含arr[i]。那么dp[i] = dp[i-1] + // 比如，arr[0...i] = {3,4,-4}，最大累加和是不包含i位置数的时候 + // + // 可能性 2) 选出的组合，只包含arr[i]。那么dp[i] = arr[i] + // 比如，arr[0...i] = {-3,-4,4}，最大累加和是只包含i位置数的时候 + // + // 可能性 3) 选出的组合，包含arr[i], 且包含arr[0...i-2]范围上的累加和。那么dp[i] = arr[i] + dp[i-2] + // 比如，arr[0...i] = {3,1,4}，最大累加和是3和4组成的7，因为相邻不能选，所以i-1位置的数要跳过 + // + // 综上所述：dp[i] = Max { dp[i-1], arr[i] , arr[i] + dp[i-2] } + public static int rob2(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int N = arr.length; + if (N == 1) { + return arr[0]; + } + if (N == 2) { + return Math.max(arr[0], arr[1]); + } + int[] dp = new int[N]; + dp[0] = arr[0]; + dp[1] = Math.max(arr[0], arr[1]); + for (int i = 2; i < N; i++) { + dp[i] = Math.max(Math.max(dp[i - 1], arr[i]), arr[i] + dp[i - 2]); + } + return dp[N - 1]; + } + +} diff --git a/大厂刷题班/class04/Code05_CandyProblem.java b/大厂刷题班/class04/Code05_CandyProblem.java new file mode 100644 index 0000000..694aab2 --- /dev/null +++ b/大厂刷题班/class04/Code05_CandyProblem.java @@ -0,0 +1,145 @@ +package class04; + +// 测试链接 : https://leetcode.com/problems/candy/ +public class Code05_CandyProblem { + + // 这是原问题的优良解 + // 时间复杂度O(N)，额外空间复杂度O(N) + public static int candy1(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int N = arr.length; + int[] left = new int[N]; + for (int i = 1; i < N; i++) { + if (arr[i - 1] < arr[i]) { + left[i] = left[i - 1] + 1; + } + } + int[] right = new int[N]; + for (int i = N - 2; i >= 0; i--) { + if (arr[i] > arr[i + 1]) { + right[i] = right[i + 1] + 1; + } + } + int ans = 0; + for (int i = 0; i < N; i++) { + ans += Math.max(left[i], right[i]); + } + return ans + N; + } + + // 这是原问题空间优化后的解 + // 时间复杂度O(N)，额外空间复杂度O(1) + public static int candy2(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int index = nextMinIndex2(arr, 0); + int res = rightCands(arr, 0, index++); + int lbase = 1; + int next = 0; + int rcands = 0; + int rbase = 0; + while (index != arr.length) { + if (arr[index] > arr[index - 1]) { + res += ++lbase; + index++; + } else if (arr[index] < arr[index - 1]) { + next = nextMinIndex2(arr, index - 1); + rcands = rightCands(arr, index - 1, next++); + rbase = next - index + 1; + res += rcands + (rbase > lbase ? -lbase : -rbase); + lbase = 1; + index = next; + } else { + res += 1; + lbase = 1; + index++; + } + } + return res; + } + + public static int nextMinIndex2(int[] arr, int start) { + for (int i = start; i != arr.length - 1; i++) { + if (arr[i] <= arr[i + 1]) { + return i; + } + } + return arr.length - 1; + } + + public static int rightCands(int[] arr, int left, int right) { + int n = right - left + 1; + return n + n * (n - 1) / 2; + } + + // 这是进阶问题的最优解，不要提交这个 + // 时间复杂度O(N), 额外空间复杂度O(1) + public static int candy3(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int index = nextMinIndex3(arr, 0); + int[] data = rightCandsAndBase(arr, 0, index++); + int res = data[0]; + int lbase = 1; + int same = 1; + int next = 0; + while (index != arr.length) { + if (arr[index] > arr[index - 1]) { + res += ++lbase; + same = 1; + index++; + } else if (arr[index] < arr[index - 1]) { + next = nextMinIndex3(arr, index - 1); + data = rightCandsAndBase(arr, index - 1, next++); + if (data[1] <= lbase) { + res += data[0] - data[1]; + } else { + res += -lbase * same + data[0] - data[1] + data[1] * same; + } + index = next; + lbase = 1; + same = 1; + } else { + res += lbase; + same++; + index++; + } + } + return res; + } + + public static int nextMinIndex3(int[] arr, int start) { + for (int i = start; i != arr.length - 1; i++) { + if (arr[i] < arr[i + 1]) { + return i; + } + } + return arr.length - 1; + } + + public static int[] rightCandsAndBase(int[] arr, int left, int right) { + int base = 1; + int cands = 1; + for (int i = right - 1; i >= left; i--) { + if (arr[i] == arr[i + 1]) { + cands += base; + } else { + cands += ++base; + } + } + return new int[] { cands, base }; + } + + public static void main(String[] args) { + int[] test1 = { 3, 0, 5, 5, 4, 4, 0 }; + System.out.println(candy2(test1)); + + int[] test2 = { 3, 0, 5, 5, 4, 4, 0 }; + System.out.println(candy3(test2)); + } + +} diff --git a/大厂刷题班/class04/Code06_MakeNo.java b/大厂刷题班/class04/Code06_MakeNo.java new file mode 100644 index 0000000..2a4af46 --- /dev/null +++ b/大厂刷题班/class04/Code06_MakeNo.java @@ -0,0 +1,59 @@ +package class04; + +public class Code06_MakeNo { + + // 生成长度为size的达标数组 + // 达标：对于任意的 i 等长奇数达标来 + // base -> 等长偶数达标来 + int[] ans = new int[size]; + int index = 0; + for (; index < halfSize; index++) { + ans[index] = base[index] * 2 - 1; + } + for (int i = 0; index < size; index++, i++) { + ans[index] = base[i] * 2; + } + return ans; + } + + // 检验函数 + public static boolean isValid(int[] arr) { + int N = arr.length; + for (int i = 0; i < N; i++) { + for (int k = i + 1; k < N; k++) { + for (int j = k + 1; j < N; j++) { + if (arr[i] + arr[j] == 2 * arr[k]) { + return false; + } + } + } + } + return true; + } + + public static void main(String[] args) { + System.out.println("test begin"); + for (int N = 1; N < 1000; N++) { + int[] arr = makeNo(N); + if (!isValid(arr)) { + System.out.println("Oops!"); + } + } + System.out.println("test end"); + System.out.println(isValid(makeNo(1042))); + System.out.println(isValid(makeNo(2981))); + } + +} diff --git a/大厂刷题班/class04/Code07_InterleavingString.java b/大厂刷题班/class04/Code07_InterleavingString.java new file mode 100644 index 0000000..f53e113 --- /dev/null +++ b/大厂刷题班/class04/Code07_InterleavingString.java @@ -0,0 +1,53 @@ +package class04; + +// 本题测试链接 : https://leetcode.com/problems/interleaving-string/ +public class Code07_InterleavingString { + + public static boolean isInterleave(String s1, String s2, String s3) { + if (s1 == null || s2 == null || s3 == null) { + return false; + } + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + char[] str3 = s3.toCharArray(); + if (str3.length != str1.length + str2.length) { + return false; + } + boolean[][] dp = new boolean[str1.length + 1][str2.length + 1]; + dp[0][0] = true; + for (int i = 1; i <= str1.length; i++) { + if (str1[i - 1] != str3[i - 1]) { + break; + } + dp[i][0] = true; + } + for (int j = 1; j <= str2.length; j++) { + if (str2[j - 1] != str3[j - 1]) { + break; + } + dp[0][j] = true; + } + for (int i = 1; i <= str1.length; i++) { + for (int j = 1; j <= str2.length; j++) { + if ( + (str1[i - 1] == str3[i + j - 1] && dp[i - 1][j]) + + + + + || + + + + (str2[j - 1] == str3[i + j - 1] && dp[i][j - 1]) + + + ) { + dp[i][j] = true; + } + } + } + return dp[str1.length][str2.length]; + } + +} diff --git a/大厂刷题班/class04/Code08_TheSkylineProblem.java b/大厂刷题班/class04/Code08_TheSkylineProblem.java new file mode 100644 index 0000000..29f3ccd --- /dev/null +++ b/大厂刷题班/class04/Code08_TheSkylineProblem.java @@ -0,0 +1,73 @@ +package class04; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.Comparator; +import java.util.List; +import java.util.Map.Entry; +import java.util.TreeMap; + +// 本题测试链接 : https://leetcode.com/problems/the-skyline-problem/ +public class Code08_TheSkylineProblem { + + public static class Node { + public int x; + public boolean isAdd; + public int h; + + public Node(int x, boolean isAdd, int h) { + this.x = x; + this.isAdd = isAdd; + this.h = h; + } + } + + public static class NodeComparator implements Comparator { + @Override + public int compare(Node o1, Node o2) { + return o1.x - o2.x; + } + } + + public static List> getSkyline(int[][] matrix) { + Node[] nodes = new Node[matrix.length * 2]; + for (int i = 0; i < matrix.length; i++) { + nodes[i * 2] = new Node(matrix[i][0], true, matrix[i][2]); + nodes[i * 2 + 1] = new Node(matrix[i][1], false, matrix[i][2]); + } + Arrays.sort(nodes, new NodeComparator()); + // key 高度 value 次数 + TreeMap mapHeightTimes = new TreeMap<>(); + TreeMap mapXHeight = new TreeMap<>(); + for (int i = 0; i < nodes.length; i++) { + if (nodes[i].isAdd) { + if (!mapHeightTimes.containsKey(nodes[i].h)) { + mapHeightTimes.put(nodes[i].h, 1); + } else { + mapHeightTimes.put(nodes[i].h, mapHeightTimes.get(nodes[i].h) + 1); + } + } else { + if (mapHeightTimes.get(nodes[i].h) == 1) { + mapHeightTimes.remove(nodes[i].h); + } else { + mapHeightTimes.put(nodes[i].h, mapHeightTimes.get(nodes[i].h) - 1); + } + } + if (mapHeightTimes.isEmpty()) { + mapXHeight.put(nodes[i].x, 0); + } else { + mapXHeight.put(nodes[i].x, mapHeightTimes.lastKey()); + } + } + List> ans = new ArrayList<>(); + for (Entry entry : mapXHeight.entrySet()) { + int curX = entry.getKey(); + int curMaxHeight = entry.getValue(); + if (ans.isEmpty() || ans.get(ans.size() - 1).get(1) != curMaxHeight) { + ans.add(new ArrayList<>(Arrays.asList(curX, curMaxHeight))); + } + } + return ans; + } + +} diff --git a/大厂刷题班/class05/Code01_ConstructBinarySearchTreeFromPreorderTraversal.java b/大厂刷题班/class05/Code01_ConstructBinarySearchTreeFromPreorderTraversal.java new file mode 100644 index 0000000..3eaf865 --- /dev/null +++ b/大厂刷题班/class05/Code01_ConstructBinarySearchTreeFromPreorderTraversal.java @@ -0,0 +1,115 @@ +package class05; + +import java.util.Stack; + +// 本题测试链接 : https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal/ +public class Code01_ConstructBinarySearchTreeFromPreorderTraversal { + + // 不用提交这个类 + public static class TreeNode { + public int val; + public TreeNode left; + public TreeNode right; + + public TreeNode() { + } + + public TreeNode(int val) { + this.val = val; + } + + public TreeNode(int val, TreeNode left, TreeNode right) { + this.val = val; + this.left = left; + this.right = right; + } + } + + // 提交下面的方法 + public static TreeNode bstFromPreorder1(int[] pre) { + if (pre == null || pre.length == 0) { + return null; + } + return process1(pre, 0, pre.length - 1); + } + + public static TreeNode process1(int[] pre, int L, int R) { + if (L > R) { + return null; + } + int firstBig = L + 1; + for (; firstBig <= R; firstBig++) { + if (pre[firstBig] > pre[L]) { + break; + } + } + TreeNode head = new TreeNode(pre[L]); + head.left = process1(pre, L + 1, firstBig - 1); + head.right = process1(pre, firstBig, R); + return head; + } + + // 已经是时间复杂度最优的方法了，但是常数项还能优化 + public static TreeNode bstFromPreorder2(int[] pre) { + if (pre == null || pre.length == 0) { + return null; + } + int N = pre.length; + int[] nearBig = new int[N]; + for (int i = 0; i < N; i++) { + nearBig[i] = -1; + } + Stack stack = new Stack<>(); + for (int i = 0; i < N; i++) { + while (!stack.isEmpty() && pre[stack.peek()] < pre[i]) { + nearBig[stack.pop()] = i; + } + stack.push(i); + } + return process2(pre, 0, N - 1, nearBig); + } + + public static TreeNode process2(int[] pre, int L, int R, int[] nearBig) { + if (L > R) { + return null; + } + int firstBig = (nearBig[L] == -1 || nearBig[L] > R) ? R + 1 : nearBig[L]; + TreeNode head = new TreeNode(pre[L]); + head.left = process2(pre, L + 1, firstBig - 1, nearBig); + head.right = process2(pre, firstBig, R, nearBig); + return head; + } + + // 最优解 + public static TreeNode bstFromPreorder3(int[] pre) { + if (pre == null || pre.length == 0) { + return null; + } + int N = pre.length; + int[] nearBig = new int[N]; + for (int i = 0; i < N; i++) { + nearBig[i] = -1; + } + int[] stack = new int[N]; + int size = 0; + for (int i = 0; i < N; i++) { + while (size != 0 && pre[stack[size - 1]] < pre[i]) { + nearBig[stack[--size]] = i; + } + stack[size++] = i; + } + return process3(pre, 0, N - 1, nearBig); + } + + public static TreeNode process3(int[] pre, int L, int R, int[] nearBig) { + if (L > R) { + return null; + } + int firstBig = (nearBig[L] == -1 || nearBig[L] > R) ? R + 1 : nearBig[L]; + TreeNode head = new TreeNode(pre[L]); + head.left = process3(pre, L + 1, firstBig - 1, nearBig); + head.right = process3(pre, firstBig, R, nearBig); + return head; + } + +} diff --git a/大厂刷题班/class05/Code02_LeftRightSameTreeNumber.java b/大厂刷题班/class05/Code02_LeftRightSameTreeNumber.java new file mode 100644 index 0000000..7b53125 --- /dev/null +++ b/大厂刷题班/class05/Code02_LeftRightSameTreeNumber.java @@ -0,0 +1,95 @@ +package class05; + +// 如果一个节点X，它左树结构和右树结构完全一样，那么我们说以X为头的子树是相等子树 +// 给定一棵二叉树的头节点head，返回head整棵树上有多少棵相等子树 +public class Code02_LeftRightSameTreeNumber { + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int v) { + value = v; + } + } + + // 时间复杂度O(N * logN) + public static int sameNumber1(Node head) { + if (head == null) { + return 0; + } + return sameNumber1(head.left) + sameNumber1(head.right) + (same(head.left, head.right) ? 1 : 0); + } + + public static boolean same(Node h1, Node h2) { + if (h1 == null ^ h2 == null) { + return false; + } + if (h1 == null && h2 == null) { + return true; + } + // 两个都不为空 + return h1.value == h2.value && same(h1.left, h2.left) && same(h1.right, h2.right); + } + + // 时间复杂度O(N) + public static int sameNumber2(Node head) { + String algorithm = "SHA-256"; + Hash hash = new Hash(algorithm); + return process(head, hash).ans; + } + + public static class Info { + public int ans; + public String str; + + public Info(int a, String s) { + ans = a; + str = s; + } + } + + public static Info process(Node head, Hash hash) { + if (head == null) { + return new Info(0, hash.hashCode("#,")); + } + Info l = process(head.left, hash); + Info r = process(head.right, hash); + int ans = (l.str.equals(r.str) ? 1 : 0) + l.ans + r.ans; + String str = hash.hashCode(String.valueOf(head.value) + "," + l.str + r.str); + return new Info(ans, str); + } + + public static Node randomBinaryTree(int restLevel, int maxValue) { + if (restLevel == 0) { + return null; + } + Node head = Math.random() < 0.2 ? null : new Node((int) (Math.random() * maxValue)); + if (head != null) { + head.left = randomBinaryTree(restLevel - 1, maxValue); + head.right = randomBinaryTree(restLevel - 1, maxValue); + } + return head; + } + + public static void main(String[] args) { + int maxLevel = 8; + int maxValue = 4; + int testTime = 100000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + Node head = randomBinaryTree(maxLevel, maxValue); + int ans1 = sameNumber1(head); + int ans2 = sameNumber2(head); + if (ans1 != ans2) { + System.out.println("出错了！"); + System.out.println(ans1); + System.out.println(ans2); + } + } + System.out.println("测试结束"); + + } + +} diff --git a/大厂刷题班/class05/Code03_EditCost.java b/大厂刷题班/class05/Code03_EditCost.java new file mode 100644 index 0000000..0079c51 --- /dev/null +++ b/大厂刷题班/class05/Code03_EditCost.java @@ -0,0 +1,89 @@ +package class05; + +public class Code03_EditCost { + + public static int minCost1(String s1, String s2, int ic, int dc, int rc) { + if (s1 == null || s2 == null) { + return 0; + } + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + int N = str1.length + 1; + int M = str2.length + 1; + int[][] dp = new int[N][M]; + // dp[0][0] = 0 + for (int i = 1; i < N; i++) { + dp[i][0] = dc * i; + } + for (int j = 1; j < M; j++) { + dp[0][j] = ic * j; + } + for (int i = 1; i < N; i++) { + for (int j = 1; j < M; j++) { + dp[i][j] = dp[i - 1][j - 1] + (str1[i - 1] == str2[j - 1] ? 0 : rc); + dp[i][j] = Math.min(dp[i][j], dp[i][j - 1] + ic); + dp[i][j] = Math.min(dp[i][j], dp[i - 1][j] + dc); + } + } + return dp[N - 1][M - 1]; + } + + public static int minCost2(String str1, String str2, int ic, int dc, int rc) { + if (str1 == null || str2 == null) { + return 0; + } + char[] chs1 = str1.toCharArray(); + char[] chs2 = str2.toCharArray(); + char[] longs = chs1.length >= chs2.length ? chs1 : chs2; + char[] shorts = chs1.length < chs2.length ? chs1 : chs2; + if (chs1.length < chs2.length) { + int tmp = ic; + ic = dc; + dc = tmp; + } + int[] dp = new int[shorts.length + 1]; + for (int i = 1; i <= shorts.length; i++) { + dp[i] = ic * i; + } + for (int i = 1; i <= longs.length; i++) { + int pre = dp[0]; + dp[0] = dc * i; + for (int j = 1; j <= shorts.length; j++) { + int tmp = dp[j]; + if (longs[i - 1] == shorts[j - 1]) { + dp[j] = pre; + } else { + dp[j] = pre + rc; + } + dp[j] = Math.min(dp[j], dp[j - 1] + ic); + dp[j] = Math.min(dp[j], tmp + dc); + pre = tmp; + } + } + return dp[shorts.length]; + } + + public static void main(String[] args) { + String str1 = "ab12cd3"; + String str2 = "abcdf"; + System.out.println(minCost1(str1, str2, 5, 3, 2)); + System.out.println(minCost2(str1, str2, 5, 3, 2)); + + str1 = "abcdf"; + str2 = "ab12cd3"; + System.out.println(minCost1(str1, str2, 3, 2, 4)); + System.out.println(minCost2(str1, str2, 3, 2, 4)); + + str1 = ""; + str2 = "ab12cd3"; + System.out.println(minCost1(str1, str2, 1, 7, 5)); + System.out.println(minCost2(str1, str2, 1, 7, 5)); + + str1 = "abcdf"; + str2 = ""; + System.out.println(minCost1(str1, str2, 2, 9, 8)); + System.out.println(minCost2(str1, str2, 2, 9, 8)); + + } + +} diff --git a/大厂刷题班/class05/Code04_DeleteMinCost.java b/大厂刷题班/class05/Code04_DeleteMinCost.java new file mode 100644 index 0000000..7aebee3 --- /dev/null +++ b/大厂刷题班/class05/Code04_DeleteMinCost.java @@ -0,0 +1,258 @@ +package class05; + +import java.util.ArrayList; +import java.util.Comparator; +import java.util.HashMap; +import java.util.List; + +public class Code04_DeleteMinCost { + + // 题目： + // 给定两个字符串s1和s2，问s2最少删除多少字符可以成为s1的子串？ + // 比如 s1 = "abcde"，s2 = "axbc" + // 返回 1 + + // 解法一 + // 求出str2所有的子序列，然后按照长度排序，长度大的排在前面。 + // 然后考察哪个子序列字符串和s1的某个子串相等(KMP)，答案就出来了。 + // 分析： + // 因为题目原本的样本数据中，有特别说明s2的长度很小。所以这么做也没有太大问题，也几乎不会超时。 + // 但是如果某一次考试给定的s2长度远大于s1，这么做就不合适了。 + public static int minCost1(String s1, String s2) { + List s2Subs = new ArrayList<>(); + process(s2.toCharArray(), 0, "", s2Subs); + s2Subs.sort(new LenComp()); + for (String str : s2Subs) { + if (s1.indexOf(str) != -1) { // indexOf底层和KMP算法代价几乎一样，也可以用KMP代替 + return s2.length() - str.length(); + } + } + return s2.length(); + } + + public static void process(char[] str2, int index, String path, List list) { + if (index == str2.length) { + list.add(path); + return; + } + process(str2, index + 1, path, list); + process(str2, index + 1, path + str2[index], list); + } + + // x字符串只通过删除的方式，变到y字符串 + // 返回至少要删几个字符 + // 如果变不成，返回Integer.Max + public static int onlyDelete(char[] x, char[] y) { + if (x.length < y.length) { + return Integer.MAX_VALUE; + } + int N = x.length; + int M = y.length; + int[][] dp = new int[N + 1][M + 1]; + for (int i = 0; i <= N; i++) { + for (int j = 0; j <= M; j++) { + dp[i][j] = Integer.MAX_VALUE; + } + } + dp[0][0] = 0; + // dp[i][j]表示前缀长度 + for (int i = 1; i <= N; i++) { + dp[i][0] = i; + } + for (int xlen = 1; xlen <= N; xlen++) { + for (int ylen = 1; ylen <= Math.min(M, xlen); ylen++) { + if (dp[xlen - 1][ylen] != Integer.MAX_VALUE) { + dp[xlen][ylen] = dp[xlen - 1][ylen] + 1; + } + if (x[xlen - 1] == y[ylen - 1] && dp[xlen - 1][ylen - 1] != Integer.MAX_VALUE) { + dp[xlen][ylen] = Math.min(dp[xlen][ylen], dp[xlen - 1][ylen - 1]); + } + } + } + return dp[N][M]; + } + + public static class LenComp implements Comparator { + + @Override + public int compare(String o1, String o2) { + return o2.length() - o1.length(); + } + + } + + // 解法二 + // 生成所有s1的子串 + // 然后考察每个子串和s2的编辑距离(假设编辑距离只有删除动作且删除一个字符的代价为1) + // 如果s1的长度较小，s2长度较大，这个方法比较合适 + public static int minCost2(String s1, String s2) { + if (s1.length() == 0 || s2.length() == 0) { + return s2.length(); + } + int ans = Integer.MAX_VALUE; + char[] str2 = s2.toCharArray(); + for (int start = 0; start < s1.length(); start++) { + for (int end = start + 1; end <= s1.length(); end++) { + // str1[start....end] + // substring -> [ 0,1 ) + ans = Math.min(ans, distance(str2, s1.substring(start, end).toCharArray())); + } + } + return ans == Integer.MAX_VALUE ? s2.length() : ans; + } + + // 求str2到s1sub的编辑距离 + // 假设编辑距离只有删除动作且删除一个字符的代价为1 + public static int distance(char[] str2, char[] s1sub) { + int row = str2.length; + int col = s1sub.length; + int[][] dp = new int[row][col]; + // dp[i][j]的含义： + // str2[0..i]仅通过删除行为变成s1sub[0..j]的最小代价 + // 可能性一： + // str2[0..i]变的过程中，不保留最后一个字符(str2[i])， + // 那么就是通过str2[0..i-1]变成s1sub[0..j]之后，再最后删掉str2[i]即可 -> dp[i][j] = dp[i-1][j] + 1 + // 可能性二： + // str2[0..i]变的过程中，想保留最后一个字符(str2[i])，然后变成s1sub[0..j]， + // 这要求str2[i] == s1sub[j]才有这种可能, 然后str2[0..i-1]变成s1sub[0..j-1]即可 + // 也就是str2[i] == s1sub[j] 的条件下，dp[i][j] = dp[i-1][j-1] + dp[0][0] = str2[0] == s1sub[0] ? 0 : Integer.MAX_VALUE; + for (int j = 1; j < col; j++) { + dp[0][j] = Integer.MAX_VALUE; + } + for (int i = 1; i < row; i++) { + dp[i][0] = (dp[i - 1][0] != Integer.MAX_VALUE || str2[i] == s1sub[0]) ? i : Integer.MAX_VALUE; + } + for (int i = 1; i < row; i++) { + for (int j = 1; j < col; j++) { + dp[i][j] = Integer.MAX_VALUE; + if (dp[i - 1][j] != Integer.MAX_VALUE) { + dp[i][j] = dp[i - 1][j] + 1; + } + if (str2[i] == s1sub[j] && dp[i - 1][j - 1] != Integer.MAX_VALUE) { + dp[i][j] = Math.min(dp[i][j], dp[i - 1][j - 1]); + } + + } + } + return dp[row - 1][col - 1]; + } + + // 解法二的优化 + public static int minCost3(String s1, String s2) { + if (s1.length() == 0 || s2.length() == 0) { + return s2.length(); + } + char[] str2 = s2.toCharArray(); + char[] str1 = s1.toCharArray(); + int M = str2.length; + int N = str1.length; + int[][] dp = new int[M][N]; + int ans = M; + for (int start = 0; start < N; start++) { // 开始的列数 + dp[0][start] = str2[0] == str1[start] ? 0 : M; + for (int row = 1; row < M; row++) { + dp[row][start] = (str2[row] == str1[start] || dp[row - 1][start] != M) ? row : M; + } + ans = Math.min(ans, dp[M - 1][start]); + // 以上已经把start列，填好 + // 以下要把dp[...][start+1....N-1]的信息填好 + // start...end end - start +2 + for (int end = start + 1; end < N && end - start < M; end++) { + // 0... first-1 行 不用管 + int first = end - start; + dp[first][end] = (str2[first] == str1[end] && dp[first - 1][end - 1] == 0) ? 0 : M; + for (int row = first + 1; row < M; row++) { + dp[row][end] = M; + if (dp[row - 1][end] != M) { + dp[row][end] = dp[row - 1][end] + 1; + } + if (dp[row - 1][end - 1] != M && str2[row] == str1[end]) { + dp[row][end] = Math.min(dp[row][end], dp[row - 1][end - 1]); + } + } + ans = Math.min(ans, dp[M - 1][end]); + } + } + return ans; + } + + // 来自学生的做法，时间复杂度O(N * M平方) + // 复杂度和方法三一样，但是思路截然不同 + public static int minCost4(String s1, String s2) { + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + HashMap> map1 = new HashMap<>(); + for (int i = 0; i < str1.length; i++) { + ArrayList list = map1.getOrDefault(str1[i], new ArrayList()); + list.add(i); + map1.put(str1[i], list); + } + int ans = 0; + // 假设删除后的str2必以i位置开头 + // 那么查找i位置在str1上一共有几个，并对str1上的每个位置开始遍历 + // 再次遍历str2一次，看存在对应str1中i后续连续子串可容纳的最长长度 + for (int i = 0; i < str2.length; i++) { + if (map1.containsKey(str2[i])) { + ArrayList keyList = map1.get(str2[i]); + for (int j = 0; j < keyList.size(); j++) { + int cur1 = keyList.get(j) + 1; + int cur2 = i + 1; + int count = 1; + for (int k = cur2; k < str2.length && cur1 < str1.length; k++) { + if (str2[k] == str1[cur1]) { + cur1++; + count++; + } + } + ans = Math.max(ans, count); + } + } + } + return s2.length() - ans; + } + + public static String generateRandomString(int l, int v) { + int len = (int) (Math.random() * l); + char[] str = new char[len]; + for (int i = 0; i < len; i++) { + str[i] = (char) ('a' + (int) (Math.random() * v)); + } + return String.valueOf(str); + } + + public static void main(String[] args) { + + char[] x = { 'a', 'b', 'c', 'd' }; + char[] y = { 'a', 'd' }; + + System.out.println(onlyDelete(x, y)); + + int str1Len = 20; + int str2Len = 10; + int v = 5; + int testTime = 10000; + boolean pass = true; + System.out.println("test begin"); + for (int i = 0; i < testTime; i++) { + String str1 = generateRandomString(str1Len, v); + String str2 = generateRandomString(str2Len, v); + int ans1 = minCost1(str1, str2); + int ans2 = minCost2(str1, str2); + int ans3 = minCost3(str1, str2); + int ans4 = minCost4(str1, str2); + if (ans1 != ans2 || ans3 != ans4 || ans1 != ans3) { + pass = false; + System.out.println(str1); + System.out.println(str2); + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + System.out.println(ans4); + break; + } + } + System.out.println("test pass : " + pass); + } + +} diff --git a/大厂刷题班/class05/Hash.java b/大厂刷题班/class05/Hash.java new file mode 100644 index 0000000..c995ac9 --- /dev/null +++ b/大厂刷题班/class05/Hash.java @@ -0,0 +1,49 @@ +package class05; + +import java.security.MessageDigest; +import java.security.NoSuchAlgorithmException; +import java.security.Security; + +// 这个包的jar，你需要自己下载一下，导入到项目中 +import javax.xml.bind.DatatypeConverter; + +public class Hash { + + private MessageDigest hash; + + public Hash(String algorithm) { + try { + hash = MessageDigest.getInstance(algorithm); + } catch (NoSuchAlgorithmException e) { + e.printStackTrace(); + } + } + + public String hashCode(String input) { + return DatatypeConverter.printHexBinary(hash.digest(input.getBytes())).toUpperCase(); + } + + public static void main(String[] args) { + System.out.println("支持的算法 : "); + for (String str : Security.getAlgorithms("MessageDigest")) { + System.out.println(str); + } + System.out.println("======="); + + String algorithm = "SHA-256"; + Hash hash = new Hash(algorithm); + + String input1 = "zuochengyunzuochengyun1"; + String input2 = "zuochengyunzuochengyun2"; + String input3 = "zuochengyunzuochengyun3"; + String input4 = "zuochengyunzuochengyun4"; + String input5 = "zuochengyunzuochengyun5"; + System.out.println(hash.hashCode(input1)); + System.out.println(hash.hashCode(input2)); + System.out.println(hash.hashCode(input3)); + System.out.println(hash.hashCode(input4)); + System.out.println(hash.hashCode(input5)); + + } + +} diff --git a/大厂刷题班/class06/Code01_MaxXOR.java b/大厂刷题班/class06/Code01_MaxXOR.java new file mode 100644 index 0000000..d6ee49d --- /dev/null +++ b/大厂刷题班/class06/Code01_MaxXOR.java @@ -0,0 +1,130 @@ +package class06; + +public class Code01_MaxXOR { + + // O(N^2) + public static int maxXorSubarray1(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + // 准备一个前缀异或和数组eor + // eor[i] = arr[0...i]的异或结果 + int[] eor = new int[arr.length]; + eor[0] = arr[0]; + // 生成eor数组，eor[i]代表arr[0..i]的异或和 + for (int i = 1; i < arr.length; i++) { + eor[i] = eor[i - 1] ^ arr[i]; + } + int max = Integer.MIN_VALUE; + for (int j = 0; j < arr.length; j++) { + for (int i = 0; i <= j; i++) { // 依次尝试arr[0..j]、arr[1..j]..arr[i..j]..arr[j..j] + max = Math.max(max, i == 0 ? eor[j] : eor[j] ^ eor[i - 1]); + } + } + return max; + } + + // 前缀树的Node结构 + // nexts[0] -> 0方向的路 + // nexts[1] -> 1方向的路 + // nexts[0] == null 0方向上没路！ + // nexts[0] != null 0方向有路，可以跳下一个节点 + // nexts[1] == null 1方向上没路！ + // nexts[1] != null 1方向有路，可以跳下一个节点 + public static class Node { + public Node[] nexts = new Node[2]; + } + + // 基于本题，定制前缀树的实现 + public static class NumTrie { + // 头节点 + public Node head = new Node(); + + public void add(int newNum) { + Node cur = head; + for (int move = 31; move >= 0; move--) { + int path = ((newNum >> move) & 1); + cur.nexts[path] = cur.nexts[path] == null ? new Node() : cur.nexts[path]; + cur = cur.nexts[path]; + } + } + + // 该结构之前收集了一票数字，并且建好了前缀树 + // num和 谁 ^ 最大的结果（把结果返回） + public int maxXor(int num) { + Node cur = head; + int ans = 0; + for (int move = 31; move >= 0; move--) { + // 取出num中第move位的状态，path只有两种值0就1，整数 + int path = (num >> move) & 1; + // 期待遇到的东西 + int best = move == 31 ? path : (path ^ 1); + // 实际遇到的东西 + best = cur.nexts[best] != null ? best : (best ^ 1); + // (path ^ best) 当前位位异或完的结果 + ans |= (path ^ best) << move; + cur = cur.nexts[best]; + } + return ans; + } + } + + // O(N) + public static int maxXorSubarray2(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int max = Integer.MIN_VALUE; + // 0~i整体异或和 + int xor = 0; + NumTrie numTrie = new NumTrie(); + numTrie.add(0); + for (int i = 0; i < arr.length; i++) { + xor ^= arr[i]; // 0 ~ i + max = Math.max(max, numTrie.maxXor(xor)); + numTrie.add(xor); + } + return max; + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random()); + } + return arr; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + int testTime = 500000; + int maxSize = 30; + int maxValue = 50; + boolean succeed = true; + for (int i = 0; i < testTime; i++) { + int[] arr = generateRandomArray(maxSize, maxValue); + int comp = maxXorSubarray1(arr); + int res = maxXorSubarray2(arr); + if (res != comp) { + succeed = false; + printArray(arr); + System.out.println(res); + System.out.println(comp); + break; + } + } + System.out.println(succeed ? "Nice!" : "Fucking fucked!"); + } +} diff --git a/大厂刷题班/class06/Code02_MaximumXorOfTwoNumbersInAnArray.java b/大厂刷题班/class06/Code02_MaximumXorOfTwoNumbersInAnArray.java new file mode 100644 index 0000000..4ee4c6e --- /dev/null +++ b/大厂刷题班/class06/Code02_MaximumXorOfTwoNumbersInAnArray.java @@ -0,0 +1,50 @@ +package class06; + +// 本题测试链接 : https://leetcode.com/problems/maximum-xor-of-two-numbers-in-an-array/ +public class Code02_MaximumXorOfTwoNumbersInAnArray { + + public static class Node { + public Node[] nexts = new Node[2]; + } + + public static class NumTrie { + public Node head = new Node(); + + public void add(int newNum) { + Node cur = head; + for (int move = 31; move >= 0; move--) { + int path = ((newNum >> move) & 1); + cur.nexts[path] = cur.nexts[path] == null ? new Node() : cur.nexts[path]; + cur = cur.nexts[path]; + } + } + + public int maxXor(int sum) { + Node cur = head; + int res = 0; + for (int move = 31; move >= 0; move--) { + int path = (sum >> move) & 1; + int best = move == 31 ? path : (path ^ 1); + best = cur.nexts[best] != null ? best : (best ^ 1); + res |= (path ^ best) << move; + cur = cur.nexts[best]; + } + return res; + } + } + + public static int findMaximumXOR(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int max = Integer.MIN_VALUE; + NumTrie numTrie = new NumTrie(); + numTrie.add(arr[0]); + for (int i = 1; i < arr.length; i++) { + max = Math.max(max, numTrie.maxXor(arr[i])); + numTrie.add(arr[i]); + } + return max; + } + +} diff --git a/大厂刷题班/class06/Code03_MaximumXorWithAnElementFromArray.java b/大厂刷题班/class06/Code03_MaximumXorWithAnElementFromArray.java new file mode 100644 index 0000000..7ad2f29 --- /dev/null +++ b/大厂刷题班/class06/Code03_MaximumXorWithAnElementFromArray.java @@ -0,0 +1,67 @@ +package class06; + +// 测试链接 : https://leetcode.com/problems/maximum-xor-with-an-element-from-array/ +public class Code03_MaximumXorWithAnElementFromArray { + + public static int[] maximizeXor(int[] nums, int[][] queries) { + int N = nums.length; + NumTrie trie = new NumTrie(); + for (int i = 0; i < N; i++) { + trie.add(nums[i]); + } + int M = queries.length; + int[] ans = new int[M]; + for (int i = 0; i < M; i++) { + ans[i] = trie.maxXorWithXBehindM(queries[i][0], queries[i][1]); + } + return ans; + } + + public static class Node { + public int min; + public Node[] nexts; + + public Node() { + min = Integer.MAX_VALUE; + nexts = new Node[2]; + } + } + + public static class NumTrie { + public Node head = new Node(); + + public void add(int num) { + Node cur = head; + head.min = Math.min(head.min, num); + for (int move = 30; move >= 0; move--) { + int path = ((num >> move) & 1); + cur.nexts[path] = cur.nexts[path] == null ? new Node() : cur.nexts[path]; + cur = cur.nexts[path]; + cur.min = Math.min(cur.min, num); + } + } + + // 这个结构中，已经收集了一票数字 + // 请返回哪个数字与X异或的结果最大，返回最大结果 + // 但是，只有<=m的数字，可以被考虑 + public int maxXorWithXBehindM(int x, int m) { + if (head.min > m) { + return -1; + } + // 一定存在某个数可以和x结合 + Node cur = head; + int ans = 0; + for (int move = 30; move >= 0; move--) { + int path = (x >> move) & 1; + // 期待遇到的东西 + int best = (path ^ 1); + best ^= (cur.nexts[best] == null || cur.nexts[best].min > m) ? 1 : 0; + // best变成了实际遇到的 + ans |= (path ^ best) << move; + cur = cur.nexts[best]; + } + return ans; + } + } + +} diff --git a/大厂刷题班/class06/Code04_MostXorZero.java b/大厂刷题班/class06/Code04_MostXorZero.java new file mode 100644 index 0000000..a73e5c5 --- /dev/null +++ b/大厂刷题班/class06/Code04_MostXorZero.java @@ -0,0 +1,122 @@ +package class06; + +import java.util.ArrayList; +import java.util.HashMap; + +public class Code04_MostXorZero { + + // 暴力方法 + public static int comparator(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int N = arr.length; + int[] eor = new int[N]; + eor[0] = arr[0]; + for (int i = 1; i < N; i++) { + eor[i] = eor[i - 1] ^ arr[i]; + } + return process(eor, 1, new ArrayList<>()); + } + + // index去决定：前一坨部分，结不结束！ + // 如果结束！就把index放入到parts里去 + // 如果不结束，就不放 + public static int process(int[] eor, int index, ArrayList parts) { + int ans = 0; + if (index == eor.length) { + parts.add(eor.length); + ans = eorZeroParts(eor, parts); + parts.remove(parts.size() - 1); + } else { + int p1 = process(eor, index + 1, parts); + parts.add(index); + int p2 = process(eor, index + 1, parts); + parts.remove(parts.size() - 1); + ans = Math.max(p1, p2); + } + return ans; + } + + public static int eorZeroParts(int[] eor, ArrayList parts) { + int L = 0; + int ans = 0; + for (Integer end : parts) { + if ((eor[end - 1] ^ (L == 0 ? 0 : eor[L - 1])) == 0) { + ans++; + } + L = end; + } + return ans; + } + + // 时间复杂度O(N)的方法 + public static int mostXor(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int N = arr.length; + int[] dp = new int[N]; + + // key 某一个前缀异或和 + // value 这个前缀异或和上次出现的位置(最晚！) + HashMap map = new HashMap<>(); + map.put(0, -1); + // 0~i整体的异或和 + int xor = 0; + for (int i = 0; i < N; i++) { + xor ^= arr[i]; + if (map.containsKey(xor)) { // 可能性2 + int pre = map.get(xor); + dp[i] = pre == -1 ? 1 : (dp[pre] + 1); + } + if (i > 0) { + dp[i] = Math.max(dp[i - 1], dp[i]); + } + map.put(xor, i); + } + return dp[N - 1]; + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) ((maxSize + 1) * Math.random())]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()); + } + return arr; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static void main(String[] args) { + int testTime = 150000; + int maxSize = 12; + int maxValue = 5; + boolean succeed = true; + for (int i = 0; i < testTime; i++) { + int[] arr = generateRandomArray(maxSize, maxValue); + int res = mostXor(arr); + int comp = comparator(arr); + if (res != comp) { + succeed = false; + printArray(arr); + System.out.println(res); + System.out.println(comp); + break; + } + } + System.out.println(succeed ? "Nice!" : "Fucking fucked!"); + } + +} diff --git a/大厂刷题班/class06/Code05_Nim.java b/大厂刷题班/class06/Code05_Nim.java new file mode 100644 index 0000000..621ecc0 --- /dev/null +++ b/大厂刷题班/class06/Code05_Nim.java @@ -0,0 +1,18 @@ +package class06; + +public class Code05_Nim { + + // 保证arr是正数数组 + public static void printWinner(int[] arr) { + int eor = 0; + for (int num : arr) { + eor ^= num; + } + if (eor == 0) { + System.out.println("后手赢"); + } else { + System.out.println("先手赢"); + } + } + +} diff --git a/大厂刷题班/class07/Code01_MaxAndValue.java b/大厂刷题班/class07/Code01_MaxAndValue.java new file mode 100644 index 0000000..fb9bda9 --- /dev/null +++ b/大厂刷题班/class07/Code01_MaxAndValue.java @@ -0,0 +1,102 @@ +package class07; + +// 给定一个正数组成的数组，长度一定大于1，求数组中哪两个数与的结果最大 +public class Code01_MaxAndValue { + + // O(N^2)的暴力解 + public static int maxAndValue1(int[] arr) { + int N = arr.length; + int max = Integer.MIN_VALUE; + for (int i = 0; i < N; i++) { + for (int j = i + 1; j < N; j++) { + max = Math.max(max, arr[i] & arr[j]); + } + } + return max; + } + + // O(N)的解 + // 因为是正数，所以不用考虑符号位(31位) + // 首先来到30位，假设剩余的数字有N个(整体)，看看这一位是1的数，有几个 + // 如果有0个、或者1个 + // 说明不管怎么在数组中选择，任何两个数&的结果在第30位上都不可能有1了 + // 答案在第30位上的状态一定是0， + // 保留剩余的N个数，继续考察第29位，谁也不淘汰(因为谁也不行，干脆接受30位上没有1的事实) + // 如果有2个， + // 说明答案就是这两个数(直接返回答案)，因为别的数在第30位都没有1，就这两个数有。 + // 如果有>2个，比如K个 + // 说明答案一定只用在这K个数中去选择某两个数，因为别的数在第30位都没有1，就这K个数有。 + // 答案在第30位上的状态一定是1， + // 只把这K个数作为剩余的数，继续考察第29位，其他数都淘汰掉 + // ..... + // 现在来到i位，假设剩余的数字有M个，看看这一位是1的数，有几个 + // 如果有0个、或者1个 + // 说明不管怎么在M个数中选择，任何两个数&的结果在第i位上都不可能有1了 + // 答案在第i位上的状态一定是0， + // 保留剩余的M个数，继续考察第i-1位 + // 如果有2个， + // 说明答案就是这两个数(直接返回答案)，因为别的数在第i位都没有1，就这两个数有。 + // 如果有>2个，比如K个 + // 说明答案一定只用在这K个数中去选择某两个数，因为别的数在第i位都没有1，就这K个数有。 + // 答案在第i位上的状态一定是1， + // 只把这K个数作为剩余的数，继续考察第i-1位，其他数都淘汰掉 + public static int maxAndValue2(int[] arr) { + // arr[0...M-1] arr[M....] + int M = arr.length; + int ans = 0; + for (int bit = 30; bit >= 0; bit--) { + // arr[0...M-1] arr[M...] + int i = 0; + int tmp = M; + while (i < M) { // arr[0...M-1] + if ((arr[i] & (1 << bit)) == 0) { + swap(arr, i, --M); + } else { + i++; + } + } + if (M == 2) { // arr[0,1] + return arr[0] & arr[1]; + } + if (M < 2) { + M = tmp; + } else { // > 2个数 bit位上有1 + ans |= (1 << bit); + } + } + return ans; + } + + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + + public static int[] randomArray(int size, int range) { + int[] arr = new int[size]; + for (int i = 0; i < size; i++) { + arr[i] = (int) (Math.random() * range) + 1; + } + return arr; + } + + public static void main(String[] args) { + int maxSize = 50; + int range = 30; + int testTime = 1000000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int size = (int) (Math.random() * maxSize) + 2; + int[] arr = randomArray(size, range); + int ans1 = maxAndValue1(arr); + int ans2 = maxAndValue2(arr); + if (ans1 != ans2) { + System.out.println("Oops!"); + } + } + System.out.println("测试结束"); + + } + +} diff --git a/大厂刷题班/class07/Code02_MinCameraCover.java b/大厂刷题班/class07/Code02_MinCameraCover.java new file mode 100644 index 0000000..ebde2ee --- /dev/null +++ b/大厂刷题班/class07/Code02_MinCameraCover.java @@ -0,0 +1,113 @@ +package class07; + +// 本题测试链接 : https://leetcode.com/problems/binary-tree-cameras/ +public class Code02_MinCameraCover { + + public static class TreeNode { + public int value; + public TreeNode left; + public TreeNode right; + } + + public static int minCameraCover1(TreeNode root) { + Info data = process1(root); + return (int) Math.min(data.uncovered + 1, Math.min(data.coveredNoCamera, data.coveredHasCamera)); + } + + // 潜台词：x是头节点，x下方的点都被覆盖的情况下 + public static class Info { + public long uncovered; // x没有被覆盖，x为头的树至少需要几个相机 + public long coveredNoCamera; // x被相机覆盖，但是x没相机，x为头的树至少需要几个相机 + public long coveredHasCamera; // x被相机覆盖了，并且x上放了相机，x为头的树至少需要几个相机 + + public Info(long un, long no, long has) { + uncovered = un; + coveredNoCamera = no; + coveredHasCamera = has; + } + } + + // 所有可能性都穷尽了 + public static Info process1(TreeNode X) { + if (X == null) { // base case + return new Info(Integer.MAX_VALUE, 0, Integer.MAX_VALUE); + } + + Info left = process1(X.left); + Info right = process1(X.right); + // x uncovered x自己不被覆盖，x下方所有节点，都被覆盖 + // 左孩子： 左孩子没被覆盖，左孩子以下的点都被覆盖 + // 左孩子被覆盖但没相机，左孩子以下的点都被覆盖 + // 左孩子被覆盖也有相机，左孩子以下的点都被覆盖 + long uncovered = left.coveredNoCamera + right.coveredNoCamera; + + // x下方的点都被covered，x也被cover，但x上没相机 + long coveredNoCamera = Math.min( + // 1) + left.coveredHasCamera + right.coveredHasCamera, + + Math.min( + // 2) + left.coveredHasCamera + right.coveredNoCamera, + + // 3) + left.coveredNoCamera + right.coveredHasCamera)); + + + + + // x下方的点都被covered，x也被cover，且x上有相机 + long coveredHasCamera = + Math.min(left.uncovered, Math.min(left.coveredNoCamera, left.coveredHasCamera)) + + + Math.min(right.uncovered, Math.min(right.coveredNoCamera, right.coveredHasCamera)) + + + 1; + + return new Info(uncovered, coveredNoCamera, coveredHasCamera); + } + + public static int minCameraCover2(TreeNode root) { + Data data = process2(root); + return data.cameras + (data.status == Status.UNCOVERED ? 1 : 0); + } + + // 以x为头，x下方的节点都是被covered，x自己的状况，分三种 + public static enum Status { + UNCOVERED, COVERED_NO_CAMERA, COVERED_HAS_CAMERA + } + + // 以x为头，x下方的节点都是被covered，得到的最优解中： + // x是什么状态，在这种状态下，需要至少几个相机 + public static class Data { + public Status status; + public int cameras; + + public Data(Status status, int cameras) { + this.status = status; + this.cameras = cameras; + } + } + + public static Data process2(TreeNode X) { + if (X == null) { + return new Data(Status.COVERED_NO_CAMERA, 0); + } + Data left = process2(X.left); + Data right = process2(X.right); + int cameras = left.cameras + right.cameras; + + // 左、或右，哪怕有一个没覆盖 + if (left.status == Status.UNCOVERED || right.status == Status.UNCOVERED) { + return new Data(Status.COVERED_HAS_CAMERA, cameras + 1); + } + + // 左右孩子，不存在没被覆盖的情况 + if (left.status == Status.COVERED_HAS_CAMERA || right.status == Status.COVERED_HAS_CAMERA) { + return new Data(Status.COVERED_NO_CAMERA, cameras); + } + // 左右孩子，不存在没被覆盖的情况，也都没有相机 + return new Data(Status.UNCOVERED, cameras); + } + +} diff --git a/大厂刷题班/class07/Code03_MaxGap.java b/大厂刷题班/class07/Code03_MaxGap.java new file mode 100644 index 0000000..486c6d3 --- /dev/null +++ b/大厂刷题班/class07/Code03_MaxGap.java @@ -0,0 +1,46 @@ +package class07; + +// 测试链接 : https://leetcode.com/problems/maximum-gap/ +public class Code03_MaxGap { + + public static int maximumGap(int[] nums) { + if (nums == null || nums.length < 2) { + return 0; + } + int len = nums.length; + int min = Integer.MAX_VALUE; + int max = Integer.MIN_VALUE; + for (int i = 0; i < len; i++) { + min = Math.min(min, nums[i]); + max = Math.max(max, nums[i]); + } + if (min == max) { + return 0; + } + boolean[] hasNum = new boolean[len + 1]; + int[] maxs = new int[len + 1]; + int[] mins = new int[len + 1]; + int bid = 0; + for (int i = 0; i < len; i++) { + bid = bucket(nums[i], len, min, max); + mins[bid] = hasNum[bid] ? Math.min(mins[bid], nums[i]) : nums[i]; + maxs[bid] = hasNum[bid] ? Math.max(maxs[bid], nums[i]) : nums[i]; + hasNum[bid] = true; + } + int res = 0; + int lastMax = maxs[0]; + int i = 1; + for (; i <= len; i++) { + if (hasNum[i]) { + res = Math.max(res, mins[i] - lastMax); + lastMax = maxs[i]; + } + } + return res; + } + + public static int bucket(long num, long len, long min, long max) { + return (int) ((num - min) * len / (max - min)); + } + +} diff --git a/大厂刷题班/class07/Code04_Power2Diffs.java b/大厂刷题班/class07/Code04_Power2Diffs.java new file mode 100644 index 0000000..ab6c553 --- /dev/null +++ b/大厂刷题班/class07/Code04_Power2Diffs.java @@ -0,0 +1,97 @@ +package class07; + +import java.util.Arrays; +import java.util.HashSet; + +public class Code04_Power2Diffs { + + /* + * 给定一个有序数组arr，其中值可能为正、负、0。 返回arr中每个数都平方之后不同的结果有多少种？ + * + * 给定一个数组arr，先递减然后递增，返回arr中有多少个绝对值不同的数字？ + * + */ + + // 时间复杂度O(N)，额外空间复杂度O(N) + public static int diff1(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + HashSet set = new HashSet<>(); + for (int cur : arr) { + set.add(cur * cur); + } + return set.size(); + } + + // 时间复杂度O(N)，额外空间复杂度O(1) + public static int diff2(int[] arr) { + int N = arr.length; + int L = 0; + int R = N - 1; + int count = 0; + int leftAbs = 0; + int rightAbs = 0; + while (L <= R) { + count++; + leftAbs = Math.abs(arr[L]); + rightAbs = Math.abs(arr[R]); + if (leftAbs < rightAbs) { + while (R >= 0 && Math.abs(arr[R]) == rightAbs) { + R--; + } + } else if (leftAbs > rightAbs) { + while (L < N && Math.abs(arr[L]) == leftAbs) { + L++; + } + } else { + while (L < N && Math.abs(arr[L]) == leftAbs) { + L++; + } + while (R >= 0 && Math.abs(arr[R]) == rightAbs) { + R--; + } + } + } + return count; + } + + // for test + public static int[] randomSortedArray(int len, int value) { + int[] ans = new int[(int) (Math.random() * len) + 1]; + for (int i = 0; i < ans.length; i++) { + ans[i] = (int) (Math.random() * value) - (int) (Math.random() * value); + } + Arrays.sort(ans); + return ans; + } + + // for test + public static void printArray(int[] arr) { + for (int cur : arr) { + System.out.print(cur + " "); + } + System.out.println(); + } + + public static void main(String[] args) { + int len = 100; + int value = 500; + int testTimes = 200000; + System.out.println("test begin"); + for (int i = 0; i < testTimes; i++) { + int[] arr = randomSortedArray(len, value); + int ans1 = diff1(arr); + int ans2 = diff2(arr); + if (ans1 != ans2) { + printArray(arr); + System.out.println(ans1); + System.out.println(ans2); + System.out.println("Oops!"); + break; + } + } + System.out.println("test finish"); + } + +} diff --git a/大厂刷题班/class07/Code05_WorldBreak.java b/大厂刷题班/class07/Code05_WorldBreak.java new file mode 100644 index 0000000..134073a --- /dev/null +++ b/大厂刷题班/class07/Code05_WorldBreak.java @@ -0,0 +1,237 @@ +package class07; + +import java.util.HashSet; + +public class Code05_WorldBreak { + /* + * + * 假设所有字符都是小写字母. 大字符串是str. arr是去重的单词表, 每个单词都不是空字符串且可以使用任意次. + * 使用arr中的单词有多少种拼接str的方式. 返回方法数. + * + */ + + public static int ways(String str, String[] arr) { + HashSet set = new HashSet<>(); + for (String candidate : arr) { + set.add(candidate); + } + return process(str, 0, set); + } + + // 所有的可分解字符串，都已经放在了set中 + // str[i....] 能够被set中的贴纸分解的话，返回分解的方法数 + public static int process(String str, int i, HashSet set) { + if (i == str.length()) { // 没字符串需要分解了！ + return 1; + } + // i....还有字符串需要分解 + int ways = 0; + // [i ... end] 前缀串 每一个前缀串 + for (int end = i; end < str.length(); end++) { + String pre = str.substring(i, end + 1);// [) + if (set.contains(pre)) { + ways += process(str, end + 1, set); + } + } + return ways; + } + + public static int ways1(String str, String[] arr) { + if (str == null || str.length() == 0 || arr == null || arr.length == 0) { + return 0; + } + HashSet map = new HashSet<>(); + for (String s : arr) { + map.add(s); + } + return f(str, map, 0); + } + + public static int f(String str, HashSet map, int index) { + if (index == str.length()) { + return 1; + } + int ways = 0; + for (int end = index; end < str.length(); end++) { + if (map.contains(str.substring(index, end + 1))) { + ways += f(str, map, end + 1); + } + } + return ways; + } + + public static int ways2(String str, String[] arr) { + if (str == null || str.length() == 0 || arr == null || arr.length == 0) { + return 0; + } + HashSet map = new HashSet<>(); + for (String s : arr) { + map.add(s); + } + int N = str.length(); + int[] dp = new int[N + 1]; + dp[N] = 1; + for (int i = N - 1; i >= 0; i--) { + for (int end = i; end < N; end++) { + if (map.contains(str.substring(i, end + 1))) { + dp[i] += dp[end + 1]; + } + } + } + return dp[0]; + } + + public static class Node { + public boolean end; + public Node[] nexts; + + public Node() { + end = false; + nexts = new Node[26]; + } + } + + public static int ways3(String str, String[] arr) { + if (str == null || str.length() == 0 || arr == null || arr.length == 0) { + return 0; + } + Node root = new Node(); + for (String s : arr) { + char[] chs = s.toCharArray(); + Node node = root; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = chs[i] - 'a'; + if (node.nexts[index] == null) { + node.nexts[index] = new Node(); + } + node = node.nexts[index]; + } + node.end = true; + } + return g(str.toCharArray(), root, 0); + } + + // str[i...] 被分解的方法数，返回 + + public static int g(char[] str, Node root, int i) { + if (i == str.length) { + return 1; + } + int ways = 0; + Node cur = root; + // i...end + for (int end = i; end < str.length; end++) { + int path = str[end] - 'a'; + if (cur.nexts[path] == null) { + break; + } + cur = cur.nexts[path]; + if (cur.end) { // i...end + ways += g(str, root, end + 1); + } + } + return ways; + } + + public static int ways4(String s, String[] arr) { + if (s == null || s.length() == 0 || arr == null || arr.length == 0) { + return 0; + } + Node root = new Node(); + for (String str : arr) { + char[] chs = str.toCharArray(); + Node node = root; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = chs[i] - 'a'; + if (node.nexts[index] == null) { + node.nexts[index] = new Node(); + } + node = node.nexts[index]; + } + node.end = true; + } + char[] str = s.toCharArray(); + int N = str.length; + int[] dp = new int[N + 1]; + dp[N] = 1; + for (int i = N - 1; i >= 0; i--) { + Node cur = root; + for (int end = i; end < N; end++) { + int path = str[end] - 'a'; + if (cur.nexts[path] == null) { + break; + } + cur = cur.nexts[path]; + if (cur.end) { + dp[i] += dp[end + 1]; + } + } + } + return dp[0]; + } + + // 以下的逻辑都是为了测试 + public static class RandomSample { + public String str; + public String[] arr; + + public RandomSample(String s, String[] a) { + str = s; + arr = a; + } + } + + // 随机样本产生器 + public static RandomSample generateRandomSample(char[] candidates, int num, int len, int joint) { + String[] seeds = randomSeeds(candidates, num, len); + HashSet set = new HashSet<>(); + for (String str : seeds) { + set.add(str); + } + String[] arr = new String[set.size()]; + int index = 0; + for (String str : set) { + arr[index++] = str; + } + StringBuilder all = new StringBuilder(); + for (int i = 0; i < joint; i++) { + all.append(arr[(int) (Math.random() * arr.length)]); + } + return new RandomSample(all.toString(), arr); + } + + public static String[] randomSeeds(char[] candidates, int num, int len) { + String[] arr = new String[(int) (Math.random() * num) + 1]; + for (int i = 0; i < arr.length; i++) { + char[] str = new char[(int) (Math.random() * len) + 1]; + for (int j = 0; j < str.length; j++) { + str[j] = candidates[(int) (Math.random() * candidates.length)]; + } + arr[i] = String.valueOf(str); + } + return arr; + } + + public static void main(String[] args) { + char[] candidates = { 'a', 'b' }; + int num = 20; + int len = 4; + int joint = 5; + int testTimes = 30000; + boolean testResult = true; + for (int i = 0; i < testTimes; i++) { + RandomSample sample = generateRandomSample(candidates, num, len, joint); + int ans1 = ways1(sample.str, sample.arr); + int ans2 = ways2(sample.str, sample.arr); + int ans3 = ways3(sample.str, sample.arr); + int ans4 = ways4(sample.str, sample.arr); + if (ans1 != ans2 || ans3 != ans4 || ans2 != ans4) { + testResult = false; + } + } + System.out.println(testTimes + "次随机测试是否通过：" + testResult); + } + +} diff --git a/大厂刷题班/class07/Code06_SplitStringMaxValue.java b/大厂刷题班/class07/Code06_SplitStringMaxValue.java new file mode 100644 index 0000000..c6e76cc --- /dev/null +++ b/大厂刷题班/class07/Code06_SplitStringMaxValue.java @@ -0,0 +1,137 @@ +package class07; + +import java.util.HashMap; + +public class Code06_SplitStringMaxValue { + + // 暴力解 + public static int maxRecord1(String str, int K, String[] parts, int[] record) { + if (str == null || str.length() == 0) { + return 0; + } + HashMap records = new HashMap<>(); + for (int i = 0; i < parts.length; i++) { + records.put(parts[i], record[i]); + } + return process(str, 0, K, records); + } + + public static int process(String str, int index, int rest, HashMap records) { + if (rest < 0) { + return -1; + } + if (index == str.length()) { + return rest == 0 ? 0 : -1; + } + int ans = -1; + for (int end = index; end < str.length(); end++) { + String first = str.substring(index, end + 1); + int next = records.containsKey(first) ? process(str, end + 1, rest - 1, records) : -1; + if (next != -1) { + ans = Math.max(ans, records.get(first) + next); + } + } + return ans; + } + + // 动态规划解 + public static int maxRecord2(String str, int K, String[] parts, int[] record) { + if (str == null || str.length() == 0) { + return 0; + } + HashMap records = new HashMap<>(); + for (int i = 0; i < parts.length; i++) { + records.put(parts[i], record[i]); + } + int N = str.length(); + int[][] dp = new int[N + 1][K + 1]; + for (int rest = 1; rest <= K; rest++) { + dp[N][rest] = -1; + } + for (int index = N - 1; index >= 0; index--) { + for (int rest = 0; rest <= K; rest++) { + int ans = -1; + for (int end = index; end < N; end++) { + String first = str.substring(index, end + 1); + int next = rest > 0 && records.containsKey(first) ? dp[end + 1][rest - 1] : -1; + if (next != -1) { + ans = Math.max(ans, records.get(first) + next); + } + } + dp[index][rest] = ans; + } + } + return dp[0][K]; + } + + // 动态规划解 + 前缀树优化 + public static int maxRecord3(String s, int K, String[] parts, int[] record) { + if (s == null || s.length() == 0) { + return 0; + } + TrieNode root = rootNode(parts, record); + char[] str = s.toCharArray(); + int N = str.length; + int[][] dp = new int[N + 1][K + 1]; + for (int rest = 1; rest <= K; rest++) { + dp[N][rest] = -1; + } + for (int index = N - 1; index >= 0; index--) { + for (int rest = 0; rest <= K; rest++) { + int ans = -1; + TrieNode cur = root; + for (int end = index; end < N; end++) { + int path = str[end] - 'a'; + if (cur.nexts[path] == null) { + break; + } + cur = cur.nexts[path]; + int next = rest > 0 && cur.value != -1 ? dp[end + 1][rest - 1] : -1; + if (next != -1) { + ans = Math.max(ans, cur.value + next); + } + } + dp[index][rest] = ans; + } + } + return dp[0][K]; + } + + public static class TrieNode { + public TrieNode[] nexts; + public int value; + + public TrieNode() { + nexts = new TrieNode[26]; + value = -1; + } + } + + public static TrieNode rootNode(String[] parts, int[] record) { + TrieNode root = new TrieNode(); + for (int i = 0; i < parts.length; i++) { + char[] str = parts[i].toCharArray(); + TrieNode cur = root; + for (int j = 0; j < str.length; j++) { + int path = str[j] - 'a'; + if (cur.nexts[path] == null) { + cur.nexts[path] = new TrieNode(); + } + cur = cur.nexts[path]; + } + cur.value = record[i]; + } + return root; + } + + public static void main(String[] args) { + String str = "abcdefg"; + int K = 3; + String[] parts = { "abc", "def", "g", "ab", "cd", "efg", "defg" }; + int[] record = { 1, 1, 1, 3, 3, 3, 2 }; + System.out.println(maxRecord1(str, K, parts, record)); + System.out.println(maxRecord2(str, K, parts, record)); + System.out.println(maxRecord3(str, K, parts, record)); + } + +} diff --git a/大厂刷题班/class08/Code01_ExpressionCompute.java b/大厂刷题班/class08/Code01_ExpressionCompute.java new file mode 100644 index 0000000..6d1192b --- /dev/null +++ b/大厂刷题班/class08/Code01_ExpressionCompute.java @@ -0,0 +1,71 @@ +package class08; + +import java.util.LinkedList; + +// 本题测试链接 : https://leetcode.com/problems/basic-calculator-iii/ +public class Code01_ExpressionCompute { + + public static int calculate(String str) { + return f(str.toCharArray(), 0)[0]; + } + + // 请从str[i...]往下算，遇到字符串终止位置或者右括号，就停止 + // 返回两个值，长度为2的数组 + // 0) 负责的这一段的结果是多少 + // 1) 负责的这一段计算到了哪个位置 + public static int[] f(char[] str, int i) { + LinkedList que = new LinkedList(); + int cur = 0; + int[] bra = null; + // 从i出发，开始撸串 + while (i < str.length && str[i] != ')') { + if (str[i] >= '0' && str[i] <= '9') { + cur = cur * 10 + str[i++] - '0'; + } else if (str[i] != '(') { // 遇到的是运算符号 + addNum(que, cur); + que.addLast(String.valueOf(str[i++])); + cur = 0; + } else { // 遇到左括号了 + bra = f(str, i + 1); + cur = bra[0]; + i = bra[1] + 1; + } + } + addNum(que, cur); + return new int[] { getNum(que), i }; + } + + public static void addNum(LinkedList que, int num) { + if (!que.isEmpty()) { + int cur = 0; + String top = que.pollLast(); + if (top.equals("+") || top.equals("-")) { + que.addLast(top); + } else { + cur = Integer.valueOf(que.pollLast()); + num = top.equals("*") ? (cur * num) : (cur / num); + } + } + que.addLast(String.valueOf(num)); + } + + public static int getNum(LinkedList que) { + int res = 0; + boolean add = true; + String cur = null; + int num = 0; + while (!que.isEmpty()) { + cur = que.pollFirst(); + if (cur.equals("+")) { + add = true; + } else if (cur.equals("-")) { + add = false; + } else { + num = Integer.valueOf(cur); + res += add ? num : (-num); + } + } + return res; + } + +} diff --git a/大厂刷题班/class08/Code02_ContainerWithMostWater.java b/大厂刷题班/class08/Code02_ContainerWithMostWater.java new file mode 100644 index 0000000..9053a24 --- /dev/null +++ b/大厂刷题班/class08/Code02_ContainerWithMostWater.java @@ -0,0 +1,32 @@ +package class08; + +// 本题测试链接 : https://leetcode.com/problems/container-with-most-water/ +public class Code02_ContainerWithMostWater { + + public static int maxArea1(int[] h) { + int max = 0; + int N = h.length; + for (int i = 0; i < N; i++) { // h[i] + for (int j = i + 1; j < N; j++) { // h[j] + max = Math.max(max, Math.min(h[i], h[j]) * (j - i)); + } + } + return max; + } + + public static int maxArea2(int[] h) { + int max = 0; + int l = 0; + int r = h.length - 1; + while (l < r) { + max = Math.max(max, Math.min(h[l], h[r]) * (r - l)); + if (h[l] > h[r]) { + r--; + } else { + l++; + } + } + return max; + } + +} diff --git a/大厂刷题班/class08/Code03_FindWordInMatrix.java b/大厂刷题班/class08/Code03_FindWordInMatrix.java new file mode 100644 index 0000000..e985c43 --- /dev/null +++ b/大厂刷题班/class08/Code03_FindWordInMatrix.java @@ -0,0 +1,159 @@ +package class08; + +/* + * 给定一个char[][] matrix，也就是char类型的二维数组，再给定一个字符串word， + * 可以从任何一个某个位置出发，可以走上下左右，能不能找到word？ + * 比如： + * char[][] m = { + * { 'a', 'b', 'z' }, + * { 'c', 'd', 'o' }, + * { 'f', 'e', 'o' }, + * }; + * word = "zooe" + * 是可以找到的 + * + * 设定1：可以走重复路的情况下，返回能不能找到 + * 比如，word = "zoooz"，是可以找到的，z -> o -> o -> o -> z，因为允许走一条路径中已经走过的字符 + * + * 设定2：不可以走重复路的情况下，返回能不能找到 + * 比如，word = "zoooz"，是不可以找到的，因为允许走一条路径中已经走过的字符不能重复走 + * + * 写出两种设定下的code + * + * */ +public class Code03_FindWordInMatrix { + + // 可以走重复的设定 + public static boolean findWord1(char[][] m, String word) { + if (word == null || word.equals("")) { + return true; + } + if (m == null || m.length == 0 || m[0] == null || m[0].length == 0) { + return false; + } + char[] w = word.toCharArray(); + int N = m.length; + int M = m[0].length; + int len = w.length; + // dp[i][j][k]表示：必须以m[i][j]这个字符结尾的情况下，能不能找到w[0...k]这个前缀串 + boolean[][][] dp = new boolean[N][M][len]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + dp[i][j][0] = m[i][j] == w[0]; + } + } + for (int k = 1; k < len; k++) { + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + dp[i][j][k] = (m[i][j] == w[k] && checkPrevious(dp, i, j, k)); + } + } + } + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + if (dp[i][j][len - 1]) { + return true; + } + } + } + return false; + } + + // 可以走重复路 + // 从m[i][j]这个字符出发，能不能找到str[k...]这个后缀串 + public static boolean canLoop(char[][] m, int i, int j, char[] str, int k) { + if (k == str.length) { + return true; + } + if (i == -1 || i == m.length || j == -1 || j == m[0].length || m[i][j] != str[k]) { + return false; + } + // 不越界！m[i][j] == str[k] 对的上的！ + // str[k+1....] + boolean ans = false; + if (canLoop(m, i + 1, j, str, k + 1) || canLoop(m, i - 1, j, str, k + 1) || canLoop(m, i, j + 1, str, k + 1) + || canLoop(m, i, j - 1, str, k + 1)) { + ans = true; + } + return ans; + } + + // 不能走重复路 + // 从m[i][j]这个字符出发，能不能找到str[k...]这个后缀串 + public static boolean noLoop(char[][] m, int i, int j, char[] str, int k) { + if (k == str.length) { + return true; + } + if (i == -1 || i == m.length || j == -1 || j == m[0].length || m[i][j] != str[k]) { + return false; + } + // 不越界！也不是回头路！m[i][j] == str[k] 也对的上！ + m[i][j] = 0; + boolean ans = false; + if (noLoop(m, i + 1, j, str, k + 1) || noLoop(m, i - 1, j, str, k + 1) || noLoop(m, i, j + 1, str, k + 1) + || noLoop(m, i, j - 1, str, k + 1)) { + ans = true; + } + m[i][j] = str[k]; + return ans; + } + + public static boolean checkPrevious(boolean[][][] dp, int i, int j, int k) { + boolean up = i > 0 ? (dp[i - 1][j][k - 1]) : false; + boolean down = i < dp.length - 1 ? (dp[i + 1][j][k - 1]) : false; + boolean left = j > 0 ? (dp[i][j - 1][k - 1]) : false; + boolean right = j < dp[0].length - 1 ? (dp[i][j + 1][k - 1]) : false; + return up || down || left || right; + } + + // 不可以走重复路的设定 + public static boolean findWord2(char[][] m, String word) { + if (word == null || word.equals("")) { + return true; + } + if (m == null || m.length == 0 || m[0] == null || m[0].length == 0) { + return false; + } + char[] w = word.toCharArray(); + for (int i = 0; i < m.length; i++) { + for (int j = 0; j < m[0].length; j++) { + if (process(m, i, j, w, 0)) { + return true; + } + } + } + return false; + } + + // 从m[i][j]这个字符出发，能不能找到w[k...]这个后缀串 + public static boolean process(char[][] m, int i, int j, char[] str, int k) { + if (k == str.length) { + return true; + } + if (i == -1 || i == m.length || j == -1 || j == m[0].length || m[i][j] == 0 || m[i][j] != str[k]) { + return false; + } + m[i][j] = 0; + boolean ans = false; + if (process(m, i + 1, j, str, k + 1) || process(m, i - 1, j, str, k + 1) || process(m, i, j + 1, str, k + 1) + || process(m, i, j - 1, str, k + 1)) { + ans = true; + } + m[i][j] = str[k]; + return ans; + } + + public static void main(String[] args) { + char[][] m = { { 'a', 'b', 'z' }, { 'c', 'd', 'o' }, { 'f', 'e', 'o' }, }; + String word1 = "zoooz"; + String word2 = "zoo"; + // 可以走重复路的设定 + System.out.println(findWord1(m, word1)); + System.out.println(findWord1(m, word2)); + // 不可以走重复路的设定 + System.out.println(findWord2(m, word1)); + System.out.println(findWord2(m, word2)); + + } + +} diff --git a/大厂刷题班/class08/Code04_SnakeGame.java b/大厂刷题班/class08/Code04_SnakeGame.java new file mode 100644 index 0000000..1aa5316 --- /dev/null +++ b/大厂刷题班/class08/Code04_SnakeGame.java @@ -0,0 +1,188 @@ +package class08; + +import java.util.Arrays; + +public class Code04_SnakeGame { + + public static int walk1(int[][] matrix) { + if (matrix == null || matrix.length == 0 || matrix[0].length == 0) { + return 0; + } + int res = Integer.MIN_VALUE; + for (int i = 0; i < matrix.length; i++) { + for (int j = 0; j < matrix[0].length; j++) { + int[] ans = process(matrix, i, j); + res = Math.max(res, Math.max(ans[0], ans[1])); + } + } + return res; + } + + public static int zuo(int[][] matrix) { + if (matrix == null || matrix.length == 0 || matrix[0].length == 0) { + return 0; + } + int ans = 0; + for (int i = 0; i < matrix.length; i++) { + for (int j = 0; j < matrix[0].length; j++) { + Info cur = f(matrix, i, j); + ans = Math.max(ans, Math.max(cur.no, cur.yes)); + } + } + return ans; + } + + public static class Info { + public int no; + public int yes; + + public Info(int n, int y) { + no = n; + yes = y; + } + } + + // 蛇从某一个最左列，且最优的空降点降落 + // 沿途走到(i,j)必须停！ + // 返回，一次能力也不用，获得的最大成长值 + // 返回，用了一次能力，获得的最大成长值 + // 如果蛇从某一个最左列，且最优的空降点降落，不用能力，怎么都到不了(i,j)，那么no = -1 + // 如果蛇从某一个最左列，且最优的空降点降落，用了一次能力，怎么都到不了(i,j)，那么yes = -1 + public static Info f(int[][] matrix, int i, int j) { + if (j == 0) { // 最左列 + int no = Math.max(matrix[i][0], -1); + int yes = Math.max(-matrix[i][0], -1); + return new Info(no, yes); + } + // j > 0 不在最左列 + int preNo = -1; + int preYes = -1; + Info pre = f(matrix, i, j - 1); + preNo = Math.max(pre.no, preNo); + preYes = Math.max(pre.yes, preYes); + if (i > 0) { + pre = f(matrix, i - 1, j - 1); + preNo = Math.max(pre.no, preNo); + preYes = Math.max(pre.yes, preYes); + } + if (i < matrix.length - 1) { + pre = f(matrix, i + 1, j - 1); + preNo = Math.max(pre.no, preNo); + preYes = Math.max(pre.yes, preYes); + } + int no = preNo == -1 ? -1 : (Math.max(-1, preNo + matrix[i][j])); + // 能力只有一次，是之前用的！ + int p1 = preYes == -1 ? -1 : (Math.max(-1, preYes + matrix[i][j])); + // 能力只有一次，就当前用！ + int p2 = preNo == -1 ? -1 : (Math.max(-1, preNo - matrix[i][j])); + int yes = Math.max(Math.max(p1, p2), -1); + return new Info(no, yes); + } + + // 从假想的最优左侧到达(i,j)的旅程中 + // 0) 在没有使用过能力的情况下，返回路径最大和，没有可能到达的话，返回负 + // 1) 在使用过能力的情况下，返回路径最大和，没有可能到达的话，返回负 + public static int[] process(int[][] m, int i, int j) { + if (j == 0) { // (i,j)就是最左侧的位置 + return new int[] { m[i][j], -m[i][j] }; + } + int[] preAns = process(m, i, j - 1); + // 所有的路中，完全不使用能力的情况下，能够到达的最好长度是多大 + int preUnuse = preAns[0]; + // 所有的路中，使用过一次能力的情况下，能够到达的最好长度是多大 + int preUse = preAns[1]; + if (i - 1 >= 0) { + preAns = process(m, i - 1, j - 1); + preUnuse = Math.max(preUnuse, preAns[0]); + preUse = Math.max(preUse, preAns[1]); + } + if (i + 1 < m.length) { + preAns = process(m, i + 1, j - 1); + preUnuse = Math.max(preUnuse, preAns[0]); + preUse = Math.max(preUse, preAns[1]); + } + // preUnuse 之前旅程，没用过能力 + // preUse 之前旅程，已经使用过能力了 + int no = -1; // 之前没使用过能力，当前位置也不使用能力，的最优解 + int yes = -1; // 不管是之前使用能力，还是当前使用了能力，请保证能力只使用一次，最优解 + if (preUnuse >= 0) { + no = m[i][j] + preUnuse; + yes = -m[i][j] + preUnuse; + } + if (preUse >= 0) { + yes = Math.max(yes, m[i][j] + preUse); + } + return new int[] { no, yes }; + } + + public static int walk2(int[][] matrix) { + if (matrix == null || matrix.length == 0 || matrix[0].length == 0) { + return 0; + } + int max = Integer.MIN_VALUE; + int[][][] dp = new int[matrix.length][matrix[0].length][2]; + for (int i = 0; i < dp.length; i++) { + dp[i][0][0] = matrix[i][0]; + dp[i][0][1] = -matrix[i][0]; + max = Math.max(max, Math.max(dp[i][0][0], dp[i][0][1])); + } + for (int j = 1; j < matrix[0].length; j++) { + for (int i = 0; i < matrix.length; i++) { + int preUnuse = dp[i][j - 1][0]; + int preUse = dp[i][j - 1][1]; + if (i - 1 >= 0) { + preUnuse = Math.max(preUnuse, dp[i - 1][j - 1][0]); + preUse = Math.max(preUse, dp[i - 1][j - 1][1]); + } + if (i + 1 < matrix.length) { + preUnuse = Math.max(preUnuse, dp[i + 1][j - 1][0]); + preUse = Math.max(preUse, dp[i + 1][j - 1][1]); + } + dp[i][j][0] = -1; + dp[i][j][1] = -1; + if (preUnuse >= 0) { + dp[i][j][0] = matrix[i][j] + preUnuse; + dp[i][j][1] = -matrix[i][j] + preUnuse; + } + if (preUse >= 0) { + dp[i][j][1] = Math.max(dp[i][j][1], matrix[i][j] + preUse); + } + max = Math.max(max, Math.max(dp[i][j][0], dp[i][j][1])); + } + } + return max; + } + + public static int[][] generateRandomArray(int row, int col, int value) { + int[][] arr = new int[row][col]; + for (int i = 0; i < arr.length; i++) { + for (int j = 0; j < arr[0].length; j++) { + arr[i][j] = (int) (Math.random() * value) * (Math.random() > 0.5 ? -1 : 1); + } + } + return arr; + } + + public static void main(String[] args) { + int N = 7; + int M = 7; + int V = 10; + int times = 1000000; + for (int i = 0; i < times; i++) { + int r = (int) (Math.random() * (N + 1)); + int c = (int) (Math.random() * (M + 1)); + int[][] matrix = generateRandomArray(r, c, V); + int ans1 = zuo(matrix); + int ans2 = walk2(matrix); + if (ans1 != ans2) { + for (int j = 0; j < matrix.length; j++) { + System.out.println(Arrays.toString(matrix[j])); + } + System.out.println("Oops ans1: " + ans1 + " ans2:" + ans2); + break; + } + } + System.out.println("finish"); + } + +} diff --git a/大厂刷题班/class09/Code01_LightProblem.java b/大厂刷题班/class09/Code01_LightProblem.java new file mode 100644 index 0000000..c42f962 --- /dev/null +++ b/大厂刷题班/class09/Code01_LightProblem.java @@ -0,0 +1,384 @@ +package class09; + +/* + * 给定一个数组arr，长度为N，arr中的值不是0就是1 + * arr[i]表示第i栈灯的状态，0代表灭灯，1代表亮灯 + * 每一栈灯都有开关，但是按下i号灯的开关，会同时改变i-1、i、i+2栈灯的状态 + * 问题一： + * 如果N栈灯排成一条直线,请问最少按下多少次开关,能让灯都亮起来 + * 排成一条直线说明： + * i为中间位置时，i号灯的开关能影响i-1、i和i+1 + * 0号灯的开关只能影响0和1位置的灯 + * N-1号灯的开关只能影响N-2和N-1位置的灯 + * + * 问题二： + * 如果N栈灯排成一个圈,请问最少按下多少次开关,能让灯都亮起来 + * 排成一个圈说明： + * i为中间位置时，i号灯的开关能影响i-1、i和i+1 + * 0号灯的开关能影响N-1、0和1位置的灯 + * N-1号灯的开关能影响N-2、N-1和0位置的灯 + * + * */ +public class Code01_LightProblem { + + // 无环改灯问题的暴力版本 + public static int noLoopRight(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + if (arr.length == 1) { + return arr[0] == 1 ? 0 : 1; + } + if (arr.length == 2) { + return arr[0] != arr[1] ? Integer.MAX_VALUE : (arr[0] ^ 1); + } + return f1(arr, 0); + } + + public static int f1(int[] arr, int i) { + if (i == arr.length) { + return valid(arr) ? 0 : Integer.MAX_VALUE; + } + int p1 = f1(arr, i + 1); + change1(arr, i); + int p2 = f1(arr, i + 1); + change1(arr, i); + p2 = (p2 == Integer.MAX_VALUE) ? p2 : (p2 + 1); + return Math.min(p1, p2); + } + + public static void change1(int[] arr, int i) { + if (i == 0) { + arr[0] ^= 1; + arr[1] ^= 1; + } else if (i == arr.length - 1) { + arr[i - 1] ^= 1; + arr[i] ^= 1; + } else { + arr[i - 1] ^= 1; + arr[i] ^= 1; + arr[i + 1] ^= 1; + } + } + + public static boolean valid(int[] arr) { + for (int i = 0; i < arr.length; i++) { + if (arr[i] == 0) { + return false; + } + } + return true; + } + + // 无环改灯问题的递归版本 + public static int noLoopMinStep1(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + if (arr.length == 1) { + return arr[0] ^ 1; + } + if (arr.length == 2) { + return arr[0] != arr[1] ? Integer.MAX_VALUE : (arr[0] ^ 1); + } + // 不变0位置的状态 + int p1 = process1(arr, 2, arr[0], arr[1]); + // 改变0位置的状态 + int p2 = process1(arr, 2, arr[0] ^ 1, arr[1] ^ 1); + if (p2 != Integer.MAX_VALUE) { + p2++; + } + return Math.min(p1, p2); + } + + // 当前在哪个位置上，做选择，nextIndex - 1 = cur ，当前！ + // cur - 1 preStatus + // cur curStatus + // 0....cur-2 全亮的！ + public static int process1(int[] arr, int nextIndex, int preStatus, int curStatus) { + if (nextIndex == arr.length) { // 当前来到最后一个开关的位置 + return preStatus != curStatus ? (Integer.MAX_VALUE) : (curStatus ^ 1); + } + // 没到最后一个按钮呢！ + // i < arr.length + if (preStatus == 0) { // 一定要改变 + curStatus ^= 1; + int cur = arr[nextIndex] ^ 1; + int next = process1(arr, nextIndex + 1, curStatus, cur); + return next == Integer.MAX_VALUE ? next : (next + 1); + } else { // 一定不能改变 + return process1(arr, nextIndex + 1, curStatus, arr[nextIndex]); + } + } + + // 无环改灯问题的迭代版本 + public static int noLoopMinStep2(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + if (arr.length == 1) { + return arr[0] == 1 ? 0 : 1; + } + if (arr.length == 2) { + return arr[0] != arr[1] ? Integer.MAX_VALUE : (arr[0] ^ 1); + } + int p1 = traceNoLoop(arr, arr[0], arr[1]); + int p2 = traceNoLoop(arr, arr[0] ^ 1, arr[1] ^ 1); + p2 = (p2 == Integer.MAX_VALUE) ? p2 : (p2 + 1); + return Math.min(p1, p2); + } + + public static int traceNoLoop(int[] arr, int preStatus, int curStatus) { + int i = 2; + int op = 0; + while (i != arr.length) { + if (preStatus == 0) { + op++; + preStatus = curStatus ^ 1; + curStatus = arr[i++] ^ 1; + } else { + preStatus = curStatus; + curStatus = arr[i++]; + } + } + return (preStatus != curStatus) ? Integer.MAX_VALUE : (op + (curStatus ^ 1)); + } + + // 有环改灯问题的暴力版本 + public static int loopRight(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + if (arr.length == 1) { + return arr[0] == 1 ? 0 : 1; + } + if (arr.length == 2) { + return arr[0] != arr[1] ? Integer.MAX_VALUE : (arr[0] ^ 1); + } + return f2(arr, 0); + } + + public static int f2(int[] arr, int i) { + if (i == arr.length) { + return valid(arr) ? 0 : Integer.MAX_VALUE; + } + int p1 = f2(arr, i + 1); + change2(arr, i); + int p2 = f2(arr, i + 1); + change2(arr, i); + p2 = (p2 == Integer.MAX_VALUE) ? p2 : (p2 + 1); + return Math.min(p1, p2); + } + + public static void change2(int[] arr, int i) { + arr[lastIndex(i, arr.length)] ^= 1; + arr[i] ^= 1; + arr[nextIndex(i, arr.length)] ^= 1; + } + + public static int lastIndex(int i, int N) { + return i == 0 ? (N - 1) : (i - 1); + } + + public static int nextIndex(int i, int N) { + return i == N - 1 ? 0 : (i + 1); + } + + // 有环改灯问题的递归版本 + public static int loopMinStep1(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + if (arr.length == 1) { + return arr[0] == 1 ? 0 : 1; + } + if (arr.length == 2) { + return arr[0] != arr[1] ? Integer.MAX_VALUE : (arr[0] ^ 1); + } + if (arr.length == 3) { + return (arr[0] != arr[1] || arr[0] != arr[2]) ? Integer.MAX_VALUE : (arr[0] ^ 1); + } + // 0不变，1不变 + int p1 = process2(arr, 3, arr[1], arr[2], arr[arr.length - 1], arr[0]); + // 0改变，1不变 + int p2 = process2(arr, 3, arr[1] ^ 1, arr[2], arr[arr.length - 1] ^ 1, arr[0] ^ 1); + // 0不变，1改变 + int p3 = process2(arr, 3, arr[1] ^ 1, arr[2] ^ 1, arr[arr.length - 1], arr[0] ^ 1); + // 0改变，1改变 + int p4 = process2(arr, 3, arr[1], arr[2] ^ 1, arr[arr.length - 1] ^ 1, arr[0]); + p2 = p2 != Integer.MAX_VALUE ? (p2 + 1) : p2; + p3 = p3 != Integer.MAX_VALUE ? (p3 + 1) : p3; + p4 = p4 != Integer.MAX_VALUE ? (p4 + 2) : p4; + return Math.min(Math.min(p1, p2), Math.min(p3, p4)); + } + + + // 下一个位置是，nextIndex + // 当前位置是，nextIndex - 1 -> curIndex + // 上一个位置是, nextIndex - 2 -> preIndex preStatus + // 当前位置是，nextIndex - 1, curStatus + // endStatus, N-1位置的状态 + // firstStatus, 0位置的状态 + // 返回，让所有灯都亮，至少按下几次按钮 + + // 当前来到的位置(nextIndex - 1)，一定不能是1！至少从2开始 + // nextIndex >= 3 + public static int process2(int[] arr, + int nextIndex, int preStatus, int curStatus, + int endStatus, int firstStatus) { + + if (nextIndex == arr.length) { // 最后一按钮！ + return (endStatus != firstStatus || endStatus != preStatus) ? Integer.MAX_VALUE : (endStatus ^ 1); + } + // 当前位置，nextIndex - 1 + // 当前的状态，叫curStatus + // 如果不按下按钮，下一步的preStatus, curStatus + // 如果按下按钮，下一步的preStatus, curStatus ^ 1 + // 如果不按下按钮，下一步的curStatus, arr[nextIndex] + // 如果按下按钮，下一步的curStatus, arr[nextIndex] ^ 1 + int noNextPreStatus = 0; + int yesNextPreStatus = 0; + int noNextCurStatus =0; + int yesNextCurStatus = 0; + int noEndStatus = 0; + int yesEndStatus = 0; + if(nextIndex < arr.length - 1) {// 当前没来到N-2位置 + noNextPreStatus = curStatus; + yesNextPreStatus = curStatus ^ 1; + noNextCurStatus = arr[nextIndex]; + yesNextCurStatus = arr[nextIndex] ^ 1; + } else if(nextIndex == arr.length - 1) {// 当前来到的就是N-2位置 + noNextPreStatus = curStatus; + yesNextPreStatus = curStatus ^ 1; + noNextCurStatus = endStatus; + yesNextCurStatus = endStatus ^ 1; + noEndStatus = endStatus; + yesEndStatus = endStatus ^ 1; + } + if(preStatus == 0) { + int next = process2(arr, nextIndex + 1, yesNextPreStatus, yesNextCurStatus, + nextIndex == arr.length - 1 ? yesEndStatus : endStatus, firstStatus); + return next == Integer.MAX_VALUE ? next : (next + 1); + }else { + return process2(arr, nextIndex + 1, noNextPreStatus, noNextCurStatus, + nextIndex == arr.length - 1 ? noEndStatus : endStatus, firstStatus); + + } +// int curStay = (nextIndex == arr.length - 1) ? endStatus : arr[nextIndex]; +// int curChange = (nextIndex == arr.length - 1) ? (endStatus ^ 1) : (arr[nextIndex] ^ 1); +// int endChange = (nextIndex == arr.length - 1) ? curChange : endStatus; +// if (preStatus == 0) { +// int next = process2(arr, nextIndex + 1, curStatus ^ 1, curChange, endChange, firstStatus); +// return next == Integer.MAX_VALUE ? next : (next + 1); +// } else { +// return process2(arr, nextIndex + 1, curStatus, curStay, endStatus, firstStatus); +// } + } + + // 有环改灯问题的迭代版本 + public static int loopMinStep2(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + if (arr.length == 1) { + return arr[0] == 1 ? 0 : 1; + } + if (arr.length == 2) { + return arr[0] != arr[1] ? Integer.MAX_VALUE : (arr[0] ^ 1); + } + if (arr.length == 3) { + return (arr[0] != arr[1] || arr[0] != arr[2]) ? Integer.MAX_VALUE : (arr[0] ^ 1); + } + // 0不变，1不变 + int p1 = traceLoop(arr, arr[1], arr[2], arr[arr.length - 1], arr[0]); + // 0改变，1不变 + int p2 = traceLoop(arr, arr[1] ^ 1, arr[2], arr[arr.length - 1] ^ 1, arr[0] ^ 1); + // 0不变，1改变 + int p3 = traceLoop(arr, arr[1] ^ 1, arr[2] ^ 1, arr[arr.length - 1], arr[0] ^ 1); + // 0改变，1改变 + int p4 = traceLoop(arr, arr[1], arr[2] ^ 1, arr[arr.length - 1] ^ 1, arr[0]); + p2 = p2 != Integer.MAX_VALUE ? (p2 + 1) : p2; + p3 = p3 != Integer.MAX_VALUE ? (p3 + 1) : p3; + p4 = p4 != Integer.MAX_VALUE ? (p4 + 2) : p4; + return Math.min(Math.min(p1, p2), Math.min(p3, p4)); + } + + public static int traceLoop(int[] arr, int preStatus, int curStatus, int endStatus, int firstStatus) { + int i = 3; + int op = 0; + while (i < arr.length - 1) { + if (preStatus == 0) { + op++; + preStatus = curStatus ^ 1; + curStatus = (arr[i++] ^ 1); + } else { + preStatus = curStatus; + curStatus = arr[i++]; + } + } + if (preStatus == 0) { + op++; + preStatus = curStatus ^ 1; + endStatus ^= 1; + curStatus = endStatus; + } else { + preStatus = curStatus; + curStatus = endStatus; + } + return (endStatus != firstStatus || endStatus != preStatus) ? Integer.MAX_VALUE : (op + (endStatus ^ 1)); + } + + // 生成长度为len的随机数组，值只有0和1两种值 + public static int[] randomArray(int len) { + int[] arr = new int[len]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) (Math.random() * 2); + } + return arr; + } + + public static void main(String[] args) { + System.out.println("如果没有任何Oops打印，说明所有方法都正确"); + System.out.println("test begin"); + int testTime = 20000; + int lenMax = 12; + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * lenMax); + int[] arr = randomArray(len); + int ans1 = noLoopRight(arr); + int ans2 = noLoopMinStep1(arr); + int ans3 = noLoopMinStep2(arr); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("1 Oops!"); + } + } + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * lenMax); + int[] arr = randomArray(len); + int ans1 = loopRight(arr); + int ans2 = loopMinStep1(arr); + int ans3 = loopMinStep2(arr); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("2 Oops!"); + } + } + System.out.println("test end"); + + int len = 100000000; + System.out.println("性能测试"); + System.out.println("数组大小：" + len); + int[] arr = randomArray(len); + long start = 0; + long end = 0; + start = System.currentTimeMillis(); + noLoopMinStep2(arr); + end = System.currentTimeMillis(); + System.out.println("noLoopMinStep2 run time: " + (end - start) + "(ms)"); + + start = System.currentTimeMillis(); + loopMinStep2(arr); + end = System.currentTimeMillis(); + System.out.println("loopMinStep2 run time: " + (end - start) + "(ms)"); + } + +} diff --git a/大厂刷题班/class09/Code02_RemoveInvalidParentheses.java b/大厂刷题班/class09/Code02_RemoveInvalidParentheses.java new file mode 100644 index 0000000..f13423e --- /dev/null +++ b/大厂刷题班/class09/Code02_RemoveInvalidParentheses.java @@ -0,0 +1,64 @@ +package class09; + +import java.util.ArrayList; +import java.util.List; + +// 测试链接 : https://leetcode.com/problems/remove-invalid-parentheses/ +public class Code02_RemoveInvalidParentheses { + + // 来自leetcode投票第一的答案，实现非常好，我们来赏析一下 + public static List removeInvalidParentheses(String s) { + List ans = new ArrayList<>(); + remove(s, ans, 0, 0, new char[] { '(', ')' }); + return ans; + } + + // modifyIndex <= checkIndex + // 只查s[checkIndex....]的部分，因为之前的一定已经调整对了 + // 但是之前的部分是怎么调整对的，调整到了哪？就是modifyIndex + // 比如： + // ( ( ) ( ) ) ) ... + // 0 1 2 3 4 5 6 + // 一开始当然checkIndex = 0，modifyIndex = 0 + // 当查到6的时候，发现不对了， + // 然后可以去掉2位置、4位置的 )，都可以 + // 如果去掉2位置的 ), 那么下一步就是 + // ( ( ( ) ) ) ... + // 0 1 2 3 4 5 6 + // checkIndex = 6 ，modifyIndex = 2 + // 如果去掉4位置的 ), 那么下一步就是 + // ( ( ) ( ) ) ... + // 0 1 2 3 4 5 6 + // checkIndex = 6 ，modifyIndex = 4 + // 也就是说， + // checkIndex和modifyIndex，分别表示查的开始 和 调的开始，之前的都不用管了 par ( ) + public static void remove(String s, List ans, int checkIndex, int deleteIndex, char[] par) { + for (int count = 0, i = checkIndex; i < s.length(); i++) { + if (s.charAt(i) == par[0]) { + count++; + } + if (s.charAt(i) == par[1]) { + count--; + } + // i check计数<0的第一个位置 + if (count < 0) { + for (int j = deleteIndex; j <= i; ++j) { + // 比如 + if (s.charAt(j) == par[1] && (j == deleteIndex || s.charAt(j - 1) != par[1])) { + remove( + s.substring(0, j) + s.substring(j + 1, s.length()), + ans, i, j, par); + } + } + return; + } + } + String reversed = new StringBuilder(s).reverse().toString(); + if (par[0] == '(') { + remove(reversed, ans, 0, 0, new char[] { ')', '(' }); + } else { + ans.add(reversed); + } + } + +} diff --git a/大厂刷题班/class09/Code03_LIS.java b/大厂刷题班/class09/Code03_LIS.java new file mode 100644 index 0000000..42e197b --- /dev/null +++ b/大厂刷题班/class09/Code03_LIS.java @@ -0,0 +1,55 @@ +package class09; + +// 本题测试链接 : https://leetcode.com/problems/longest-increasing-subsequence +public class Code03_LIS { + + public static int lengthOfLIS(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + // ends数组 + // ends[i]表示 : 目前所有长度为i+1的递增子序列的最小结尾 + int[] ends = new int[arr.length]; + // 根据含义, 一开始ends[0] = arr[0] + ends[0] = arr[0]; + // ends有效区范围是0...right，right往右为无效区 + // 所以一开始right = 0, 表示有效区只有0...0范围 + int right = 0; + // 最长递增子序列的长度 + // 全局变量，抓取每一步的答案，取最大的结果 + int max = 1; + for (int i = 1; i < arr.length; i++) { + int l = 0; + int r = right; + // 在ends[l...r]范围上二分 + // 如果 当前数(arr[i]) > ends[m]，砍掉左侧 + // 如果 当前数(arr[i]) <= ends[m]，砍掉右侧 + // 整个二分就是在ends里寻找 >= 当前数(arr[i])的最左位置 + // 就是从while里面出来时，l所在的位置。 + // 如果ends中不存在 >= 当前数(arr[i])的情况，将返回有效区的越界位置 + // 也就是从while里面出来时，l所在的位置，是有效区的越界位置 + // 比如 : ends = { 3, 5, 9, 12, 再往右无效} + // 如果当前数为8, 从while里面出来时，l将来到2位置 + // 比如 : ends = { 3, 5, 9, 12, 再往右无效} + // 如果当前数为13, 从while里面出来时，l将来到有效区的越界位置，4位置 + while (l <= r) { + int m = (l + r) / 2; + if (arr[i] > ends[m]) { + l = m + 1; + } else { + r = m - 1; + } + } + // 从while里面出来，看l的位置 + // 如果l比right大，说明扩充了有效区，那么right变量要随之变大 + // 如果l不比right大，说明l没有来到有效区的越界位置，right不变 + right = Math.max(right, l); + // l的位置，就是当前数应该填到ends数组里的位置 + ends[l] = arr[i]; + // 更新全局变量 + max = Math.max(max, l + 1); + } + return max; + } + +} \ No newline at end of file diff --git a/大厂刷题班/class09/Code04_EnvelopesProblem.java b/大厂刷题班/class09/Code04_EnvelopesProblem.java new file mode 100644 index 0000000..ea22dd4 --- /dev/null +++ b/大厂刷题班/class09/Code04_EnvelopesProblem.java @@ -0,0 +1,60 @@ +package class09; + +import java.util.Arrays; +import java.util.Comparator; + +// 本题测试链接 : https://leetcode.com/problems/russian-doll-envelopes/ +public class Code04_EnvelopesProblem { + + public static int maxEnvelopes(int[][] matrix) { + Envelope[] arr = sort(matrix); + int[] ends = new int[matrix.length]; + ends[0] = arr[0].h; + int right = 0; + int l = 0; + int r = 0; + int m = 0; + for (int i = 1; i < arr.length; i++) { + l = 0; + r = right; + while (l <= r) { + m = (l + r) / 2; + if (arr[i].h > ends[m]) { + l = m + 1; + } else { + r = m - 1; + } + } + right = Math.max(right, l); + ends[l] = arr[i].h; + } + return right + 1; + } + + public static class Envelope { + public int l; + public int h; + + public Envelope(int weight, int hight) { + l = weight; + h = hight; + } + } + + public static class EnvelopeComparator implements Comparator { + @Override + public int compare(Envelope o1, Envelope o2) { + return o1.l != o2.l ? o1.l - o2.l : o2.h - o1.h; + } + } + + public static Envelope[] sort(int[][] matrix) { + Envelope[] res = new Envelope[matrix.length]; + for (int i = 0; i < matrix.length; i++) { + res[i] = new Envelope(matrix[i][0], matrix[i][1]); + } + Arrays.sort(res, new EnvelopeComparator()); + return res; + } + +} diff --git a/大厂刷题班/class09/Code05_IsStepSum.java b/大厂刷题班/class09/Code05_IsStepSum.java new file mode 100644 index 0000000..5bdd807 --- /dev/null +++ b/大厂刷题班/class09/Code05_IsStepSum.java @@ -0,0 +1,58 @@ +package class09; + +import java.util.HashMap; + +public class Code05_IsStepSum { + + public static boolean isStepSum(int stepSum) { + int L = 0; + int R = stepSum; + int M = 0; + int cur = 0; + while (L <= R) { + M = L + ((R - L) >> 1); + cur = stepSum(M); + if (cur == stepSum) { + return true; + } else if (cur < stepSum) { + L = M + 1; + } else { + R = M - 1; + } + } + return false; + } + + public static int stepSum(int num) { + int sum = 0; + while (num != 0) { + sum += num; + num /= 10; + } + return sum; + } + + // for test + public static HashMap generateStepSumNumberMap(int numMax) { + HashMap map = new HashMap<>(); + for (int i = 0; i <= numMax; i++) { + map.put(stepSum(i), i); + } + return map; + } + + // for test + public static void main(String[] args) { + int max = 1000000; + int maxStepSum = stepSum(max); + HashMap ans = generateStepSumNumberMap(max); + System.out.println("测试开始"); + for (int i = 0; i <= maxStepSum; i++) { + if (isStepSum(i) ^ ans.containsKey(i)) { + System.out.println("出错了！"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class10/Code01_JumpGame.java b/大厂刷题班/class10/Code01_JumpGame.java new file mode 100644 index 0000000..bcb2fe1 --- /dev/null +++ b/大厂刷题班/class10/Code01_JumpGame.java @@ -0,0 +1,23 @@ +package class10; + +// 本题测试链接 : https://leetcode.com/problems/jump-game-ii/ +public class Code01_JumpGame { + + public static int jump(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int step = 0; + int cur = 0; + int next = 0; + for (int i = 0; i < arr.length; i++) { + if (cur < i) { + step++; + cur = next; + } + next = Math.max(next, i + arr[i]); + } + return step; + } + +} diff --git a/大厂刷题班/class10/Code02_TopK.java b/大厂刷题班/class10/Code02_TopK.java new file mode 100644 index 0000000..d563da6 --- /dev/null +++ b/大厂刷题班/class10/Code02_TopK.java @@ -0,0 +1,154 @@ +package class10; + +import java.util.ArrayList; +import java.util.Comparator; +import java.util.HashMap; +import java.util.List; +import java.util.TreeSet; + +// 本题测试链接：https://www.lintcode.com/problem/top-k-frequent-words-ii/ +// 以上的代码不要粘贴, 把以下的代码粘贴进java环境编辑器 +// 把类名和构造方法名改成TopK, 可以直接通过 +public class Code02_TopK { + + private Node[] heap; + private int heapSize; + // 词频表 key abc value (abc,7) + private HashMap strNodeMap; + private HashMap nodeIndexMap; + private NodeHeapComp comp; + private TreeSet treeSet; + + public Code02_TopK(int K) { + heap = new Node[K]; + heapSize = 0; + strNodeMap = new HashMap(); + nodeIndexMap = new HashMap(); + comp = new NodeHeapComp(); + treeSet = new TreeSet<>(new NodeTreeSetComp()); + } + + public static class Node { + public String str; + public int times; + + public Node(String s, int t) { + str = s; + times = t; + } + } + + public static class NodeHeapComp implements Comparator { + + @Override + public int compare(Node o1, Node o2) { + return o1.times != o2.times ? (o1.times - o2.times) : (o2.str.compareTo(o1.str)); + } + + } + + public static class NodeTreeSetComp implements Comparator { + + @Override + public int compare(Node o1, Node o2) { + return o1.times != o2.times ? (o2.times - o1.times) : (o1.str.compareTo(o2.str)); + } + + } + + public void add(String str) { + if (heap.length == 0) { + return; + } + // str 找到对应节点 curNode + Node curNode = null; + // 对应节点 curNode 在堆上的位置 + int preIndex = -1; + if (!strNodeMap.containsKey(str)) { + curNode = new Node(str, 1); + strNodeMap.put(str, curNode); + nodeIndexMap.put(curNode, -1); + } else { + curNode = strNodeMap.get(str); + // 要在time++之前，先在treeSet中删掉 + // 原因是因为一但times++，curNode在treeSet中的排序就失效了 + // 这种失效会导致整棵treeSet出现问题 + if (treeSet.contains(curNode)) { + treeSet.remove(curNode); + } + curNode.times++; + preIndex = nodeIndexMap.get(curNode); + } + if (preIndex == -1) { + if (heapSize == heap.length) { + if (comp.compare(heap[0], curNode) < 0) { + treeSet.remove(heap[0]); + treeSet.add(curNode); + nodeIndexMap.put(heap[0], -1); + nodeIndexMap.put(curNode, 0); + heap[0] = curNode; + heapify(0, heapSize); + } + } else { + treeSet.add(curNode); + nodeIndexMap.put(curNode, heapSize); + heap[heapSize] = curNode; + heapInsert(heapSize++); + } + } else { + treeSet.add(curNode); + heapify(preIndex, heapSize); + } + } + + public List topk() { + ArrayList ans = new ArrayList<>(); + for (Node node : treeSet) { + ans.add(node.str); + } + return ans; + } + + private void heapInsert(int index) { + while (index != 0) { + int parent = (index - 1) / 2; + if (comp.compare(heap[index], heap[parent]) < 0) { + swap(parent, index); + index = parent; + } else { + break; + } + } + } + + private void heapify(int index, int heapSize) { + int l = index * 2 + 1; + int r = index * 2 + 2; + int smallest = index; + while (l < heapSize) { + if (comp.compare(heap[l], heap[index]) < 0) { + smallest = l; + } + if (r < heapSize && comp.compare(heap[r], heap[smallest]) < 0) { + smallest = r; + } + if (smallest != index) { + swap(smallest, index); + } else { + break; + } + index = smallest; + l = index * 2 + 1; + r = index * 2 + 2; + } + } + + private void swap(int index1, int index2) { + nodeIndexMap.put(heap[index1], index2); + nodeIndexMap.put(heap[index2], index1); + Node tmp = heap[index1]; + heap[index1] = heap[index2]; + heap[index2] = tmp; + } + +} \ No newline at end of file diff --git a/大厂刷题班/class10/Code03_KInversePairs.java b/大厂刷题班/class10/Code03_KInversePairs.java new file mode 100644 index 0000000..79a6775 --- /dev/null +++ b/大厂刷题班/class10/Code03_KInversePairs.java @@ -0,0 +1,43 @@ +package class10; + +// 测试链接 : https://leetcode.com/problems/k-inverse-pairs-array/ +public class Code03_KInversePairs { + + public static int kInversePairs(int n, int k) { + if (n < 1 || k < 0) { + return 0; + } + int[][] dp = new int[n + 1][k + 1]; + dp[0][0] = 1; + int mod = 1000000007; + for (int i = 1; i <= n; i++) { + dp[i][0] = 1; + for (int j = 1; j <= k; j++) { + dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % mod; + if (j >= i) { + dp[i][j] = (dp[i][j] - dp[i - 1][j - i] + mod) % mod; + } + } + } + return dp[n][k]; + } + + public static int kInversePairs2(int n, int k) { + if (n < 1 || k < 0) { + return 0; + } + int[][] dp = new int[n + 1][k + 1]; + dp[0][0] = 1; + for (int i = 1; i <= n; i++) { + dp[i][0] = 1; + for (int j = 1; j <= k; j++) { + dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; + if (j >= i) { + dp[i][j] -= dp[i - 1][j - i]; + } + } + } + return dp[n][k]; + } + +} diff --git a/大厂刷题班/class10/Code04_BSTtoDoubleLinkedList.java b/大厂刷题班/class10/Code04_BSTtoDoubleLinkedList.java new file mode 100644 index 0000000..b937b66 --- /dev/null +++ b/大厂刷题班/class10/Code04_BSTtoDoubleLinkedList.java @@ -0,0 +1,57 @@ +package class10; + +// 本题测试链接 : https://leetcode.com/problems/convert-binary-search-tree-to-sorted-doubly-linked-list/ +public class Code04_BSTtoDoubleLinkedList { + + // 提交时不要提交这个类 + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int data) { + this.value = data; + } + } + + // 提交下面的代码 + public static Node treeToDoublyList(Node head) { + if (head == null) { + return null; + } + Info allInfo = process(head); + allInfo.end.right = allInfo.start; + allInfo.start.left = allInfo.end; + return allInfo.start; + } + + public static class Info { + public Node start; + public Node end; + + public Info(Node start, Node end) { + this.start = start; + this.end = end; + } + } + + public static Info process(Node X) { + if (X == null) { + return new Info(null, null); + } + Info lInfo = process(X.left); + Info rInfo = process(X.right); + if (lInfo.end != null) { + lInfo.end.right = X; + } + X.left = lInfo.end; + X.right = rInfo.start; + if (rInfo.start != null) { + rInfo.start.left = X; + } + // 整体链表的头 lInfo.start != null ? lInfo.start : X + // 整体链表的尾 rInfo.end != null ? rInfo.end : X + return new Info(lInfo.start != null ? lInfo.start : X, rInfo.end != null ? rInfo.end : X); + } + +} \ No newline at end of file diff --git a/大厂刷题班/class10/Code05_BooleanEvaluation.java b/大厂刷题班/class10/Code05_BooleanEvaluation.java new file mode 100644 index 0000000..9219a79 --- /dev/null +++ b/大厂刷题班/class10/Code05_BooleanEvaluation.java @@ -0,0 +1,156 @@ +package class10; + +// 本题测试链接 : https://leetcode-cn.com/problems/boolean-evaluation-lcci/ +public class Code05_BooleanEvaluation { + + public static int countEval0(String express, int desired) { + if (express == null || express.equals("")) { + return 0; + } + char[] exp = express.toCharArray(); + int N = exp.length; + Info[][] dp = new Info[N][N]; + Info allInfo = func(exp, 0, exp.length - 1, dp); + return desired == 1 ? allInfo.t : allInfo.f; + } + + public static class Info { + public int t; + public int f; + + public Info(int tr, int fa) { + t = tr; + f = fa; + } + } + + // 限制: + // L...R上，一定有奇数个字符 + // L位置的字符和R位置的字符，非0即1，不能是逻辑符号！ + // 返回str[L...R]这一段，为true的方法数，和false的方法数 + public static Info func(char[] str, int L, int R, Info[][] dp) { + if (dp[L][R] != null) { + return dp[L][R]; + } + int t = 0; + int f = 0; + if (L == R) { + t = str[L] == '1' ? 1 : 0; + f = str[L] == '0' ? 1 : 0; + } else { // L..R >=3 + // 每一个种逻辑符号，split枚举的东西 + // 都去试试最后结合 + for (int split = L + 1; split < R; split += 2) { + Info leftInfo = func(str, L, split - 1, dp); + Info rightInfo = func(str, split + 1, R, dp); + int a = leftInfo.t; + int b = leftInfo.f; + int c = rightInfo.t; + int d = rightInfo.f; + switch (str[split]) { + case '&': + t += a * c; + f += b * c + b * d + a * d; + break; + case '|': + t += a * c + a * d + b * c; + f += b * d; + break; + case '^': + t += a * d + b * c; + f += a * c + b * d; + break; + } + } + + } + dp[L][R] = new Info(t, f); + return dp[L][R]; + } + + public static int countEval1(String express, int desired) { + if (express == null || express.equals("")) { + return 0; + } + char[] exp = express.toCharArray(); + return f(exp, desired, 0, exp.length - 1); + } + + public static int f(char[] str, int desired, int L, int R) { + if (L == R) { + if (str[L] == '1') { + return desired; + } else { + return desired ^ 1; + } + } + int res = 0; + if (desired == 1) { + for (int i = L + 1; i < R; i += 2) { + switch (str[i]) { + case '&': + res += f(str, 1, L, i - 1) * f(str, 1, i + 1, R); + break; + case '|': + res += f(str, 1, L, i - 1) * f(str, 0, i + 1, R); + res += f(str, 0, L, i - 1) * f(str, 1, i + 1, R); + res += f(str, 1, L, i - 1) * f(str, 1, i + 1, R); + break; + case '^': + res += f(str, 1, L, i - 1) * f(str, 0, i + 1, R); + res += f(str, 0, L, i - 1) * f(str, 1, i + 1, R); + break; + } + } + } else { + for (int i = L + 1; i < R; i += 2) { + switch (str[i]) { + case '&': + res += f(str, 0, L, i - 1) * f(str, 1, i + 1, R); + res += f(str, 1, L, i - 1) * f(str, 0, i + 1, R); + res += f(str, 0, L, i - 1) * f(str, 0, i + 1, R); + break; + case '|': + res += f(str, 0, L, i - 1) * f(str, 0, i + 1, R); + break; + case '^': + res += f(str, 1, L, i - 1) * f(str, 1, i + 1, R); + res += f(str, 0, L, i - 1) * f(str, 0, i + 1, R); + break; + } + } + } + return res; + } + + public static int countEval2(String express, int desired) { + if (express == null || express.equals("")) { + return 0; + } + char[] exp = express.toCharArray(); + int N = exp.length; + int[][][] dp = new int[2][N][N]; + dp[0][0][0] = exp[0] == '0' ? 1 : 0; + dp[1][0][0] = dp[0][0][0] ^ 1; + for (int i = 2; i < exp.length; i += 2) { + dp[0][i][i] = exp[i] == '1' ? 0 : 1; + dp[1][i][i] = exp[i] == '0' ? 0 : 1; + for (int j = i - 2; j >= 0; j -= 2) { + for (int k = j; k < i; k += 2) { + if (exp[k + 1] == '&') { + dp[1][j][i] += dp[1][j][k] * dp[1][k + 2][i]; + dp[0][j][i] += (dp[0][j][k] + dp[1][j][k]) * dp[0][k + 2][i] + dp[0][j][k] * dp[1][k + 2][i]; + } else if (exp[k + 1] == '|') { + dp[1][j][i] += (dp[0][j][k] + dp[1][j][k]) * dp[1][k + 2][i] + dp[1][j][k] * dp[0][k + 2][i]; + dp[0][j][i] += dp[0][j][k] * dp[0][k + 2][i]; + } else { + dp[1][j][i] += dp[0][j][k] * dp[1][k + 2][i] + dp[1][j][k] * dp[0][k + 2][i]; + dp[0][j][i] += dp[0][j][k] * dp[0][k + 2][i] + dp[1][j][k] * dp[1][k + 2][i]; + } + } + } + } + return dp[desired][0][N - 1]; + } + +} diff --git a/大厂刷题班/class11/Code01_MinimumInsertionStepsToMakeAStringPalindrome.java b/大厂刷题班/class11/Code01_MinimumInsertionStepsToMakeAStringPalindrome.java new file mode 100644 index 0000000..94e3bd8 --- /dev/null +++ b/大厂刷题班/class11/Code01_MinimumInsertionStepsToMakeAStringPalindrome.java @@ -0,0 +1,173 @@ +package class11; + +import java.util.ArrayList; +import java.util.List; + +// 本题测试链接 : https://leetcode.com/problems/minimum-insertion-steps-to-make-a-string-palindrome/ +public class Code01_MinimumInsertionStepsToMakeAStringPalindrome { + + // 测试链接只测了本题的第一问，直接提交可以通过 + public static int minInsertions(String s) { + if (s == null || s.length() < 2) { + return 0; + } + char[] str = s.toCharArray(); + int N = str.length; + int[][] dp = new int[N][N]; + for (int i = 0; i < N - 1; i++) { + dp[i][i + 1] = str[i] == str[i + 1] ? 0 : 1; + } + for (int i = N - 3; i >= 0; i--) { + for (int j = i + 2; j < N; j++) { + dp[i][j] = Math.min(dp[i][j - 1], dp[i + 1][j]) + 1; + if (str[i] == str[j]) { + dp[i][j] = Math.min(dp[i][j], dp[i + 1][j - 1]); + } + } + } + return dp[0][N - 1]; + } + + // 本题第二问，返回其中一种结果 + public static String minInsertionsOneWay(String s) { + if (s == null || s.length() < 2) { + return s; + } + char[] str = s.toCharArray(); + int N = str.length; + int[][] dp = new int[N][N]; + for (int i = 0; i < N - 1; i++) { + dp[i][i + 1] = str[i] == str[i + 1] ? 0 : 1; + } + for (int i = N - 3; i >= 0; i--) { + for (int j = i + 2; j < N; j++) { + dp[i][j] = Math.min(dp[i][j - 1], dp[i + 1][j]) + 1; + if (str[i] == str[j]) { + dp[i][j] = Math.min(dp[i][j], dp[i + 1][j - 1]); + } + } + } + + int L = 0; + int R = N - 1; + char[] ans = new char[N + dp[L][R]]; + int ansl = 0; + int ansr = ans.length - 1; + while (L < R) { + if (dp[L][R - 1] == dp[L][R] - 1) { + ans[ansl++] = str[R]; + ans[ansr--] = str[R--]; + } else if (dp[L + 1][R] == dp[L][R] - 1) { + ans[ansl++] = str[L]; + ans[ansr--] = str[L++]; + } else { + ans[ansl++] = str[L++]; + ans[ansr--] = str[R--]; + } + } + if (L == R) { + ans[ansl] = str[L]; + } + return String.valueOf(ans); + } + + // 本题第三问，返回所有可能的结果 + public static List minInsertionsAllWays(String s) { + List ans = new ArrayList<>(); + if (s == null || s.length() < 2) { + ans.add(s); + } else { + char[] str = s.toCharArray(); + int N = str.length; + int[][] dp = new int[N][N]; + for (int i = 0; i < N - 1; i++) { + dp[i][i + 1] = str[i] == str[i + 1] ? 0 : 1; + } + for (int i = N - 3; i >= 0; i--) { + for (int j = i + 2; j < N; j++) { + dp[i][j] = Math.min(dp[i][j - 1], dp[i + 1][j]) + 1; + if (str[i] == str[j]) { + dp[i][j] = Math.min(dp[i][j], dp[i + 1][j - 1]); + } + } + } + int M = N + dp[0][N - 1]; + char[] path = new char[M]; + process(str, dp, 0, N - 1, path, 0, M - 1, ans); + } + return ans; + } + + // 当前来到的动态规划中的格子，(L,R) + // path .... [pl....pr] .... + public static void process(char[] str, int[][] dp, int L, int R, char[] path, int pl, int pr, List ans) { + if (L >= R) { // L > R L==R + if (L == R) { + path[pl] = str[L]; + } + ans.add(String.valueOf(path)); + } else { + if (dp[L][R - 1] == dp[L][R] - 1) { + path[pl] = str[R]; + path[pr] = str[R]; + process(str, dp, L, R - 1, path, pl + 1, pr - 1, ans); + } + if (dp[L + 1][R] == dp[L][R] - 1) { + path[pl] = str[L]; + path[pr] = str[L]; + process(str, dp, L + 1, R, path, pl + 1, pr - 1, ans); + } + if (str[L] == str[R] && (L == R - 1 || dp[L + 1][R - 1] == dp[L][R])) { + path[pl] = str[L]; + path[pr] = str[R]; + process(str, dp, L + 1, R - 1, path, pl + 1, pr - 1, ans); + } + } + } + + public static void main(String[] args) { + String s = null; + String ans2 = null; + List ans3 = null; + + System.out.println("本题第二问，返回其中一种结果测试开始"); + s = "mbadm"; + ans2 = minInsertionsOneWay(s); + System.out.println(ans2); + + s = "leetcode"; + ans2 = minInsertionsOneWay(s); + System.out.println(ans2); + + s = "aabaa"; + ans2 = minInsertionsOneWay(s); + System.out.println(ans2); + System.out.println("本题第二问，返回其中一种结果测试结束"); + + System.out.println(); + + System.out.println("本题第三问，返回所有可能的结果测试开始"); + s = "mbadm"; + ans3 = minInsertionsAllWays(s); + for (String way : ans3) { + System.out.println(way); + } + System.out.println(); + + s = "leetcode"; + ans3 = minInsertionsAllWays(s); + for (String way : ans3) { + System.out.println(way); + } + System.out.println(); + + s = "aabaa"; + ans3 = minInsertionsAllWays(s); + for (String way : ans3) { + System.out.println(way); + } + System.out.println(); + System.out.println("本题第三问，返回所有可能的结果测试结束"); + } + +} diff --git a/大厂刷题班/class11/Code02_PalindromePartitioningII.java b/大厂刷题班/class11/Code02_PalindromePartitioningII.java new file mode 100644 index 0000000..08633db --- /dev/null +++ b/大厂刷题班/class11/Code02_PalindromePartitioningII.java @@ -0,0 +1,207 @@ +package class11; + +import java.util.ArrayList; +import java.util.List; + +// 本题测试链接 : https://leetcode.com/problems/palindrome-partitioning-ii/ +public class Code02_PalindromePartitioningII { + + // 测试链接只测了本题的第一问，直接提交可以通过 + public static int minCut(String s) { + if (s == null || s.length() < 2) { + return 0; + } + char[] str = s.toCharArray(); + int N = str.length; + boolean[][] checkMap = createCheckMap(str, N); + int[] dp = new int[N + 1]; + dp[N] = 0; + for (int i = N - 1; i >= 0; i--) { + if (checkMap[i][N - 1]) { + dp[i] = 1; + } else { + int next = Integer.MAX_VALUE; + for (int j = i; j < N; j++) { + if (checkMap[i][j]) { + next = Math.min(next, dp[j + 1]); + } + } + dp[i] = 1 + next; + } + } + return dp[0] - 1; + } + + public static boolean[][] createCheckMap(char[] str, int N) { + boolean[][] ans = new boolean[N][N]; + for (int i = 0; i < N - 1; i++) { + ans[i][i] = true; + ans[i][i + 1] = str[i] == str[i + 1]; + } + ans[N - 1][N - 1] = true; + for (int i = N - 3; i >= 0; i--) { + for (int j = i + 2; j < N; j++) { + ans[i][j] = str[i] == str[j] && ans[i + 1][j - 1]; + } + } + return ans; + } + + // 本题第二问，返回其中一种结果 + public static List minCutOneWay(String s) { + List ans = new ArrayList<>(); + if (s == null || s.length() < 2) { + ans.add(s); + } else { + char[] str = s.toCharArray(); + int N = str.length; + boolean[][] checkMap = createCheckMap(str, N); + int[] dp = new int[N + 1]; + dp[N] = 0; + for (int i = N - 1; i >= 0; i--) { + if (checkMap[i][N - 1]) { + dp[i] = 1; + } else { + int next = Integer.MAX_VALUE; + for (int j = i; j < N; j++) { + if (checkMap[i][j]) { + next = Math.min(next, dp[j + 1]); + } + } + dp[i] = 1 + next; + } + } + // dp[i] (0....5) 回文！ dp[0] == dp[6] + 1 + // (0....5) 6 + for (int i = 0, j = 1; j <= N; j++) { + if (checkMap[i][j - 1] && dp[i] == dp[j] + 1) { + ans.add(s.substring(i, j)); + i = j; + } + } + } + return ans; + } + + // 本题第三问，返回所有结果 + public static List> minCutAllWays(String s) { + List> ans = new ArrayList<>(); + if (s == null || s.length() < 2) { + List cur = new ArrayList<>(); + cur.add(s); + ans.add(cur); + } else { + char[] str = s.toCharArray(); + int N = str.length; + boolean[][] checkMap = createCheckMap(str, N); + int[] dp = new int[N + 1]; + dp[N] = 0; + for (int i = N - 1; i >= 0; i--) { + if (checkMap[i][N - 1]) { + dp[i] = 1; + } else { + int next = Integer.MAX_VALUE; + for (int j = i; j < N; j++) { + if (checkMap[i][j]) { + next = Math.min(next, dp[j + 1]); + } + } + dp[i] = 1 + next; + } + } + process(s, 0, 1, checkMap, dp, new ArrayList<>(), ans); + } + return ans; + } + + // s[0....i-1] 存到path里去了 + // s[i..j-1]考察的分出来的第一份 + public static void process(String s, int i, int j, boolean[][] checkMap, int[] dp, + List path, + List> ans) { + if (j == s.length()) { // s[i...N-1] + if (checkMap[i][j - 1] && dp[i] == dp[j] + 1) { + path.add(s.substring(i, j)); + ans.add(copyStringList(path)); + path.remove(path.size() - 1); + } + } else {// s[i...j-1] + if (checkMap[i][j - 1] && dp[i] == dp[j] + 1) { + path.add(s.substring(i, j)); + process(s, j, j + 1, checkMap, dp, path, ans); + path.remove(path.size() - 1); + } + process(s, i, j + 1, checkMap, dp, path, ans); + } + } + + public static List copyStringList(List list) { + List ans = new ArrayList<>(); + for (String str : list) { + ans.add(str); + } + return ans; + } + + public static void main(String[] args) { + String s = null; + List ans2 = null; + List> ans3 = null; + + System.out.println("本题第二问，返回其中一种结果测试开始"); + s = "abacbc"; + ans2 = minCutOneWay(s); + for (String str : ans2) { + System.out.print(str + " "); + } + System.out.println(); + + s = "aabccbac"; + ans2 = minCutOneWay(s); + for (String str : ans2) { + System.out.print(str + " "); + } + System.out.println(); + + s = "aabaa"; + ans2 = minCutOneWay(s); + for (String str : ans2) { + System.out.print(str + " "); + } + System.out.println(); + System.out.println("本题第二问，返回其中一种结果测试结束"); + System.out.println(); + System.out.println("本题第三问，返回所有可能结果测试开始"); + s = "cbbbcbc"; + ans3 = minCutAllWays(s); + for (List way : ans3) { + for (String str : way) { + System.out.print(str + " "); + } + System.out.println(); + } + System.out.println(); + + s = "aaaaaa"; + ans3 = minCutAllWays(s); + for (List way : ans3) { + for (String str : way) { + System.out.print(str + " "); + } + System.out.println(); + } + System.out.println(); + + s = "fcfffcffcc"; + ans3 = minCutAllWays(s); + for (List way : ans3) { + for (String str : way) { + System.out.print(str + " "); + } + System.out.println(); + } + System.out.println(); + System.out.println("本题第三问，返回所有可能结果测试结束"); + } + +} diff --git a/大厂刷题班/class12/Code01_ContainAllCharExactly.java b/大厂刷题班/class12/Code01_ContainAllCharExactly.java new file mode 100644 index 0000000..1774deb --- /dev/null +++ b/大厂刷题班/class12/Code01_ContainAllCharExactly.java @@ -0,0 +1,125 @@ +package class12; + +import java.util.Arrays; + +// 本题测试链接 : https://leetcode.com/problems/permutation-in-string/ +public class Code01_ContainAllCharExactly { + + public static int containExactly1(String s, String a) { + if (s == null || a == null || s.length() < a.length()) { + return -1; + } + char[] aim = a.toCharArray(); + Arrays.sort(aim); + String aimSort = String.valueOf(aim); + for (int L = 0; L < s.length(); L++) { + for (int R = L; R < s.length(); R++) { + char[] cur = s.substring(L, R + 1).toCharArray(); + Arrays.sort(cur); + String curSort = String.valueOf(cur); + if (curSort.equals(aimSort)) { + return L; + } + } + } + return -1; + } + + public static int containExactly2(String s, String a) { + if (s == null || a == null || s.length() < a.length()) { + return -1; + } + char[] str = s.toCharArray(); + char[] aim = a.toCharArray(); + for (int L = 0; L <= str.length - aim.length; L++) { + if (isCountEqual(str, L, aim)) { + return L; + } + } + return -1; + } + + public static boolean isCountEqual(char[] str, int L, char[] aim) { + int[] count = new int[256]; + for (int i = 0; i < aim.length; i++) { + count[aim[i]]++; + } + for (int i = 0; i < aim.length; i++) { + if (count[str[L + i]]-- == 0) { + return false; + } + } + return true; + } + + public static int containExactly3(String s1, String s2) { + if (s1 == null || s2 == null || s1.length() < s2.length()) { + return -1; + } + char[] str2 = s2.toCharArray(); + int M = str2.length; + int[] count = new int[256]; + for (int i = 0; i < M; i++) { + count[str2[i]]++; + } + int all = M; + char[] str1 = s1.toCharArray(); + int R = 0; + // 0~M-1 + for (; R < M; R++) { // 最早的M个字符，让其窗口初步形成 + if (count[str1[R]]-- > 0) { + all--; + } + } + // 窗口初步形成了，并没有判断有效无效，决定下一个位置一上来判断 + // 接下来的过程，窗口右进一个，左吐一个 + for (; R < str1.length; R++) { + if (all == 0) { // R-1 + return R - M; + } + if (count[str1[R]]-- > 0) { + all--; + } + if (count[str1[R - M]]++ >= 0) { + all++; + } + } + return all == 0 ? R - M : -1; + } + + // for test + public static String getRandomString(int possibilities, int maxSize) { + char[] ans = new char[(int) (Math.random() * maxSize) + 1]; + for (int i = 0; i < ans.length; i++) { + ans[i] = (char) ((int) (Math.random() * possibilities) + 'a'); + } + return String.valueOf(ans); + } + + public static void main(String[] args) { + int possibilities = 5; + int strMaxSize = 20; + int aimMaxSize = 10; + int testTimes = 500000; + System.out.println("test begin, test time : " + testTimes); + for (int i = 0; i < testTimes; i++) { + String str = getRandomString(possibilities, strMaxSize); + String aim = getRandomString(possibilities, aimMaxSize); + int ans1 = containExactly1(str, aim); + int ans2 = containExactly2(str, aim); + int ans3 = containExactly3(str, aim); + if (ans1 != ans2 || ans2 != ans3) { + System.out.println("Oops!"); + System.out.println(str); + System.out.println(aim); + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + break; + } + } + System.out.println("test finish"); + + } + +} diff --git a/大厂刷题班/class12/Code03_FindKthMinNumber.java b/大厂刷题班/class12/Code03_FindKthMinNumber.java new file mode 100644 index 0000000..c36cbda --- /dev/null +++ b/大厂刷题班/class12/Code03_FindKthMinNumber.java @@ -0,0 +1,103 @@ +package class12; + +// 本题测试链接 : https://leetcode.com/problems/median-of-two-sorted-arrays/ +public class Code03_FindKthMinNumber { + + public double findMedianSortedArrays(int[] nums1, int[] nums2) { + int size = nums1.length + nums2.length; + boolean even = (size & 1) == 0; + if (nums1.length != 0 && nums2.length != 0) { + if (even) { + return (double) (findKthNum(nums1, nums2, size / 2) + findKthNum(nums1, nums2, size / 2 + 1)) / 2D; + } else { + return findKthNum(nums1, nums2, size / 2 + 1); + } + } else if (nums1.length != 0) { + if (even) { + return (double) (nums1[(size - 1) / 2] + nums1[size / 2]) / 2; + } else { + return nums1[size / 2]; + } + } else if (nums2.length != 0) { + if (even) { + return (double) (nums2[(size - 1) / 2] + nums2[size / 2]) / 2; + } else { + return nums2[size / 2]; + } + } else { + return 0; + } + } + + // 进阶问题 : 在两个都有序的数组中，找整体第K小的数 + // 可以做到O(log(Min(M,N))) + public static int findKthNum(int[] arr1, int[] arr2, int kth) { + int[] longs = arr1.length >= arr2.length ? arr1 : arr2; + int[] shorts = arr1.length < arr2.length ? arr1 : arr2; + int l = longs.length; + int s = shorts.length; + if (kth <= s) { // 1) + return getUpMedian(shorts, 0, kth - 1, longs, 0, kth - 1); + } + if (kth > l) { // 3) + if (shorts[kth - l - 1] >= longs[l - 1]) { + return shorts[kth - l - 1]; + } + if (longs[kth - s - 1] >= shorts[s - 1]) { + return longs[kth - s - 1]; + } + return getUpMedian(shorts, kth - l, s - 1, longs, kth - s, l - 1); + } + // 2) s < k <= l + if (longs[kth - s - 1] >= shorts[s - 1]) { + return longs[kth - s - 1]; + } + return getUpMedian(shorts, 0, s - 1, longs, kth - s, kth - 1); + } + + + + + // A[s1...e1] + // B[s2...e2] + // 一定等长！ + // 返回整体的，上中位数！8（4） 10（5） 12（6） + public static int getUpMedian(int[] A, int s1, int e1, int[] B, int s2, int e2) { + int mid1 = 0; + int mid2 = 0; + while (s1 < e1) { + // mid1 = s1 + (e1 - s1) >> 1 + mid1 = (s1 + e1) / 2; + mid2 = (s2 + e2) / 2; + if (A[mid1] == B[mid2]) { + return A[mid1]; + } + // 两个中点一定不等！ + if (((e1 - s1 + 1) & 1) == 1) { // 奇数长度 + if (A[mid1] > B[mid2]) { + if (B[mid2] >= A[mid1 - 1]) { + return B[mid2]; + } + e1 = mid1 - 1; + s2 = mid2 + 1; + } else { // A[mid1] < B[mid2] + if (A[mid1] >= B[mid2 - 1]) { + return A[mid1]; + } + e2 = mid2 - 1; + s1 = mid1 + 1; + } + } else { // 偶数长度 + if (A[mid1] > B[mid2]) { + e1 = mid1; + s2 = mid2 + 1; + } else { + e2 = mid2; + s1 = mid1 + 1; + } + } + } + return Math.min(A[s1], B[s2]); + } + +} diff --git a/大厂刷题班/class12/Code03_LongestConsecutive.java b/大厂刷题班/class12/Code03_LongestConsecutive.java new file mode 100644 index 0000000..8beb48d --- /dev/null +++ b/大厂刷题班/class12/Code03_LongestConsecutive.java @@ -0,0 +1,65 @@ +package class12; + +import java.util.HashMap; +import java.util.HashSet; + +// 本题测试链接 : https://leetcode.com/problems/longest-consecutive-sequence/ +public class Code03_LongestConsecutive { + + // 课上讲的解法 + public static int longestConsecutive(int[] nums) { + HashMap map = new HashMap<>(); + int len = 0; + for (int num : nums) { + if (!map.containsKey(num)) { + map.put(num, 1); + int preLen = map.containsKey(num - 1) ? map.get(num - 1) : 0; + int posLen = map.containsKey(num + 1) ? map.get(num + 1) : 0; + int all = preLen + posLen + 1; + map.put(num - preLen, all); + map.put(num + posLen, all); + len = Math.max(len, all); + } + } + return len; + } + + // 补充一个两张表：头表、尾表。非常好理解的方法 + // 不是最优解，但是好理解 + public static int longestConsecutive2(int[] nums) { + HashMap headMap = new HashMap(); + HashMap tailMap = new HashMap(); + HashSet visited = new HashSet<>(); + for (int num : nums) { + if (!visited.contains(num)) { + visited.add(num); + headMap.put(num, 1); + tailMap.put(num, 1); + if (tailMap.containsKey(num - 1)) { + int preLen = tailMap.get(num - 1); + int preHead = num - preLen; + headMap.put(preHead, preLen + 1); + tailMap.put(num, preLen + 1); + headMap.remove(num); + tailMap.remove(num - 1); + } + if (headMap.containsKey(num + 1)) { + int preLen = tailMap.get(num); + int preHead = num - preLen + 1; + int postLen = headMap.get(num + 1); + int postTail = num + postLen; + headMap.put(preHead, preLen + postLen); + tailMap.put(postTail, preLen + postLen); + headMap.remove(num + 1); + tailMap.remove(num); + } + } + } + int ans = 0; + for (int len : headMap.values()) { + ans = Math.max(ans, len); + } + return ans; + } + +} diff --git a/大厂刷题班/class12/Code04_RegularExpressionMatch.java b/大厂刷题班/class12/Code04_RegularExpressionMatch.java new file mode 100644 index 0000000..c4b35b7 --- /dev/null +++ b/大厂刷题班/class12/Code04_RegularExpressionMatch.java @@ -0,0 +1,135 @@ +package class12; + +// 测试链接 : https://leetcode.com/problems/regular-expression-matching/ +public class Code04_RegularExpressionMatch { + + public static boolean isValid(char[] s, char[] e) { + // s中不能有'.' or '*' + for (int i = 0; i < s.length; i++) { + if (s[i] == '*' || s[i] == '.') { + return false; + } + } + // 开头的e[0]不能是'*'，没有相邻的'*' + for (int i = 0; i < e.length; i++) { + if (e[i] == '*' && (i == 0 || e[i - 1] == '*')) { + return false; + } + } + return true; + } + + // 初始尝试版本，不包含斜率优化 + public static boolean isMatch1(String str, String exp) { + if (str == null || exp == null) { + return false; + } + char[] s = str.toCharArray(); + char[] e = exp.toCharArray(); + return isValid(s, e) && process(s, e, 0, 0); + } + + // str[si.....] 能不能被 exp[ei.....]配出来！ true false + public static boolean process(char[] s, char[] e, int si, int ei) { + if (ei == e.length) { // exp 没了 str？ + return si == s.length; + } + // exp[ei]还有字符 + // ei + 1位置的字符，不是* + if (ei + 1 == e.length || e[ei + 1] != '*') { + // ei + 1 不是* + // str[si] 必须和 exp[ei] 能配上！ + return si != s.length && (e[ei] == s[si] || e[ei] == '.') && process(s, e, si + 1, ei + 1); + } + // exp[ei]还有字符 + // ei + 1位置的字符，是*! + while (si != s.length && (e[ei] == s[si] || e[ei] == '.')) { + if (process(s, e, si, ei + 2)) { + return true; + } + si++; + } + return process(s, e, si, ei + 2); + } + + // 改记忆化搜索+斜率优化 + public static boolean isMatch2(String str, String exp) { + if (str == null || exp == null) { + return false; + } + char[] s = str.toCharArray(); + char[] e = exp.toCharArray(); + if (!isValid(s, e)) { + return false; + } + int[][] dp = new int[s.length + 1][e.length + 1]; + // dp[i][j] = 0, 没算过！ + // dp[i][j] = -1 算过，返回值是false + // dp[i][j] = 1 算过，返回值是true + return isValid(s, e) && process2(s, e, 0, 0, dp); + } + + public static boolean process2(char[] s, char[] e, int si, int ei, int[][] dp) { + if (dp[si][ei] != 0) { + return dp[si][ei] == 1; + } + boolean ans = false; + if (ei == e.length) { + ans = si == s.length; + } else { + if (ei + 1 == e.length || e[ei + 1] != '*') { + ans = si != s.length && (e[ei] == s[si] || e[ei] == '.') && process2(s, e, si + 1, ei + 1, dp); + } else { + if (si == s.length) { // ei ei+1 * + ans = process2(s, e, si, ei + 2, dp); + } else { // si没结束 + if (s[si] != e[ei] && e[ei] != '.') { + ans = process2(s, e, si, ei + 2, dp); + } else { // s[si] 可以和 e[ei]配上 + ans = process2(s, e, si, ei + 2, dp) || process2(s, e, si + 1, ei, dp); + } + } + } + } + dp[si][ei] = ans ? 1 : -1; + return ans; + } + + // 动态规划版本 + 斜率优化 + public static boolean isMatch3(String str, String pattern) { + if (str == null || pattern == null) { + return false; + } + char[] s = str.toCharArray(); + char[] p = pattern.toCharArray(); + if (!isValid(s, p)) { + return false; + } + int N = s.length; + int M = p.length; + boolean[][] dp = new boolean[N + 1][M + 1]; + dp[N][M] = true; + for (int j = M - 1; j >= 0; j--) { + dp[N][j] = (j + 1 < M && p[j + 1] == '*') && dp[N][j + 2]; + } + // dp[0..N-2][M-1]都等于false，只有dp[N-1][M-1]需要讨论 + if (N > 0 && M > 0) { + dp[N - 1][M - 1] = (s[N - 1] == p[M - 1] || p[M - 1] == '.'); + } + for (int i = N - 1; i >= 0; i--) { + for (int j = M - 2; j >= 0; j--) { + if (p[j + 1] != '*') { + dp[i][j] = ((s[i] == p[j]) || (p[j] == '.')) && dp[i + 1][j + 1]; + } else { + if ((s[i] == p[j] || p[j] == '.') && dp[i + 1][j]) { + dp[i][j] = true; + } else { + dp[i][j] = dp[i][j + 2]; + } + } + } + } + return dp[0][0]; + } + +} diff --git a/大厂刷题班/class13/Code01_NCardsABWin.java b/大厂刷题班/class13/Code01_NCardsABWin.java new file mode 100644 index 0000000..4a2b832 --- /dev/null +++ b/大厂刷题班/class13/Code01_NCardsABWin.java @@ -0,0 +1,169 @@ +package class13; + +import java.text.DecimalFormat; + +public class Code01_NCardsABWin { + + // 谷歌面试题 + // 面值为1~10的牌组成一组， + // 每次你从组里等概率的抽出1~10中的一张 + // 下次抽会换一个新的组，有无限组 + // 当累加和<17时，你将一直抽牌 + // 当累加和>=17且<21时，你将获胜 + // 当累加和>=21时，你将失败 + // 返回获胜的概率 + public static double f1() { + return p1(0); + } + + // 游戏的规则，如上 + // 当你来到cur这个累加和的时候，获胜概率是多少返回！ + public static double p1(int cur) { + if (cur >= 17 && cur < 21) { + return 1.0; + } + if (cur >= 21) { + return 0.0; + } + double w = 0.0; + for (int i = 1; i <= 10; i++) { + w += p1(cur + i); + } + return w / 10; + } + + // 谷歌面试题扩展版 + // 面值为1~N的牌组成一组， + // 每次你从组里等概率的抽出1~N中的一张 + // 下次抽会换一个新的组，有无限组 + // 当累加和=a且=b时，你将失败 + // 返回获胜的概率，给定的参数为N，a，b + public static double f2(int N, int a, int b) { + if (N < 1 || a >= b || a < 0 || b < 0) { + return 0.0; + } + if (b - a >= N) { + return 1.0; + } + // 所有参数都合法，并且b-a < N + return p2(0, N, a, b); + } + + // 游戏规则，如上，int N, int a, int b，固定参数！ + // cur，目前到达了cur的累加和 + // 返回赢的概率 + public static double p2(int cur, int N, int a, int b) { + if (cur >= a && cur < b) { + return 1.0; + } + if (cur >= b) { + return 0.0; + } + double w = 0.0; + for (int i = 1; i <= N; i++) { + w += p2(cur + i, N, a, b); + } + return w / N; + } + + // f2的改进版本，用到了观察位置优化枚举的技巧 + // 可以课上讲一下 + public static double f3(int N, int a, int b) { + if (N < 1 || a >= b || a < 0 || b < 0) { + return 0.0; + } + if (b - a >= N) { + return 1.0; + } + return p3(0, N, a, b); + } + + public static double p3(int cur, int N, int a, int b) { + if (cur >= a && cur < b) { + return 1.0; + } + if (cur >= b) { + return 0.0; + } + if (cur == a - 1) { + return 1.0 * (b - a) / N; + } + double w = p3(cur + 1, N, a, b) + p3(cur + 1, N, a, b) * N; + if (cur + 1 + N < b) { + w -= p3(cur + 1 + N, N, a, b); + } + return w / N; + } + + // f3的改进版本的动态规划 + // 可以课上讲一下 + public static double f4(int N, int a, int b) { + if (N < 1 || a >= b || a < 0 || b < 0) { + return 0.0; + } + if (b - a >= N) { + return 1.0; + } + double[] dp = new double[b]; + for (int i = a; i < b; i++) { + dp[i] = 1.0; + } + if (a - 1 >= 0) { + dp[a - 1] = 1.0 * (b - a) / N; + } + for (int cur = a - 2; cur >= 0; cur--) { + double w = dp[cur + 1] + dp[cur + 1] * N; + if (cur + 1 + N < b) { + w -= dp[cur + 1 + N]; + } + dp[cur] = w / N; + } + return dp[0]; + } + + public static void main(String[] args) { + int N = 10; + int a = 17; + int b = 21; + System.out.println("N = " + N + ", a = " + a + ", b = " + b); + System.out.println(f1()); + System.out.println(f2(N, a, b)); + System.out.println(f3(N, a, b)); + System.out.println(f4(N, a, b)); + + int maxN = 15; + int maxM = 20; + int testTime = 100000; + System.out.println("测试开始"); + System.out.println("比对double类型答案可能会有精度对不准的问题"); + System.out.println("所以答案一律只保留小数点后四位进行比对"); + System.out.println("如果没有错误提示, 说明验证通过"); + DecimalFormat df = new DecimalFormat("#.####"); + for (int i = 0; i < testTime; i++) { + N = (int) (Math.random() * maxN); + a = (int) (Math.random() * maxM); + b = (int) (Math.random() * maxM); + double ans2 = Double.valueOf(df.format(f2(N, a, b))); + double ans3 = Double.valueOf(df.format(f3(N, a, b))); + double ans4 = Double.valueOf(df.format(f4(N, a, b))); + if (ans2 != ans3 || ans2 != ans4) { + System.out.println("Oops!"); + System.out.println(N + " , " + a + " , " + b); + System.out.println(ans2); + System.out.println(ans3); + System.out.println(ans4); + } + } + System.out.println("测试结束"); + + N = 10000; + a = 67834; + b = 72315; + System.out.println("N = " + N + ", a = " + a + ", b = " + b + "时, 除了方法4外都超时"); + System.out.print("方法4答案: "); + System.out.println(f4(N, a, b)); + } + +} diff --git a/大厂刷题班/class13/Code02_SuperWashingMachines.java b/大厂刷题班/class13/Code02_SuperWashingMachines.java new file mode 100644 index 0000000..bd23ae4 --- /dev/null +++ b/大厂刷题班/class13/Code02_SuperWashingMachines.java @@ -0,0 +1,34 @@ +package class13; + +// 本题测试链接 : https://leetcode.com/problems/super-washing-machines/ +public class Code02_SuperWashingMachines { + + public static int findMinMoves(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int size = arr.length; + int sum = 0; + for (int i = 0; i < size; i++) { + sum += arr[i]; + } + if (sum % size != 0) { + return -1; + } + int avg = sum / size; + int leftSum = 0; + int ans = 0; + for (int i = 0; i < arr.length; i++) { + int leftRest = leftSum - i * avg; + int rightRest = (sum - leftSum - arr[i]) - (size - i - 1) * avg; + if (leftRest < 0 && rightRest < 0) { + ans = Math.max(ans, Math.abs(leftRest) + Math.abs(rightRest)); + } else { + ans = Math.max(ans, Math.max(Math.abs(leftRest), Math.abs(rightRest))); + } + leftSum += arr[i]; + } + return ans; + } + +} diff --git a/大厂刷题班/class13/Code03_ScrambleString.java b/大厂刷题班/class13/Code03_ScrambleString.java new file mode 100644 index 0000000..49244c4 --- /dev/null +++ b/大厂刷题班/class13/Code03_ScrambleString.java @@ -0,0 +1,186 @@ +package class13; + +// 本题测试链接 : https://leetcode.com/problems/scramble-string/ +public class Code03_ScrambleString { + + public static boolean isScramble0(String s1, String s2) { + if ((s1 == null && s2 != null) || (s1 != null && s2 == null)) { + return false; + } + if (s1 == null && s2 == null) { + return true; + } + if (s1.equals(s2)) { + return true; + } + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + if (!sameTypeSameNumber(str1, str2)) { + return false; + } + return process0(str1, 0, str1.length - 1, str2, 0, str2.length - 1); + } + + // str1[L1...R1] str2[L2...R2] 是否互为玄变串 + // 一定保证这两段是等长的！ + public static boolean process0(char[] str1, int L1, int R1, char[] str2, int L2, int R2) { + if (L1 == R1) { + return str1[L1] == str2[L2]; + } + for (int leftEnd = L1; leftEnd < R1; leftEnd++) { + boolean p1 = process0(str1, L1, leftEnd, str2, L2, L2 + leftEnd - L1) + && process0(str1, leftEnd + 1, R1, str2, L2 + leftEnd - L1 + 1, R2); + boolean p2 = process0(str1, L1, leftEnd, str2, R2 - (leftEnd - L1), R2) + && process0(str1, leftEnd + 1, R1, str2, L2, R2 - (leftEnd - L1) - 1); + if (p1 || p2) { + return true; + } + } + return false; + } + + public static boolean sameTypeSameNumber(char[] str1, char[] str2) { + if (str1.length != str2.length) { + return false; + } + int[] map = new int[256]; + for (int i = 0; i < str1.length; i++) { + map[str1[i]]++; + } + for (int i = 0; i < str2.length; i++) { + if (--map[str2[i]] < 0) { + return false; + } + } + return true; + } + + public static boolean isScramble1(String s1, String s2) { + if ((s1 == null && s2 != null) || (s1 != null && s2 == null)) { + return false; + } + if (s1 == null && s2 == null) { + return true; + } + if (s1.equals(s2)) { + return true; + } + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + if (!sameTypeSameNumber(str1, str2)) { + return false; + } + int N = s1.length(); + return process1(str1, str2, 0, 0, N); + } + + // 返回str1[从L1开始往右长度为size的子串]和str2[从L2开始往右长度为size的子串]是否互为旋变字符串 + // 在str1中的这一段和str2中的这一段一定是等长的，所以只用一个参数size + public static boolean process1(char[] str1, char[] str2, int L1, int L2, int size) { + if (size == 1) { + return str1[L1] == str2[L2]; + } + // 枚举每一种情况，有一个计算出互为旋变就返回true。都算不出来最后返回false + for (int leftPart = 1; leftPart < size; leftPart++) { + if ( + // 如果1左对2左，并且1右对2右 + (process1(str1, str2, L1, L2, leftPart) + && process1(str1, str2, L1 + leftPart, L2 + leftPart, size - leftPart)) || + // 如果1左对2右，并且1右对2左 + (process1(str1, str2, L1, L2 + size - leftPart, leftPart) + && process1(str1, str2, L1 + leftPart, L2, size - leftPart))) { + return true; + } + } + return false; + } + + public static boolean isScramble2(String s1, String s2) { + if ((s1 == null && s2 != null) || (s1 != null && s2 == null)) { + return false; + } + if (s1 == null && s2 == null) { + return true; + } + if (s1.equals(s2)) { + return true; + } + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + if (!sameTypeSameNumber(str1, str2)) { + return false; + } + int N = s1.length(); + int[][][] dp = new int[N][N][N + 1]; + // dp[i][j][k] = 0 processDP(i,j,k)状态之前没有算过的 + // dp[i][j][k] = -1 processDP(i,j,k)状态之前算过的,返回值是false + // dp[i][j][k] = 1 processDP(i,j,k)状态之前算过的,返回值是true + return process2(str1, str2, 0, 0, N, dp); + } + + public static boolean process2(char[] str1, char[] str2, int L1, int L2, int size, int[][][] dp) { + if (dp[L1][L2][size] != 0) { + return dp[L1][L2][size] == 1; + } + boolean ans = false; + if (size == 1) { + ans = str1[L1] == str2[L2]; + } else { + for (int leftPart = 1; leftPart < size; leftPart++) { + if ((process2(str1, str2, L1, L2, leftPart, dp) + && process2(str1, str2, L1 + leftPart, L2 + leftPart, size - leftPart, dp)) + || (process2(str1, str2, L1, L2 + size - leftPart, leftPart, dp) + && process2(str1, str2, L1 + leftPart, L2, size - leftPart, dp))) { + ans = true; + break; + } + } + } + dp[L1][L2][size] = ans ? 1 : -1; + return ans; + } + + public static boolean isScramble3(String s1, String s2) { + if ((s1 == null && s2 != null) || (s1 != null && s2 == null)) { + return false; + } + if (s1 == null && s2 == null) { + return true; + } + if (s1.equals(s2)) { + return true; + } + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + if (!sameTypeSameNumber(str1, str2)) { + return false; + } + int N = s1.length(); + boolean[][][] dp = new boolean[N][N][N + 1]; + for (int L1 = 0; L1 < N; L1++) { + for (int L2 = 0; L2 < N; L2++) { + dp[L1][L2][1] = str1[L1] == str2[L2]; + } + } + // 第一层for循环含义是：依次填size=2层、size=3层..size=N层，每一层都是一个二维平面 + // 第二、三层for循环含义是：在具体的一层，整个面都要填写，所以用两个for循环去填一个二维面 + // L1的取值氛围是[0,N-size]，因为从L1出发往右长度为size的子串，L1是不能从N-size+1出发的，这样往右就不够size个字符了 + // L2的取值范围同理 + // 第4层for循环完全是递归函数怎么写，这里就怎么改的 + for (int size = 2; size <= N; size++) { + for (int L1 = 0; L1 <= N - size; L1++) { + for (int L2 = 0; L2 <= N - size; L2++) { + for (int leftPart = 1; leftPart < size; leftPart++) { + if ((dp[L1][L2][leftPart] && dp[L1 + leftPart][L2 + leftPart][size - leftPart]) + || (dp[L1][L2 + size - leftPart][leftPart] && dp[L1 + leftPart][L2][size - leftPart])) { + dp[L1][L2][size] = true; + break; + } + } + } + } + } + return dp[0][0][N]; + } + +} diff --git a/大厂刷题班/class13/Code04_BricksFallingWhenHit.java b/大厂刷题班/class13/Code04_BricksFallingWhenHit.java new file mode 100644 index 0000000..657095c --- /dev/null +++ b/大厂刷题班/class13/Code04_BricksFallingWhenHit.java @@ -0,0 +1,149 @@ +package class13; + +// 本题测试链接 : https://leetcode.com/problems/bricks-falling-when-hit/ +public class Code04_BricksFallingWhenHit { + + public static int[] hitBricks(int[][] grid, int[][] hits) { + for (int i = 0; i < hits.length; i++) { + if (grid[hits[i][0]][hits[i][1]] == 1) { + grid[hits[i][0]][hits[i][1]] = 2; + } + } + UnionFind unionFind = new UnionFind(grid); + int[] ans = new int[hits.length]; + for (int i = hits.length - 1; i >= 0; i--) { + if (grid[hits[i][0]][hits[i][1]] == 2) { + ans[i] = unionFind.finger(hits[i][0], hits[i][1]); + } + } + return ans; + } + + // 并查集 + public static class UnionFind { + private int N; + private int M; + // 有多少块砖，连到了天花板上 + private int cellingAll; + // 原始矩阵，因为炮弹的影响，1 -> 2 + private int[][] grid; + // cellingSet[i] = true; i 是头节点，所在的集合是天花板集合 + private boolean[] cellingSet; + private int[] fatherMap; + private int[] sizeMap; + private int[] stack; + + public UnionFind(int[][] matrix) { + initSpace(matrix); + initConnect(); + } + + private void initSpace(int[][] matrix) { + grid = matrix; + N = grid.length; + M = grid[0].length; + int all = N * M; + cellingAll = 0; + cellingSet = new boolean[all]; + fatherMap = new int[all]; + sizeMap = new int[all]; + stack = new int[all]; + for (int row = 0; row < N; row++) { + for (int col = 0; col < M; col++) { + if (grid[row][col] == 1) { + int index = row * M + col; + fatherMap[index] = index; + sizeMap[index] = 1; + if (row == 0) { + cellingSet[index] = true; + cellingAll++; + } + } + } + } + } + + private void initConnect() { + for (int row = 0; row < N; row++) { + for (int col = 0; col < M; col++) { + union(row, col, row - 1, col); + union(row, col, row + 1, col); + union(row, col, row, col - 1); + union(row, col, row, col + 1); + } + } + } + + private int find(int row, int col) { + int stackSize = 0; + int index = row * M + col; + while (index != fatherMap[index]) { + stack[stackSize++] = index; + index = fatherMap[index]; + } + while (stackSize != 0) { + fatherMap[stack[--stackSize]] = index; + } + return index; + } + + private void union(int r1, int c1, int r2, int c2) { + if (valid(r1, c1) && valid(r2, c2)) { + int father1 = find(r1, c1); + int father2 = find(r2, c2); + if (father1 != father2) { + int size1 = sizeMap[father1]; + int size2 = sizeMap[father2]; + boolean status1 = cellingSet[father1]; + boolean status2 = cellingSet[father2]; + if (size1 <= size2) { + fatherMap[father1] = father2; + sizeMap[father2] = size1 + size2; + if (status1 ^ status2) { + cellingSet[father2] = true; + cellingAll += status1 ? size2 : size1; + } + } else { + fatherMap[father2] = father1; + sizeMap[father1] = size1 + size2; + if (status1 ^ status2) { + cellingSet[father1] = true; + cellingAll += status1 ? size2 : size1; + } + } + } + } + } + + private boolean valid(int row, int col) { + return row >= 0 && row < N && col >= 0 && col < M && grid[row][col] == 1; + } + + public int cellingNum() { + return cellingAll; + } + + public int finger(int row, int col) { + grid[row][col] = 1; + int cur = row * M + col; + if (row == 0) { + cellingSet[cur] = true; + cellingAll++; + } + fatherMap[cur] = cur; + sizeMap[cur] = 1; + int pre = cellingAll; + union(row, col, row - 1, col); + union(row, col, row + 1, col); + union(row, col, row, col - 1); + union(row, col, row, col + 1); + int now = cellingAll; + if (row == 0) { + return now - pre; + } else { + return now == pre ? 0 : now - pre - 1; + } + } + } + +} diff --git a/大厂刷题班/class14/Code01_Parentheses.java b/大厂刷题班/class14/Code01_Parentheses.java new file mode 100644 index 0000000..acd810f --- /dev/null +++ b/大厂刷题班/class14/Code01_Parentheses.java @@ -0,0 +1,97 @@ +package class14; + +public class Code01_Parentheses { + + public static boolean valid(String s) { + char[] str = s.toCharArray(); + int count = 0; + for (int i = 0; i < str.length; i++) { + count += str[i] == '(' ? 1 : -1; + if (count < 0) { + return false; + } + } + return count == 0; + } + + public static int needParentheses(String s) { + char[] str = s.toCharArray(); + int count = 0; + int need = 0; + for (int i = 0; i < str.length; i++) { + if (str[i] == '(') { + count++; + } else { // 遇到的是')' + if (count == 0) { + need++; + } else { + count--; + } + } + } + return count + need; + } + + public static boolean isValid(char[] str) { + if (str == null || str.length == 0) { + return false; + } + int status = 0; + for (int i = 0; i < str.length; i++) { + if (str[i] != ')' && str[i] != '(') { + return false; + } + if (str[i] == ')' && --status < 0) { + return false; + } + if (str[i] == '(') { + status++; + } + } + return status == 0; + } + + public static int deep(String s) { + char[] str = s.toCharArray(); + if (!isValid(str)) { + return 0; + } + int count = 0; + int max = 0; + for (int i = 0; i < str.length; i++) { + if (str[i] == '(') { + max = Math.max(max, ++count); + } else { + count--; + } + } + return max; + } + + // s只由(和)组成 + // 求最长有效括号子串长度 + // 本题测试链接 : https://leetcode.com/problems/longest-valid-parentheses/ + public static int longestValidParentheses(String s) { + if (s == null || s.length() < 2) { + return 0; + } + char[] str = s.toCharArray(); + // dp[i] : 子串必须以i位置结尾的情况下，往左最远能扩出多长的有效区域 + int[] dp = new int[str.length]; + // dp[0] = 0; （ ） + int pre = 0; + int ans = 0; + for (int i = 1; i < str.length; i++) { + if (str[i] == ')') { + // 当前谁和i位置的)，去配！ + pre = i - dp[i - 1] - 1; // 与str[i]配对的左括号的位置 pre + if (pre >= 0 && str[pre] == '(') { + dp[i] = dp[i - 1] + 2 + (pre > 0 ? dp[pre - 1] : 0); + } + } + ans = Math.max(ans, dp[i]); + } + return ans; + } + +} diff --git a/大厂刷题班/class14/Code02_MaxSubArraySumLessOrEqualK.java b/大厂刷题班/class14/Code02_MaxSubArraySumLessOrEqualK.java new file mode 100644 index 0000000..0456e4b --- /dev/null +++ b/大厂刷题班/class14/Code02_MaxSubArraySumLessOrEqualK.java @@ -0,0 +1,28 @@ +package class14; + +import java.util.TreeSet; + +public class Code02_MaxSubArraySumLessOrEqualK { + + // 请返回arr中，求个子数组的累加和，是<=K的，并且是最大的。 + // 返回这个最大的累加和 + public static int getMaxLessOrEqualK(int[] arr, int K) { + // 记录i之前的，前缀和，按照有序表组织 + TreeSet set = new TreeSet(); + // 一个数也没有的时候，就已经有一个前缀和是0了 + set.add(0); + int max = Integer.MIN_VALUE; + int sum = 0; + // 每一步的i，都求子数组必须以i结尾的情况下，求个子数组的累加和，是<=K的，并且是最大的 + for (int i = 0; i < arr.length; i++) { + sum += arr[i]; // sum -> arr[0..i]; + if (set.ceiling(sum - K) != null) { + max = Math.max(max, sum - set.ceiling(sum - K)); + } + set.add(sum); // 当前的前缀和加入到set中去 + } + return max; + + } + +} diff --git a/大厂刷题班/class14/Code03_BiggestBSTTopologyInTree.java b/大厂刷题班/class14/Code03_BiggestBSTTopologyInTree.java new file mode 100644 index 0000000..c45d2f8 --- /dev/null +++ b/大厂刷题班/class14/Code03_BiggestBSTTopologyInTree.java @@ -0,0 +1,65 @@ +package class14; + +// 本题测试链接 : https://www.nowcoder.com/practice/e13bceaca5b14860b83cbcc4912c5d4a +// 提交以下的代码，并把主类名改成Main +// 可以直接通过 +import java.util.Scanner; + +public class Code03_BiggestBSTTopologyInTree { + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + int n = sc.nextInt(); + int h = sc.nextInt(); + int[][] tree = new int[n + 1][3]; + for (int i = 1; i <= n; i++) { + int c = sc.nextInt(); + int l = sc.nextInt(); + int r = sc.nextInt(); + tree[l][0] = c; + tree[r][0] = c; + tree[c][1] = l; + tree[c][2] = r; + } + System.out.println(maxBSTTopology(h, tree, new int[n + 1])); + sc.close(); + } + + // h: 代表当前的头节点 + // t: 代表树 t[i][0]是i节点的父节点、t[i][1]是i节点的左孩子、t[i][2]是i节点的右孩子 + // m: i节点为头的最大bst拓扑结构大小 -> m[i] + // 返回: 以h为头的整棵树上，最大bst拓扑结构的大小 + public static int maxBSTTopology(int h, int[][] t, int[] m) { + if (h == 0) { + return 0; + } + int l = t[h][1]; + int r = t[h][2]; + int c = 0; + int p1 = maxBSTTopology(l, t, m); + int p2 = maxBSTTopology(r, t, m); + while (l < h && m[l] != 0) { + l = t[l][2]; + } + if (m[l] != 0) { + c = m[l]; + while (l != h) { + m[l] -= c; + l = t[l][0]; + } + } + while (r > h && m[r] != 0) { + r = t[r][1]; + } + if (m[r] != 0) { + c = m[r]; + while (r != h) { + m[r] -= c; + r = t[r][0]; + } + } + m[h] = m[t[h][1]] + m[t[h][2]] + 1; + return Math.max(Math.max(p1, p2), m[h]); + } + +} diff --git a/大厂刷题班/class14/Code04_CompleteTreeNodeNumber.java b/大厂刷题班/class14/Code04_CompleteTreeNodeNumber.java new file mode 100644 index 0000000..5e0c09b --- /dev/null +++ b/大厂刷题班/class14/Code04_CompleteTreeNodeNumber.java @@ -0,0 +1,45 @@ +package class14; + +//本题测试链接 : https://leetcode.cn/problems/count-complete-tree-nodes/ +public class Code04_CompleteTreeNodeNumber { + + // 提交时不要提交这个类 + public class TreeNode { + int val; + TreeNode left; + TreeNode right; + } + + // 提交如下的方法 + public static int countNodes(TreeNode head) { + if (head == null) { + return 0; + } + return bs(head, 1, mostLeftLevel(head, 1)); + } + + // 当前来到node节点，node节点在level层，总层数是h + // 返回node为头的子树(必是完全二叉树)，有多少个节点 + public static int bs(TreeNode node, int Level, int h) { + if (Level == h) { + return 1; + } + if (mostLeftLevel(node.right, Level + 1) == h) { + return (1 << (h - Level)) + bs(node.right, Level + 1, h); + } else { + return (1 << (h - Level - 1)) + bs(node.left, Level + 1, h); + } + } + + // 如果node在第level层， + // 求以node为头的子树，最大深度是多少 + // node为头的子树，一定是完全二叉树 + public static int mostLeftLevel(TreeNode node, int level) { + while (node != null) { + level++; + node = node.left; + } + return level - 1; + } + +} diff --git a/大厂刷题班/class14/Code05_RecoverBinarySearchTree.java b/大厂刷题班/class14/Code05_RecoverBinarySearchTree.java new file mode 100644 index 0000000..2ee6e62 --- /dev/null +++ b/大厂刷题班/class14/Code05_RecoverBinarySearchTree.java @@ -0,0 +1,511 @@ +package class14; + +import java.util.Stack; + +// 本题测试链接 : https://leetcode.com/problems/recover-binary-search-tree/ +public class Code05_RecoverBinarySearchTree { + + // 不要提交这个类 + public static class TreeNode { + public int val; + public TreeNode left; + public TreeNode right; + + public TreeNode(int v) { + val = v; + } + } + + // 如果能过leetcode，只需要提交这个方法即可 + // 但其实recoverTree2才是正路，只不过leetcode没有那么考 + public static void recoverTree(TreeNode root) { + TreeNode[] errors = twoErrors(root); + if (errors[0] != null && errors[1] != null) { + int tmp = errors[0].val; + errors[0].val = errors[1].val; + errors[1].val = tmp; + } + } + + public static TreeNode[] twoErrors(TreeNode head) { + TreeNode[] ans = new TreeNode[2]; + if (head == null) { + return ans; + } + TreeNode cur = head; + TreeNode mostRight = null; + TreeNode pre = null; + TreeNode e1 = null; + TreeNode e2 = null; + while (cur != null) { + mostRight = cur.left; + if (mostRight != null) { + while (mostRight.right != null && mostRight.right != cur) { + mostRight = mostRight.right; + } + if (mostRight.right == null) { + mostRight.right = cur; + cur = cur.left; + continue; + } else { + mostRight.right = null; + } + } + if (pre != null && pre.val >= cur.val) { + e1 = e1 == null ? pre : e1; + e2 = cur; + } + pre = cur; + cur = cur.right; + } + ans[0] = e1; + ans[1] = e2; + return ans; + } + + // 以下的方法，提交leetcode是通过不了的，但那是因为leetcode的验证方式有问题 + // 但其实！以下的方法，才是正路！在结构上彻底交换两个节点，而不是值交换 + public static TreeNode recoverTree2(TreeNode head) { + TreeNode[] errs = getTwoErrNodes(head); + TreeNode[] parents = getTwoErrParents(head, errs[0], errs[1]); + TreeNode e1 = errs[0]; + TreeNode e1P = parents[0]; + TreeNode e1L = e1.left; + TreeNode e1R = e1.right; + TreeNode e2 = errs[1]; + TreeNode e2P = parents[1]; + TreeNode e2L = e2.left; + TreeNode e2R = e2.right; + if (e1 == head) { + if (e1 == e2P) { + e1.left = e2L; + e1.right = e2R; + e2.right = e1; + e2.left = e1L; + } else if (e2P.left == e2) { + e2P.left = e1; + e2.left = e1L; + e2.right = e1R; + e1.left = e2L; + e1.right = e2R; + } else { + e2P.right = e1; + e2.left = e1L; + e2.right = e1R; + e1.left = e2L; + e1.right = e2R; + } + head = e2; + } else if (e2 == head) { + if (e2 == e1P) { + e2.left = e1L; + e2.right = e1R; + e1.left = e2; + e1.right = e2R; + } else if (e1P.left == e1) { + e1P.left = e2; + e1.left = e2L; + e1.right = e2R; + e2.left = e1L; + e2.right = e1R; + } else { + e1P.right = e2; + e1.left = e2L; + e1.right = e2R; + e2.left = e1L; + e2.right = e1R; + } + head = e1; + } else { + if (e1 == e2P) { + if (e1P.left == e1) { + e1P.left = e2; + e1.left = e2L; + e1.right = e2R; + e2.left = e1L; + e2.right = e1; + } else { + e1P.right = e2; + e1.left = e2L; + e1.right = e2R; + e2.left = e1L; + e2.right = e1; + } + } else if (e2 == e1P) { + if (e2P.left == e2) { + e2P.left = e1; + e2.left = e1L; + e2.right = e1R; + e1.left = e2; + e1.right = e2R; + } else { + e2P.right = e1; + e2.left = e1L; + e2.right = e1R; + e1.left = e2; + e1.right = e2R; + } + } else { + if (e1P.left == e1) { + if (e2P.left == e2) { + e1.left = e2L; + e1.right = e2R; + e2.left = e1L; + e2.right = e1R; + e1P.left = e2; + e2P.left = e1; + } else { + e1.left = e2L; + e1.right = e2R; + e2.left = e1L; + e2.right = e1R; + e1P.left = e2; + e2P.right = e1; + } + } else { + if (e2P.left == e2) { + e1.left = e2L; + e1.right = e2R; + e2.left = e1L; + e2.right = e1R; + e1P.right = e2; + e2P.left = e1; + } else { + e1.left = e2L; + e1.right = e2R; + e2.left = e1L; + e2.right = e1R; + e1P.right = e2; + e2P.right = e1; + } + } + } + } + return head; + } + + public static TreeNode[] getTwoErrNodes(TreeNode head) { + TreeNode[] errs = new TreeNode[2]; + if (head == null) { + return errs; + } + Stack stack = new Stack(); + TreeNode pre = null; + while (!stack.isEmpty() || head != null) { + if (head != null) { + stack.push(head); + head = head.left; + } else { + head = stack.pop(); + if (pre != null && pre.val > head.val) { + errs[0] = errs[0] == null ? pre : errs[0]; + errs[1] = head; + } + pre = head; + head = head.right; + } + } + return errs; + } + + public static TreeNode[] getTwoErrParents(TreeNode head, TreeNode e1, TreeNode e2) { + TreeNode[] parents = new TreeNode[2]; + if (head == null) { + return parents; + } + Stack stack = new Stack(); + while (!stack.isEmpty() || head != null) { + if (head != null) { + stack.push(head); + head = head.left; + } else { + head = stack.pop(); + if (head.left == e1 || head.right == e1) { + parents[0] = head; + } + if (head.left == e2 || head.right == e2) { + parents[1] = head; + } + head = head.right; + } + } + return parents; + } + + // for test -- print tree + public static void printTree(TreeNode head) { + System.out.println("Binary Tree:"); + printInOrder(head, 0, "H", 17); + System.out.println(); + } + + public static void printInOrder(TreeNode head, int height, String to, int len) { + if (head == null) { + return; + } + printInOrder(head.right, height + 1, "v", len); + String val = to + head.val + to; + int lenM = val.length(); + int lenL = (len - lenM) / 2; + int lenR = len - lenM - lenL; + val = getSpace(lenL) + val + getSpace(lenR); + System.out.println(getSpace(height * len) + val); + printInOrder(head.left, height + 1, "^", len); + } + + public static String getSpace(int num) { + String space = " "; + StringBuffer buf = new StringBuffer(""); + for (int i = 0; i < num; i++) { + buf.append(space); + } + return buf.toString(); + } + + // 为了测试 + public static boolean isBST(TreeNode head) { + if (head == null) { + return false; + } + Stack stack = new Stack(); + TreeNode pre = null; + while (!stack.isEmpty() || head != null) { + if (head != null) { + stack.push(head); + head = head.left; + } else { + head = stack.pop(); + if (pre != null && pre.val > head.val) { + return false; + } + pre = head; + head = head.right; + } + } + return true; + } + + public static void main(String[] args) { + TreeNode head = new TreeNode(5); + head.left = new TreeNode(3); + head.right = new TreeNode(7); + head.left.left = new TreeNode(2); + head.left.right = new TreeNode(4); + head.right.left = new TreeNode(6); + head.right.right = new TreeNode(8); + head.left.left.left = new TreeNode(1); + printTree(head); + System.out.println(isBST(head)); + + System.out.println("situation 1"); + TreeNode head1 = new TreeNode(7); + head1.left = new TreeNode(3); + head1.right = new TreeNode(5); + head1.left.left = new TreeNode(2); + head1.left.right = new TreeNode(4); + head1.right.left = new TreeNode(6); + head1.right.right = new TreeNode(8); + head1.left.left.left = new TreeNode(1); + printTree(head1); + System.out.println(isBST(head1)); + TreeNode res1 = recoverTree2(head1); + printTree(res1); + System.out.println(isBST(res1)); + + System.out.println("situation 2"); + TreeNode head2 = new TreeNode(6); + head2.left = new TreeNode(3); + head2.right = new TreeNode(7); + head2.left.left = new TreeNode(2); + head2.left.right = new TreeNode(4); + head2.right.left = new TreeNode(5); + head2.right.right = new TreeNode(8); + head2.left.left.left = new TreeNode(1); + printTree(head2); + System.out.println(isBST(head2)); + TreeNode res2 = recoverTree2(head2); + printTree(res2); + System.out.println(isBST(res2)); + + System.out.println("situation 3"); + TreeNode head3 = new TreeNode(8); + head3.left = new TreeNode(3); + head3.right = new TreeNode(7); + head3.left.left = new TreeNode(2); + head3.left.right = new TreeNode(4); + head3.right.left = new TreeNode(6); + head3.right.right = new TreeNode(5); + head3.left.left.left = new TreeNode(1); + printTree(head3); + System.out.println(isBST(head3)); + TreeNode res3 = recoverTree2(head3); + printTree(res3); + System.out.println(isBST(res3)); + + System.out.println("situation 4"); + TreeNode head4 = new TreeNode(3); + head4.left = new TreeNode(5); + head4.right = new TreeNode(7); + head4.left.left = new TreeNode(2); + head4.left.right = new TreeNode(4); + head4.right.left = new TreeNode(6); + head4.right.right = new TreeNode(8); + head4.left.left.left = new TreeNode(1); + printTree(head4); + System.out.println(isBST(head4)); + TreeNode res4 = recoverTree2(head4); + printTree(res4); + System.out.println(isBST(res4)); + + System.out.println("situation 5"); + TreeNode head5 = new TreeNode(2); + head5.left = new TreeNode(3); + head5.right = new TreeNode(7); + head5.left.left = new TreeNode(5); + head5.left.right = new TreeNode(4); + head5.right.left = new TreeNode(6); + head5.right.right = new TreeNode(8); + head5.left.left.left = new TreeNode(1); + printTree(head5); + System.out.println(isBST(head5)); + TreeNode res5 = recoverTree2(head5); + printTree(res5); + System.out.println(isBST(res5)); + + System.out.println("situation 6"); + TreeNode head6 = new TreeNode(4); + head6.left = new TreeNode(3); + head6.right = new TreeNode(7); + head6.left.left = new TreeNode(2); + head6.left.right = new TreeNode(5); + head6.right.left = new TreeNode(6); + head6.right.right = new TreeNode(8); + head6.left.left.left = new TreeNode(1); + printTree(head6); + System.out.println(isBST(head6)); + TreeNode res6 = recoverTree2(head6); + printTree(res6); + System.out.println(isBST(res6)); + + System.out.println("situation 7"); + TreeNode head7 = new TreeNode(5); + head7.left = new TreeNode(4); + head7.right = new TreeNode(7); + head7.left.left = new TreeNode(2); + head7.left.right = new TreeNode(3); + head7.right.left = new TreeNode(6); + head7.right.right = new TreeNode(8); + head7.left.left.left = new TreeNode(1); + printTree(head7); + System.out.println(isBST(head7)); + TreeNode res7 = recoverTree2(head7); + printTree(res7); + System.out.println(isBST(res7)); + + System.out.println("situation 8"); + TreeNode head8 = new TreeNode(5); + head8.left = new TreeNode(3); + head8.right = new TreeNode(8); + head8.left.left = new TreeNode(2); + head8.left.right = new TreeNode(4); + head8.right.left = new TreeNode(6); + head8.right.right = new TreeNode(7); + head8.left.left.left = new TreeNode(1); + printTree(head8); + System.out.println(isBST(head8)); + TreeNode res8 = recoverTree2(head8); + printTree(res8); + System.out.println(isBST(res8)); + + System.out.println("situation 9"); + TreeNode head9 = new TreeNode(5); + head9.left = new TreeNode(2); + head9.right = new TreeNode(7); + head9.left.left = new TreeNode(3); + head9.left.right = new TreeNode(4); + head9.right.left = new TreeNode(6); + head9.right.right = new TreeNode(8); + head9.left.left.left = new TreeNode(1); + printTree(head9); + System.out.println(isBST(head9)); + TreeNode res9 = recoverTree2(head9); + printTree(res9); + System.out.println(isBST(res9)); + + System.out.println("situation 10"); + TreeNode head10 = new TreeNode(5); + head10.left = new TreeNode(3); + head10.right = new TreeNode(6); + head10.left.left = new TreeNode(2); + head10.left.right = new TreeNode(4); + head10.right.left = new TreeNode(7); + head10.right.right = new TreeNode(8); + head10.left.left.left = new TreeNode(1); + printTree(head10); + System.out.println(isBST(head10)); + TreeNode res10 = recoverTree2(head10); + printTree(res10); + System.out.println(isBST(res10)); + + System.out.println("situation 11"); + TreeNode head11 = new TreeNode(5); + head11.left = new TreeNode(3); + head11.right = new TreeNode(7); + head11.left.left = new TreeNode(6); + head11.left.right = new TreeNode(4); + head11.right.left = new TreeNode(2); + head11.right.right = new TreeNode(8); + head11.left.left.left = new TreeNode(1); + printTree(head11); + System.out.println(isBST(head11)); + TreeNode res11 = recoverTree2(head11); + printTree(res11); + System.out.println(isBST(res11)); + + System.out.println("situation 12"); + TreeNode head12 = new TreeNode(5); + head12.left = new TreeNode(3); + head12.right = new TreeNode(7); + head12.left.left = new TreeNode(8); + head12.left.right = new TreeNode(4); + head12.right.left = new TreeNode(6); + head12.right.right = new TreeNode(2); + head12.left.left.left = new TreeNode(1); + printTree(head12); + System.out.println(isBST(head12)); + TreeNode res12 = recoverTree2(head12); + printTree(res12); + System.out.println(isBST(res12)); + + System.out.println("situation 13"); + TreeNode head13 = new TreeNode(5); + head13.left = new TreeNode(3); + head13.right = new TreeNode(7); + head13.left.left = new TreeNode(2); + head13.left.right = new TreeNode(6); + head13.right.left = new TreeNode(4); + head13.right.right = new TreeNode(8); + head13.left.left.left = new TreeNode(1); + printTree(head13); + System.out.println(isBST(head13)); + TreeNode res13 = recoverTree2(head13); + printTree(res13); + System.out.println(isBST(res13)); + + System.out.println("situation 14"); + TreeNode head14 = new TreeNode(5); + head14.left = new TreeNode(3); + head14.right = new TreeNode(7); + head14.left.left = new TreeNode(2); + head14.left.right = new TreeNode(8); + head14.right.left = new TreeNode(6); + head14.right.right = new TreeNode(4); + head14.left.left.left = new TreeNode(1); + printTree(head14); + System.out.println(isBST(head14)); + TreeNode res14 = recoverTree2(head14); + printTree(res14); + System.out.println(isBST(res14)); + } + +} diff --git a/大厂刷题班/class14/Code06_MissingNumber.java b/大厂刷题班/class14/Code06_MissingNumber.java new file mode 100644 index 0000000..9abca16 --- /dev/null +++ b/大厂刷题班/class14/Code06_MissingNumber.java @@ -0,0 +1,29 @@ +package class14; + +// 测试链接：https://leetcode.com/problems/first-missing-positive/ +public class Code06_MissingNumber { + + public static int firstMissingPositive(int[] arr) { + // l是盯着的位置 + // 0 ~ L-1有效区 + int L = 0; + int R = arr.length; + while (L != R) { + if (arr[L] == L + 1) { + L++; + } else if (arr[L] <= L || arr[L] > R || arr[arr[L] - 1] == arr[L]) { // 垃圾的情况 + swap(arr, L, --R); + } else { + swap(arr, L, arr[L] - 1); + } + } + return L + 1; + } + + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + +} diff --git a/大厂刷题班/class15/Code01_BestTimeToBuyAndSellStock.java b/大厂刷题班/class15/Code01_BestTimeToBuyAndSellStock.java new file mode 100644 index 0000000..3660c43 --- /dev/null +++ b/大厂刷题班/class15/Code01_BestTimeToBuyAndSellStock.java @@ -0,0 +1,21 @@ +package class15; + +// leetcode 121 +public class Code01_BestTimeToBuyAndSellStock { + + public static int maxProfit(int[] prices) { + if (prices == null || prices.length == 0) { + return 0; + } + // 必须在0时刻卖掉，[0] - [0] + int ans = 0; + // arr[0...0] + int min = prices[0]; + for (int i = 1; i < prices.length; i++) { + min = Math.min(min, prices[i]); + ans = Math.max(ans, prices[i] - min); + } + return ans; + } + +} diff --git a/大厂刷题班/class15/Code02_BestTimeToBuyAndSellStockII.java b/大厂刷题班/class15/Code02_BestTimeToBuyAndSellStockII.java new file mode 100644 index 0000000..c751f54 --- /dev/null +++ b/大厂刷题班/class15/Code02_BestTimeToBuyAndSellStockII.java @@ -0,0 +1,17 @@ +package class15; + +//leetcode 122 +public class Code02_BestTimeToBuyAndSellStockII { + + public static int maxProfit(int[] prices) { + if (prices == null || prices.length == 0) { + return 0; + } + int ans = 0; + for (int i = 1; i < prices.length; i++) { + ans += Math.max(prices[i] - prices[i-1], 0); + } + return ans; + } + +} diff --git a/大厂刷题班/class15/Code03_BestTimeToBuyAndSellStockIII.java b/大厂刷题班/class15/Code03_BestTimeToBuyAndSellStockIII.java new file mode 100644 index 0000000..a1aadde --- /dev/null +++ b/大厂刷题班/class15/Code03_BestTimeToBuyAndSellStockIII.java @@ -0,0 +1,23 @@ +package class15; + +//leetcode 123 +public class Code03_BestTimeToBuyAndSellStockIII { + + public static int maxProfit(int[] prices) { + if (prices == null || prices.length < 2) { + return 0; + } + int ans = 0; + int doneOnceMinusBuyMax = -prices[0]; + int doneOnceMax = 0; + int min = prices[0]; + for (int i = 1; i < prices.length; i++) { + min = Math.min(min, prices[i]); + ans = Math.max(ans, doneOnceMinusBuyMax + prices[i]); + doneOnceMax = Math.max(doneOnceMax, prices[i] - min); + doneOnceMinusBuyMax = Math.max(doneOnceMinusBuyMax, doneOnceMax - prices[i]); + } + return ans; + } + +} diff --git a/大厂刷题班/class15/Code04_BestTimeToBuyAndSellStockIV.java b/大厂刷题班/class15/Code04_BestTimeToBuyAndSellStockIV.java new file mode 100644 index 0000000..7b8bd72 --- /dev/null +++ b/大厂刷题班/class15/Code04_BestTimeToBuyAndSellStockIV.java @@ -0,0 +1,65 @@ +package class15; + +//leetcode 188 +public class Code04_BestTimeToBuyAndSellStockIV { + + public static int maxProfit(int K, int[] prices) { + if (prices == null || prices.length == 0) { + return 0; + } + int N = prices.length; + if (K >= N / 2) { + return allTrans(prices); + } + int[][] dp = new int[K + 1][N]; + int ans = 0; + for (int tran = 1; tran <= K; tran++) { + int pre = dp[tran][0]; + int best = pre - prices[0]; + for (int index = 1; index < N; index++) { + pre = dp[tran - 1][index]; + dp[tran][index] = Math.max(dp[tran][index - 1], prices[index] + best); + best = Math.max(best, pre - prices[index]); + ans = Math.max(dp[tran][index], ans); + } + } + return ans; + } + + public static int allTrans(int[] prices) { + int ans = 0; + for (int i = 1; i < prices.length; i++) { + ans += Math.max(prices[i] - prices[i - 1], 0); + } + return ans; + } + + // 课上写的版本，对了 + public static int maxProfit2(int K, int[] arr) { + if (arr == null || arr.length == 0 || K < 1) { + return 0; + } + int N = arr.length; + if (K >= N / 2) { + return allTrans(arr); + } + int[][] dp = new int[N][K + 1]; + // dp[...][0] = 0 + // dp[0][...] = arr[0.0] 0 + for (int j = 1; j <= K; j++) { + // dp[1][j] + int p1 = dp[0][j]; + int best = Math.max(dp[1][j - 1] - arr[1], dp[0][j - 1] - arr[0]); + dp[1][j] = Math.max(p1, best + arr[1]); + // dp[1][j] 准备好一些枚举，接下来准备好的枚举 + for (int i = 2; i < N; i++) { + p1 = dp[i - 1][j]; + int newP = dp[i][j - 1] - arr[i]; + best = Math.max(newP, best); + dp[i][j] = Math.max(p1, best + arr[i]); + } + } + return dp[N - 1][K]; + } + +} diff --git a/大厂刷题班/class15/Code05_BestTimeToBuyAndSellStockWithCooldown.java b/大厂刷题班/class15/Code05_BestTimeToBuyAndSellStockWithCooldown.java new file mode 100644 index 0000000..0dea85d --- /dev/null +++ b/大厂刷题班/class15/Code05_BestTimeToBuyAndSellStockWithCooldown.java @@ -0,0 +1,101 @@ +package class15; + +//leetcode 309 +public class Code05_BestTimeToBuyAndSellStockWithCooldown { + + // 该方法是对的，提交之后大量的测试用例通过，最后几个会超时 + // 如果把这个方法改成动态规划，是可以通过的，但这个尝试不是最优解 + public static int maxProfit1(int[] prices) { + return process1(prices, false, 0, 0); + } + + // buy == false 目前可以交易，而且当前没有购买行为 + // buy == true 已经买了，买入价buyPrices，待卖出 + public static int process1(int[] prices, boolean buy, int index, int buyPrices) { + if (index >= prices.length) { + return 0; + } + if (buy) { + int noSell = process1(prices, true, index + 1, buyPrices); + int yesSell = prices[index] - buyPrices + process1(prices, false, index + 2, 0); + return Math.max(noSell, yesSell); + } else { + int noBuy = process1(prices, false, index + 1, 0); + int yesBuy = process1(prices, true, index + 1, prices[index]); + return Math.max(noBuy, yesBuy); + } + } + + // 最优尝试如下： + // buy[i] : 在0...i范围上，最后一次操作是buy动作， + // 这最后一次操作有可能发生在i位置，也可能发生在i之前 + // buy[i]值的含义是：max{ 所有可能性[之前交易获得的最大收益 - 最后buy动作的收购价格] } + // 比如：arr[0...i]假设为[1,3,4,6,2,7,1...i之后的数字不用管] + // 什么叫，所有可能性[之前交易获得的最大收益 - 最后buy动作的收购价格]？ + // 比如其中一种可能性： + // 假设最后一次buy动作发生在2这个数的时候，那么之前的交易只能在[1,3,4]上结束，因为6要cooldown的， + // 此时最大收益是多少呢？是4-1==3。那么，之前交易获得的最大收益 - 最后buy动作的收购价格 = 3 - 2 = 1 + // 另一种可能性： + // 再比如最后一次buy动作发生在最后的1这个数的时候，那么之前的交易只能在[1,3,4,6,2]上发生，因为7要cooldown的， + // 此时最大收益是多少呢？是6-1==5。那么，之前交易获得的最大收益 - 最后buy动作的收购价格 = 5 - 1 = 4 + // 除了上面两种可能性之外，还有很多可能性，你可以假设每个数字都是最后buy动作的时候， + // 所有可能性中，(之前交易获得的最大收益 - 最后buy动作的收购价格)的最大值，就是buy[i]的含义 + // 为啥buy[i]要算之前的最大收益 - 最后一次收购价格？尤其是最后为什么要减这么一下？ + // 因为这样一来，当你之后以X价格做成一笔交易的时候，当前最好的总收益直接就是 X + buy[i]了 + // + // sell[i] :0...i范围上，最后一次操作是sell动作，这最后一次操作有可能发生在i位置，也可能发生在之前 + // sell[i]值的含义：0...i范围上，最后一次动作是sell的情况下，最好的收益 + // + // 于是通过分析，能得到以下的转移方程： + // buy[i] = Math.max(buy[i - 1], sell[i - 2] - prices[i]) + + + + + // 如果i位置没有发生buy行为，说明有没有i位置都一样，那么buy[i] = buy[i-1]，这显而易见 + // 如果i位置发生了buy行为, 那么buy[i] = sell[i - 2] - prices[i]， + // 因为你想在i位置买的话，你必须保证之前交易行为发生在0...i-2上， + // 因为如果i-1位置有可能参与交易的话，i位置就要cooldown了， + // 而且之前交易行为必须以sell结束，你才能buy，而且要保证之前交易尽可能得到最好的利润， + // 这正好是sell[i - 2]所代表的含义，并且根据buy[i]的定义，最后一定要 - prices[i] + // + // sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i]) + // 如果i位置没有发生sell行为，那么sell[i] = sell[i-1]，这显而易见 + // 如果i位置发生了sell行为，那么我们一定要找到 {之前获得尽可能好的收益 - 最后一次的收购价格尽可能低}， + // 而这正好是buy[i - 1]的含义！之前所有的"尽可能"中，最好的一个！ + public static int maxProfit2(int[] prices) { + if (prices.length < 2) { + return 0; + } + int N = prices.length; + int[] buy = new int[N]; + int[] sell = new int[N]; + // buy[0] 不需要设置 buy[0] = -arr[0] + // sell[0] = 0 + buy[1] = Math.max(-prices[0], -prices[1]); + sell[1] = Math.max(0, prices[1] - prices[0]); + for (int i = 2; i < N; i++) { + buy[i] = Math.max(buy[i - 1], sell[i - 2] - prices[i]); + sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i]); + } + return sell[N - 1]; + } + + // 最优解就是方法2的空间优化而已 + public static int maxProfit3(int[] prices) { + if (prices.length < 2) { + return 0; + } + int buy1 = Math.max(-prices[0], -prices[1]); + int sell1 = Math.max(0, prices[1] - prices[0]); + int sell2 = 0; + for (int i = 2; i < prices.length; i++) { + int tmp = sell1; + sell1 = Math.max(sell1, buy1 + prices[i]); + buy1 = Math.max(buy1, sell2 - prices[i]); + sell2 = tmp; + } + return sell1; + } + +} diff --git a/大厂刷题班/class15/Code06_BestTimeToBuyAndSellStockWithTransactionFee.java b/大厂刷题班/class15/Code06_BestTimeToBuyAndSellStockWithTransactionFee.java new file mode 100644 index 0000000..121e8a6 --- /dev/null +++ b/大厂刷题班/class15/Code06_BestTimeToBuyAndSellStockWithTransactionFee.java @@ -0,0 +1,27 @@ +package class15; + +//leetcode 714 +public class Code06_BestTimeToBuyAndSellStockWithTransactionFee { + + public static int maxProfit(int[] arr, int fee) { + if (arr == null || arr.length < 2) { + return 0; + } + int N = arr.length; + // 0..0 0 -[0] - fee + int bestbuy = -arr[0] - fee; + // 0..0 卖 0 + int bestsell = 0; + for (int i = 1; i < N; i++) { + // 来到i位置了！ + // 如果在i必须买 收入 - 批发价 - fee + int curbuy = bestsell - arr[i] - fee; + // 如果在i必须卖 整体最优（收入 - 良好批发价 - fee） + int cursell = bestbuy + arr[i]; + bestbuy = Math.max(bestbuy, curbuy); + bestsell = Math.max(bestsell, cursell); + } + return bestsell; + } + +} diff --git a/大厂刷题班/class16/Code01_IsSum.java b/大厂刷题班/class16/Code01_IsSum.java new file mode 100644 index 0000000..210c277 --- /dev/null +++ b/大厂刷题班/class16/Code01_IsSum.java @@ -0,0 +1,206 @@ +package class16; + +import java.util.HashMap; +import java.util.HashSet; + +// 这道题是一个小小的补充，课上没有讲 +// 但是如果你听过体系学习班动态规划专题和本节课的话 +// 这道题就是一道水题 +public class Code01_IsSum { + + // arr中的值可能为正，可能为负，可能为0 + // 自由选择arr中的数字，能不能累加得到sum + // 暴力递归方法 + public static boolean isSum1(int[] arr, int sum) { + if (sum == 0) { + return true; + } + if (arr == null || arr.length == 0) { + return false; + } + return process1(arr, arr.length - 1, sum); + } + + // 可以自由使用arr[0...i]上的数字，能不能累加得到sum + public static boolean process1(int[] arr, int i, int sum) { + if (sum == 0) { + return true; + } + if (i == -1) { + return false; + } + return process1(arr, i - 1, sum) || process1(arr, i - 1, sum - arr[i]); + } + + // arr中的值可能为正，可能为负，可能为0 + // 自由选择arr中的数字，能不能累加得到sum + // 记忆化搜索方法 + // 从暴力递归方法来，加了记忆化缓存，就是动态规划了 + public static boolean isSum2(int[] arr, int sum) { + if (sum == 0) { + return true; + } + if (arr == null || arr.length == 0) { + return false; + } + return process2(arr, arr.length - 1, sum, new HashMap<>()); + } + + public static boolean process2(int[] arr, int i, int sum, HashMap> dp) { + if (dp.containsKey(i) && dp.get(i).containsKey(sum)) { + return dp.get(i).get(sum); + } + boolean ans = false; + if (sum == 0) { + ans = true; + } else if (i != -1) { + ans = process2(arr, i - 1, sum, dp) || process2(arr, i - 1, sum - arr[i], dp); + } + if (!dp.containsKey(i)) { + dp.put(i, new HashMap<>()); + } + dp.get(i).put(sum, ans); + return ans; + } + + // arr中的值可能为正，可能为负，可能为0 + // 自由选择arr中的数字，能不能累加得到sum + // 经典动态规划 + public static boolean isSum3(int[] arr, int sum) { + if (sum == 0) { + return true; + } + // sum != 0 + if (arr == null || arr.length == 0) { + return false; + } + // arr有数，sum不为0 + int min = 0; + int max = 0; + for (int num : arr) { + min += num < 0 ? num : 0; + max += num > 0 ? num : 0; + } + // min~max + if (sum < min || sum > max) { + return false; + } + + // min <= sum <= max + int N = arr.length; + // dp[i][j] + // + // 0 1 2 3 4 5 6 7 (实际的对应) + // -7 -6 -5 -4 -3 -2 -1 0（想象中） + // + // dp[0][-min] -> dp[0][7] -> dp[0][0] + boolean[][] dp = new boolean[N][max - min + 1]; + // dp[0][0] = true + dp[0][-min] = true; + // dp[0][arr[0]] = true + dp[0][arr[0] - min] = true; + for (int i = 1; i < N; i++) { + for (int j = min; j <= max; j++) { + // dp[i][j] = dp[i-1][j] + dp[i][j - min] = dp[i - 1][j - min]; + // dp[i][j] |= dp[i-1][j - arr[i]] + int next = j - min - arr[i]; + dp[i][j - min] |= (next >= 0 && next <= max - min && dp[i - 1][next]); + } + } + // dp[N-1][sum] + return dp[N - 1][sum - min]; + } + + // arr中的值可能为正，可能为负，可能为0 + // 自由选择arr中的数字，能不能累加得到sum + // 分治的方法 + // 如果arr中的数值特别大，动态规划方法依然会很慢 + // 此时如果arr的数字个数不算多(40以内)，哪怕其中的数值很大，分治的方法也将是最优解 + public static boolean isSum4(int[] arr, int sum) { + if (sum == 0) { + return true; + } + if (arr == null || arr.length == 0) { + return false; + } + if (arr.length == 1) { + return arr[0] == sum; + } + int N = arr.length; + int mid = N >> 1; + HashSet leftSum = new HashSet<>(); + HashSet rightSum = new HashSet<>(); + // 0...mid-1 + process4(arr, 0, mid, 0, leftSum); + // mid..N-1 + process4(arr, mid, N, 0, rightSum); + // 单独查看，只使用左部分，能不能搞出sum + // 单独查看，只使用右部分，能不能搞出sum + // 左+右，联合能不能搞出sum + // 左部分搞出所有累加和的时候，包含左部分一个数也没有，这种情况的，leftsum表里，0 + // 17 17 + for (int l : leftSum) { + if (rightSum.contains(sum - l)) { + return true; + } + } + return false; + } + + // arr[0...i-1]决定已经做完了！形成的累加和是pre + // arr[i...end - 1] end(终止) 所有数字随意选择， + // arr[0...end-1]所有可能的累加和存到ans里去 + public static void process4(int[] arr, int i, int end, int pre, HashSet ans) { + if (i == end) { + ans.add(pre); + } else { + process4(arr, i + 1, end, pre, ans); + process4(arr, i + 1, end, pre + arr[i], ans); + } + } + + // 为了测试 + // 生成长度为len的随机数组 + // 值在[-max, max]上随机 + public static int[] randomArray(int len, int max) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * ((max << 1) + 1)) - max; + } + return arr; + } + + // 对数器验证所有方法 + public static void main(String[] args) { + int N = 20; + int M = 100; + int testTime = 100000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int size = (int) (Math.random() * (N + 1)); + int[] arr = randomArray(size, M); + int sum = (int) (Math.random() * ((M << 1) + 1)) - M; + boolean ans1 = isSum1(arr, sum); + boolean ans2 = isSum2(arr, sum); + boolean ans3 = isSum3(arr, sum); + boolean ans4 = isSum4(arr, sum); + if (ans1 ^ ans2 || ans3 ^ ans4 || ans1 ^ ans3) { + System.out.println("出错了！"); + System.out.print("arr : "); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println("sum : " + sum); + System.out.println("方法一答案 : " + ans1); + System.out.println("方法二答案 : " + ans2); + System.out.println("方法三答案 : " + ans3); + System.out.println("方法四答案 : " + ans4); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class16/Code02_SmallestUnFormedSum.java b/大厂刷题班/class16/Code02_SmallestUnFormedSum.java new file mode 100644 index 0000000..645b3af --- /dev/null +++ b/大厂刷题班/class16/Code02_SmallestUnFormedSum.java @@ -0,0 +1,123 @@ +package class16; + +import java.util.Arrays; +import java.util.HashSet; + +public class Code02_SmallestUnFormedSum { + + public static int unformedSum1(int[] arr) { + if (arr == null || arr.length == 0) { + return 1; + } + HashSet set = new HashSet(); + process(arr, 0, 0, set); + int min = Integer.MAX_VALUE; + for (int i = 0; i != arr.length; i++) { + min = Math.min(min, arr[i]); + } + for (int i = min + 1; i != Integer.MIN_VALUE; i++) { + if (!set.contains(i)) { + return i; + } + } + return 0; + } + + public static void process(int[] arr, int i, int sum, HashSet set) { + if (i == arr.length) { + set.add(sum); + return; + } + process(arr, i + 1, sum, set); + process(arr, i + 1, sum + arr[i], set); + } + + public static int unformedSum2(int[] arr) { + if (arr == null || arr.length == 0) { + return 1; + } + int sum = 0; + int min = Integer.MAX_VALUE; + for (int i = 0; i != arr.length; i++) { + sum += arr[i]; + min = Math.min(min, arr[i]); + } + // boolean[][] dp ... + int N = arr.length; + boolean[][] dp = new boolean[N][sum + 1]; + for (int i = 0; i < N; i++) {// arr[0..i] 0 + dp[i][0] = true; + } + dp[0][arr[0]] = true; + for (int i = 1; i < N; i++) { + for (int j = 1; j <= sum; j++) { + dp[i][j] = dp[i - 1][j] || ((j - arr[i] >= 0) ? dp[i - 1][j - arr[i]] : false); + } + } + for (int j = min; j <= sum; j++) { + if (!dp[N - 1][j]) { + return j; + } + } + return sum + 1; + } + + // 已知arr中肯定有1这个数 + public static int unformedSum3(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + Arrays.sort(arr); // O (N * logN) + int range = 1; + // arr[0] == 1 + for (int i = 1; i != arr.length; i++) { + if (arr[i] > range + 1) { + return range + 1; + } else { + range += arr[i]; + } + } + return range + 1; + } + + public static int[] generateArray(int len, int maxValue) { + int[] res = new int[len]; + for (int i = 0; i != res.length; i++) { + res[i] = (int) (Math.random() * maxValue) + 1; + } + return res; + } + + public static void printArray(int[] arr) { + for (int i = 0; i != arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + public static void main(String[] args) { + int len = 27; + int max = 30; + int[] arr = generateArray(len, max); + printArray(arr); + long start = System.currentTimeMillis(); + System.out.println(unformedSum1(arr)); + long end = System.currentTimeMillis(); + System.out.println("cost time: " + (end - start) + " ms"); + System.out.println("======================================"); + + start = System.currentTimeMillis(); + System.out.println(unformedSum2(arr)); + end = System.currentTimeMillis(); + System.out.println("cost time: " + (end - start) + " ms"); + System.out.println("======================================"); + + System.out.println("set arr[0] to 1"); + arr[0] = 1; + start = System.currentTimeMillis(); + System.out.println(unformedSum3(arr)); + end = System.currentTimeMillis(); + System.out.println("cost time: " + (end - start) + " ms"); + + } +} diff --git a/大厂刷题班/class16/Code03_MinPatches.java b/大厂刷题班/class16/Code03_MinPatches.java new file mode 100644 index 0000000..de5eb5d --- /dev/null +++ b/大厂刷题班/class16/Code03_MinPatches.java @@ -0,0 +1,82 @@ +package class16; + +import java.util.Arrays; + +public class Code03_MinPatches { + + // arr请保证有序，且正数 1~aim + public static int minPatches(int[] arr, int aim) { + int patches = 0; // 缺多少个数字 + long range = 0; // 已经完成了1 ~ range的目标 + Arrays.sort(arr); + for (int i = 0; i != arr.length; i++) { + // arr[i] + // 要求：1 ~ arr[i]-1 范围被搞定！ + while (arr[i] - 1 > range) { // arr[i] 1 ~ arr[i]-1 + range += range + 1; // range + 1 是缺的数字 + patches++; + if (range >= aim) { + return patches; + } + } + // 要求被满足了！ + range += arr[i]; + if (range >= aim) { + return patches; + } + } + while (aim >= range + 1) { + range += range + 1; + patches++; + } + return patches; + } + + // 嘚瑟 + public static int minPatches2(int[] arr, int K) { + int patches = 0; // 缺多少个数字 + int range = 0; // 已经完成了1 ~ range的目标 + for (int i = 0; i != arr.length; i++) { + // 1~range + // 1 ~ arr[i]-1 + while (arr[i] > range + 1) { // arr[i] 1 ~ arr[i]-1 + + if (range > Integer.MAX_VALUE - range - 1) { + return patches + 1; + } + + range += range + 1; // range + 1 是缺的数字 + patches++; + if (range >= K) { + return patches; + } + } + if (range > Integer.MAX_VALUE - arr[i]) { + return patches; + } + range += arr[i]; + if (range >= K) { + return patches; + } + } + while (K >= range + 1) { + if (K == range && K == Integer.MAX_VALUE) { + return patches; + } + if (range > Integer.MAX_VALUE - range - 1) { + return patches + 1; + } + range += range + 1; + patches++; + } + return patches; + } + + public static void main(String[] args) { + int[] test = { 1, 2, 31, 33 }; + int n = 2147483647; + System.out.println(minPatches(test, n)); + + } + +} diff --git a/大厂刷题班/class16/Code04_MergeRecord.java b/大厂刷题班/class16/Code04_MergeRecord.java new file mode 100644 index 0000000..a7b1798 --- /dev/null +++ b/大厂刷题班/class16/Code04_MergeRecord.java @@ -0,0 +1,220 @@ +package class16; + +public class Code04_MergeRecord { + + /* + * 腾讯原题 + * + * 给定整数power，给定一个数组arr，给定一个数组reverse。含义如下： + * arr的长度一定是2的power次方，reverse中的每个值一定都在0~power范围。 + * 例如power = 2, arr = {3, 1, 4, 2}，reverse = {0, 1, 0, 2} + * 任何一个在前的数字可以和任何一个在后的数组，构成一对数。可能是升序关系、相等关系或者降序关系。 + * 比如arr开始时有如下的降序对：(3,1)、(3,2)、(4,2)，一共3个。 + * 接下来根据reverse对arr进行调整： + * reverse[0] = 0, 表示在arr中，划分每1(2的0次方)个数一组，然后每个小组内部逆序，那么arr变成 + * [3,1,4,2]，此时有3个逆序对。 + * reverse[1] = 1, 表示在arr中，划分每2(2的1次方)个数一组，然后每个小组内部逆序，那么arr变成 + * [1,3,2,4]，此时有1个逆序对 + * reverse[2] = 0, 表示在arr中，划分每1(2的0次方)个数一组，然后每个小组内部逆序，那么arr变成 + * [1,3,2,4]，此时有1个逆序对。 + * reverse[3] = 2, 表示在arr中，划分每4(2的2次方)个数一组，然后每个小组内部逆序，那么arr变成 + * [4,2,3,1]，此时有5个逆序对。 + * 所以返回[3,1,1,5]，表示每次调整之后的逆序对数量。 + * + * 输入数据状况： + * power的范围[0,20] + * arr长度范围[1,10的7次方] + * reverse长度范围[1,10的6次方] + * + * */ + + public static int[] reversePair1(int[] originArr, int[] reverseArr, int power) { + int[] ans = new int[reverseArr.length]; + for (int i = 0; i < reverseArr.length; i++) { + reverseArray(originArr, 1 << (reverseArr[i])); + ans[i] = countReversePair(originArr); + } + return ans; + } + + public static void reverseArray(int[] originArr, int teamSize) { + if (teamSize < 2) { + return; + } + for (int i = 0; i < originArr.length; i += teamSize) { + reversePart(originArr, i, i + teamSize - 1); + } + } + + public static void reversePart(int[] arr, int L, int R) { + while (L < R) { + int tmp = arr[L]; + arr[L++] = arr[R]; + arr[R--] = tmp; + } + } + + public static int countReversePair(int[] originArr) { + int ans = 0; + for (int i = 0; i < originArr.length; i++) { + for (int j = i + 1; j < originArr.length; j++) { + if (originArr[i] > originArr[j]) { + ans++; + } + } + } + return ans; + } + + public static int[] reversePair2(int[] originArr, int[] reverseArr, int power) { + int[] reverse = copyArray(originArr); + reversePart(reverse, 0, reverse.length - 1); + int[] recordDown = new int[power + 1]; + int[] recordUp = new int[power + 1]; + process(originArr, 0, originArr.length - 1, power, recordDown); + process(reverse, 0, reverse.length - 1, power, recordUp); + int[] ans = new int[reverseArr.length]; + for (int i = 0; i < reverseArr.length; i++) { + int curPower = reverseArr[i]; + for (int p = 1; p <= curPower; p++) { + int tmp = recordDown[p]; + recordDown[p] = recordUp[p]; + recordUp[p] = tmp; + } + for (int p = 1; p <= power; p++) { + ans[i] += recordDown[p]; + } + } + return ans; + } + + // originArr[L...R]完成排序！ + // L...M左 M...R右 merge + // L...R 2的power次方 + + public static void process(int[] originArr, int L, int R, int power, int[] record) { + if (L == R) { + return; + } + int mid = L + ((R - L) >> 1); + process(originArr, L, mid, power - 1, record); + process(originArr, mid + 1, R, power - 1, record); + record[power] += merge(originArr, L, mid, R); + } + + public static int merge(int[] arr, int L, int m, int r) { + int[] help = new int[r - L + 1]; + int i = 0; + int p1 = L; + int p2 = m + 1; + int ans = 0; + while (p1 <= m && p2 <= r) { + ans += arr[p1] > arr[p2] ? (m - p1 + 1) : 0; + help[i++] = arr[p1] <= arr[p2] ? arr[p1++] : arr[p2++]; + } + while (p1 <= m) { + help[i++] = arr[p1++]; + } + while (p2 <= r) { + help[i++] = arr[p2++]; + } + for (i = 0; i < help.length; i++) { + arr[L + i] = help[i]; + } + return ans; + } + + // for test + public static int[] generateRandomOriginArray(int power, int value) { + int[] ans = new int[1 << power]; + for (int i = 0; i < ans.length; i++) { + ans[i] = (int) (Math.random() * value); + } + return ans; + } + + // for test + public static int[] generateRandomReverseArray(int len, int power) { + int[] ans = new int[len]; + for (int i = 0; i < ans.length; i++) { + ans[i] = (int) (Math.random() * (power + 1)); + } + return ans; + } + + // for test + public static void printArray(int[] arr) { + System.out.println("arr size : " + arr.length); + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + public static void main(String[] args) { + int powerMax = 8; + int msizeMax = 10; + int value = 30; + int testTime = 50000; + System.out.println("test begin"); + for (int i = 0; i < testTime; i++) { + int power = (int) (Math.random() * powerMax) + 1; + int msize = (int) (Math.random() * msizeMax) + 1; + int[] originArr = generateRandomOriginArray(power, value); + int[] originArr1 = copyArray(originArr); + int[] originArr2 = copyArray(originArr); + int[] reverseArr = generateRandomReverseArray(msize, power); + int[] reverseArr1 = copyArray(reverseArr); + int[] reverseArr2 = copyArray(reverseArr); + int[] ans1 = reversePair1(originArr1, reverseArr1, power); + int[] ans2 = reversePair2(originArr2, reverseArr2, power); + if (!isEqual(ans1, ans2)) { + System.out.println("Oops!"); + } + } + System.out.println("test finish!"); + + powerMax = 20; + msizeMax = 1000000; + value = 1000; + int[] originArr = generateRandomOriginArray(powerMax, value); + int[] reverseArr = generateRandomReverseArray(msizeMax, powerMax); + // int[] ans1 = reversePair1(originArr1, reverseArr1, powerMax); + long start = System.currentTimeMillis(); + reversePair2(originArr, reverseArr, powerMax); + long end = System.currentTimeMillis(); + System.out.println("run time : " + (end - start) + " ms"); + } + +} diff --git a/大厂刷题班/class16/Code05_JosephusProblem.java b/大厂刷题班/class16/Code05_JosephusProblem.java new file mode 100644 index 0000000..0aaebd8 --- /dev/null +++ b/大厂刷题班/class16/Code05_JosephusProblem.java @@ -0,0 +1,120 @@ +package class16; + +// 本题测试链接 : https://leetcode-cn.com/problems/yuan-quan-zhong-zui-hou-sheng-xia-de-shu-zi-lcof/ +public class Code05_JosephusProblem { + + // 提交直接通过 + // 给定的编号是0~n-1的情况下，数到m就杀 + // 返回谁会活？ + public int lastRemaining1(int n, int m) { + return getLive(n, m) - 1; + } + + // 课上题目的设定是，给定的编号是1~n的情况下，数到m就杀 + // 返回谁会活？ + public static int getLive(int n, int m) { + if (n == 1) { + return 1; + } + return (getLive(n - 1, m) + m - 1) % n + 1; + } + + // 提交直接通过 + // 给定的编号是0~n-1的情况下，数到m就杀 + // 返回谁会活？ + // 这个版本是迭代版 + public int lastRemaining2(int n, int m) { + int ans = 1; + int r = 1; + while (r <= n) { + ans = (ans + m - 1) % (r++) + 1; + } + return ans - 1; + } + + // 以下的code针对单链表，不要提交 + public static class Node { + public int value; + public Node next; + + public Node(int data) { + this.value = data; + } + } + + public static Node josephusKill1(Node head, int m) { + if (head == null || head.next == head || m < 1) { + return head; + } + Node last = head; + while (last.next != head) { + last = last.next; + } + int count = 0; + while (head != last) { + if (++count == m) { + last.next = head.next; + count = 0; + } else { + last = last.next; + } + head = last.next; + } + return head; + } + + public static Node josephusKill2(Node head, int m) { + if (head == null || head.next == head || m < 1) { + return head; + } + Node cur = head.next; + int size = 1; // tmp -> list size + while (cur != head) { + size++; + cur = cur.next; + } + int live = getLive(size, m); // tmp -> service node position + while (--live != 0) { + head = head.next; + } + head.next = head; + return head; + } + + public static void printCircularList(Node head) { + if (head == null) { + return; + } + System.out.print("Circular List: " + head.value + " "); + Node cur = head.next; + while (cur != head) { + System.out.print(cur.value + " "); + cur = cur.next; + } + System.out.println("-> " + head.value); + } + + public static void main(String[] args) { + Node head1 = new Node(1); + head1.next = new Node(2); + head1.next.next = new Node(3); + head1.next.next.next = new Node(4); + head1.next.next.next.next = new Node(5); + head1.next.next.next.next.next = head1; + printCircularList(head1); + head1 = josephusKill1(head1, 3); + printCircularList(head1); + + Node head2 = new Node(1); + head2.next = new Node(2); + head2.next.next = new Node(3); + head2.next.next.next = new Node(4); + head2.next.next.next.next = new Node(5); + head2.next.next.next.next.next = head2; + printCircularList(head2); + head2 = josephusKill2(head2, 3); + printCircularList(head2); + + } + +} \ No newline at end of file diff --git a/大厂刷题班/class17/Code01_FindNumInSortedMatrix.java b/大厂刷题班/class17/Code01_FindNumInSortedMatrix.java new file mode 100644 index 0000000..16aea90 --- /dev/null +++ b/大厂刷题班/class17/Code01_FindNumInSortedMatrix.java @@ -0,0 +1,34 @@ +package class17; + +public class Code01_FindNumInSortedMatrix { + + public static boolean isContains(int[][] matrix, int K) { + int row = 0; + int col = matrix[0].length - 1; + while (row < matrix.length && col > -1) { + if (matrix[row][col] == K) { + return true; + } else if (matrix[row][col] > K) { + col--; + } else { + row++; + } + } + return false; + } + + public static void main(String[] args) { + int[][] matrix = new int[][] { { 0, 1, 2, 3, 4, 5, 6 }, // 0 + { 10, 12, 13, 15, 16, 17, 18 }, // 1 + { 23, 24, 25, 26, 27, 28, 29 }, // 2 + { 44, 45, 46, 47, 48, 49, 50 }, // 3 + { 65, 66, 67, 68, 69, 70, 71 }, // 4 + { 96, 97, 98, 99, 100, 111, 122 }, // 5 + { 166, 176, 186, 187, 190, 195, 200 }, // 6 + { 233, 243, 321, 341, 356, 370, 380 } // 7 + }; + int K = 233; + System.out.println(isContains(matrix, K)); + } + +} diff --git a/大厂刷题班/class17/Code02_KthSmallestElementInSortedMatrix.java b/大厂刷题班/class17/Code02_KthSmallestElementInSortedMatrix.java new file mode 100644 index 0000000..0dcd8e9 --- /dev/null +++ b/大厂刷题班/class17/Code02_KthSmallestElementInSortedMatrix.java @@ -0,0 +1,110 @@ +package class17; + +import java.util.Comparator; +import java.util.PriorityQueue; + +// 本题测试链接 : https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/ +public class Code02_KthSmallestElementInSortedMatrix { + + // 堆的方法 + public static int kthSmallest1(int[][] matrix, int k) { + int N = matrix.length; + int M = matrix[0].length; + PriorityQueue heap = new PriorityQueue<>(new NodeComparator()); + boolean[][] set = new boolean[N][M]; + heap.add(new Node(matrix[0][0], 0, 0)); + set[0][0] = true; + int count = 0; + Node ans = null; + while (!heap.isEmpty()) { + ans = heap.poll(); + if (++count == k) { + break; + } + int row = ans.row; + int col = ans.col; + if (row + 1 < N && !set[row + 1][col]) { + heap.add(new Node(matrix[row + 1][col], row + 1, col)); + set[row + 1][col] = true; + } + if (col + 1 < M && !set[row][col + 1]) { + heap.add(new Node(matrix[row][col + 1], row, col + 1)); + set[row][col + 1] = true; + } + } + return ans.value; + } + + public static class Node { + public int value; + public int row; + public int col; + + public Node(int v, int r, int c) { + value = v; + row = r; + col = c; + } + + } + + public static class NodeComparator implements Comparator { + + @Override + public int compare(Node o1, Node o2) { + return o1.value - o2.value; + } + + } + + // 二分的方法 + public static int kthSmallest2(int[][] matrix, int k) { + int N = matrix.length; + int M = matrix[0].length; + int left = matrix[0][0]; + int right = matrix[N - 1][M - 1]; + int ans = 0; + while (left <= right) { + int mid = left + ((right - left) >> 1); + // <=mid 有几个 <= mid 在矩阵中真实出现的数，谁最接近mid + Info info = noMoreNum(matrix, mid); + if (info.num < k) { + left = mid + 1; + } else { + ans = info.near; + right = mid - 1; + } + } + return ans; + } + + public static class Info { + public int near; + public int num; + + public Info(int n1, int n2) { + near = n1; + num = n2; + } + } + + public static Info noMoreNum(int[][] matrix, int value) { + int near = Integer.MIN_VALUE; + int num = 0; + int N = matrix.length; + int M = matrix[0].length; + int row = 0; + int col = M - 1; + while (row < N && col >= 0) { + if (matrix[row][col] <= value) { + near = Math.max(near, matrix[row][col]); + num += col + 1; + row++; + } else { + col--; + } + } + return new Info(near, num); + } + +} diff --git a/大厂刷题班/class17/Code03_PalindromePairs.java b/大厂刷题班/class17/Code03_PalindromePairs.java new file mode 100644 index 0000000..b3d110b --- /dev/null +++ b/大厂刷题班/class17/Code03_PalindromePairs.java @@ -0,0 +1,104 @@ +package class17; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.List; + +// 测试链接 : https://leetcode.com/problems/palindrome-pairs/ +public class Code03_PalindromePairs { + + public static List> palindromePairs(String[] words) { + HashMap wordset = new HashMap<>(); + for (int i = 0; i < words.length; i++) { + wordset.put(words[i], i); + } + List> res = new ArrayList<>(); + //{ [6,23] 、 [7,13] } + for (int i = 0; i < words.length; i++) { + // i words[i] + // findAll(字符串，在i位置，wordset) 返回所有生成的结果返回 + res.addAll(findAll(words[i], i, wordset)); + } + return res; + } + + public static List> findAll(String word, int index, HashMap words) { + List> res = new ArrayList<>(); + String reverse = reverse(word); + Integer rest = words.get(""); + if (rest != null && rest != index && word.equals(reverse)) { + addRecord(res, rest, index); + addRecord(res, index, rest); + } + int[] rs = manacherss(word); + int mid = rs.length >> 1; + for (int i = 1; i < mid; i++) { + if (i - rs[i] == -1) { + rest = words.get(reverse.substring(0, mid - i)); + if (rest != null && rest != index) { + addRecord(res, rest, index); + } + } + } + for (int i = mid + 1; i < rs.length; i++) { + if (i + rs[i] == rs.length) { + rest = words.get(reverse.substring((mid << 1) - i)); + if (rest != null && rest != index) { + addRecord(res, index, rest); + } + } + } + return res; + } + + public static void addRecord(List> res, int left, int right) { + List newr = new ArrayList<>(); + newr.add(left); + newr.add(right); + res.add(newr); + } + + public static int[] manacherss(String word) { + char[] mchs = manachercs(word); + int[] rs = new int[mchs.length]; + int center = -1; + int pr = -1; + for (int i = 0; i != mchs.length; i++) { + rs[i] = pr > i ? Math.min(rs[(center << 1) - i], pr - i) : 1; + while (i + rs[i] < mchs.length && i - rs[i] > -1) { + if (mchs[i + rs[i]] != mchs[i - rs[i]]) { + break; + } + rs[i]++; + } + if (i + rs[i] > pr) { + pr = i + rs[i]; + center = i; + } + } + return rs; + } + + public static char[] manachercs(String word) { + char[] chs = word.toCharArray(); + char[] mchs = new char[chs.length * 2 + 1]; + int index = 0; + for (int i = 0; i != mchs.length; i++) { + mchs[i] = (i & 1) == 0 ? '#' : chs[index++]; + } + return mchs; + } + + public static String reverse(String str) { + char[] chs = str.toCharArray(); + int l = 0; + int r = chs.length - 1; + while (l < r) { + char tmp = chs[l]; + chs[l++] = chs[r]; + chs[r--] = tmp; + } + return String.valueOf(chs); + } + +} \ No newline at end of file diff --git a/大厂刷题班/class17/Code04_DistinctSubseq.java b/大厂刷题班/class17/Code04_DistinctSubseq.java new file mode 100644 index 0000000..ff047e4 --- /dev/null +++ b/大厂刷题班/class17/Code04_DistinctSubseq.java @@ -0,0 +1,101 @@ +package class17; + +// 测试链接 : https://leetcode-cn.com/problems/21dk04/ +public class Code04_DistinctSubseq { + + public static int numDistinct1(String S, String T) { + char[] s = S.toCharArray(); + char[] t = T.toCharArray(); + return process(s, t, s.length, t.length); + } + + public static int process(char[] s, char[] t, int i, int j) { + if (j == 0) { + return 1; + } + if (i == 0) { + return 0; + } + int res = process(s, t, i - 1, j); + if (s[i - 1] == t[j - 1]) { + res += process(s, t, i - 1, j - 1); + } + return res; + } + + // S[...i]的所有子序列中，包含多少个字面值等于T[...j]这个字符串的子序列 + // 记为dp[i][j] + // 可能性1）S[...i]的所有子序列中，都不以s[i]结尾，则dp[i][j]肯定包含dp[i-1][j] + // 可能性2）S[...i]的所有子序列中，都必须以s[i]结尾， + // 这要求S[i] == T[j]，则dp[i][j]包含dp[i-1][j-1] + public static int numDistinct2(String S, String T) { + char[] s = S.toCharArray(); + char[] t = T.toCharArray(); + // dp[i][j] : s[0..i] T[0...j] + + // dp[i][j] : s只拿前i个字符做子序列，有多少个子序列，字面值等于T的前j个字符的前缀串 + int[][] dp = new int[s.length + 1][t.length + 1]; + // dp[0][0] + // dp[0][j] = s只拿前0个字符做子序列, T前j个字符 + for (int j = 0; j <= t.length; j++) { + dp[0][j] = 0; + } + for (int i = 0; i <= s.length; i++) { + dp[i][0] = 1; + } + for (int i = 1; i <= s.length; i++) { + for (int j = 1; j <= t.length; j++) { + dp[i][j] = dp[i - 1][j] + (s[i - 1] == t[j - 1] ? dp[i - 1][j - 1] : 0); + } + } + return dp[s.length][t.length]; + } + + public static int numDistinct3(String S, String T) { + char[] s = S.toCharArray(); + char[] t = T.toCharArray(); + int[] dp = new int[t.length + 1]; + dp[0] = 1; + for (int j = 1; j <= t.length; j++) { + dp[j] = 0; + } + for (int i = 1; i <= s.length; i++) { + for (int j = t.length; j >= 1; j--) { + dp[j] += s[i - 1] == t[j - 1] ? dp[j - 1] : 0; + } + } + return dp[t.length]; + } + + public static int dp(String S, String T) { + char[] s = S.toCharArray(); + char[] t = T.toCharArray(); + int N = s.length; + int M = t.length; + int[][] dp = new int[N][M]; + // s[0..0] T[0..0] dp[0][0] + dp[0][0] = s[0] == t[0] ? 1 : 0; + for (int i = 1; i < N; i++) { + dp[i][0] = s[i] == t[0] ? (dp[i - 1][0] + 1) : dp[i - 1][0]; + } + for (int i = 1; i < N; i++) { + for (int j = 1; j <= Math.min(i, M - 1); j++) { + dp[i][j] = dp[i - 1][j]; + if (s[i] == t[j]) { + dp[i][j] += dp[i - 1][j - 1]; + } + } + } + return dp[N - 1][M - 1]; + } + + public static void main(String[] args) { + String S = "1212311112121132"; + String T = "13"; + + System.out.println(numDistinct3(S, T)); + System.out.println(dp(S, T)); + + } + +} diff --git a/大厂刷题班/class17/Code05_DistinctSubseqValue.java b/大厂刷题班/class17/Code05_DistinctSubseqValue.java new file mode 100644 index 0000000..17f00d5 --- /dev/null +++ b/大厂刷题班/class17/Code05_DistinctSubseqValue.java @@ -0,0 +1,50 @@ +package class17; + +import java.util.HashMap; + +// 本题测试链接 : https://leetcode.com/problems/distinct-subsequences-ii/ +public class Code05_DistinctSubseqValue { + + public static int distinctSubseqII(String s) { + if (s == null || s.length() == 0) { + return 0; + } + int m = 1000000007; + char[] str = s.toCharArray(); + int[] count = new int[26]; + int all = 1; // 算空集 + for (char x : str) { + int add = (all - count[x - 'a'] + m) % m; + all = (all + add) % m; + count[x - 'a'] = (count[x - 'a'] + add) % m; + } + return all - 1; + } + + public static int zuo(String s) { + if (s == null || s.length() == 0) { + return 0; + } + int m = 1000000007; + char[] str = s.toCharArray(); + HashMap map = new HashMap<>(); + int all = 1; // 一个字符也没遍历的时候，有空集 + for (char x : str) { + int newAdd = all; +// int curAll = all + newAdd - (map.containsKey(x) ? map.get(x) : 0); + int curAll = all; + curAll = (curAll + newAdd) % m; + curAll = (curAll - (map.containsKey(x) ? map.get(x) : 0) + m) % m; + all = curAll; + map.put(x, newAdd); + } + return all; + } + + public static void main(String[] args) { + String s = "bccaccbaabbc"; + System.out.println(distinctSubseqII(s) + 1); + System.out.println(zuo(s)); + } + +} diff --git a/大厂刷题班/class18/Code01_HanoiProblem.java b/大厂刷题班/class18/Code01_HanoiProblem.java new file mode 100644 index 0000000..588377e --- /dev/null +++ b/大厂刷题班/class18/Code01_HanoiProblem.java @@ -0,0 +1,97 @@ +package class18; + +public class Code01_HanoiProblem { + + public static int step1(int[] arr) { + if (arr == null || arr.length == 0) { + return -1; + } + return process(arr, arr.length - 1, 1, 2, 3); + } + + // 目标是: 把0~i的圆盘，从from全部挪到to上 + // 返回，根据arr中的状态arr[0..i]，它是最优解的第几步？ + // f(i, 3 , 2, 1) f(i, 1, 3, 2) f(i, 3, 1, 2) + public static int process(int[] arr, int i, int from, int other, int to) { + if (i == -1) { + return 0; + } + if (arr[i] != from && arr[i] != to) { + return -1; + } + if (arr[i] == from) { // 第一大步没走完 + return process(arr, i - 1, from, to, other); + } else { // arr[i] == to + // 已经走完1，2两步了， + int rest = process(arr, i - 1, other, from, to); // 第三大步完成的程度 + if (rest == -1) { + return -1; + } + return (1 << i) + rest; + } + } + + public static int step2(int[] arr) { + if (arr == null || arr.length == 0) { + return -1; + } + int from = 1; + int mid = 2; + int to = 3; + int i = arr.length - 1; + int res = 0; + int tmp = 0; + while (i >= 0) { + if (arr[i] != from && arr[i] != to) { + return -1; + } + if (arr[i] == to) { + res += 1 << i; + tmp = from; + from = mid; + } else { + tmp = to; + to = mid; + } + mid = tmp; + i--; + } + return res; + } + + public static int kth(int[] arr) { + int N = arr.length; + return step(arr, N - 1, 1, 3, 2); + } + + // 0...index这些圆盘，arr[0..index] index+1层塔 + // 在哪？from 去哪？to 另一个是啥？other + // arr[0..index]这些状态，是index+1层汉诺塔问题的，最优解第几步 + public static int step(int[] arr, int index, int from, int to, int other) { + if (index == -1) { + return 0; + } + if (arr[index] == other) { + return -1; + } + // arr[index] == from arr[index] == to; + if (arr[index] == from) { + return step(arr, index - 1, from, other, to); + } else { + int p1 = (1 << index) - 1; + int p2 = 1; + int p3 = step(arr, index - 1, other, to, from); + if (p3 == -1) { + return -1; + } + return p1 + p2 + p3; + } + } + + public static void main(String[] args) { + int[] arr = { 3, 3, 2, 1 }; + System.out.println(step1(arr)); + System.out.println(step2(arr)); + System.out.println(kth(arr)); + } +} diff --git a/大厂刷题班/class18/Code02_ShortestBridge.java b/大厂刷题班/class18/Code02_ShortestBridge.java new file mode 100644 index 0000000..cac7b20 --- /dev/null +++ b/大厂刷题班/class18/Code02_ShortestBridge.java @@ -0,0 +1,96 @@ +package class18; + +// 本题测试链接 : https://leetcode.com/problems/shortest-bridge/ +public class Code02_ShortestBridge { + + public static int shortestBridge(int[][] m) { + int N = m.length; + int M = m[0].length; + int all = N * M; + int island = 0; + int[] curs = new int[all]; + int[] nexts = new int[all]; + int[][] records = new int[2][all]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + if (m[i][j] == 1) { // 当前位置发现了1！ + // 把这一片的1，都变成2，同时，抓上来了，这一片1组成的初始队列 + // curs, 把这一片的1到自己的距离，都设置成1了，records + int queueSize = infect(m, i, j, N, M, curs, 0, records[island]); + int V = 1; + while (queueSize != 0) { + V++; + // curs里面的点，上下左右，records[点]==0， nexts + queueSize = bfs(N, M, all, V, curs, queueSize, nexts, records[island]); + int[] tmp = curs; + curs = nexts; + nexts = tmp; + } + island++; + } + } + } + int min = Integer.MAX_VALUE; + for (int i = 0; i < all; i++) { + min = Math.min(min, records[0][i] + records[1][i]); + } + return min - 3; + } + + // 当前来到m[i][j] , 总行数是N，总列数是M + // m[i][j]感染出去(找到这一片岛所有的1),把每一个1的坐标，放入到int[] curs队列！ + // 1 (a,b) -> curs[index++] = (a * M + b) + // 1 (c,d) -> curs[index++] = (c * M + d) + // 二维已经变成一维了， 1 (a,b) -> a * M + b + // 设置距离record[a * M +b ] = 1 + public static int infect(int[][] m, int i, int j, int N, int M, int[] curs, int index, int[] record) { + if (i < 0 || i == N || j < 0 || j == M || m[i][j] != 1) { + return index; + } + // m[i][j] 不越界，且m[i][j] == 1 + m[i][j] = 2; + int p = i * M + j; + record[p] = 1; + // 收集到不同的1 + curs[index++] = p; + index = infect(m, i - 1, j, N, M, curs, index, record); + index = infect(m, i + 1, j, N, M, curs, index, record); + index = infect(m, i, j - 1, N, M, curs, index, record); + index = infect(m, i, j + 1, N, M, curs, index, record); + return index; + } + + // 二维原始矩阵中，N总行数，M总列数 + // all 总 all = N * M + // V 要生成的是第几层 curs V-1 nexts V + // record里面拿距离 + public static int bfs(int N, int M, int all, int V, + int[] curs, int size, int[] nexts, int[] record) { + int nexti = 0; // 我要生成的下一层队列成长到哪了？ + for (int i = 0; i < size; i++) { + // curs[i] -> 一个位置 + int up = curs[i] < M ? -1 : curs[i] - M; + int down = curs[i] + M >= all ? -1 : curs[i] + M; + int left = curs[i] % M == 0 ? -1 : curs[i] - 1; + int right = curs[i] % M == M - 1 ? -1 : curs[i] + 1; + if (up != -1 && record[up] == 0) { + record[up] = V; + nexts[nexti++] = up; + } + if (down != -1 && record[down] == 0) { + record[down] = V; + nexts[nexti++] = down; + } + if (left != -1 && record[left] == 0) { + record[left] = V; + nexts[nexti++] = left; + } + if (right != -1 && record[right] == 0) { + record[right] = V; + nexts[nexti++] = right; + } + } + return nexti; + } + +} diff --git a/大厂刷题班/class18/Code03_CherryPickup.java b/大厂刷题班/class18/Code03_CherryPickup.java new file mode 100644 index 0000000..07b9c54 --- /dev/null +++ b/大厂刷题班/class18/Code03_CherryPickup.java @@ -0,0 +1,70 @@ +package class18; + +// 牛客的测试链接： +// https://www.nowcoder.com/questionTerminal/8ecfe02124674e908b2aae65aad4efdf +// 把如下的全部代码拷贝进java编辑器 +// 把文件大类名字改成Main，可以直接通过 + +import java.util.Scanner; + +public class Code03_CherryPickup { + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + int N = sc.nextInt(); + int M = sc.nextInt(); + int[][] matrix = new int[N][M]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + matrix[i][j] = sc.nextInt(); + } + } + int ans = cherryPickup(matrix); + System.out.println(ans); + sc.close(); + } + + public static int cherryPickup(int[][] grid) { + int N = grid.length; + int M = grid[0].length; + int[][][] dp = new int[N][M][N]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + for (int k = 0; k < N; k++) { + dp[i][j][k] = Integer.MIN_VALUE; + } + } + } + int ans = process(grid, 0, 0, 0, dp); + return ans < 0 ? 0 : ans; + } + + public static int process(int[][] grid, int x1, int y1, int x2, int[][][] dp) { + if (x1 == grid.length || y1 == grid[0].length || x2 == grid.length || x1 + y1 - x2 == grid[0].length) { + return Integer.MIN_VALUE; + } + if (dp[x1][y1][x2] != Integer.MIN_VALUE) { + return dp[x1][y1][x2]; + } + if (x1 == grid.length - 1 && y1 == grid[0].length - 1) { + dp[x1][y1][x2] = grid[x1][y1]; + return dp[x1][y1][x2]; + } + int next = Integer.MIN_VALUE; + next = Math.max(next, process(grid, x1 + 1, y1, x2 + 1, dp)); + next = Math.max(next, process(grid, x1 + 1, y1, x2, dp)); + next = Math.max(next, process(grid, x1, y1 + 1, x2, dp)); + next = Math.max(next, process(grid, x1, y1 + 1, x2 + 1, dp)); + if (grid[x1][y1] == -1 || grid[x2][x1 + y1 - x2] == -1 || next == -1) { + dp[x1][y1][x2] = -1; + return dp[x1][y1][x2]; + } + if (x1 == x2) { + dp[x1][y1][x2] = grid[x1][y1] + next; + return dp[x1][y1][x2]; + } + dp[x1][y1][x2] = grid[x1][y1] + grid[x2][x1 + y1 - x2] + next; + return dp[x1][y1][x2]; + } + +} \ No newline at end of file diff --git a/大厂刷题班/class18/Code04_TopKSumCrossTwoArrays.java b/大厂刷题班/class18/Code04_TopKSumCrossTwoArrays.java new file mode 100644 index 0000000..9820986 --- /dev/null +++ b/大厂刷题班/class18/Code04_TopKSumCrossTwoArrays.java @@ -0,0 +1,93 @@ +package class18; + +// 牛客的测试链接： +// https://www.nowcoder.com/practice/7201cacf73e7495aa5f88b223bbbf6d1 +// 不要提交包信息，把import底下的类名改成Main，提交下面的代码可以直接通过 +// 因为测试平台会卡空间，所以把set换成了动态加和减的结构 + +import java.util.Scanner; +import java.util.Comparator; +import java.util.HashSet; +import java.util.PriorityQueue; + +public class Code04_TopKSumCrossTwoArrays { + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + int N = sc.nextInt(); + int K = sc.nextInt(); + int[] arr1 = new int[N]; + int[] arr2 = new int[N]; + for (int i = 0; i < N; i++) { + arr1[i] = sc.nextInt(); + } + for (int i = 0; i < N; i++) { + arr2[i] = sc.nextInt(); + } + int[] topK = topKSum(arr1, arr2, K); + for (int i = 0; i < K; i++) { + System.out.print(topK[i] + " "); + } + System.out.println(); + sc.close(); + } + + // 放入大根堆中的结构 + public static class Node { + public int index1;// arr1中的位置 + public int index2;// arr2中的位置 + public int sum;// arr1[index1] + arr2[index2]的值 + + public Node(int i1, int i2, int s) { + index1 = i1; + index2 = i2; + sum = s; + } + } + + // 生成大根堆的比较器 + public static class MaxHeapComp implements Comparator { + @Override + public int compare(Node o1, Node o2) { + return o2.sum - o1.sum; + } + } + + public static int[] topKSum(int[] arr1, int[] arr2, int topK) { + if (arr1 == null || arr2 == null || topK < 1) { + return null; + } + int N = arr1.length; + int M = arr2.length; + topK = Math.min(topK, N * M); + int[] res = new int[topK]; + int resIndex = 0; + PriorityQueue maxHeap = new PriorityQueue<>(new MaxHeapComp()); + HashSet set = new HashSet<>(); + int i1 = N - 1; + int i2 = M - 1; + maxHeap.add(new Node(i1, i2, arr1[i1] + arr2[i2])); + set.add(x(i1, i2, M)); + while (resIndex != topK) { + Node curNode = maxHeap.poll(); + res[resIndex++] = curNode.sum; + i1 = curNode.index1; + i2 = curNode.index2; + set.remove(x(i1, i2, M)); + if (i1 - 1 >= 0 && !set.contains(x(i1 - 1, i2, M))) { + set.add(x(i1 - 1, i2, M)); + maxHeap.add(new Node(i1 - 1, i2, arr1[i1 - 1] + arr2[i2])); + } + if (i2 - 1 >= 0 && !set.contains(x(i1, i2 - 1, M))) { + set.add(x(i1, i2 - 1, M)); + maxHeap.add(new Node(i1, i2 - 1, arr1[i1] + arr2[i2 - 1])); + } + } + return res; + } + + public static long x(int i1, int i2, int M) { + return (long) i1 * (long) M + (long) i2; + } + +} diff --git a/大厂刷题班/class19/Code01_LRUCache.java b/大厂刷题班/class19/Code01_LRUCache.java new file mode 100644 index 0000000..4cae760 --- /dev/null +++ b/大厂刷题班/class19/Code01_LRUCache.java @@ -0,0 +1,142 @@ +package class19; + +import java.util.HashMap; + +// 本题测试链接 : https://leetcode.com/problems/lru-cache/ +// 提交时把类名和构造方法名改成 : LRUCache +public class Code01_LRUCache { + + public Code01_LRUCache(int capacity) { + cache = new MyCache<>(capacity); + } + + private MyCache cache; + + public int get(int key) { + Integer ans = cache.get(key); + return ans == null ? -1 : ans; + } + + public void put(int key, int value) { + cache.set(key, value); + } + + public static class Node { + public K key; + public V value; + public Node last; + public Node next; + + public Node(K key, V value) { + this.key = key; + this.value = value; + } + } + + public static class NodeDoubleLinkedList { + private Node head; + private Node tail; + + public NodeDoubleLinkedList() { + head = null; + tail = null; + } + + // 现在来了一个新的node，请挂到尾巴上去 + public void addNode(Node newNode) { + if (newNode == null) { + return; + } + if (head == null) { + head = newNode; + tail = newNode; + } else { + tail.next = newNode; + newNode.last = tail; + tail = newNode; + } + } + + // node 入参，一定保证！node在双向链表里！ + // node原始的位置，左右重新连好，然后把node分离出来 + // 挂到整个链表的尾巴上 + public void moveNodeToTail(Node node) { + if (tail == node) { + return; + } + if (head == node) { + head = node.next; + head.last = null; + } else { + node.last.next = node.next; + node.next.last = node.last; + } + node.last = tail; + node.next = null; + tail.next = node; + tail = node; + } + + public Node removeHead() { + if (head == null) { + return null; + } + Node res = head; + if (head == tail) { + head = null; + tail = null; + } else { + head = res.next; + res.next = null; + head.last = null; + } + return res; + } + + } + + public static class MyCache { + private HashMap> keyNodeMap; + private NodeDoubleLinkedList nodeList; + private final int capacity; + + public MyCache(int cap) { + keyNodeMap = new HashMap>(); + nodeList = new NodeDoubleLinkedList(); + capacity = cap; + } + + public V get(K key) { + if (keyNodeMap.containsKey(key)) { + Node res = keyNodeMap.get(key); + nodeList.moveNodeToTail(res); + return res.value; + } + return null; + } + + // set(Key, Value) + // 新增 更新value的操作 + public void set(K key, V value) { + if (keyNodeMap.containsKey(key)) { + Node node = keyNodeMap.get(key); + node.value = value; + nodeList.moveNodeToTail(node); + } else { // 新增！ + Node newNode = new Node(key, value); + keyNodeMap.put(key, newNode); + nodeList.addNode(newNode); + if (keyNodeMap.size() == capacity + 1) { + removeMostUnusedCache(); + } + } + } + + private void removeMostUnusedCache() { + Node removeNode = nodeList.removeHead(); + keyNodeMap.remove(removeNode.key); + } + + } + +} diff --git a/大厂刷题班/class19/Code02_LFUCache.java b/大厂刷题班/class19/Code02_LFUCache.java new file mode 100644 index 0000000..30ed43a --- /dev/null +++ b/大厂刷题班/class19/Code02_LFUCache.java @@ -0,0 +1,204 @@ +package class19; + +import java.util.HashMap; + +// 本题测试链接 : https://leetcode.com/problems/lfu-cache/ +// 提交时把类名和构造方法名改为 : LFUCache +public class Code02_LFUCache { + + public Code02_LFUCache(int K) { + capacity = K; + size = 0; + records = new HashMap<>(); + heads = new HashMap<>(); + headList = null; + } + + private int capacity; // 缓存的大小限制，即K + private int size; // 缓存目前有多少个节点 + private HashMap records;// 表示key(Integer)由哪个节点(Node)代表 + private HashMap heads; // 表示节点(Node)在哪个桶(NodeList)里 + private NodeList headList; // 整个结构中位于最左的桶 + + // 节点的数据结构 + public static class Node { + public Integer key; + public Integer value; + public Integer times; // 这个节点发生get或者set的次数总和 + public Node up; // 节点之间是双向链表所以有上一个节点 + public Node down;// 节点之间是双向链表所以有下一个节点 + + public Node(int k, int v, int t) { + key = k; + value = v; + times = t; + } + } + + // 桶结构 + public static class NodeList { + public Node head; // 桶的头节点 + public Node tail; // 桶的尾节点 + public NodeList last; // 桶之间是双向链表所以有前一个桶 + public NodeList next; // 桶之间是双向链表所以有后一个桶 + + public NodeList(Node node) { + head = node; + tail = node; + } + + // 把一个新的节点加入这个桶，新的节点都放在顶端变成新的头部 + public void addNodeFromHead(Node newHead) { + newHead.down = head; + head.up = newHead; + head = newHead; + } + + // 判断这个桶是不是空的 + public boolean isEmpty() { + return head == null; + } + + // 删除node节点并保证node的上下环境重新连接 + public void deleteNode(Node node) { + if (head == tail) { + head = null; + tail = null; + } else { + if (node == head) { + head = node.down; + head.up = null; + } else if (node == tail) { + tail = node.up; + tail.down = null; + } else { + node.up.down = node.down; + node.down.up = node.up; + } + } + node.up = null; + node.down = null; + } + } + + // removeNodeList：刚刚减少了一个节点的桶 + // 这个函数的功能是，判断刚刚减少了一个节点的桶是不是已经空了。 + // 1）如果不空，什么也不做 + // + // 2)如果空了，removeNodeList还是整个缓存结构最左的桶(headList)。 + // 删掉这个桶的同时也要让最左的桶变成removeNodeList的下一个。 + // + // 3)如果空了，removeNodeList不是整个缓存结构最左的桶(headList)。 + // 把这个桶删除，并保证上一个的桶和下一个桶之间还是双向链表的连接方式 + // + // 函数的返回值表示刚刚减少了一个节点的桶是不是已经空了，空了返回true；不空返回false + private boolean modifyHeadList(NodeList removeNodeList) { + if (removeNodeList.isEmpty()) { + if (headList == removeNodeList) { + headList = removeNodeList.next; + if (headList != null) { + headList.last = null; + } + } else { + removeNodeList.last.next = removeNodeList.next; + if (removeNodeList.next != null) { + removeNodeList.next.last = removeNodeList.last; + } + } + return true; + } + return false; + } + + // 函数的功能 + // node这个节点的次数+1了，这个节点原来在oldNodeList里。 + // 把node从oldNodeList删掉，然后放到次数+1的桶中 + // 整个过程既要保证桶之间仍然是双向链表，也要保证节点之间仍然是双向链表 + private void move(Node node, NodeList oldNodeList) { + oldNodeList.deleteNode(node); + // preList表示次数+1的桶的前一个桶是谁 + // 如果oldNodeList删掉node之后还有节点，oldNodeList就是次数+1的桶的前一个桶 + // 如果oldNodeList删掉node之后空了，oldNodeList是需要删除的，所以次数+1的桶的前一个桶，是oldNodeList的前一个 + NodeList preList = modifyHeadList(oldNodeList) ? oldNodeList.last : oldNodeList; + // nextList表示次数+1的桶的后一个桶是谁 + NodeList nextList = oldNodeList.next; + if (nextList == null) { + NodeList newList = new NodeList(node); + if (preList != null) { + preList.next = newList; + } + newList.last = preList; + if (headList == null) { + headList = newList; + } + heads.put(node, newList); + } else { + if (nextList.head.times.equals(node.times)) { + nextList.addNodeFromHead(node); + heads.put(node, nextList); + } else { + NodeList newList = new NodeList(node); + if (preList != null) { + preList.next = newList; + } + newList.last = preList; + newList.next = nextList; + nextList.last = newList; + if (headList == nextList) { + headList = newList; + } + heads.put(node, newList); + } + } + } + + public void put(int key, int value) { + if (capacity == 0) { + return; + } + if (records.containsKey(key)) { + Node node = records.get(key); + node.value = value; + node.times++; + NodeList curNodeList = heads.get(node); + move(node, curNodeList); + } else { + if (size == capacity) { + Node node = headList.tail; + headList.deleteNode(node); + modifyHeadList(headList); + records.remove(node.key); + heads.remove(node); + size--; + } + Node node = new Node(key, value, 1); + if (headList == null) { + headList = new NodeList(node); + } else { + if (headList.head.times.equals(node.times)) { + headList.addNodeFromHead(node); + } else { + NodeList newList = new NodeList(node); + newList.next = headList; + headList.last = newList; + headList = newList; + } + } + records.put(key, node); + heads.put(node, headList); + size++; + } + } + + public int get(int key) { + if (!records.containsKey(key)) { + return -1; + } + Node node = records.get(key); + node.times++; + NodeList curNodeList = heads.get(node); + move(node, curNodeList); + return node.value; + } + +} \ No newline at end of file diff --git a/大厂刷题班/class19/Code03_OneNumber.java b/大厂刷题班/class19/Code03_OneNumber.java new file mode 100644 index 0000000..36f2cd6 --- /dev/null +++ b/大厂刷题班/class19/Code03_OneNumber.java @@ -0,0 +1,83 @@ +package class19; + +public class Code03_OneNumber { + + public static int solution1(int num) { + if (num < 1) { + return 0; + } + int count = 0; + for (int i = 1; i != num + 1; i++) { + count += get1Nums(i); + } + return count; + } + + public static int get1Nums(int num) { + int res = 0; + while (num != 0) { + if (num % 10 == 1) { + res++; + } + num /= 10; + } + return res; + } + + + // 1 ~ num 这个范围上，画了几道1 + public static int solution2(int num) { + if (num < 1) { + return 0; + } + // num -> 13625 + // len = 5位数 + int len = getLenOfNum(num); + if (len == 1) { + return 1; + } + // num 13625 + // tmp1 10000 + // + // num 7872328738273 + // tmp1 1000000000000 + int tmp1 = powerBaseOf10(len - 1); + // num最高位 num / tmp1 + int first = num / tmp1; + // 最高1 N % tmp1 + 1 + // 最高位first tmp1 + int firstOneNum = first == 1 ? num % tmp1 + 1 : tmp1; + // 除去最高位之外，剩下1的数量 + // 最高位1 10(k-2次方) * (k-1) * 1 + // 最高位first 10(k-2次方) * (k-1) * first + int otherOneNum = first * (len - 1) * (tmp1 / 10); + return firstOneNum + otherOneNum + solution2(num % tmp1); + } + + public static int getLenOfNum(int num) { + int len = 0; + while (num != 0) { + len++; + num /= 10; + } + return len; + } + + public static int powerBaseOf10(int base) { + return (int) Math.pow(10, base); + } + + public static void main(String[] args) { + int num = 50000000; + long start1 = System.currentTimeMillis(); + System.out.println(solution1(num)); + long end1 = System.currentTimeMillis(); + System.out.println("cost time: " + (end1 - start1) + " ms"); + + long start2 = System.currentTimeMillis(); + System.out.println(solution2(num)); + long end2 = System.currentTimeMillis(); + System.out.println("cost time: " + (end2 - start2) + " ms"); + + } +} diff --git a/大厂刷题班/class19/Code04_SmallestRangeCoveringElementsfromKLists.java b/大厂刷题班/class19/Code04_SmallestRangeCoveringElementsfromKLists.java new file mode 100644 index 0000000..5a49728 --- /dev/null +++ b/大厂刷题班/class19/Code04_SmallestRangeCoveringElementsfromKLists.java @@ -0,0 +1,58 @@ +package class19; + +import java.util.Comparator; +import java.util.List; +import java.util.TreeSet; + +// 本题测试链接 : https://leetcode.com/problems/smallest-range-covering-elements-from-k-lists/ +public class Code04_SmallestRangeCoveringElementsfromKLists { + + public static class Node { + public int value; + public int arrid; + public int index; + + public Node(int v, int ai, int i) { + value = v; + arrid = ai; + index = i; + } + } + + public static class NodeComparator implements Comparator { + + @Override + public int compare(Node o1, Node o2) { + return o1.value != o2.value ? o1.value - o2.value : o1.arrid - o2.arrid; + } + + } + + public static int[] smallestRange(List> nums) { + int N = nums.size(); + TreeSet orderSet = new TreeSet<>(new NodeComparator()); + for (int i = 0; i < N; i++) { + orderSet.add(new Node(nums.get(i).get(0), i, 0)); + } + boolean set = false; + int a = 0; + int b = 0; + while (orderSet.size() == N) { + Node min = orderSet.first(); + Node max = orderSet.last(); + if (!set || (max.value - min.value < b - a)) { + set = true; + a = min.value; + b = max.value; + } + min = orderSet.pollFirst(); + int arrid = min.arrid; + int index = min.index + 1; + if (index != nums.get(arrid).size()) { + orderSet.add(new Node(nums.get(arrid).get(index), arrid, index)); + } + } + return new int[] { a, b }; + } + +} diff --git a/大厂刷题班/class19/Code05_CardsProblem.java b/大厂刷题班/class19/Code05_CardsProblem.java new file mode 100644 index 0000000..5b9092c --- /dev/null +++ b/大厂刷题班/class19/Code05_CardsProblem.java @@ -0,0 +1,166 @@ +package class19; + +import java.util.LinkedList; + +/* + * 一张扑克有3个属性，每种属性有3种值（A、B、C） + * 比如"AAA"，第一个属性值A，第二个属性值A，第三个属性值A + * 比如"BCA"，第一个属性值B，第二个属性值C，第三个属性值A + * 给定一个字符串类型的数组cards[]，每一个字符串代表一张扑克 + * 从中挑选三张扑克，一个属性达标的条件是：这个属性在三张扑克中全一样，或全不一样 + * 挑选的三张扑克达标的要求是：每种属性都满足上面的条件 + * 比如："ABC"、"CBC"、"BBC" + * 第一张第一个属性为"A"、第二张第一个属性为"C"、第三张第一个属性为"B"，全不一样 + * 第一张第二个属性为"B"、第二张第二个属性为"B"、第三张第二个属性为"B"，全一样 + * 第一张第三个属性为"C"、第二张第三个属性为"C"、第三张第三个属性为"C"，全一样 + * 每种属性都满足在三张扑克中全一样，或全不一样，所以这三张扑克达标 + * 返回在cards[]中任意挑选三张扑克，达标的方法数 + * + * */ +public class Code05_CardsProblem { + + public static int ways1(String[] cards) { + LinkedList picks = new LinkedList<>(); + return process1(cards, 0, picks); + } + + public static int process1(String[] cards, int index, LinkedList picks) { + if (picks.size() == 3) { + return getWays1(picks); + } + if (index == cards.length) { + return 0; + } + int ways = process1(cards, index + 1, picks); + picks.addLast(cards[index]); + ways += process1(cards, index + 1, picks); + picks.pollLast(); + return ways; + } + + public static int getWays1(LinkedList picks) { + char[] s1 = picks.get(0).toCharArray(); + char[] s2 = picks.get(1).toCharArray(); + char[] s3 = picks.get(2).toCharArray(); + for (int i = 0; i < 3; i++) { + if ((s1[i] != s2[i] && s1[i] != s3[i] && s2[i] != s3[i]) || (s1[i] == s2[i] && s1[i] == s3[i])) { + continue; + } + return 0; + } + return 1; + } + + public static int ways2(String[] cards) { + int[] counts = new int[27]; + for (String s : cards) { + char[] str = s.toCharArray(); + counts[(str[0] - 'A') * 9 + (str[1] - 'A') * 3 + (str[2] - 'A') * 1]++; + } + int ways = 0; + for (int status = 0; status < 27; status++) { + int n = counts[status]; + if (n > 2) { + ways += n == 3 ? 1 : (n * (n - 1) * (n - 2) / 6); + } + } + LinkedList path = new LinkedList<>(); + for (int i = 0; i < 27; i++) { + if (counts[i] != 0) { + path.addLast(i); + ways += process2(counts, i, path); + path.pollLast(); + } + } + return ways; + } + + // 之前的牌面，拿了一些 ABC BBB ... + // pre = BBB + // ABC ... + // pre = ABC + // ABC BBB CAB + // pre = CAB + // 牌面一定要依次变大，所有形成的有效牌面，把方法数返回 + public static int process2(int[] counts, int pre, LinkedList path) { + if (path.size() == 3) { + return getWays2(counts, path); + } + int ways = 0; + for (int next = pre + 1; next < 27; next++) { + if (counts[next] != 0) { + path.addLast(next); + ways += process2(counts, next, path); + path.pollLast(); + } + } + return ways; + } + + public static int getWays2(int[] counts, LinkedList path) { + int v1 = path.get(0); + int v2 = path.get(1); + int v3 = path.get(2); + for (int i = 9; i > 0; i /= 3) { + int cur1 = v1 / i; + int cur2 = v2 / i; + int cur3 = v3 / i; + v1 %= i; + v2 %= i; + v3 %= i; + if ((cur1 != cur2 && cur1 != cur3 && cur2 != cur3) || (cur1 == cur2 && cur1 == cur3)) { + continue; + } + return 0; + } + v1 = path.get(0); + v2 = path.get(1); + v3 = path.get(2); + return counts[v1] * counts[v2] * counts[v3]; + } + + // for test + public static String[] generateCards(int size) { + int n = (int) (Math.random() * size) + 3; + String[] ans = new String[n]; + for (int i = 0; i < n; i++) { + char cha0 = (char) ((int) (Math.random() * 3) + 'A'); + char cha1 = (char) ((int) (Math.random() * 3) + 'A'); + char cha2 = (char) ((int) (Math.random() * 3) + 'A'); + ans[i] = String.valueOf(cha0) + String.valueOf(cha1) + String.valueOf(cha2); + } + return ans; + } + + // for test + public static void main(String[] args) { + int size = 20; + int testTime = 100000; + System.out.println("test begin"); + for (int i = 0; i < testTime; i++) { + String[] arr = generateCards(size); + int ans1 = ways1(arr); + int ans2 = ways2(arr); + if (ans1 != ans2) { + for (String str : arr) { + System.out.println(str); + } + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("test finish"); + + long start = 0; + long end = 0; + String[] arr = generateCards(10000000); + System.out.println("arr size : " + arr.length + " runtime test begin"); + start = System.currentTimeMillis(); + ways2(arr); + end = System.currentTimeMillis(); + System.out.println("run time : " + (end - start) + " ms"); + System.out.println("runtime test end"); + } + +} diff --git a/大厂刷题班/class20/Code01_PreAndInArrayToPosArray.java b/大厂刷题班/class20/Code01_PreAndInArrayToPosArray.java new file mode 100644 index 0000000..abc2bcc --- /dev/null +++ b/大厂刷题班/class20/Code01_PreAndInArrayToPosArray.java @@ -0,0 +1,233 @@ +package class20; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.HashSet; + +public class Code01_PreAndInArrayToPosArray { + + public static int[] zuo(int[] pre, int[] in) { + if (pre == null || in == null || pre.length != in.length) { + return null; + } + int N = pre.length; + HashMap inMap = new HashMap<>(); + for (int i = 0; i < N; i++) { + inMap.put(in[i], i); + } + int[] pos = new int[N]; + func(pre, 0, N - 1, in, 0, N - 1, pos, 0, N - 1, inMap); + return pos; + } + + public static void func(int[] pre, int L1, int R1, int[] in, int L2, int R2, int[] pos, int L3, int R3, + HashMap inMap) { + if (L1 > R1) { + return; + } + if (L1 == R1) { + pos[L3] = pre[L1]; + } else { + pos[R3] = pre[L1]; + int index = inMap.get(pre[L1]); + func(pre, L1 + 1, L1 + index - L2, in, L2, index - 1, pos, L3, L3 + index - L2 - 1, inMap); + func(pre, L1 + index - L2 + 1, R1, in, index + 1, R2, pos, L3 + index - L2, R3 - 1, inMap); + } + } + + public static class Node { + public int value; + public Node left; + public Node right; + + public Node(int v) { + value = v; + } + } + + public static int[] preInToPos1(int[] pre, int[] in) { + if (pre == null || in == null || pre.length != in.length) { + return null; + } + int N = pre.length; + int[] pos = new int[N]; + process1(pre, 0, N - 1, in, 0, N - 1, pos, 0, N - 1); + return pos; + } + + // L1...R1 L2...R2 L3...R3 + public static void process1(int[] pre, int L1, int R1, int[] in, int L2, int R2, int[] pos, int L3, int R3) { + if (L1 > R1) { + return; + } + if (L1 == R1) { + pos[L3] = pre[L1]; + return; + } + pos[R3] = pre[L1]; + int mid = L2; + for (; mid <= R2; mid++) { + if (in[mid] == pre[L1]) { + break; + } + } + int leftSize = mid - L2; + process1(pre, L1 + 1, L1 + leftSize, in, L2, mid - 1, pos, L3, L3 + leftSize - 1); + process1(pre, L1 + leftSize + 1, R1, in, mid + 1, R2, pos, L3 + leftSize, R3 - 1); + } + + public static int[] preInToPos2(int[] pre, int[] in) { + if (pre == null || in == null || pre.length != in.length) { + return null; + } + int N = pre.length; + HashMap inMap = new HashMap<>(); + for (int i = 0; i < N; i++) { + inMap.put(in[i], i); + } + int[] pos = new int[N]; + process2(pre, 0, N - 1, in, 0, N - 1, pos, 0, N - 1, inMap); + return pos; + } + + public static void process2(int[] pre, int L1, int R1, int[] in, int L2, int R2, int[] pos, int L3, int R3, + HashMap inMap) { + if (L1 > R1) { + return; + } + if (L1 == R1) { + pos[L3] = pre[L1]; + return; + } + pos[R3] = pre[L1]; + int mid = inMap.get(pre[L1]); + int leftSize = mid - L2; + process2(pre, L1 + 1, L1 + leftSize, in, L2, mid - 1, pos, L3, L3 + leftSize - 1, inMap); + process2(pre, L1 + leftSize + 1, R1, in, mid + 1, R2, pos, L3 + leftSize, R3 - 1, inMap); + } + + // for test + public static int[] getPreArray(Node head) { + ArrayList arr = new ArrayList<>(); + fillPreArray(head, arr); + int[] ans = new int[arr.size()]; + for (int i = 0; i < ans.length; i++) { + ans[i] = arr.get(i); + } + return ans; + } + + // for test + public static void fillPreArray(Node head, ArrayList arr) { + if (head == null) { + return; + } + arr.add(head.value); + fillPreArray(head.left, arr); + fillPreArray(head.right, arr); + } + + // for test + public static int[] getInArray(Node head) { + ArrayList arr = new ArrayList<>(); + fillInArray(head, arr); + int[] ans = new int[arr.size()]; + for (int i = 0; i < ans.length; i++) { + ans[i] = arr.get(i); + } + return ans; + } + + // for test + public static void fillInArray(Node head, ArrayList arr) { + if (head == null) { + return; + } + fillInArray(head.left, arr); + arr.add(head.value); + fillInArray(head.right, arr); + } + + // for test + public static int[] getPosArray(Node head) { + ArrayList arr = new ArrayList<>(); + fillPostArray(head, arr); + int[] ans = new int[arr.size()]; + for (int i = 0; i < ans.length; i++) { + ans[i] = arr.get(i); + } + return ans; + } + + // for test + public static void fillPostArray(Node head, ArrayList arr) { + if (head == null) { + return; + } + fillPostArray(head.left, arr); + fillPostArray(head.right, arr); + arr.add(head.value); + } + + // for test + public static Node generateRandomTree(int value, int maxLevel) { + HashSet hasValue = new HashSet(); + return createTree(value, 1, maxLevel, hasValue); + } + + // for test + public static Node createTree(int value, int level, int maxLevel, HashSet hasValue) { + if (level > maxLevel) { + return null; + } + int cur = 0; + do { + cur = (int) (Math.random() * value) + 1; + } while (hasValue.contains(cur)); + hasValue.add(cur); + Node head = new Node(cur); + head.left = createTree(value, level + 1, maxLevel, hasValue); + head.right = createTree(value, level + 1, maxLevel, hasValue); + return head; + } + + // for test + public static boolean isEqual(int[] arr1, int[] arr2) { + if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) { + return false; + } + if (arr1 == null && arr2 == null) { + return true; + } + if (arr1.length != arr2.length) { + return false; + } + for (int i = 0; i < arr1.length; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + public static void main(String[] args) { + System.out.println("test begin"); + int maxLevel = 5; + int value = 1000; + int testTime = 100000; + for (int i = 0; i < testTime; i++) { + Node head = generateRandomTree(value, maxLevel); + int[] pre = getPreArray(head); + int[] in = getInArray(head); + int[] pos = getPosArray(head); + int[] ans1 = preInToPos1(pre, in); + int[] ans2 = preInToPos2(pre, in); + int[] classAns = zuo(pre, in); + if (!isEqual(pos, ans1) || !isEqual(ans1, ans2) || !isEqual(pos, classAns)) { + System.out.println("Oops!"); + } + } + System.out.println("test end"); + + } +} diff --git a/大厂刷题班/class20/Code02_LargestComponentSizebyCommonFactor.java b/大厂刷题班/class20/Code02_LargestComponentSizebyCommonFactor.java new file mode 100644 index 0000000..bb746cb --- /dev/null +++ b/大厂刷题班/class20/Code02_LargestComponentSizebyCommonFactor.java @@ -0,0 +1,109 @@ +package class20; + +import java.util.HashMap; + +// 本题为leetcode原题 +// 测试链接：https://leetcode.com/problems/largest-component-size-by-common-factor/ +// 方法1会超时，但是方法2直接通过 +public class Code02_LargestComponentSizebyCommonFactor { + + public static int largestComponentSize1(int[] arr) { + int N = arr.length; + UnionFind set = new UnionFind(N); + for (int i = 0; i < N; i++) { + for (int j = i + 1; j < N; j++) { + if (gcd(arr[i], arr[j]) != 1) { + set.union(i, j); + } + } + } + return set.maxSize(); + } + + public static int largestComponentSize2(int[] arr) { + int N = arr.length; + // arr中，N个位置，在并查集初始时，每个位置自己是一个集合 + UnionFind unionFind = new UnionFind(N); + // key 某个因子 value 哪个位置拥有这个因子 + HashMap fatorsMap = new HashMap<>(); + for (int i = 0; i < N; i++) { + int num = arr[i]; + // 求出根号N， -> limit + int limit = (int) Math.sqrt(num); + for (int j = 1; j <= limit; j++) { + if (num % j == 0) { + if (j != 1) { + if (!fatorsMap.containsKey(j)) { + fatorsMap.put(j, i); + } else { + unionFind.union(fatorsMap.get(j), i); + } + } + int other = num / j; + if (other != 1) { + if (!fatorsMap.containsKey(other)) { + fatorsMap.put(other, i); + } else { + unionFind.union(fatorsMap.get(other), i); + } + } + } + } + } + return unionFind.maxSize(); + } + + // O(1) + // m,n 要是正数，不能有任何一个等于0 + public static int gcd(int a, int b) { + return b == 0 ? a : gcd(b, a % b); + } + + public static class UnionFind { + private int[] parents; + private int[] sizes; + private int[] help; + + public UnionFind(int N) { + parents = new int[N]; + sizes = new int[N]; + help = new int[N]; + for (int i = 0; i < N; i++) { + parents[i] = i; + sizes[i] = 1; + } + } + + public int maxSize() { + int ans = 0; + for (int size : sizes) { + ans = Math.max(ans, size); + } + return ans; + } + + private int find(int i) { + int hi = 0; + while (i != parents[i]) { + help[hi++] = i; + i = parents[i]; + } + for (hi--; hi >= 0; hi--) { + parents[help[hi]] = i; + } + return i; + } + + public void union(int i, int j) { + int f1 = find(i); + int f2 = find(j); + if (f1 != f2) { + int big = sizes[f1] >= sizes[f2] ? f1 : f2; + int small = big == f1 ? f2 : f1; + parents[small] = big; + sizes[big] = sizes[f1] + sizes[f2]; + } + } + } + +} diff --git a/大厂刷题班/class20/Code03_ShuffleProblem.java b/大厂刷题班/class20/Code03_ShuffleProblem.java new file mode 100644 index 0000000..fabd281 --- /dev/null +++ b/大厂刷题班/class20/Code03_ShuffleProblem.java @@ -0,0 +1,154 @@ +package class20; + +import java.util.Arrays; + +public class Code03_ShuffleProblem { + + // 数组的长度为len，调整前的位置是i，返回调整之后的位置 + // 下标不从0开始，从1开始 + public static int modifyIndex1(int i, int len) { + if (i <= len / 2) { + return 2 * i; + } else { + return 2 * (i - (len / 2)) - 1; + } + } + + // 数组的长度为len，调整前的位置是i，返回调整之后的位置 + // 下标不从0开始，从1开始 + public static int modifyIndex2(int i, int len) { + return (2 * i) % (len + 1); + } + + // 主函数 + // 数组必须不为空，且长度为偶数 + public static void shuffle(int[] arr) { + if (arr != null && arr.length != 0 && (arr.length & 1) == 0) { + shuffle(arr, 0, arr.length - 1); + } + } + + // 在arr[L..R]上做完美洗牌的调整（arr[L..R]范围上一定要是偶数个数字） + public static void shuffle(int[] arr, int L, int R) { + while (R - L + 1 > 0) { // 切成一块一块的解决，每一块的长度满足(3^k)-1 + int len = R - L + 1; + int base = 3; + int k = 1; + // 计算小于等于len并且是离len最近的，满足(3^k)-1的数 + // 也就是找到最大的k，满足3^k <= len+1 + while (base <= (len + 1) / 3) { // base > (N+1)/3 + base *= 3; + k++; + } + // 3^k -1 + // 当前要解决长度为base-1的块，一半就是再除2 + int half = (base - 1) / 2; + // [L..R]的中点位置 + int mid = (L + R) / 2; + // 要旋转的左部分为[L+half...mid], 右部分为arr[mid+1..mid+half] + // 注意在这里，arr下标是从0开始的 + rotate(arr, L + half, mid, mid + half); + // 旋转完成后，从L开始算起，长度为base-1的部分进行下标连续推 + cycles(arr, L, base - 1, k); + // 解决了前base-1的部分，剩下的部分继续处理 + L = L + base - 1; // L -> [] [+1...R] + } + } + + // 从start位置开始，往右len的长度这一段，做下标连续推 + // 出发位置依次为1,3,9... + public static void cycles(int[] arr, int start, int len, int k) { + // 找到每一个出发位置trigger，一共k个 + // 每一个trigger都进行下标连续推 + // 出发位置是从1开始算的，而数组下标是从0开始算的。 + for (int i = 0, trigger = 1; i < k; i++, trigger *= 3) { + int preValue = arr[trigger + start - 1]; + int cur = modifyIndex2(trigger, len); + while (cur != trigger) { + int tmp = arr[cur + start - 1]; + arr[cur + start - 1] = preValue; + preValue = tmp; + cur = modifyIndex2(cur, len); + } + arr[cur + start - 1] = preValue; + } + } + + // [L..M]为左部分，[M+1..R]为右部分，左右两部分互换 + public static void rotate(int[] arr, int L, int M, int R) { + reverse(arr, L, M); + reverse(arr, M + 1, R); + reverse(arr, L, R); + } + + // [L..R]做逆序调整 + public static void reverse(int[] arr, int L, int R) { + while (L < R) { + int tmp = arr[L]; + arr[L++] = arr[R]; + arr[R--] = tmp; + } + } + + public static void wiggleSort(int[] arr) { + if (arr == null || arr.length == 0) { + return; + } + // 假设这个排序是额外空间复杂度O(1)的，当然系统提供的排序并不是，你可以自己实现一个堆排序 + Arrays.sort(arr); + if ((arr.length & 1) == 1) { + shuffle(arr, 1, arr.length - 1); + } else { + shuffle(arr, 0, arr.length - 1); + for (int i = 0; i < arr.length; i += 2) { + int tmp = arr[i]; + arr[i] = arr[i + 1]; + arr[i + 1] = tmp; + } + } + } + + // for test + public static boolean isValidWiggle(int[] arr) { + for (int i = 1; i < arr.length; i++) { + if ((i & 1) == 1 && arr[i] < arr[i - 1]) { + return false; + } + if ((i & 1) == 0 && arr[i] > arr[i - 1]) { + return false; + } + } + return true; + } + + // for test + public static void printArray(int[] arr) { + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + // for test + public static int[] generateArray() { + int len = (int) (Math.random() * 10) * 2; + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * 100); + } + return arr; + } + + public static void main(String[] args) { + for (int i = 0; i < 5000000; i++) { + int[] arr = generateArray(); + wiggleSort(arr); + if (!isValidWiggle(arr)) { + System.out.println("ooops!"); + printArray(arr); + break; + } + } + } + +} diff --git a/大厂刷题班/class20/Code04_PalindromeWays.java b/大厂刷题班/class20/Code04_PalindromeWays.java new file mode 100644 index 0000000..0608e0c --- /dev/null +++ b/大厂刷题班/class20/Code04_PalindromeWays.java @@ -0,0 +1,91 @@ +package class20; + +public class Code04_PalindromeWays { + + public static int ways1(String str) { + if (str == null || str.length() == 0) { + return 0; + } + char[] s = str.toCharArray(); + char[] path = new char[s.length]; + return process(str.toCharArray(), 0, path, 0); + } + + public static int process(char[] s, int si, char[] path, int pi) { + if (si == s.length) { + return isP(path, pi) ? 1 : 0; + } + int ans = process(s, si + 1, path, pi); + path[pi] = s[si]; + ans += process(s, si + 1, path, pi + 1); + return ans; + } + + public static boolean isP(char[] path, int pi) { + if (pi == 0) { + return false; + } + int L = 0; + int R = pi - 1; + while (L < R) { + if (path[L++] != path[R--]) { + return false; + } + } + return true; + } + + public static int ways2(String str) { + if (str == null || str.length() == 0) { + return 0; + } + char[] s = str.toCharArray(); + int n = s.length; + int[][] dp = new int[n][n]; + for (int i = 0; i < n; i++) { + dp[i][i] = 1; + } + for (int i = 0; i < n - 1; i++) { + dp[i][i + 1] = s[i] == s[i + 1] ? 3 : 2; + } + for (int L = n - 3; L >= 0; L--) { + for (int R = L + 2; R < n; R++) { + dp[L][R] = dp[L + 1][R] + dp[L][R - 1] - dp[L + 1][R - 1]; + if (s[L] == s[R]) { + dp[L][R] += dp[L + 1][R - 1] + 1; + } + } + } + return dp[0][n - 1]; + } + + public static String randomString(int len, int types) { + char[] str = new char[len]; + for (int i = 0; i < str.length; i++) { + str[i] = (char) ('a' + (int) (Math.random() * types)); + } + return String.valueOf(str); + } + + public static void main(String[] args) { + int N = 10; + int types = 5; + int testTimes = 100000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + int len = (int) (Math.random() * N); + String str = randomString(len, types); + int ans1 = ways1(str); + int ans2 = ways2(str); + if (ans1 != ans2) { + System.out.println(str); + System.out.println(ans1); + System.out.println(ans2); + System.out.println("Oops!"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class21/TreeChainPartition.java b/大厂刷题班/class21/TreeChainPartition.java new file mode 100644 index 0000000..94ff19c --- /dev/null +++ b/大厂刷题班/class21/TreeChainPartition.java @@ -0,0 +1,416 @@ +package class21; + +import java.util.HashMap; + +public class TreeChainPartition { + + public static class TreeChain { + // 时间戳 0 1 2 3 4 + private int tim; + // 节点个数是n，节点编号是1~n + private int n; + // 谁是头 + private int h; + // 朴素树结构 + private int[][] tree; + // 权重数组 原始的0节点权重是6 -> val[1] = 6 + private int[] val; + // father数组一个平移，因为标号要+1 + private int[] fa; + // 深度数组！ + private int[] dep; + // son[i] = 0 i这个节点，没有儿子 + // son[i] != 0 j i这个节点，重儿子是j + private int[] son; + // siz[i] i这个节点为头的子树，有多少个节点 + private int[] siz; + // top[i] = j i这个节点，所在的重链，头是j + private int[] top; + // dfn[i] = j i这个节点，在dfs序中是第j个 + private int[] dfn; + // 如果原来的节点a，权重是10 + // 如果a节点在dfs序中是第5个节点, tnw[5] = 10 + private int[] tnw; + // 线段树，在tnw上，玩连续的区间查询或者更新 + private SegmentTree seg; + + public TreeChain(int[] father, int[] values) { + // 原始的树 tree，弄好了，可以从i这个点，找到下级的直接孩子 + // 上面的一大堆结构，准备好了空间，values -> val + // 找到头部点 + initTree(father, values); + // fa; + // dep; + // son; + // siz; + dfs1(h, 0); + // top; + // dfn; + // tnw; + dfs2(h, h); + seg = new SegmentTree(tnw); + seg.build(1, n, 1); + } + + private void initTree(int[] father, int[] values) { + tim = 0; + n = father.length + 1; + tree = new int[n][]; + val = new int[n]; + fa = new int[n]; + dep = new int[n]; + son = new int[n]; + siz = new int[n]; + top = new int[n]; + dfn = new int[n]; + tnw = new int[n--]; + int[] cnum = new int[n]; + for (int i = 0; i < n; i++) { + val[i + 1] = values[i]; + } + for (int i = 0; i < n; i++) { + if (father[i] == i) { + h = i + 1; + } else { + cnum[father[i]]++; + } + } + tree[0] = new int[0]; + for (int i = 0; i < n; i++) { + tree[i + 1] = new int[cnum[i]]; + } + for (int i = 0; i < n; i++) { + if (i + 1 != h) { + tree[father[i] + 1][--cnum[father[i]]] = i + 1; + } + } + } + + // u 当前节点 + // f u的父节点 + private void dfs1(int u, int f) { + fa[u] = f; + dep[u] = dep[f] + 1; + siz[u] = 1; + int maxSize = -1; + for (int v : tree[u]) { // 遍历u节点，所有的直接孩子 + dfs1(v, u); + siz[u] += siz[v]; + if (siz[v] > maxSize) { + maxSize = siz[v]; + son[u] = v; + } + } + } + + // u当前节点 + // t是u所在重链的头部 + private void dfs2(int u, int t) { + dfn[u] = ++tim; + top[u] = t; + tnw[tim] = val[u]; + if (son[u] != 0) { // 如果u有儿子 siz[u] > 1 + dfs2(son[u], t); + for (int v : tree[u]) { + if (v != son[u]) { + dfs2(v, v); + } + } + } + } + + // head为头的子树上，所有节点值+value + // 因为节点经过平移，所以head(原始节点) -> head(平移节点) + public void addSubtree(int head, int value) { + // 原始点编号 -> 平移编号 + head++; + // 平移编号 -> dfs编号 dfn[head] + seg.add(dfn[head], dfn[head] + siz[head] - 1, value, 1, n, 1); + } + + public int querySubtree(int head) { + head++; + return seg.query(dfn[head], dfn[head] + siz[head] - 1, 1, n, 1); + } + + public void addChain(int a, int b, int v) { + a++; + b++; + while (top[a] != top[b]) { + if (dep[top[a]] > dep[top[b]]) { + seg.add(dfn[top[a]], dfn[a], v, 1, n, 1); + a = fa[top[a]]; + } else { + seg.add(dfn[top[b]], dfn[b], v, 1, n, 1); + b = fa[top[b]]; + } + } + if (dep[a] > dep[b]) { + seg.add(dfn[b], dfn[a], v, 1, n, 1); + } else { + seg.add(dfn[a], dfn[b], v, 1, n, 1); + } + } + + public int queryChain(int a, int b) { + a++; + b++; + int ans = 0; + while (top[a] != top[b]) { + if (dep[top[a]] > dep[top[b]]) { + ans += seg.query(dfn[top[a]], dfn[a], 1, n, 1); + a = fa[top[a]]; + } else { + ans += seg.query(dfn[top[b]], dfn[b], 1, n, 1); + b = fa[top[b]]; + } + } + if (dep[a] > dep[b]) { + ans += seg.query(dfn[b], dfn[a], 1, n, 1); + } else { + ans += seg.query(dfn[a], dfn[b], 1, n, 1); + } + return ans; + } + } + + public static class SegmentTree { + private int MAXN; + private int[] arr; + private int[] sum; + private int[] lazy; + + public SegmentTree(int[] origin) { + MAXN = origin.length; + arr = origin; + sum = new int[MAXN << 2]; + lazy = new int[MAXN << 2]; + } + + private void pushUp(int rt) { + sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; + } + + private void pushDown(int rt, int ln, int rn) { + if (lazy[rt] != 0) { + lazy[rt << 1] += lazy[rt]; + sum[rt << 1] += lazy[rt] * ln; + lazy[rt << 1 | 1] += lazy[rt]; + sum[rt << 1 | 1] += lazy[rt] * rn; + lazy[rt] = 0; + } + } + + public void build(int l, int r, int rt) { + if (l == r) { + sum[rt] = arr[l]; + return; + } + int mid = (l + r) >> 1; + build(l, mid, rt << 1); + build(mid + 1, r, rt << 1 | 1); + pushUp(rt); + } + + public void add(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + sum[rt] += C * (r - l + 1); + lazy[rt] += C; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + add(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + add(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + public int query(int L, int R, int l, int r, int rt) { + if (L <= l && r <= R) { + return sum[rt]; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + int ans = 0; + if (L <= mid) { + ans += query(L, R, l, mid, rt << 1); + } + if (R > mid) { + ans += query(L, R, mid + 1, r, rt << 1 | 1); + } + return ans; + } + + } + + // 为了测试，这个结构是暴力但正确的方法 + public static class Right { + private int n; + private int[][] tree; + private int[] fa; + private int[] val; + private HashMap path; + + public Right(int[] father, int[] value) { + n = father.length; + tree = new int[n][]; + fa = new int[n]; + val = new int[n]; + for (int i = 0; i < n; i++) { + fa[i] = father[i]; + val[i] = value[i]; + } + int[] help = new int[n]; + int h = 0; + for (int i = 0; i < n; i++) { + if (father[i] == i) { + h = i; + } else { + help[father[i]]++; + } + } + for (int i = 0; i < n; i++) { + tree[i] = new int[help[i]]; + } + for (int i = 0; i < n; i++) { + if (i != h) { + tree[father[i]][--help[father[i]]] = i; + } + } + path = new HashMap<>(); + } + + public void addSubtree(int head, int value) { + val[head] += value; + for (int next : tree[head]) { + addSubtree(next, value); + } + } + + public int querySubtree(int head) { + int ans = val[head]; + for (int next : tree[head]) { + ans += querySubtree(next); + } + return ans; + } + + public void addChain(int a, int b, int v) { + path.clear(); + path.put(a, null); + while (a != fa[a]) { + path.put(fa[a], a); + a = fa[a]; + } + while (!path.containsKey(b)) { + val[b] += v; + b = fa[b]; + } + val[b] += v; + while (path.get(b) != null) { + b = path.get(b); + val[b] += v; + } + } + + public int queryChain(int a, int b) { + path.clear(); + path.put(a, null); + while (a != fa[a]) { + path.put(fa[a], a); + a = fa[a]; + } + int ans = 0; + while (!path.containsKey(b)) { + ans += val[b]; + b = fa[b]; + } + ans += val[b]; + while (path.get(b) != null) { + b = path.get(b); + ans += val[b]; + } + return ans; + } + + } + + // 为了测试 + // 随机生成N个节点树，可能是多叉树，并且一定不是森林 + // 输入参数N要大于0 + public static int[] generateFatherArray(int N) { + int[] order = new int[N]; + for (int i = 0; i < N; i++) { + order[i] = i; + } + for (int i = N - 1; i >= 0; i--) { + swap(order, i, (int) (Math.random() * (i + 1))); + } + int[] ans = new int[N]; + ans[order[0]] = order[0]; + for (int i = 1; i < N; i++) { + ans[order[i]] = order[(int) (Math.random() * i)]; + } + return ans; + } + + // 为了测试 + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + + // 为了测试 + public static int[] generateValueArray(int N, int V) { + int[] ans = new int[N]; + for (int i = 0; i < N; i++) { + ans[i] = (int) (Math.random() * V) + 1; + } + return ans; + } + + // 对数器 + public static void main(String[] args) { + int N = 50000; + int V = 100000; + int[] father = generateFatherArray(N); + int[] values = generateValueArray(N, V); + TreeChain tc = new TreeChain(father, values); + Right right = new Right(father, values); + int testTime = 1000000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + double decision = Math.random(); + if (decision < 0.25) { + int head = (int) (Math.random() * N); + int value = (int) (Math.random() * V); + tc.addSubtree(head, value); + right.addSubtree(head, value); + } else if (decision < 0.5) { + int head = (int) (Math.random() * N); + if (tc.querySubtree(head) != right.querySubtree(head)) { + System.out.println("出错了!"); + } + } else if (decision < 0.75) { + int a = (int) (Math.random() * N); + int b = (int) (Math.random() * N); + int value = (int) (Math.random() * V); + tc.addChain(a, b, value); + right.addChain(a, b, value); + } else { + int a = (int) (Math.random() * N); + int b = (int) (Math.random() * N); + if (tc.queryChain(a, b) != right.queryChain(a, b)) { + System.out.println("出错了!"); + } + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class22/Code01_MaximumSumof3NonOverlappingSubarrays.java b/大厂刷题班/class22/Code01_MaximumSumof3NonOverlappingSubarrays.java new file mode 100644 index 0000000..22f2a37 --- /dev/null +++ b/大厂刷题班/class22/Code01_MaximumSumof3NonOverlappingSubarrays.java @@ -0,0 +1,102 @@ +package class22; + +// 本题测试链接 : https://leetcode.com/problems/maximum-sum-of-3-non-overlapping-subarrays/ +public class Code01_MaximumSumof3NonOverlappingSubarrays { + +// public static int[] maxSumArray1(int[] arr) { +// int N = arr.length; +// int[] help = new int[N]; +// // help[i] 子数组必须以i位置结尾的情况下，累加和最大是多少？ +// help[0] = arr[0]; +// for (int i = 1; i < N; i++) { +// int p1 = arr[i]; +// int p2 = arr[i] + help[i - 1]; +// help[i] = Math.max(p1, p2); +// } +// // dp[i] 在0~i范围上，随意选一个子数组，累加和最大是多少？ +// int[] dp = new int[N]; +// dp[0] = help[0]; +// for (int i = 1; i < N; i++) { +// int p1 = help[i]; +// int p2 = dp[i - 1]; +// dp[i] = Math.max(p1, p2); +// } +// return dp; +// } +// +// public static int maxSumLenK(int[] arr, int k) { +// int N = arr.length; +// // 子数组必须以i位置的数结尾，长度一定要是K，累加和最大是多少？ +// // help[0] help[k-2] ... +// int sum = 0; +// for (int i = 0; i < k - 1; i++) { +// sum += arr[i]; +// } +// // 0...k-2 k-1 sum +// int[] help = new int[N]; +// for (int i = k - 1; i < N; i++) { +// // 0..k-2 +// // 01..k-1 +// sum += arr[i]; +// help[i] = sum; +// // i == k-1 +// sum -= arr[i - k + 1]; +// } +// // help[i] - > dp[i] 0-..i K +// +// } + + public static int[] maxSumOfThreeSubarrays(int[] nums, int k) { + int N = nums.length; + int[] range = new int[N]; + int[] left = new int[N]; + int sum = 0; + for (int i = 0; i < k; i++) { + sum += nums[i]; + } + range[0] = sum; + left[k - 1] = 0; + int max = sum; + for (int i = k; i < N; i++) { + sum = sum - nums[i - k] + nums[i]; + range[i - k + 1] = sum; + left[i] = left[i - 1]; + if (sum > max) { + max = sum; + left[i] = i - k + 1; + } + } + sum = 0; + for (int i = N - 1; i >= N - k; i--) { + sum += nums[i]; + } + max = sum; + int[] right = new int[N]; + right[N - k] = N - k; + for (int i = N - k - 1; i >= 0; i--) { + sum = sum - nums[i + k] + nums[i]; + right[i] = right[i + 1]; + if (sum >= max) { + max = sum; + right[i] = i; + } + } + int a = 0; + int b = 0; + int c = 0; + max = 0; + for (int i = k; i < N - 2 * k + 1; i++) { // 中间一块的起始点 (0...k-1)选不了 i == N-1 + int part1 = range[left[i - 1]]; + int part2 = range[i]; + int part3 = range[right[i + k]]; + if (part1 + part2 + part3 > max) { + max = part1 + part2 + part3; + a = left[i - 1]; + b = i; + c = right[i + k]; + } + } + return new int[] { a, b, c }; + } + +} diff --git a/大厂刷题班/class22/Code02_TrappingRainWater.java b/大厂刷题班/class22/Code02_TrappingRainWater.java new file mode 100644 index 0000000..e022a12 --- /dev/null +++ b/大厂刷题班/class22/Code02_TrappingRainWater.java @@ -0,0 +1,28 @@ +package class22; + +// 本题测试链接 : https://leetcode.com/problems/trapping-rain-water/ +public class Code02_TrappingRainWater { + + public static int trap(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int N = arr.length; + int L = 1; + int leftMax = arr[0]; + int R = N - 2; + int rightMax = arr[N - 1]; + int water = 0; + while (L <= R) { + if (leftMax <= rightMax) { + water += Math.max(0, leftMax - arr[L]); + leftMax = Math.max(leftMax, arr[L++]); + } else { + water += Math.max(0, rightMax - arr[R]); + rightMax = Math.max(rightMax, arr[R--]); + } + } + return water; + } + +} diff --git a/大厂刷题班/class22/Code03_TrappingRainWaterII.java b/大厂刷题班/class22/Code03_TrappingRainWaterII.java new file mode 100644 index 0000000..144dbac --- /dev/null +++ b/大厂刷题班/class22/Code03_TrappingRainWaterII.java @@ -0,0 +1,76 @@ +package class22; + +import java.util.PriorityQueue; + +// 本题测试链接 : https://leetcode.com/problems/trapping-rain-water-ii/ +public class Code03_TrappingRainWaterII { + + public static class Node { + public int value; + public int row; + public int col; + + public Node(int v, int r, int c) { + value = v; + row = r; + col = c; + } + + } + + public static int trapRainWater(int[][] heightMap) { + if (heightMap == null || heightMap.length == 0 || heightMap[0] == null || heightMap[0].length == 0) { + return 0; + } + int N = heightMap.length; + int M = heightMap[0].length; + boolean[][] isEnter = new boolean[N][M]; + PriorityQueue heap = new PriorityQueue<>((a, b) -> a.value - b.value); + for (int col = 0; col < M - 1; col++) { + isEnter[0][col] = true; + heap.add(new Node(heightMap[0][col], 0, col)); + } + for (int row = 0; row < N - 1; row++) { + isEnter[row][M - 1] = true; + heap.add(new Node(heightMap[row][M - 1], row, M - 1)); + } + for (int col = M - 1; col > 0; col--) { + isEnter[N - 1][col] = true; + heap.add(new Node(heightMap[N - 1][col], N - 1, col)); + } + for (int row = N - 1; row > 0; row--) { + isEnter[row][0] = true; + heap.add(new Node(heightMap[row][0], row, 0)); + } + int water = 0; + int max = 0; + while (!heap.isEmpty()) { + Node cur = heap.poll(); + max = Math.max(max, cur.value); + int r = cur.row; + int c = cur.col; + if (r > 0 && !isEnter[r - 1][c]) { + water += Math.max(0, max - heightMap[r - 1][c]); + isEnter[r - 1][c] = true; + heap.add(new Node(heightMap[r - 1][c], r - 1, c)); + } + if (r < N - 1 && !isEnter[r + 1][c]) { + water += Math.max(0, max - heightMap[r + 1][c]); + isEnter[r + 1][c] = true; + heap.add(new Node(heightMap[r + 1][c], r + 1, c)); + } + if (c > 0 && !isEnter[r][c - 1]) { + water += Math.max(0, max - heightMap[r][c - 1]); + isEnter[r][c - 1] = true; + heap.add(new Node(heightMap[r][c - 1], r, c - 1)); + } + if (c < M - 1 && !isEnter[r][c + 1]) { + water += Math.max(0, max - heightMap[r][c + 1]); + isEnter[r][c + 1] = true; + heap.add(new Node(heightMap[r][c + 1], r, c + 1)); + } + } + return water; + } + +} diff --git a/大厂刷题班/class22/Code04_VisibleMountains.java b/大厂刷题班/class22/Code04_VisibleMountains.java new file mode 100644 index 0000000..24dc0be --- /dev/null +++ b/大厂刷题班/class22/Code04_VisibleMountains.java @@ -0,0 +1,194 @@ +package class22; + +import java.util.HashSet; +import java.util.Stack; + +public class Code04_VisibleMountains { + + // 栈中放的记录， + // value就是指，times是收集的个数 + public static class Record { + public int value; + public int times; + + public Record(int value) { + this.value = value; + this.times = 1; + } + } + + public static int getVisibleNum(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int N = arr.length; + int maxIndex = 0; + // 先在环中找到其中一个最大值的位置，哪一个都行 + for (int i = 0; i < N; i++) { + maxIndex = arr[maxIndex] < arr[i] ? i : maxIndex; + } + Stack stack = new Stack(); + // 先把(最大值,1)这个记录放入stack中 + stack.push(new Record(arr[maxIndex])); + // 从最大值位置的下一个位置开始沿next方向遍历 + int index = nextIndex(maxIndex, N); + // 用“小找大”的方式统计所有可见山峰对 + int res = 0; + // 遍历阶段开始，当index再次回到maxIndex的时候，说明转了一圈，遍历阶段就结束 + while (index != maxIndex) { + // 当前数要进入栈，判断会不会破坏第一维的数字从顶到底依次变大 + // 如果破坏了，就依次弹出栈顶记录，并计算山峰对数量 + while (stack.peek().value < arr[index]) { + int k = stack.pop().times; + // 弹出记录为(X,K)，如果K==1，产生2对; 如果K>1，产生2*K + C(2,K)对。 + res += getInternalSum(k) + 2 * k; + } + // 当前数字arr[index]要进入栈了，如果和当前栈顶数字一样就合并 + // 不一样就把记录(arr[index],1)放入栈中 + if (stack.peek().value == arr[index]) { + stack.peek().times++; + } else { // > + stack.push(new Record(arr[index])); + } + index = nextIndex(index, N); + } + // 清算阶段开始了 + // 清算阶段的第1小阶段 + while (stack.size() > 2) { + int times = stack.pop().times; + res += getInternalSum(times) + 2 * times; + } + // 清算阶段的第2小阶段 + if (stack.size() == 2) { + int times = stack.pop().times; + res += getInternalSum(times) + + (stack.peek().times == 1 ? times : 2 * times); + } + // 清算阶段的第3小阶段 + res += getInternalSum(stack.pop().times); + return res; + } + + // 如果k==1返回0，如果k>1返回C(2,k) + public static int getInternalSum(int k) { + return k == 1 ? 0 : (k * (k - 1) / 2); + } + + // 环形数组中当前位置为i，数组长度为size，返回i的下一个位置 + public static int nextIndex(int i, int size) { + return i < (size - 1) ? (i + 1) : 0; + } + + // 环形数组中当前位置为i，数组长度为size，返回i的上一个位置 + public static int lastIndex(int i, int size) { + return i > 0 ? (i - 1) : (size - 1); + } + + // for test, O(N^2)的解法，绝对正确 + public static int rightWay(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int res = 0; + HashSet equalCounted = new HashSet<>(); + for (int i = 0; i < arr.length; i++) { + // 枚举从每一个位置出发，根据“小找大”原则能找到多少对儿，并且保证不重复找 + res += getVisibleNumFromIndex(arr, i, equalCounted); + } + return res; + } + + // for test + // 根据“小找大”的原则返回从index出发能找到多少对 + // 相等情况下，比如arr[1]==3，arr[5]==3 + // 之前如果从位置1找过位置5，那么等到从位置5出发时就不再找位置1（去重） + // 之前找过的、所有相等情况的山峰对，都保存在了equalCounted中 + public static int getVisibleNumFromIndex(int[] arr, int index, + HashSet equalCounted) { + int res = 0; + for (int i = 0; i < arr.length; i++) { + if (i != index) { // 不找自己 + if (arr[i] == arr[index]) { + String key = Math.min(index, i) + "_" + Math.max(index, i); + // 相等情况下，确保之前没找过这一对 + if (equalCounted.add(key) && isVisible(arr, index, i)) { + res++; + } + } else if (isVisible(arr, index, i)) { // 不相等的情况下直接找 + res++; + } + } + } + return res; + } + + // for test + // 调用该函数的前提是，lowIndex和highIndex一定不是同一个位置 + // 在“小找大”的策略下，从lowIndex位置能不能看到highIndex位置 + // next方向或者last方向有一个能走通，就返回true，否则返回false + public static boolean isVisible(int[] arr, int lowIndex, int highIndex) { + if (arr[lowIndex] > arr[highIndex]) { // “大找小”的情况直接返回false + return false; + } + int size = arr.length; + boolean walkNext = true; + int mid = nextIndex(lowIndex, size); + // lowIndex通过next方向走到highIndex，沿途不能出现比arr[lowIndex]大的数 + while (mid != highIndex) { + if (arr[mid] > arr[lowIndex]) { + walkNext = false;// next方向失败 + break; + } + mid = nextIndex(mid, size); + } + boolean walkLast = true; + mid = lastIndex(lowIndex, size); + // lowIndex通过last方向走到highIndex，沿途不能出现比arr[lowIndex]大的数 + while (mid != highIndex) { + if (arr[mid] > arr[lowIndex]) { + walkLast = false; // last方向失败 + break; + } + mid = lastIndex(mid, size); + } + return walkNext || walkLast; // 有一个成功就是能相互看见 + } + + // for test + public static int[] getRandomArray(int size, int max) { + int[] arr = new int[(int) (Math.random() * size)]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) (Math.random() * max); + } + return arr; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null || arr.length == 0) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + public static void main(String[] args) { + int size = 10; + int max = 10; + int testTimes = 3000000; + System.out.println("test begin!"); + for (int i = 0; i < testTimes; i++) { + int[] arr = getRandomArray(size, max); + if (rightWay(arr) != getVisibleNum(arr)) { + printArray(arr); + System.out.println(rightWay(arr)); + System.out.println(getVisibleNum(arr)); + break; + } + } + System.out.println("test end!"); + } + +} \ No newline at end of file diff --git a/大厂刷题班/class22/Code05_TallestBillboard.java b/大厂刷题班/class22/Code05_TallestBillboard.java new file mode 100644 index 0000000..ede1d89 --- /dev/null +++ b/大厂刷题班/class22/Code05_TallestBillboard.java @@ -0,0 +1,38 @@ +package class22; + +import java.util.HashMap; + +// 本题测试链接 : https://leetcode.com/problems/tallest-billboard/ +public class Code05_TallestBillboard { + + public int tallestBillboard(int[] rods) { + // key 集合对的某个差 + // value 满足差值为key的集合对中，最好的一对，较小集合的累加和 + // 较大 -> value + key + HashMap dp = new HashMap<>(), cur; + dp.put(0, 0);// 空集 和 空集 + for (int num : rods) { + if (num != 0) { + // cur 内部数据完全和dp一样 + cur = new HashMap<>(dp); // 考虑x之前的集合差值状况拷贝下来 + for (int d : cur.keySet()) { + int diffMore = cur.get(d); // 最好的一对，较小集合的累加和 + // x决定放入，比较大的那个 + dp.put(d + num, Math.max(diffMore, dp.getOrDefault(num + d, 0))); + // x决定放入，比较小的那个 + // 新的差值 Math.abs(x - d) + // 之前差值为Math.abs(x - d)，的那一对，就要和这一对，决策一下 + // 之前那一对，较小集合的累加和diffXD + int diffXD = dp.getOrDefault(Math.abs(num - d), 0); + if (d >= num) { // x决定放入比较小的那个, 但是放入之后，没有超过这一对较大的那个 + dp.put(d - num, Math.max(diffMore + num, diffXD)); + } else { // x决定放入比较小的那个, 但是放入之后，没有超过这一对较大的那个 + dp.put(num - d, Math.max(diffMore + d, diffXD)); + } + } + } + } + return dp.get(0); + } + +} diff --git a/大厂刷题班/class23/Code01_LCATarjanAndTreeChainPartition.java b/大厂刷题班/class23/Code01_LCATarjanAndTreeChainPartition.java new file mode 100644 index 0000000..43154cb --- /dev/null +++ b/大厂刷题班/class23/Code01_LCATarjanAndTreeChainPartition.java @@ -0,0 +1,352 @@ +package class23; + +import java.util.HashSet; + +public class Code01_LCATarjanAndTreeChainPartition { + + // 给定数组tree大小为N，表示一共有N个节点 + // tree[i] = j 表示点i的父亲是点j，tree一定是一棵树而不是森林 + // queries是二维数组，大小为M*2，每一个长度为2的数组都表示一条查询 + // [4,9], 表示想查询4和9之间的最低公共祖先 + // [3,7], 表示想查询3和7之间的最低公共祖先 + // tree和queries里面的所有值，都一定在0~N-1之间 + // 返回一个数组ans，大小为M，ans[i]表示第i条查询的答案 + + // 暴力方法 + public static int[] query1(int[] father, int[][] queries) { + int M = queries.length; + int[] ans = new int[M]; + HashSet path = new HashSet<>(); + for (int i = 0; i < M; i++) { + int jump = queries[i][0]; + while (father[jump] != jump) { + path.add(jump); + jump = father[jump]; + } + path.add(jump); + jump = queries[i][1]; + while (!path.contains(jump)) { + jump = father[jump]; + } + ans[i] = jump; + path.clear(); + } + return ans; + } + + // 离线批量查询最优解 -> Tarjan + 并查集 + // 如果有M条查询，时间复杂度O(N + M) + // 但是必须把M条查询一次给全，不支持在线查询 + public static int[] query2(int[] father, int[][] queries) { + int N = father.length; + int M = queries.length; + int[] help = new int[N]; + int h = 0; + for (int i = 0; i < N; i++) { + if (father[i] == i) { + h = i; + } else { + help[father[i]]++; + } + } + int[][] mt = new int[N][]; + for (int i = 0; i < N; i++) { + mt[i] = new int[help[i]]; + } + for (int i = 0; i < N; i++) { + if (i != h) { + mt[father[i]][--help[father[i]]] = i; + } + } + for (int i = 0; i < M; i++) { + if (queries[i][0] != queries[i][1]) { + help[queries[i][0]]++; + help[queries[i][1]]++; + } + } + int[][] mq = new int[N][]; + int[][] mi = new int[N][]; + for (int i = 0; i < N; i++) { + mq[i] = new int[help[i]]; + mi[i] = new int[help[i]]; + } + for (int i = 0; i < M; i++) { + if (queries[i][0] != queries[i][1]) { + mq[queries[i][0]][--help[queries[i][0]]] = queries[i][1]; + mi[queries[i][0]][help[queries[i][0]]] = i; + mq[queries[i][1]][--help[queries[i][1]]] = queries[i][0]; + mi[queries[i][1]][help[queries[i][1]]] = i; + } + } + int[] ans = new int[M]; + UnionFind uf = new UnionFind(N); + process(h, mt, mq, mi, uf, ans); + for (int i = 0; i < M; i++) { + if (queries[i][0] == queries[i][1]) { + ans[i] = queries[i][0]; + } + } + return ans; + } + + // 当前来到head点 + // mt是整棵树 head下方有哪些点 mt[head] = {a,b,c,d} head的孩子是abcd + // mq问题列表 head有哪些问题 mq[head] = {x,y,z} (head，x) (head，y) (head z) + // mi得到问题的答案，填在ans的什么地方 {6,12,34} + // uf 并查集 + public static void process(int head, int[][] mt, int[][] mq, int[][] mi, UnionFind uf, int[] ans) { + for (int next : mt[head]) { // head有哪些孩子，都遍历去吧！ + process(next, mt, mq, mi, uf, ans); + uf.union(head, next); + uf.setTag(head, head); + } + // 解决head的问题！ + int[] q = mq[head]; + int[] i = mi[head]; + for (int k = 0; k < q.length; k++) { + // head和谁有问题 q[k] 答案填哪 i[k] + int tag = uf.getTag(q[k]); + if (tag != -1) { + ans[i[k]] = tag; + } + } + } + + public static class UnionFind { + private int[] f; // father -> 并查集里面father信息，i -> i的father + private int[] s; // size[] -> 集合 --> i size[i] + private int[] t; // tag[] -> 集合 ---> tag[i] = ? + private int[] h; // 栈？并查集搞扁平化 + + public UnionFind(int N) { + f = new int[N]; + s = new int[N]; + t = new int[N]; + h = new int[N]; + for (int i = 0; i < N; i++) { + f[i] = i; + s[i] = 1; + t[i] = -1; + } + } + + private int find(int i) { + int j = 0; + // i -> j -> k -> s -> a -> a + while (i != f[i]) { + h[j++] = i; + i = f[i]; + } + // i -> a + // j -> a + // k -> a + // s -> a + while (j > 0) { + h[--j] = i; + } + // a + return i; + } + + public void union(int i, int j) { + int fi = find(i); + int fj = find(j); + if (fi != fj) { + int si = s[fi]; + int sj = s[fj]; + int m = si >= sj ? fi : fj; + int l = m == fi ? fj : fi; + f[l] = m; + s[m] += s[l]; + } + } + + // 集合的某个元素是i，请把整个集合打上统一的标签，tag + public void setTag(int i, int tag) { + t[find(i)] = tag; + } + + // 集合的某个元素是i，请把整个集合的tag信息返回 + public int getTag(int i) { + return t[find(i)]; + } + + } + + // 在线查询最优解 -> 树链剖分 + // 空间复杂度O(N), 支持在线查询，单次查询时间复杂度O(logN) + // 如果有M次查询，时间复杂度O(N + M * logN) + public static int[] query3(int[] father, int[][] queries) { + TreeChain tc = new TreeChain(father); + int M = queries.length; + int[] ans = new int[M]; + for (int i = 0; i < M; i++) { + // x y ? + // x x x + if (queries[i][0] == queries[i][1]) { + ans[i] = queries[i][0]; + } else { + ans[i] = tc.lca(queries[i][0], queries[i][1]); + } + } + return ans; + } + + public static class TreeChain { + private int n; + private int h; + private int[][] tree; + private int[] fa; + private int[] dep; + private int[] son; + private int[] siz; + private int[] top; + + public TreeChain(int[] father) { + initTree(father); + dfs1(h, 0); + dfs2(h, h); + } + + private void initTree(int[] father) { + n = father.length + 1; + tree = new int[n][]; + fa = new int[n]; + dep = new int[n]; + son = new int[n]; + siz = new int[n]; + top = new int[n--]; + int[] cnum = new int[n]; + for (int i = 0; i < n; i++) { + if (father[i] == i) { + h = i + 1; + } else { + cnum[father[i]]++; + } + } + tree[0] = new int[0]; + for (int i = 0; i < n; i++) { + tree[i + 1] = new int[cnum[i]]; + } + for (int i = 0; i < n; i++) { + if (i + 1 != h) { + tree[father[i] + 1][--cnum[father[i]]] = i + 1; + } + } + } + + private void dfs1(int u, int f) { + fa[u] = f; + dep[u] = dep[f] + 1; + siz[u] = 1; + int maxSize = -1; + for (int v : tree[u]) { + dfs1(v, u); + siz[u] += siz[v]; + if (siz[v] > maxSize) { + maxSize = siz[v]; + son[u] = v; + } + } + } + + private void dfs2(int u, int t) { + top[u] = t; + if (son[u] != 0) { + dfs2(son[u], t); + for (int v : tree[u]) { + if (v != son[u]) { + dfs2(v, v); + } + } + } + } + + public int lca(int a, int b) { + a++; + b++; + while (top[a] != top[b]) { + if (dep[top[a]] > dep[top[b]]) { + a = fa[top[a]]; + } else { + b = fa[top[b]]; + } + } + return (dep[a] < dep[b] ? a : b) - 1; + } + } + + // 为了测试 + // 随机生成N个节点树，可能是多叉树，并且一定不是森林 + // 输入参数N要大于0 + public static int[] generateFatherArray(int N) { + int[] order = new int[N]; + for (int i = 0; i < N; i++) { + order[i] = i; + } + for (int i = N - 1; i >= 0; i--) { + swap(order, i, (int) (Math.random() * (i + 1))); + } + int[] ans = new int[N]; + ans[order[0]] = order[0]; + for (int i = 1; i < N; i++) { + ans[order[i]] = order[(int) (Math.random() * i)]; + } + return ans; + } + + // 为了测试 + // 随机生成M条查询，点有N个，点的编号在0~N-1之间 + // 输入参数M和N都要大于0 + public static int[][] generateQueries(int M, int N) { + int[][] ans = new int[M][2]; + for (int i = 0; i < M; i++) { + ans[i][0] = (int) (Math.random() * N); + ans[i][1] = (int) (Math.random() * N); + } + return ans; + } + + // 为了测试 + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + + // 为了测试 + public static boolean equal(int[] a, int[] b) { + if (a.length != b.length) { + return false; + } + for (int i = 0; i < a.length; i++) { + if (a[i] != b[i]) { + return false; + } + } + return true; + } + + // 为了测试 + public static void main(String[] args) { + int N = 1000; + int M = 200; + int testTime = 50000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int size = (int) (Math.random() * N) + 1; + int ques = (int) (Math.random() * M) + 1; + int[] father = generateFatherArray(size); + int[][] queries = generateQueries(ques, size); + int[] ans1 = query1(father, queries); + int[] ans2 = query2(father, queries); + int[] ans3 = query3(father, queries); + if (!equal(ans1, ans2) || !equal(ans1, ans3)) { + System.out.println("出错了！"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class23/Code02_MaxABSBetweenLeftAndRight.java b/大厂刷题班/class23/Code02_MaxABSBetweenLeftAndRight.java new file mode 100644 index 0000000..af3cd49 --- /dev/null +++ b/大厂刷题班/class23/Code02_MaxABSBetweenLeftAndRight.java @@ -0,0 +1,63 @@ +package class23; + +public class Code02_MaxABSBetweenLeftAndRight { + + public static int maxABS1(int[] arr) { + int res = Integer.MIN_VALUE; + int maxLeft = 0; + int maxRight = 0; + for (int i = 0; i != arr.length - 1; i++) { + maxLeft = Integer.MIN_VALUE; + for (int j = 0; j != i + 1; j++) { + maxLeft = Math.max(arr[j], maxLeft); + } + maxRight = Integer.MIN_VALUE; + for (int j = i + 1; j != arr.length; j++) { + maxRight = Math.max(arr[j], maxRight); + } + res = Math.max(Math.abs(maxLeft - maxRight), res); + } + return res; + } + + public static int maxABS2(int[] arr) { + int[] lArr = new int[arr.length]; + int[] rArr = new int[arr.length]; + lArr[0] = arr[0]; + rArr[arr.length - 1] = arr[arr.length - 1]; + for (int i = 1; i < arr.length; i++) { + lArr[i] = Math.max(lArr[i - 1], arr[i]); + } + for (int i = arr.length - 2; i > -1; i--) { + rArr[i] = Math.max(rArr[i + 1], arr[i]); + } + int max = 0; + for (int i = 0; i < arr.length - 1; i++) { + max = Math.max(max, Math.abs(lArr[i] - rArr[i + 1])); + } + return max; + } + + public static int maxABS3(int[] arr) { + int max = Integer.MIN_VALUE; + for (int i = 0; i < arr.length; i++) { + max = Math.max(arr[i], max); + } + return max - Math.min(arr[0], arr[arr.length - 1]); + } + + public static int[] generateRandomArray(int length) { + int[] arr = new int[length]; + for (int i = 0; i != arr.length; i++) { + arr[i] = (int) (Math.random() * 1000) - 499; + } + return arr; + } + + public static void main(String[] args) { + int[] arr = generateRandomArray(200); + System.out.println(maxABS1(arr)); + System.out.println(maxABS2(arr)); + System.out.println(maxABS3(arr)); + } +} diff --git a/大厂刷题班/class23/Code03_LongestIntegratedLength.java b/大厂刷题班/class23/Code03_LongestIntegratedLength.java new file mode 100644 index 0000000..8ae031c --- /dev/null +++ b/大厂刷题班/class23/Code03_LongestIntegratedLength.java @@ -0,0 +1,106 @@ +package class23; + +import java.util.Arrays; +import java.util.HashSet; + +public class Code03_LongestIntegratedLength { + + public static int maxLen(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int N = arr.length; + HashSet set = new HashSet<>(); + int ans = 1; + for (int L = 0; L < N; L++) { + set.clear(); + int min = arr[L]; + int max = arr[L]; + set.add(arr[L]); + // L..R + for (int R = L + 1; R < N; R++) { + // L....R + if(set.contains(arr[R])) { + break; + } + set.add(arr[R]); + min = Math.min(min, arr[R]); + max = Math.max(max, arr[R]); + if(max - min == R - L) { + ans = Math.max(ans, R - L + 1); + } + } + } + return ans; + + } + + public static int getLIL1(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int len = 0; + // O(N^3 * log N) + for (int start = 0; start < arr.length; start++) { // 子数组所有可能的开头 + for (int end = start; end < arr.length; end++) { // 开头为start的情况下，所有可能的结尾 + // arr[s ... e] + // O(N * logN) + if (isIntegrated(arr, start, end)) { + len = Math.max(len, end - start + 1); + } + } + } + return len; + } + + public static boolean isIntegrated(int[] arr, int left, int right) { + int[] newArr = Arrays.copyOfRange(arr, left, right + 1); // O(N) + Arrays.sort(newArr); // O(N*logN) + for (int i = 1; i < newArr.length; i++) { + if (newArr[i - 1] != newArr[i] - 1) { + return false; + } + } + return true; + } + + public static int getLIL2(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int len = 0; + int max = 0; + int min = 0; + HashSet set = new HashSet(); + for (int L = 0; L < arr.length; L++) { // L 左边界 + // L ....... + set.clear(); + max = Integer.MIN_VALUE; + min = Integer.MAX_VALUE; + for (int R = L; R < arr.length; R++) { // R 右边界 + // arr[L..R]这个子数组在验证 l...R L...r+1 l...r+2 + if (set.contains(arr[R])) { + // arr[L..R]上开始 出现重复值了，arr[L..R往后]不需要验证了， + // 一定不是可整合的 + break; + } + // arr[L..R]上无重复值 + set.add(arr[R]); + max = Math.max(max, arr[R]); + min = Math.min(min, arr[R]); + if (max - min == R - L) { // L..R 是可整合的 + len = Math.max(len, R - L + 1); + } + } + } + return len; + } + + public static void main(String[] args) { + int[] arr = { 5, 5, 3, 2, 6, 4, 3 }; + System.out.println(getLIL1(arr)); + System.out.println(getLIL2(arr)); + + } + +} diff --git a/大厂刷题班/class23/Code04_FindKMajority.java b/大厂刷题班/class23/Code04_FindKMajority.java new file mode 100644 index 0000000..91ea4c5 --- /dev/null +++ b/大厂刷题班/class23/Code04_FindKMajority.java @@ -0,0 +1,113 @@ +package class23; + +import java.util.HashMap; +import java.util.LinkedList; +import java.util.List; +import java.util.Map.Entry; + +public class Code04_FindKMajority { + + public static void printHalfMajor(int[] arr) { + int cand = 0; + int HP = 0; + for (int i = 0; i < arr.length; i++) { + if (HP == 0) { + cand = arr[i]; + HP = 1; + } else if (arr[i] == cand) { + HP++; + } else { + HP--; + } + } + if(HP == 0) { + System.out.println("no such number."); + return; + } + HP = 0; + for (int i = 0; i < arr.length; i++) { + if (arr[i] == cand) { + HP++; + } + } + if (HP > arr.length / 2) { + System.out.println(cand); + } else { + System.out.println("no such number."); + } + } + + public static void printKMajor(int[] arr, int K) { + if (K < 2) { + System.out.println("the value of K is invalid."); + return; + } + // 攒候选，cands，候选表，最多K-1条记录！ > N / K次的数字，最多有K-1个 + HashMap cands = new HashMap(); + for (int i = 0; i != arr.length; i++) { + if (cands.containsKey(arr[i])) { + cands.put(arr[i], cands.get(arr[i]) + 1); + } else { // arr[i] 不是候选 + if (cands.size() == K - 1) { // 当前数肯定不要！，每一个候选付出1点血量，血量变成0的候选，要删掉！ + allCandsMinusOne(cands); + } else { + cands.put(arr[i], 1); + } + } + } + // 所有可能的候选，都在cands表中！遍历一遍arr，每个候选收集真实次数 + + + + + HashMap reals = getReals(arr, cands); + boolean hasPrint = false; + for (Entry set : cands.entrySet()) { + Integer key = set.getKey(); + if (reals.get(key) > arr.length / K) { + hasPrint = true; + System.out.print(key + " "); + } + } + System.out.println(hasPrint ? "" : "no such number."); + } + + public static void allCandsMinusOne(HashMap map) { + List removeList = new LinkedList(); + for (Entry set : map.entrySet()) { + Integer key = set.getKey(); + Integer value = set.getValue(); + if (value == 1) { + removeList.add(key); + } + map.put(key, value - 1); + } + for (Integer removeKey : removeList) { + map.remove(removeKey); + } + } + + public static HashMap getReals(int[] arr, + HashMap cands) { + HashMap reals = new HashMap(); + for (int i = 0; i != arr.length; i++) { + int curNum = arr[i]; + if (cands.containsKey(curNum)) { + if (reals.containsKey(curNum)) { + reals.put(curNum, reals.get(curNum) + 1); + } else { + reals.put(curNum, 1); + } + } + } + return reals; + } + + public static void main(String[] args) { + int[] arr = { 1, 2, 3, 1, 1, 2, 1 }; + printHalfMajor(arr); + int K = 4; + printKMajor(arr, K); + } + +} diff --git a/大厂刷题班/class23/Code05_MinimumCostToMergeStones.java b/大厂刷题班/class23/Code05_MinimumCostToMergeStones.java new file mode 100644 index 0000000..45f6c84 --- /dev/null +++ b/大厂刷题班/class23/Code05_MinimumCostToMergeStones.java @@ -0,0 +1,159 @@ +package class23; + +// 本题测试链接 : https://leetcode.com/problems/minimum-cost-to-merge-stones/ +public class Code05_MinimumCostToMergeStones { + +// // arr[L...R]一定要整出P份，合并的最小代价，返回！ +// public static int f(int[] arr, int K, int L, int R, int P) { +// if(从L到R根本不可能弄出P份) { +// return Integer.MAX_VALUE; +// } +// // 真的有可能的！ +// if(P == 1) { +// return L == R ? 0 : (f(arr, K, L, R, K) + 最后一次大合并的代价); +// } +// int ans = Integer.MAX_VALUE; +// // 真的有可能，P > 1 +// for(int i = L; i < R;i++) { +// // L..i(1份) i+1...R(P-1份) +// int left = f(arr, K, L, i, 1); +// if(left == Integer.MAX_VALUE) { +// continue; +// } +// int right = f(arr, K, i+1,R,P-1); +// if(right == Integer.MAX_VALUE) { +// continue; +// } +// int all = left + right; +// ans = Math.min(ans, all); +// } +// return ans; +// } + + public static int mergeStones1(int[] stones, int K) { + int n = stones.length; + if ((n - 1) % (K - 1) > 0) { + return -1; + } + int[] presum = new int[n + 1]; + for (int i = 0; i < n; i++) { + presum[i + 1] = presum[i] + stones[i]; + } + return process1(0, n - 1, 1, stones, K, presum); + } + + // part >= 1 + // arr[L..R] 一定要弄出part份，返回最低代价 + // arr、K、presum（前缀累加和数组，求i..j的累加和，就是O(1)了） + public static int process1(int L, int R, int P, int[] arr, int K, int[] presum) { + if (L == R) { // arr[L..R] + return P == 1 ? 0 : -1; + } + // L ... R 不只一个数 + if (P == 1) { + int next = process1(L, R, K, arr, K, presum); + if (next == -1) { + return -1; + } else { + return next + presum[R + 1] - presum[L]; + } + } else { // P > 1 + int ans = Integer.MAX_VALUE; + // L...mid是第1块，剩下的是part-1块 + for (int mid = L; mid < R; mid += K - 1) { + // L..mid(一份) mid+1...R(part - 1) + int next1 = process1(L, mid, 1, arr, K, presum); + int next2 = process1(mid + 1, R, P - 1, arr, K, presum); + if (next1 != -1 && next2 != -1) { + ans = Math.min(ans, next1 + next2); + } + } + return ans; + } + } + + public static int mergeStones2(int[] stones, int K) { + int n = stones.length; + if ((n - 1) % (K - 1) > 0) { // n个数，到底能不能K个相邻的数合并，最终变成1个数！ + return -1; + } + int[] presum = new int[n + 1]; + for (int i = 0; i < n; i++) { + presum[i + 1] = presum[i] + stones[i]; + } + int[][][] dp = new int[n][n][K + 1]; + return process2(0, n - 1, 1, stones, K, presum, dp); + } + + public static int process2(int L, int R, int P, int[] arr, int K, int[] presum, int[][][] dp) { + if (dp[L][R][P] != 0) { + return dp[L][R][P]; + } + if (L == R) { + return P == 1 ? 0 : -1; + } + if (P == 1) { + int next = process2(L, R, K, arr, K, presum, dp); + if (next == -1) { + dp[L][R][P] = -1; + return -1; + } else { + dp[L][R][P] = next + presum[R + 1] - presum[L]; + return next + presum[R + 1] - presum[L]; + } + } else { + int ans = Integer.MAX_VALUE; + // i...mid是第1块，剩下的是part-1块 + for (int mid = L; mid < R; mid += K - 1) { + int next1 = process2(L, mid, 1, arr, K, presum, dp); + int next2 = process2(mid + 1, R, P - 1, arr, K, presum, dp); + if (next1 == -1 || next2 == -1) { + dp[L][R][P] = -1; + return -1; + } else { + ans = Math.min(ans, next1 + next2); + } + } + dp[L][R][P] = ans; + return ans; + } + } + + // for test + public static int[] generateRandomArray(int maxSize, int maxValue) { + int[] arr = new int[(int) (maxSize * Math.random()) + 1]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) ((maxValue + 1) * Math.random()); + } + return arr; + } + + // for test + public static void printArray(int[] arr) { + if (arr == null) { + return; + } + for (int i = 0; i < arr.length; i++) { + System.out.print(arr[i] + " "); + } + System.out.println(); + } + + public static void main(String[] args) { + int maxSize = 12; + int maxValue = 100; + System.out.println("Test begin"); + for (int testTime = 0; testTime < 100000; testTime++) { + int[] arr = generateRandomArray(maxSize, maxValue); + int K = (int) (Math.random() * 7) + 2; + int ans1 = mergeStones1(arr, K); + int ans2 = mergeStones2(arr, K); + if (ans1 != ans2) { + System.out.println(ans1); + System.out.println(ans2); + } + } + + } + +} diff --git a/大厂刷题班/class24/Code01_Split4Parts.java b/大厂刷题班/class24/Code01_Split4Parts.java new file mode 100644 index 0000000..8ac489e --- /dev/null +++ b/大厂刷题班/class24/Code01_Split4Parts.java @@ -0,0 +1,89 @@ +package class24; + +import java.util.HashMap; +import java.util.HashSet; + +public class Code01_Split4Parts { + + public static boolean canSplits1(int[] arr) { + if (arr == null || arr.length < 7) { + return false; + } + HashSet set = new HashSet(); + int sum = 0; + for (int i = 0; i < arr.length; i++) { + sum += arr[i]; + } + int leftSum = arr[0]; + for (int i = 1; i < arr.length - 1; i++) { + set.add(String.valueOf(leftSum) + "_" + String.valueOf(sum - leftSum - arr[i])); + leftSum += arr[i]; + } + int l = 1; + int lsum = arr[0]; + int r = arr.length - 2; + int rsum = arr[arr.length - 1]; + while (l < r - 3) { + if (lsum == rsum) { + String lkey = String.valueOf(lsum * 2 + arr[l]); + String rkey = String.valueOf(rsum * 2 + arr[r]); + if (set.contains(lkey + "_" + rkey)) { + return true; + } + lsum += arr[l++]; + } else if (lsum < rsum) { + lsum += arr[l++]; + } else { + rsum += arr[r--]; + } + } + return false; + } + + public static boolean canSplits2(int[] arr) { + if (arr == null || arr.length < 7) { + return false; + } + // key 某一个累加和， value出现的位置 + HashMap map = new HashMap(); + int sum = arr[0]; + for (int i = 1; i < arr.length; i++) { + map.put(sum, i); + sum += arr[i]; + } + int lsum = arr[0]; // 第一刀左侧的累加和 + for (int s1 = 1; s1 < arr.length - 5; s1++) { // s1是第一刀的位置 + int checkSum = lsum * 2 + arr[s1]; // 100 x 100 100*2 + x + if (map.containsKey(checkSum)) { + int s2 = map.get(checkSum); // j -> y + checkSum += lsum + arr[s2]; + if (map.containsKey(checkSum)) { // 100 * 3 + x + y + int s3 = map.get(checkSum); // k -> z + if (checkSum + arr[s3] + lsum == sum) { + return true; + } + } + } + lsum += arr[s1]; + } + return false; + } + + public static int[] generateRondomArray() { + int[] res = new int[(int) (Math.random() * 10) + 7]; + for (int i = 0; i < res.length; i++) { + res[i] = (int) (Math.random() * 10) + 1; + } + return res; + } + + public static void main(String[] args) { + int testTime = 3000000; + for (int i = 0; i < testTime; i++) { + int[] arr = generateRondomArray(); + if (canSplits1(arr) ^ canSplits2(arr)) { + System.out.println("Error"); + } + } + } +} diff --git a/大厂刷题班/class24/Code02_KthMinPair.java b/大厂刷题班/class24/Code02_KthMinPair.java new file mode 100644 index 0000000..ebd3d96 --- /dev/null +++ b/大厂刷题班/class24/Code02_KthMinPair.java @@ -0,0 +1,180 @@ +package class24; + +import java.util.Arrays; +import java.util.Comparator; + +public class Code02_KthMinPair { + + public static class Pair { + public int x; + public int y; + + Pair(int a, int b) { + x = a; + y = b; + } + } + + public static class PairComparator implements Comparator { + + @Override + public int compare(Pair arg0, Pair arg1) { + return arg0.x != arg1.x ? arg0.x - arg1.x : arg0.y - arg1.y; + } + + } + + // O(N^2 * log (N^2))的复杂度，你肯定过不了 + // 返回的int[] 长度是2，{3,1} int[2] = [3,1] + public static int[] kthMinPair1(int[] arr, int k) { + int N = arr.length; + if (k > N * N) { + return null; + } + Pair[] pairs = new Pair[N * N]; + int index = 0; + for (int i = 0; i < N; i++) { + for (int j = 0; j < N; j++) { + pairs[index++] = new Pair(arr[i], arr[j]); + } + } + Arrays.sort(pairs, new PairComparator()); + return new int[] { pairs[k - 1].x, pairs[k - 1].y }; + } + + // O(N*logN)的复杂度，你肯定过了 + public static int[] kthMinPair2(int[] arr, int k) { + int N = arr.length; + if (k > N * N) { + return null; + } + // O(N*logN) + Arrays.sort(arr); + // 第K小的数值对，第一维数字，是什么 是arr中 + int fristNum = arr[(k - 1) / N]; + int lessFristNumSize = 0;// 数出比fristNum小的数有几个 + int fristNumSize = 0; // 数出==fristNum的数有几个 + // <= fristNum + for (int i = 0; i < N && arr[i] <= fristNum; i++) { + if (arr[i] < fristNum) { + lessFristNumSize++; + } else { + fristNumSize++; + } + } + int rest = k - (lessFristNumSize * N); + return new int[] { fristNum, arr[(rest - 1) / fristNumSize] }; + } + + // O(N)的复杂度，你肯定蒙了 + public static int[] kthMinPair3(int[] arr, int k) { + int N = arr.length; + if (k > N * N) { + return null; + } + // 在无序数组中，找到第K小的数，返回值 + // 第K小，以1作为开始 + int fristNum = getMinKth(arr, (k - 1) / N); + // 第1维数字 + int lessFristNumSize = 0; + int fristNumSize = 0; + for (int i = 0; i < N; i++) { + if (arr[i] < fristNum) { + lessFristNumSize++; + } + if (arr[i] == fristNum) { + fristNumSize++; + } + } + int rest = k - (lessFristNumSize * N); + return new int[] { fristNum, getMinKth(arr, (rest - 1) / fristNumSize) }; + } + + // 改写快排，时间复杂度O(N) + // 在无序数组arr中，找到，如果排序的话，arr[index]的数是什么？ + public static int getMinKth(int[] arr, int index) { + int L = 0; + int R = arr.length - 1; + int pivot = 0; + int[] range = null; + while (L < R) { + pivot = arr[L + (int) (Math.random() * (R - L + 1))]; + range = partition(arr, L, R, pivot); + if (index < range[0]) { + R = range[0] - 1; + } else if (index > range[1]) { + L = range[1] + 1; + } else { + return pivot; + } + } + return arr[L]; + } + + public static int[] partition(int[] arr, int L, int R, int pivot) { + int less = L - 1; + int more = R + 1; + int cur = L; + while (cur < more) { + if (arr[cur] < pivot) { + swap(arr, ++less, cur++); + } else if (arr[cur] > pivot) { + swap(arr, cur, --more); + } else { + cur++; + } + } + return new int[] { less + 1, more - 1 }; + } + + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + + // 为了测试，生成值也随机，长度也随机的随机数组 + public static int[] getRandomArray(int max, int len) { + int[] arr = new int[(int) (Math.random() * len) + 1]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) (Math.random() * max) - (int) (Math.random() * max); + } + return arr; + } + + // 为了测试 + public static int[] copyArray(int[] arr) { + if (arr == null) { + return null; + } + int[] res = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + res[i] = arr[i]; + } + return res; + } + + // 随机测试了百万组，保证三种方法都是对的 + public static void main(String[] args) { + int max = 100; + int len = 30; + int testTimes = 100000; + System.out.println("test bagin, test times : " + testTimes); + for (int i = 0; i < testTimes; i++) { + int[] arr = getRandomArray(max, len); + int[] arr1 = copyArray(arr); + int[] arr2 = copyArray(arr); + int[] arr3 = copyArray(arr); + int N = arr.length * arr.length; + int k = (int) (Math.random() * N) + 1; + int[] ans1 = kthMinPair1(arr1, k); + int[] ans2 = kthMinPair2(arr2, k); + int[] ans3 = kthMinPair3(arr3, k); + if (ans1[0] != ans2[0] || ans2[0] != ans3[0] || ans1[1] != ans2[1] || ans2[1] != ans3[1]) { + System.out.println("Oops!"); + } + } + System.out.println("test finish"); + } + +} diff --git a/大厂刷题班/class24/Code03_NotContains4.java b/大厂刷题班/class24/Code03_NotContains4.java new file mode 100644 index 0000000..0a52911 --- /dev/null +++ b/大厂刷题班/class24/Code03_NotContains4.java @@ -0,0 +1,144 @@ +package class24; + +// 里程表不能含有4，给定一个数num，返回他是里程表里的第几个 +public class Code03_NotContains4 { + + // num中一定没有4这个数字 + public static long notContains4Nums1(long num) { + long count = 0; + for (long i = 1; i <= num; i++) { + if (isNot4(i)) { + count++; + } + } + return count; + } + + public static boolean isNot4(long num) { + while (num != 0) { + if (num % 10 == 4) { + return false; + } + num /= 10; + } + return true; + } + + // arr[1] : 比1位数还少1位，有几个数(数字里可以包含0但是不可以包含4)？0个 + // arr[2] : 比2位数还少1位，有几个数(数字里可以包含0但是不可以包含4)？9个 + // 1 -> 0 1 2 3 - 5 6 7 8 9 = 9 + // arr[3] :比3位数还少1位，有几个数(数字里可以包含0但是不可以包含4)？81个 + // 1 : 0 (0 1 2 3 - 5 6 7 8 9) = 9 + // 2 : + // 1 (0 1 2 3 - 5 6 7 8 9) = 9 + // 2 (0 1 2 3 - 5 6 7 8 9) = 9 + // 3 (0 1 2 3 - 5 6 7 8 9) = 9 + // 5 (0 1 2 3 - 5 6 7 8 9) = 9 + // ... + // 9 (0 1 2 3 - 5 6 7 8 9) = 9 + public static long[] arr = { 0L, 1L, 9L, 81L, 729L, 6561L, 59049L, 531441L, 4782969L, 43046721L, 387420489L, + 3486784401L, 31381059609L, 282429536481L, 2541865828329L, 22876792454961L, 205891132094649L, + 1853020188851841L, 16677181699666569L, 150094635296999121L, 1350851717672992089L }; + + // num中一定没有4这个数字 + public static long notContains4Nums2(long num) { + if (num <= 0) { + return 0; + } + // num的十进制位数，len + int len = decimalLength(num); + // 35621 + // 10000 + long offset = offset(len); + + // 第一位数字 + long first = num / offset; + return arr[len] - 1 + (first - (first < 4 ? 1 : 2)) * arr[len] + process(num % offset, offset / 10, len - 1); + } + + // num之前，有开头！ + // 在算0的情况下，num是第几个数字 + // num 76217 + // 10000 + // 6217 + // 1000 + // len + public static long process(long num, long offset, int len) { + if (len == 0) { + return 1; + } + long first = num / offset; + return (first < 4 ? first : (first - 1)) * arr[len] + process(num % offset, offset / 10, len - 1); + } + + // num的十进制位数 + // num = 7653210 + // 7 + public static int decimalLength(long num) { + int len = 0; + while (num != 0) { + len++; + num /= 10; + } + return len; + } + + // len = 6 + // 100000 + // len = 4 + // 1000 + // 3452168 + // 1000000 + // 3 + public static long offset(int len) { + long offset = 1; + for (int i = 1; i < len; i++) { + offset *= 10L; + } + return offset; + } + + // 讲完之后想到了课上同学的留言 + // 突然意识到，这道题的本质是一个9进制的数转成10进制的数 + // 不过好在课上的解法有实际意义，就是这种求解的方式，很多题目都这么弄 + // 还有课上的时间复杂度和"9进制的数转成10进制的数"的做法，时间复杂度都是O(lg N) + // 不过"9进制的数转成10进制的数"毫无疑问是最优解 + public static long notContains4Nums3(long num) { + if (num <= 0) { + return 0; + } + long ans = 0; + for (long base = 1, cur = 0; num != 0; num /= 10, base *= 9) { + cur = num % 10; + ans += (cur < 4 ? cur : cur - 1) * base; + } + return ans; + } + + public static void main(String[] args) { + long max = 88888888L; + System.out.println("功能测试开始，验证 0 ~ " + max + " 以内所有的结果"); + for (long i = 0; i <= max; i++) { + // 测试的时候，输入的数字i里不能含有4，这是题目的规定！ + if (isNot4(i) && notContains4Nums2(i) != notContains4Nums3(i)) { + System.out.println("Oops!"); + } + } + System.out.println("如果没有打印Oops说明验证通过"); + + long num = 8173528638135L; + long start; + long end; + System.out.println("性能测试开始，计算 num = " + num + " 的答案"); + start = System.currentTimeMillis(); + long ans2 = notContains4Nums2(num); + end = System.currentTimeMillis(); + System.out.println("方法二答案 : " + ans2 + ", 运行时间 : " + (end - start) + " ms"); + + start = System.currentTimeMillis(); + long ans3 = notContains4Nums3(num); + end = System.currentTimeMillis(); + System.out.println("方法三答案 : " + ans3 + ", 运行时间 : " + (end - start) + " ms"); + } + +} diff --git a/大厂刷题班/class24/Code04_Painting.java b/大厂刷题班/class24/Code04_Painting.java new file mode 100644 index 0000000..229a906 --- /dev/null +++ b/大厂刷题班/class24/Code04_Painting.java @@ -0,0 +1,75 @@ +package class24; + +import java.util.ArrayList; +import java.util.List; + +public class Code04_Painting { + // N * M的棋盘 + // 每种颜色的格子数必须相同的 + // 相邻格子染的颜色必须不同 + // 所有格子必须染色 + // 返回至少多少种颜色可以完成任务 + + public static int minColors(int N, int M) { + // 颜色数量是i + for (int i = 2; i < N * M; i++) { + int[][] matrix = new int[N][M]; + // 下面这一句可知，需要的最少颜色数i，一定是N*M的某个因子 + if ((N * M) % i == 0 && can(matrix, N, M, i)) { + return i; + } + } + return N * M; + } + + // 在matrix上染色，返回只用pNum种颜色是否可以做到要求 + public static boolean can(int[][] matrix, int N, int M, int pNum) { + int all = N * M; + int every = all / pNum; + ArrayList rest = new ArrayList<>(); + rest.add(0); + for (int i = 1; i <= pNum; i++) { + rest.add(every); + } + return process(matrix, N, M, pNum, 0, 0, rest); + } + + public static boolean process(int[][] matrix, int N, int M, int pNum, int row, int col, List rest) { + if (row == N) { + return true; + } + if (col == M) { + return process(matrix, N, M, pNum, row + 1, 0, rest); + } + int left = col == 0 ? 0 : matrix[row][col - 1]; + int up = row == 0 ? 0 : matrix[row - 1][col]; + for (int color = 1; color <= pNum; color++) { + if (color != left && color != up && rest.get(color) > 0) { + int count = rest.get(color); + rest.set(color, count - 1); + matrix[row][col] = color; + if (process(matrix, N, M, pNum, row, col + 1, rest)) { + return true; + } + rest.set(color, count); + matrix[row][col] = 0; + } + } + return false; + } + + public static void main(String[] args) { + // 根据代码16行的提示，打印出答案，看看是答案是哪个因子 + for (int N = 2; N < 10; N++) { + for (int M = 2; M < 10; M++) { + System.out.println("N = " + N); + System.out.println("M = " + M); + System.out.println("ans = " + minColors(N, M)); + System.out.println("==========="); + } + } + // 打印答案，分析可知，是N*M最小的质数因子，原因不明，也不重要 + // 反正打表法猜出来了 + } + +} diff --git a/大厂刷题班/class24/Code05_MinWindowLength.java b/大厂刷题班/class24/Code05_MinWindowLength.java new file mode 100644 index 0000000..60508cd --- /dev/null +++ b/大厂刷题班/class24/Code05_MinWindowLength.java @@ -0,0 +1,80 @@ +package class24; + +public class Code05_MinWindowLength { + + public static int minLength(String s1, String s2) { + if (s1 == null || s2 == null || s1.length() < s2.length()) { + return Integer.MAX_VALUE; + } + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + int[] map = new int[256]; // map[37] = 4 37 4次 + for (int i = 0; i != str2.length; i++) { + map[str2[i]]++; + } + int all = str2.length; + + // [L,R-1] R + // [L,R) -> [0,0) + int L = 0; + int R = 0; + int minLen = Integer.MAX_VALUE; + while (R != str1.length) { + map[str1[R]]--; + if (map[str1[R]] >= 0) { + all--; + } + if (all == 0) { // 还完了 + while (map[str1[L]] < 0) { + map[str1[L++]]++; + } + // [L..R] + minLen = Math.min(minLen, R - L + 1); + all++; + map[str1[L++]]++; + } + R++; + } + return minLen == Integer.MAX_VALUE ? 0 : minLen; + } + + // 测试链接 : https://leetcode.com/problems/minimum-window-substring/ + public static String minWindow(String s, String t) { + if (s.length() < t.length()) { + return ""; + } + char[] str = s.toCharArray(); + char[] target = t.toCharArray(); + int[] map = new int[256]; + for (char cha : target) { + map[cha]++; + } + int all = target.length; + int L = 0; + int R = 0; + int minLen = Integer.MAX_VALUE; + int ansl = -1; + int ansr = -1; + while (R != str.length) { + map[str[R]]--; + if (map[str[R]] >= 0) { + all--; + } + if (all == 0) { + while (map[str[L]] < 0) { + map[str[L++]]++; + } + if (minLen > R - L + 1) { + minLen = R - L + 1; + ansl = L; + ansr = R; + } + all++; + map[str[L++]]++; + } + R++; + } + return minLen == Integer.MAX_VALUE ? "" : s.substring(ansl, ansr + 1); + } + +} diff --git a/大厂刷题班/class24/Code06_RemoveDuplicateLettersLessLexi.java b/大厂刷题班/class24/Code06_RemoveDuplicateLettersLessLexi.java new file mode 100644 index 0000000..2672411 --- /dev/null +++ b/大厂刷题班/class24/Code06_RemoveDuplicateLettersLessLexi.java @@ -0,0 +1,74 @@ +package class24; + +// 本题测试链接 : https://leetcode.com/problems/remove-duplicate-letters/ +public class Code06_RemoveDuplicateLettersLessLexi { + + // 在str中，每种字符都要保留一个，让最后的结果，字典序最小 ，并返回 + public static String removeDuplicateLetters1(String str) { + if (str == null || str.length() < 2) { + return str; + } + int[] map = new int[256]; + for (int i = 0; i < str.length(); i++) { + map[str.charAt(i)]++; + } + int minACSIndex = 0; + for (int i = 0; i < str.length(); i++) { + minACSIndex = str.charAt(minACSIndex) > str.charAt(i) ? i : minACSIndex; + if (--map[str.charAt(i)] == 0) { + break; + } + } + // 0...break(之前) minACSIndex + // str[minACSIndex] 剩下的字符串str[minACSIndex+1...] -> 去掉str[minACSIndex]字符 -> s' + // s'... + return String.valueOf(str.charAt(minACSIndex)) + removeDuplicateLetters1( + str.substring(minACSIndex + 1).replaceAll(String.valueOf(str.charAt(minACSIndex)), "")); + } + + public static String removeDuplicateLetters2(String s) { + char[] str = s.toCharArray(); + // 小写字母ascii码值范围[97~122]，所以用长度为26的数组做次数统计 + // 如果map[i] > -1，则代表ascii码值为i的字符的出现次数 + // 如果map[i] == -1，则代表ascii码值为i的字符不再考虑 + int[] map = new int[26]; + for (int i = 0; i < str.length; i++) { + map[str[i] - 'a']++; + } + char[] res = new char[26]; + int index = 0; + int L = 0; + int R = 0; + while (R != str.length) { + // 如果当前字符是不再考虑的，直接跳过 + // 如果当前字符的出现次数减1之后，后面还能出现，直接跳过 + if (map[str[R] - 'a'] == -1 || --map[str[R] - 'a'] > 0) { + R++; + } else { // 当前字符需要考虑并且之后不会再出现了 + // 在str[L..R]上所有需要考虑的字符中，找到ascii码最小字符的位置 + int pick = -1; + for (int i = L; i <= R; i++) { + if (map[str[i] - 'a'] != -1 && (pick == -1 || str[i] < str[pick])) { + pick = i; + } + } + // 把ascii码最小的字符放到挑选结果中 + res[index++] = str[pick]; + // 在上一个的for循环中，str[L..R]范围上每种字符的出现次数都减少了 + // 需要把str[pick + 1..R]上每种字符的出现次数加回来 + for (int i = pick + 1; i <= R; i++) { + if (map[str[i] - 'a'] != -1) { // 只增加以后需要考虑字符的次数 + map[str[i] - 'a']++; + } + } + // 选出的ascii码最小的字符，以后不再考虑了 + map[str[pick] - 'a'] = -1; + // 继续在str[pick + 1......]上重复这个过程 + L = pick + 1; + R = L; + } + } + return String.valueOf(res, 0, index); + } + +} diff --git a/大厂刷题班/class25/Code01_IPToCIDR.java b/大厂刷题班/class25/Code01_IPToCIDR.java new file mode 100644 index 0000000..5cc59bd --- /dev/null +++ b/大厂刷题班/class25/Code01_IPToCIDR.java @@ -0,0 +1,90 @@ +package class25; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.List; + +// 本题测试链接 : https://leetcode.com/problems/ip-to-cidr/ +public class Code01_IPToCIDR { + + public static List ipToCIDR(String ip, int n) { + // ip -> 32位整数 + int cur = status(ip); + // cur这个状态，最右侧的1，能表示下2的几次方 + int maxPower = 0; + // 已经解决了多少ip了 + int solved = 0; + // 临时变量 + int power = 0; + List ans = new ArrayList<>(); + while (n > 0) { + // cur最右侧的1，能搞定2的几次方个ip + // cur : 000...000 01001000 + // 3 + maxPower = mostRightPower(cur); + // cur : 0000....0000 00001000 -> 2的3次方的问题 + // sol : 0000....0000 00000001 -> 1 2的0次方 + // sol : 0000....0000 00000010 -> 2 2的1次方 + // sol : 0000....0000 00000100 -> 4 2的2次方 + // sol : 0000....0000 00001000 -> 8 2的3次方 + solved = 1; + power = 0; + // 怕溢出 + // solved + while ((solved << 1) <= n && (power + 1) <= maxPower) { + solved <<= 1; + power++; + } + ans.add(content(cur, power)); + n -= solved; + cur += solved; + } + return ans; + } + + // ip -> int(32位状态) + public static int status(String ip) { + int ans = 0; + int move = 24; + for (String str : ip.split("\\.")) { + // 17.23.16.5 "17" "23" "16" "5" + // "17" -> 17 << 24 + // "23" -> 23 << 16 + // "16" -> 16 << 8 + // "5" -> 5 << 0 + ans |= Integer.valueOf(str) << move; + move -= 8; + } + return ans; + } + + public static HashMap map = new HashMap<>(); + // 1 000000....000000 -> 2的32次方 + + public static int mostRightPower(int num) { + // map只会生成1次，以后直接用 + if (map.isEmpty()) { + map.put(0, 32); + for (int i = 0; i < 32; i++) { + // 00...0000 00000001 2的0次方 + // 00...0000 00000010 2的1次方 + // 00...0000 00000100 2的2次方 + // 00...0000 00001000 2的3次方 + map.put(1 << i, i); + } + } + // num & (-num) -> num & (~num+1) -> 提取出最右侧的1 + return map.get(num & (-num)); + } + + public static String content(int status, int power) { + StringBuilder builder = new StringBuilder(); + for (int move = 24; move >= 0; move -= 8) { + builder.append(((status & (255 << move)) >>> move) + "."); + } + builder.setCharAt(builder.length() - 1, '/'); + builder.append(32 - power); + return builder.toString(); + } + +} diff --git a/大厂刷题班/class25/Code02_3Sum.java b/大厂刷题班/class25/Code02_3Sum.java new file mode 100644 index 0000000..363b4e3 --- /dev/null +++ b/大厂刷题班/class25/Code02_3Sum.java @@ -0,0 +1,70 @@ +package class25; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.List; + +// 本题测试链接 : https://leetcode.com/problems/3sum/ +public class Code02_3Sum { + + public static List> threeSum(int[] nums) { + Arrays.sort(nums); + int N = nums.length; + List> ans = new ArrayList<>(); + for (int i = N - 1; i > 1; i--) { // 三元组最后一个数，是arr[i] 之前....二元组 + arr[i] + if (i == N - 1 || nums[i] != nums[i + 1]) { + List> nexts = twoSum(nums, i - 1, -nums[i]); + for (List cur : nexts) { + cur.add(nums[i]); + ans.add(cur); + } + } + } + return ans; + } + + // nums[0...end]这个范围上，有多少个不同二元组，相加==target，全返回 + // {-1,5} K = 4 + // {1, 3} + public static List> twoSum(int[] nums, int end, int target) { + int L = 0; + int R = end; + List> ans = new ArrayList<>(); + while (L < R) { + if (nums[L] + nums[R] > target) { + R--; + } else if (nums[L] + nums[R] < target) { + L++; + } else { // nums[L] + nums[R] == target + if (L == 0 || nums[L - 1] != nums[L]) { + List cur = new ArrayList<>(); + cur.add(nums[L]); + cur.add(nums[R]); + ans.add(cur); + } + L++; + } + } + return ans; + } + + public static int findPairs(int[] nums, int k) { + Arrays.sort(nums); + int left = 0, right = 1; + int result = 0; + while (left < nums.length && right < nums.length) { + if (left == right || nums[right] - nums[left] < k) { + right++; + } else if (nums[right] - nums[left] > k) { + left++; + } else { + left++; + result++; + while (left < nums.length && nums[left] == nums[left - 1]) + left++; + } + } + return result; + } + +} diff --git a/大厂刷题班/class25/Code03_MaxPointsOnALine.java b/大厂刷题班/class25/Code03_MaxPointsOnALine.java new file mode 100644 index 0000000..e1599c7 --- /dev/null +++ b/大厂刷题班/class25/Code03_MaxPointsOnALine.java @@ -0,0 +1,68 @@ +package class25; + +import java.util.HashMap; +import java.util.Map; + +// 本题测试链接: https://leetcode.com/problems/max-points-on-a-line/ +public class Code03_MaxPointsOnALine { + + // [ + // [1,3] + // [4,9] + // [5,7] + // ] + + public static int maxPoints(int[][] points) { + if (points == null) { + return 0; + } + if (points.length <= 2) { + return points.length; + } + // key = 3 + // value = {7 , 10} -> 斜率为3/7的点 有10个 + // {5, 15} -> 斜率为3/5的点 有15个 + Map> map = new HashMap>(); + int result = 0; + for (int i = 0; i < points.length; i++) { + map.clear(); + int samePosition = 1; + int sameX = 0; + int sameY = 0; + int line = 0; + for (int j = i + 1; j < points.length; j++) { // i号点，和j号点，的斜率关系 + int x = points[j][0] - points[i][0]; + int y = points[j][1] - points[i][1]; + if (x == 0 && y == 0) { + samePosition++; + } else if (x == 0) { + sameX++; + } else if (y == 0) { + sameY++; + } else { // 普通斜率 + int gcd = gcd(x, y); + x /= gcd; + y /= gcd; + // x / y + if (!map.containsKey(x)) { + map.put(x, new HashMap()); + } + if (!map.get(x).containsKey(y)) { + map.get(x).put(y, 0); + } + map.get(x).put(y, map.get(x).get(y) + 1); + line = Math.max(line, map.get(x).get(y)); + } + } + result = Math.max(result, Math.max(Math.max(sameX, sameY), line) + samePosition); + } + return result; + } + + // 保证初始调用的时候，a和b不等于0 + // O(1) + public static int gcd(int a, int b) { + return b == 0 ? a : gcd(b, a % b); + } + +} diff --git a/大厂刷题班/class25/Code04_GasStation.java b/大厂刷题班/class25/Code04_GasStation.java new file mode 100644 index 0000000..cc044e6 --- /dev/null +++ b/大厂刷题班/class25/Code04_GasStation.java @@ -0,0 +1,104 @@ +package class25; + +// 本题测试链接 : https://leetcode.com/problems/gas-station/ +// 注意本题的实现比leetcode上的问法更加通用 +// leetcode只让返回其中一个良好出发点的位置 +// 本题是返回结果数组，每一个出发点是否是良好出发点都求出来了 +// 得到结果数组的过程，时间复杂度O(N)，额外空间复杂度O(1) +public class Code04_GasStation { + + public static int canCompleteCircuit(int[] gas, int[] cost) { + if (gas == null || gas.length == 0) { + return -1; + } + if (gas.length == 1) { + return gas[0] < cost[0] ? -1 : 0; + } + boolean[] good = stations(cost, gas); + for (int i = 0; i < gas.length; i++) { + if (good[i]) { + return i; + } + } + return -1; + } + + public static boolean[] stations(int[] cost, int[] gas) { + if (cost == null || gas == null || cost.length < 2 || cost.length != gas.length) { + return null; + } + int init = changeDisArrayGetInit(cost, gas); + return init == -1 ? new boolean[cost.length] : enlargeArea(cost, init); + } + + public static int changeDisArrayGetInit(int[] dis, int[] oil) { + int init = -1; + for (int i = 0; i < dis.length; i++) { + dis[i] = oil[i] - dis[i]; + if (dis[i] >= 0) { + init = i; + } + } + return init; + } + + public static boolean[] enlargeArea(int[] dis, int init) { + boolean[] res = new boolean[dis.length]; + int start = init; + int end = nextIndex(init, dis.length); + int need = 0; + int rest = 0; + do { + // 当前来到的start已经在连通区域中，可以确定后续的开始点一定无法转完一圈 + if (start != init && start == lastIndex(end, dis.length)) { + break; + } + // 当前来到的start不在连通区域中，就扩充连通区域 + // start(5) -> 联通区的头部(7) -> 2 + // start(-2) -> 联通区的头部(7) -> 9 + if (dis[start] < need) { // 当前start无法接到连通区的头部 + need -= dis[start]; + } else { // 当前start可以接到连通区的头部，开始扩充连通区域的尾巴 + // start(7) -> 联通区的头部(5) + rest += dis[start] - need; + need = 0; + while (rest >= 0 && end != start) { + rest += dis[end]; + end = nextIndex(end, dis.length); + } + // 如果连通区域已经覆盖整个环，当前的start是良好出发点，进入2阶段 + if (rest >= 0) { + res[start] = true; + connectGood(dis, lastIndex(start, dis.length), init, res); + break; + } + } + start = lastIndex(start, dis.length); + } while (start != init); + return res; + } + + // 已知start的next方向上有一个良好出发点 + // start如果可以达到这个良好出发点，那么从start出发一定可以转一圈 + public static void connectGood(int[] dis, int start, int init, boolean[] res) { + int need = 0; + while (start != init) { + if (dis[start] < need) { + need -= dis[start]; + } else { + res[start] = true; + need = 0; + } + start = lastIndex(start, dis.length); + } + } + + public static int lastIndex(int index, int size) { + return index == 0 ? (size - 1) : index - 1; + } + + public static int nextIndex(int index, int size) { + return index == size - 1 ? 0 : (index + 1); + } + +} diff --git a/大厂刷题班/class26/Code01_MinRange.java b/大厂刷题班/class26/Code01_MinRange.java new file mode 100644 index 0000000..537cc65 --- /dev/null +++ b/大厂刷题班/class26/Code01_MinRange.java @@ -0,0 +1,119 @@ +package class26; + +import java.util.Arrays; +import java.util.Comparator; +import java.util.List; +import java.util.TreeSet; + +public class Code01_MinRange { + + // 本题为求最小包含区间 + // 测试链接 : + // https://leetcode.com/problems/smallest-range-covering-elements-from-k-lists/ + public static class Node { + public int val; + public int arr; + public int idx; + + public Node(int value, int arrIndex, int index) { + val = value; + arr = arrIndex; + idx = index; + } + } + + public static class NodeComparator implements Comparator { + + @Override + public int compare(Node a, Node b) { + return a.val != b.val ? (a.val - b.val) : (a.arr - b.arr); + } + + } + + public static int[] smallestRange(List> nums) { + int N = nums.size(); + TreeSet set = new TreeSet<>(new NodeComparator()); + for (int i = 0; i < N; i++) { + set.add(new Node(nums.get(i).get(0), i, 0)); + } + int r = Integer.MAX_VALUE; + int a = 0; + int b = 0; + while (set.size() == N) { + Node max = set.last(); + Node min = set.pollFirst(); + if (max.val - min.val < r) { + r = max.val - min.val; + a = min.val; + b = max.val; + } + if (min.idx < nums.get(min.arr).size() - 1) { + set.add(new Node(nums.get(min.arr).get(min.idx + 1), min.arr, min.idx + 1)); + } + } + return new int[] { a, b }; + } + + // 以下为课堂题目，在main函数里有对数器 + public static int minRange1(int[][] m) { + int min = Integer.MAX_VALUE; + for (int i = 0; i < m[0].length; i++) { + for (int j = 0; j < m[1].length; j++) { + for (int k = 0; k < m[2].length; k++) { + min = Math.min(min, + Math.abs(m[0][i] - m[1][j]) + Math.abs(m[1][j] - m[2][k]) + Math.abs(m[2][k] - m[0][i])); + } + } + } + return min; + } + + public static int minRange2(int[][] matrix) { + int N = matrix.length; + TreeSet set = new TreeSet<>(new NodeComparator()); + for (int i = 0; i < N; i++) { + set.add(new Node(matrix[i][0], i, 0)); + } + int min = Integer.MAX_VALUE; + while (set.size() == N) { + min = Math.min(min, set.last().val - set.first().val); + Node cur = set.pollFirst(); + if (cur.idx < matrix[cur.arr].length - 1) { + set.add(new Node(matrix[cur.arr][cur.idx + 1], cur.arr, cur.idx + 1)); + } + } + return min << 1; + } + + public static int[][] generateRandomMatrix(int n, int v) { + int[][] m = new int[3][]; + int s = 0; + for (int i = 0; i < 3; i++) { + s = (int) (Math.random() * n) + 1; + m[i] = new int[s]; + for (int j = 0; j < s; j++) { + m[i][j] = (int) (Math.random() * v); + } + Arrays.sort(m[i]); + } + return m; + } + + public static void main(String[] args) { + int n = 20; + int v = 200; + int t = 1000000; + System.out.println("测试开始"); + for (int i = 0; i < t; i++) { + int[][] m = generateRandomMatrix(n, v); + int ans1 = minRange1(m); + int ans2 = minRange2(m); + if (ans1 != ans2) { + System.out.println("出错了!"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class26/Code02_WordSearchII.java b/大厂刷题班/class26/Code02_WordSearchII.java new file mode 100644 index 0000000..fcfaeec --- /dev/null +++ b/大厂刷题班/class26/Code02_WordSearchII.java @@ -0,0 +1,122 @@ +package class26; + +import java.util.ArrayList; +import java.util.HashSet; +import java.util.LinkedList; +import java.util.List; + +// 本题测试链接 : https://leetcode.com/problems/word-search-ii/ +public class Code02_WordSearchII { + + public static class TrieNode { + public TrieNode[] nexts; + public int pass; + public boolean end; + + public TrieNode() { + nexts = new TrieNode[26]; + pass = 0; + end = false; + } + + } + + public static void fillWord(TrieNode head, String word) { + head.pass++; + char[] chs = word.toCharArray(); + int index = 0; + TrieNode node = head; + for (int i = 0; i < chs.length; i++) { + index = chs[i] - 'a'; + if (node.nexts[index] == null) { + node.nexts[index] = new TrieNode(); + } + node = node.nexts[index]; + node.pass++; + } + node.end = true; + } + + public static String generatePath(LinkedList path) { + char[] str = new char[path.size()]; + int index = 0; + for (Character cha : path) { + str[index++] = cha; + } + return String.valueOf(str); + } + + public static List findWords(char[][] board, String[] words) { + TrieNode head = new TrieNode(); // 前缀树最顶端的头 + HashSet set = new HashSet<>(); + for (String word : words) { + if (!set.contains(word)) { + fillWord(head, word); + set.add(word); + } + } + // 答案 + List ans = new ArrayList<>(); + // 沿途走过的字符，收集起来，存在path里 + LinkedList path = new LinkedList<>(); + for (int row = 0; row < board.length; row++) { + for (int col = 0; col < board[0].length; col++) { + // 枚举在board中的所有位置 + // 每一个位置出发的情况下，答案都收集 + process(board, row, col, path, head, ans); + } + } + return ans; + } + + // 从board[row][col]位置的字符出发， + // 之前的路径上，走过的字符，记录在path里 + // cur还没有登上，有待检查能不能登上去的前缀树的节点 + // 如果找到words中的某个str，就记录在 res里 + // 返回值，从row,col 出发，一共找到了多少个str + public static int process( + char[][] board, int row, int col, + LinkedList path, TrieNode cur, + List res) { + char cha = board[row][col]; + if (cha == 0) { // 这个row col位置是之前走过的位置 + return 0; + } + // (row,col) 不是回头路 cha 有效 + + int index = cha - 'a'; + // 如果没路，或者这条路上最终的字符串之前加入过结果里 + if (cur.nexts[index] == null || cur.nexts[index].pass == 0) { + return 0; + } + // 没有走回头路且能登上去 + cur = cur.nexts[index]; + path.addLast(cha);// 当前位置的字符加到路径里去 + int fix = 0; // 从row和col位置出发，后续一共搞定了多少答案 + // 当我来到row col位置，如果决定不往后走了。是不是已经搞定了某个字符串了 + if (cur.end) { + res.add(generatePath(path)); + cur.end = false; + fix++; + } + // 往上、下、左、右，四个方向尝试 + board[row][col] = 0; + if (row > 0) { + fix += process(board, row - 1, col, path, cur, res); + } + if (row < board.length - 1) { + fix += process(board, row + 1, col, path, cur, res); + } + if (col > 0) { + fix += process(board, row, col - 1, path, cur, res); + } + if (col < board[0].length - 1) { + fix += process(board, row, col + 1, path, cur, res); + } + board[row][col] = cha; + path.pollLast(); + cur.pass -= fix; + return fix; + } + +} diff --git a/大厂刷题班/class26/Code03_ExpressionAddOperators.java b/大厂刷题班/class26/Code03_ExpressionAddOperators.java new file mode 100644 index 0000000..2422591 --- /dev/null +++ b/大厂刷题班/class26/Code03_ExpressionAddOperators.java @@ -0,0 +1,65 @@ +package class26; + +import java.util.LinkedList; +import java.util.List; + +// 本题测试链接 : https://leetcode.com/problems/expression-add-operators/ +public class Code03_ExpressionAddOperators { + + public static List addOperators(String num, int target) { + List ret = new LinkedList<>(); + if (num.length() == 0) { + return ret; + } + // 沿途的数字拷贝和+ - * 的决定，放在path里 + char[] path = new char[num.length() * 2 - 1]; + // num -> char[] + char[] digits = num.toCharArray(); + long n = 0; + for (int i = 0; i < digits.length; i++) { // 尝试0~i前缀作为第一部分 + n = n * 10 + digits[i] - '0'; + path[i] = digits[i]; + dfs(ret, path, i + 1, 0, n, digits, i + 1, target); // 后续过程 + if (n == 0) { + break; + } + } + return ret; + } + + // char[] digits 固定参数，字符类型数组，等同于num + // int target 目标 + // char[] path 之前做的决定，已经从左往右依次填写的字符在其中，可能含有'0'~'9' 与 * - + + // int len path[0..len-1]已经填写好，len是终止 + // int pos 字符类型数组num, 使用到了哪 + // left -> 前面固定的部分 cur -> 前一块 + // 默认 left + cur ... + public static void dfs(List res, char[] path, int len, + long left, long cur, + char[] num, int index, int aim) { + if (index == num.length) { + if (left + cur == aim) { + res.add(new String(path, 0, len)); + } + return; + } + long n = 0; + int j = len + 1; + for (int i = index; i < num.length; i++) { // pos ~ i + // 试每一个可能的前缀，作为第一个数字！ + // num[index...i] 作为第一个数字！ + n = n * 10 + num[i] - '0'; + path[j++] = num[i]; + path[len] = '+'; + dfs(res, path, j, left + cur, n, num, i + 1, aim); + path[len] = '-'; + dfs(res, path, j, left + cur, -n, num, i + 1, aim); + path[len] = '*'; + dfs(res, path, j, left, cur * n, num, i + 1, aim); + if (num[index] == '0') { + break; + } + } + } + +} diff --git a/大厂刷题班/class26/Code04_WordLadderII.java b/大厂刷题班/class26/Code04_WordLadderII.java new file mode 100644 index 0000000..d55ad1e --- /dev/null +++ b/大厂刷题班/class26/Code04_WordLadderII.java @@ -0,0 +1,100 @@ +package class26; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.HashSet; +import java.util.LinkedList; +import java.util.List; +import java.util.Queue; + +// 本题测试链接 : https://leetcode.com/problems/word-ladder-ii/ +public class Code04_WordLadderII { + + public static List> findLadders(String start, String end, List list) { + list.add(start); + HashMap> nexts = getNexts(list); + HashMap fromDistances = getDistances(start, nexts); + List> res = new ArrayList<>(); + if (!fromDistances.containsKey(end)) { + return res; + } + HashMap toDistances = getDistances(end, nexts); + LinkedList pathList = new LinkedList<>(); + getShortestPaths(start, end, nexts, fromDistances, toDistances, pathList, res); + return res; + } + + public static HashMap> getNexts(List words) { + HashSet dict = new HashSet<>(words); + HashMap> nexts = new HashMap<>(); + for (int i = 0; i < words.size(); i++) { + nexts.put(words.get(i), getNext(words.get(i), dict)); + } + return nexts; + } + + // word, 在表中，有哪些邻居，把邻居们，生成list返回 + public static List getNext(String word, HashSet dict) { + ArrayList res = new ArrayList(); + char[] chs = word.toCharArray(); + for (char cur = 'a'; cur <= 'z'; cur++) { + for (int i = 0; i < chs.length; i++) { + if (chs[i] != cur) { + char tmp = chs[i]; + chs[i] = cur; + if (dict.contains(String.valueOf(chs))) { + res.add(String.valueOf(chs)); + } + chs[i] = tmp; + } + } + } + return res; + } + + // 生成距离表，从start开始，根据邻居表，宽度优先遍历，对于能够遇到的所有字符串，生成(字符串，距离)这条记录，放入距离表中 + public static HashMap getDistances(String start, HashMap> nexts) { + HashMap distances = new HashMap<>(); + distances.put(start, 0); + Queue queue = new LinkedList<>(); + queue.add(start); + HashSet set = new HashSet<>(); + set.add(start); + while (!queue.isEmpty()) { + String cur = queue.poll(); + for (String next : nexts.get(cur)) { + if (!set.contains(next)) { + distances.put(next, distances.get(cur) + 1); + queue.add(next); + set.add(next); + } + } + } + return distances; + } + + // cur 当前来到的字符串 可变 + // to 目标，固定参数 + // nexts 每一个字符串的邻居表 + // cur 到开头距离5 -> 到开头距离是6的支路 fromDistances距离表 + // cur 到结尾距离5 -> 到开头距离是4的支路 toDistances距离表 + // path : 来到cur之前，深度优先遍历之前的历史是什么 + // res : 当遇到cur，把历史，放入res，作为一个结果 + public static void getShortestPaths(String cur, String to, HashMap> nexts, + HashMap fromDistances, HashMap toDistances, LinkedList path, + List> res) { + path.add(cur); + if (to.equals(cur)) { + res.add(new LinkedList(path)); + } else { + for (String next : nexts.get(cur)) { + if (fromDistances.get(next) == fromDistances.get(cur) + 1 + && toDistances.get(next) == toDistances.get(cur) - 1) { + getShortestPaths(next, to, nexts, fromDistances, toDistances, path, res); + } + } + } + path.pollLast(); + } + +} diff --git a/大厂刷题班/class27/Code01_PickBands.java b/大厂刷题班/class27/Code01_PickBands.java new file mode 100644 index 0000000..eaaa435 --- /dev/null +++ b/大厂刷题班/class27/Code01_PickBands.java @@ -0,0 +1,221 @@ +package class27; + +import java.util.Arrays; + +public class Code01_PickBands { + + // 每一个项目都有三个数，[a,b,c]表示这个项目a和b乐队参演，花费为c + // 每一个乐队可能在多个项目里都出现了，但是只能挑一次 + // nums是可以挑选的项目数量，所以一定会有nums*2只乐队被挑选出来 + // 乐队的全部数量一定是nums*2，且标号一定是0 ~ nums*2-1 + // 返回一共挑nums轮(也就意味着一定请到所有的乐队)，最少花费是多少？ + public static int minCost(int[][] programs, int nums) { + if (nums == 0 || programs == null || programs.length == 0) { + return 0; + } + int size = clean(programs); + int[] map1 = init(1 << (nums << 1)); + int[] map2 = null; + if ((nums & 1) == 0) { + // nums = 8 , 4 + f(programs, size, 0, 0, 0, nums >> 1, map1); + map2 = map1; + } else { + // nums == 7 4 -> map1 3 -> map2 + f(programs, size, 0, 0, 0, nums >> 1, map1); + map2 = init(1 << (nums << 1)); + f(programs, size, 0, 0, 0, nums - (nums >> 1), map2); + } + // 16 mask : 0..00 1111.1111(16个) + // 12 mask : 0..00 1111.1111(12个) + int mask = (1 << (nums << 1)) - 1; + int ans = Integer.MAX_VALUE; + for (int i = 0; i < map1.length; i++) { + if (map1[i] != Integer.MAX_VALUE && map2[mask & (~i)] != Integer.MAX_VALUE) { + ans = Math.min(ans, map1[i] + map2[mask & (~i)]); + } + } + return ans == Integer.MAX_VALUE ? -1 : ans; + } + + // [ + // [9, 1, 100] + // [2, 9 , 50] + // ... + // ] + public static int clean(int[][] programs) { + int x = 0; + int y = 0; + for (int[] p : programs) { + x = Math.min(p[0], p[1]); + y = Math.max(p[0], p[1]); + p[0] = x; + p[1] = y; + } + Arrays.sort(programs, (a, b) -> a[0] != b[0] ? (a[0] - b[0]) : (a[1] != b[1] ? (a[1] - b[1]) : (a[2] - b[2]))); + x = programs[0][0]; + y = programs[0][1]; + int n = programs.length; + // (0, 1, ? ) + for (int i = 1; i < n; i++) { + if (programs[i][0] == x && programs[i][1] == y) { + programs[i] = null; + } else { + x = programs[i][0]; + y = programs[i][1]; + } + } + int size = 1; + for (int i = 1; i < n; i++) { + if (programs[i] != null) { + programs[size++] = programs[i]; + } + } + // programs[0...size-1] + return size; + } + + public static int[] init(int size) { + int[] arr = new int[size]; + for (int i = 0; i < size; i++) { + arr[i] = Integer.MAX_VALUE; + } + return arr; + } + + public static void f(int[][] programs, int size, int index, int status, int cost, int rest, int[] map) { + if (rest == 0) { + map[status] = Math.min(map[status], cost); + } else { + if (index != size) { + f(programs, size, index + 1, status, cost, rest, map); + int pick = 0 | (1 << programs[index][0]) | (1 << programs[index][1]); + if ((pick & status) == 0) { + f(programs, size, index + 1, status | pick, cost + programs[index][2], rest - 1, map); + } + } + } + } + +// // 如果nums，2 * nums 只乐队 +// // (1 << (nums << 1)) - 1 +// // programs 洗数据 size +// // nums = 8 16只乐队 +// +// // process(programs, size, (1 << (nums << 1)) - 1, 0, 4, 0, 0) +// +// public static int minCost = Integer.MAX_VALUE; +// +// public static int[] map = new int[1 << 16]; // map初始化全变成系统最大 +// +// +// +// public static void process(int[][] programs, int size, int index, int rest, int pick, int cost) { +// if (rest == 0) { +// +// map[pick] = Math.min(map[pick], cost); +// +// } else { // 还有项目可挑 +// if (index != size) { +// // 不考虑当前的项目！programs[index]; +// process(programs, size, index + 1, rest, pick, cost); +// // 考虑当前的项目！programs[index]; +// int x = programs[index][0]; +// int y = programs[index][1]; +// int cur = (1 << x) | (1 << y); +// if ((pick & cur) == 0) { // 终于可以考虑了！ +// process(programs, size, index + 1, rest - 1, pick | cur, cost + programs[index][2]); +// } +// } +// } +// } +// +// +// public static void zuo(int[] arr, int index, int rest) { +// if(rest == 0) { +// 停止 +// } +// if(index != arr.length) { +// zuo(arr, index + 1, rest); +// zuo(arr, index + 1, rest - 1); +// } +// } + + // 为了测试 + public static int right(int[][] programs, int nums) { + min = Integer.MAX_VALUE; + r(programs, 0, nums, 0, 0); + return min == Integer.MAX_VALUE ? -1 : min; + } + + public static int min = Integer.MAX_VALUE; + + public static void r(int[][] programs, int index, int rest, int pick, int cost) { + if (rest == 0) { + min = Math.min(min, cost); + } else { + if (index < programs.length) { + r(programs, index + 1, rest, pick, cost); + int cur = (1 << programs[index][0]) | (1 << programs[index][1]); + if ((pick & cur) == 0) { + r(programs, index + 1, rest - 1, pick | cur, cost + programs[index][2]); + } + } + } + } + + // 为了测试 + public static int[][] randomPrograms(int N, int V) { + int nums = N << 1; + int n = nums * (nums - 1); + int[][] programs = new int[n][3]; + for (int i = 0; i < n; i++) { + int a = (int) (Math.random() * nums); + int b = 0; + do { + b = (int) (Math.random() * nums); + } while (b == a); + programs[i][0] = a; + programs[i][1] = b; + programs[i][2] = (int) (Math.random() * V) + 1; + } + return programs; + } + + // 为了测试 + public static void main(String[] args) { + int N = 4; + int V = 100; + int T = 10000; + System.out.println("测试开始"); + for (int i = 0; i < T; i++) { + int nums = (int) (Math.random() * N) + 1; + int[][] programs = randomPrograms(nums, V); + int ans1 = right(programs, nums); + int ans2 = minCost(programs, nums); + if (ans1 != ans2) { + System.out.println("Oops!"); + break; + } + } + System.out.println("测试结束"); + + long start; + long end; + int[][] programs; + + programs = randomPrograms(7, V); + start = System.currentTimeMillis(); + right(programs, 7); + end = System.currentTimeMillis(); + System.out.println("right方法，在nums=7时候的运行时间(毫秒) : " + (end - start)); + + programs = randomPrograms(10, V); + start = System.currentTimeMillis(); + minCost(programs, 10); + end = System.currentTimeMillis(); + System.out.println("minCost方法，在nums=10时候的运行时间(毫秒) : " + (end - start)); + + } + +} diff --git a/大厂刷题班/class27/Code02_MinPeople.java b/大厂刷题班/class27/Code02_MinPeople.java new file mode 100644 index 0000000..64dc565 --- /dev/null +++ b/大厂刷题班/class27/Code02_MinPeople.java @@ -0,0 +1,61 @@ +package class27; + +import java.util.Arrays; + +public class Code02_MinPeople { + +// 三道题如下： +// 微信面试题 +// 题目一： +// 股民小 A 有一天穿越回到了 n 天前，他能记住某只股票连续 n 天的价格；他手上有足够多的启动资金，他可以在这 n 天内多次交易，但是有个限制 +// 如果已经购买了一个股票，在卖出它之前就不能再继续购买股票了。 +// 到 n 天之后，小 A 不能再持有股票。 +// 求问 这 n 天内，小 A 能够获得的利润的最大值 +// 如果不需要手续费的话，求最大的利润 +// function(number) { +// return number +// } +// 输入: prices = [3, 1, 2, 8, 5, 9] +// 输出: 11 +// +// 题目二： +// 企鹅厂每年都会发文化衫，文化衫有很多种，厂庆的时候，企鹅们都需要穿文化衫来拍照 +// 一次采访中，记者随机遇到的企鹅，企鹅会告诉记者还有多少企鹅跟他穿一种文化衫 +// 我们将这些回答放在 answers 数组里，返回鹅厂中企鹅的最少数量。 +// 输入: answers = [1] 输出：2 +// 输入: answers = [1, 1, 2] 输出：5 +// Leetcode题目：https://leetcode.com/problems/rabbits-in-forest/ +// +// 题目三： +// WXG 的秘书有一堆的文件袋，现在秘书需要把文件袋嵌套收纳起来。请你帮他计算下，最大嵌套数量。 +// 给你一个二维整数数组 folders ，其中 folders[i] = [wi, hi] ，表示第 i 个文件袋的宽度和长度 +// 当某一个文件袋的宽度和长度都比这个另外一个文件袋大的时候，前者就能把后者装起来，称之为一组文件袋。 +// 请计算，最大的一组文件袋的数量是多少。 +// 实例 +// 输入：[[6,10],[11,14],[6,1],[16,14],[13,2]] +// 输出： 3 + + // 题目一，股票系列专题，大厂刷题班15节 + // 题目三，最长递增子序列专题，大厂刷题班第9节 + // 我们来讲一下题目二 + public static int numRabbits(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + Arrays.sort(arr); + int x = arr[0]; + int c = 1; + int ans = 0; + for (int i = 1; i < arr.length; i++) { + if (x != arr[i]) { + ans += ((c + x) / (x + 1)) * (x + 1); + x = arr[i]; + c = 1; + } else { + c++; + } + } + return ans + ((c + x) / (x + 1)) * (x + 1); + } + +} diff --git a/大厂刷题班/class27/Problem_0001_TwoSum.java b/大厂刷题班/class27/Problem_0001_TwoSum.java new file mode 100644 index 0000000..d1c4619 --- /dev/null +++ b/大厂刷题班/class27/Problem_0001_TwoSum.java @@ -0,0 +1,19 @@ +package class27; + +import java.util.HashMap; + +public class Problem_0001_TwoSum { + + public static int[] twoSum(int[] nums, int target) { + // key 某个之前的数 value 这个数出现的位置 + HashMap map = new HashMap<>(); + for (int i = 0; i < nums.length; i++) { + if (map.containsKey(target - nums[i])) { + return new int[] { map.get(target - nums[i]), i }; + } + map.put(nums[i], i); + } + return new int[] { -1, -1 }; + } + +} diff --git a/大厂刷题班/class27/Problem_0007_ReverseInteger.java b/大厂刷题班/class27/Problem_0007_ReverseInteger.java new file mode 100644 index 0000000..a26a5dc --- /dev/null +++ b/大厂刷题班/class27/Problem_0007_ReverseInteger.java @@ -0,0 +1,21 @@ +package class27; + +public class Problem_0007_ReverseInteger { + + public static int reverse(int x) { + boolean neg = ((x >>> 31) & 1) == 1; + x = neg ? x : -x; + int m = Integer.MIN_VALUE / 10; + int o = Integer.MIN_VALUE % 10; + int res = 0; + while (x != 0) { + if (res < m || (res == m && x % 10 < o)) { + return 0; + } + res = res * 10 + x % 10; + x /= 10; + } + return neg ? res : Math.abs(res); + } + +} diff --git a/大厂刷题班/class27/说明 b/大厂刷题班/class27/说明 new file mode 100644 index 0000000..dc0c1f5 --- /dev/null +++ b/大厂刷题班/class27/说明 @@ -0,0 +1,12 @@ +leetcode高频题 +leetcode全题目列表 : https://leetcode.com/problemset/all/ +在全题目列表的右侧栏，点击Top Interview Questions +或者直接进入右侧链接 : https://leetcode.com/problemset/all/?listId=wpwgkgt +即可看到leetcode高频题全列表 +本节课解决leetcode高频题列表中的如下题目 : +0001 : 大厂刷题班, 第27节, 本节 +0002 : 算法新手班, 第4节第5题 +0003 : 大厂刷题班, 第3节第1题 +0004 : 大厂刷题班, 第12节第3题 +0005 : 体系学习班, Manacher算法 +0007 : 大厂刷题班, 第27节, 本节 \ No newline at end of file diff --git a/大厂刷题班/class28/Problem_0008_StringToInteger.java b/大厂刷题班/class28/Problem_0008_StringToInteger.java new file mode 100644 index 0000000..ecab645 --- /dev/null +++ b/大厂刷题班/class28/Problem_0008_StringToInteger.java @@ -0,0 +1,80 @@ +package class28; + +public class Problem_0008_StringToInteger { + + public static int myAtoi(String s) { + if (s == null || s.equals("")) { + return 0; + } + s = removeHeadZero(s.trim()); + if (s == null || s.equals("")) { + return 0; + } + char[] str = s.toCharArray(); + if (!isValid(str)) { + return 0; + } + // str 是符合日常书写的，正经整数形式 + boolean posi = str[0] == '-' ? false : true; + int minq = Integer.MIN_VALUE / 10; + int minr = Integer.MIN_VALUE % 10; + int res = 0; + int cur = 0; + for (int i = (str[0] == '-' || str[0] == '+') ? 1 : 0; i < str.length; i++) { + // 3 cur = -3 '5' cur = -5 '0' cur = 0 + cur = '0' - str[i]; + if ((res < minq) || (res == minq && cur < minr)) { + return posi ? Integer.MAX_VALUE : Integer.MIN_VALUE; + } + res = res * 10 + cur; + } + // res 负 + if (posi && res == Integer.MIN_VALUE) { + return Integer.MAX_VALUE; + } + return posi ? -res : res; + } + + public static String removeHeadZero(String str) { + boolean r = (str.startsWith("+") || str.startsWith("-")); + int s = r ? 1 : 0; + for (; s < str.length(); s++) { + if (str.charAt(s) != '0') { + break; + } + } + // s 到了第一个不是'0'字符的位置 + int e = -1; + // 左<-右 + for (int i = str.length() - 1; i >= (r ? 1 : 0); i--) { + if (str.charAt(i) < '0' || str.charAt(i) > '9') { + e = i; + } + } + // e 到了最左的 不是数字字符的位置 + return (r ? String.valueOf(str.charAt(0)) : "") + str.substring(s, e == -1 ? str.length() : e); + } + + public static boolean isValid(char[] chas) { + if (chas[0] != '-' && chas[0] != '+' && (chas[0] < '0' || chas[0] > '9')) { + return false; + } + if ((chas[0] == '-' || chas[0] == '+') && chas.length == 1) { + return false; + } + // 0 +... -... num + for (int i = 1; i < chas.length; i++) { + if (chas[i] < '0' || chas[i] > '9') { + return false; + } + } + return true; + } + + public static void main(String[] args) { + System.out.println(Integer.MAX_VALUE); + } + + + +} diff --git a/大厂刷题班/class28/Problem_0012_IntegerToRoman.java b/大厂刷题班/class28/Problem_0012_IntegerToRoman.java new file mode 100644 index 0000000..d495efc --- /dev/null +++ b/大厂刷题班/class28/Problem_0012_IntegerToRoman.java @@ -0,0 +1,20 @@ +package class28; + +public class Problem_0012_IntegerToRoman { + + public static String intToRoman(int num) { + String[][] c = { + { "", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX" }, + { "", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC" }, + { "", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM" }, + { "", "M", "MM", "MMM" } }; + StringBuilder roman = new StringBuilder(); + roman + .append(c[3][num / 1000 % 10]) + .append(c[2][num / 100 % 10]) + .append(c[1][num / 10 % 10]) + .append(c[0][num % 10]); + return roman.toString(); + } + +} diff --git a/大厂刷题班/class28/Problem_0013_RomanToInteger.java b/大厂刷题班/class28/Problem_0013_RomanToInteger.java new file mode 100644 index 0000000..d8ae1bc --- /dev/null +++ b/大厂刷题班/class28/Problem_0013_RomanToInteger.java @@ -0,0 +1,45 @@ +package class28; + +public class Problem_0013_RomanToInteger { + + public static int romanToInt(String s) { + // C M X C I X + // 100 1000 10 100 1 10 + int nums[] = new int[s.length()]; + for (int i = 0; i < s.length(); i++) { + switch (s.charAt(i)) { + case 'M': + nums[i] = 1000; + break; + case 'D': + nums[i] = 500; + break; + case 'C': + nums[i] = 100; + break; + case 'L': + nums[i] = 50; + break; + case 'X': + nums[i] = 10; + break; + case 'V': + nums[i] = 5; + break; + case 'I': + nums[i] = 1; + break; + } + } + int sum = 0; + for (int i = 0; i < nums.length - 1; i++) { + if (nums[i] < nums[i + 1]) { + sum -= nums[i]; + } else { + sum += nums[i]; + } + } + return sum + nums[nums.length - 1]; + } + +} diff --git a/大厂刷题班/class28/Problem_0014_LongestCommonPrefix.java b/大厂刷题班/class28/Problem_0014_LongestCommonPrefix.java new file mode 100644 index 0000000..0ee811c --- /dev/null +++ b/大厂刷题班/class28/Problem_0014_LongestCommonPrefix.java @@ -0,0 +1,28 @@ +package class28; + +public class Problem_0014_LongestCommonPrefix { + + public static String longestCommonPrefix(String[] strs) { + if (strs == null || strs.length == 0) { + return ""; + } + char[] chs = strs[0].toCharArray(); + int min = Integer.MAX_VALUE; + for (String str : strs) { + char[] tmp = str.toCharArray(); + int index = 0; + while (index < tmp.length && index < chs.length) { + if (chs[index] != tmp[index]) { + break; + } + index++; + } + min = Math.min(index, min); + if (min == 0) { + return ""; + } + } + return strs[0].substring(0, min); + } + +} diff --git a/大厂刷题班/class28/Problem_0017_LetterCombinationsOfAPhoneNumber.java b/大厂刷题班/class28/Problem_0017_LetterCombinationsOfAPhoneNumber.java new file mode 100644 index 0000000..14fb701 --- /dev/null +++ b/大厂刷题班/class28/Problem_0017_LetterCombinationsOfAPhoneNumber.java @@ -0,0 +1,43 @@ +package class28; + +import java.util.ArrayList; +import java.util.List; + +public class Problem_0017_LetterCombinationsOfAPhoneNumber { + + public static char[][] phone = { + { 'a', 'b', 'c' }, // 2 0 + { 'd', 'e', 'f' }, // 3 1 + { 'g', 'h', 'i' }, // 4 2 + { 'j', 'k', 'l' }, // 5 3 + { 'm', 'n', 'o' }, // 6 + { 'p', 'q', 'r', 's' }, // 7 + { 't', 'u', 'v' }, // 8 + { 'w', 'x', 'y', 'z' }, // 9 + }; + + // "23" + public static List letterCombinations(String digits) { + List ans = new ArrayList<>(); + if (digits == null || digits.length() == 0) { + return ans; + } + char[] str = digits.toCharArray(); + char[] path = new char[str.length]; + process(str, 0, path, ans); + return ans; + } + + public static void process(char[] str, int index, char[] path, List ans) { + if (index == str.length) { + ans.add(String.valueOf(path)); + } else { + char[] cands = phone[str[index] - '2']; + for (char cur : cands) { + path[index] = cur; + process(str, index + 1, path, ans); + } + } + } + +} diff --git a/大厂刷题班/class28/Problem_0019_RemoveNthNodeFromEndofList.java b/大厂刷题班/class28/Problem_0019_RemoveNthNodeFromEndofList.java new file mode 100644 index 0000000..97cdc29 --- /dev/null +++ b/大厂刷题班/class28/Problem_0019_RemoveNthNodeFromEndofList.java @@ -0,0 +1,33 @@ +package class28; + +public class Problem_0019_RemoveNthNodeFromEndofList { + + public static class ListNode { + public int val; + public ListNode next; + } + + public static ListNode removeNthFromEnd(ListNode head, int n) { + ListNode cur = head; + ListNode pre = null; + while (cur != null) { + n--; + if (n == -1) { + pre = head; + } + if (n < -1) { + pre = pre.next; + } + cur = cur.next; + } + if (n > 0) { + return head; + } + if (pre == null) { + return head.next; + } + pre.next = pre.next.next; + return head; + } + +} diff --git a/大厂刷题班/class28/Problem_0020_ValidParentheses.java b/大厂刷题班/class28/Problem_0020_ValidParentheses.java new file mode 100644 index 0000000..1914ead --- /dev/null +++ b/大厂刷题班/class28/Problem_0020_ValidParentheses.java @@ -0,0 +1,30 @@ +package class28; + +public class Problem_0020_ValidParentheses { + + public static boolean isValid(String s) { + if (s == null || s.length() == 0) { + return true; + } + char[] str = s.toCharArray(); + int N = str.length; + char[] stack = new char[N]; + int size = 0; + for (int i = 0; i < N; i++) { + char cha = str[i]; + if (cha == '(' || cha == '[' || cha == '{') { + stack[size++] = cha == '(' ? ')' : (cha == '[' ? ']' : '}'); + } else { + if (size == 0) { + return false; + } + char last = stack[--size]; + if (cha != last) { + return false; + } + } + } + return size == 0; + } + +} diff --git a/大厂刷题班/class28/Problem_0022_GenerateParentheses.java b/大厂刷题班/class28/Problem_0022_GenerateParentheses.java new file mode 100644 index 0000000..7ca9625 --- /dev/null +++ b/大厂刷题班/class28/Problem_0022_GenerateParentheses.java @@ -0,0 +1,71 @@ +package class28; + +import java.util.ArrayList; +import java.util.List; + +public class Problem_0022_GenerateParentheses { + + public static List generateParenthesis(int n) { + char[] path = new char[n << 1]; + List ans = new ArrayList<>(); + process(path, 0, 0, n, ans); + return ans; + } + + + // path 做的决定 path[0....index-1]做完决定的！ + // path[index.....] 还没做决定，当前轮到index位置做决定！ + + public static void process(char[] path, int index, int leftMinusRight, int leftRest, List ans) { + if (index == path.length) { + ans.add(String.valueOf(path)); + } else { + // index ( ) + if (leftRest > 0) { + path[index] = '('; + process(path, index + 1, leftMinusRight + 1, leftRest - 1, ans); + } + if (leftMinusRight > 0) { + path[index] = ')'; + process(path, index + 1, leftMinusRight - 1, leftRest, ans); + } + } + } + + // 不剪枝的做法 + public static List generateParenthesis2(int n) { + char[] path = new char[n << 1]; + List ans = new ArrayList<>(); + process2(path, 0, ans); + return ans; + } + + public static void process2(char[] path, int index, List ans) { + if (index == path.length) { + if (isValid(path)) { + ans.add(String.valueOf(path)); + } + } else { + path[index] = '('; + process2(path, index + 1, ans); + path[index] = ')'; + process2(path, index + 1, ans); + } + } + + public static boolean isValid(char[] path) { + int count = 0; + for (char cha : path) { + if (cha == '(') { + count++; + } else { + count--; + } + if (count < 0) { + return false; + } + } + return count == 0; + } + +} diff --git a/大厂刷题班/class28/Problem_0026_RemoveDuplicatesFromSortedArray.java b/大厂刷题班/class28/Problem_0026_RemoveDuplicatesFromSortedArray.java new file mode 100644 index 0000000..fc2e419 --- /dev/null +++ b/大厂刷题班/class28/Problem_0026_RemoveDuplicatesFromSortedArray.java @@ -0,0 +1,21 @@ +package class28; + +public class Problem_0026_RemoveDuplicatesFromSortedArray { + + public static int removeDuplicates(int[] nums) { + if (nums == null) { + return 0; + } + if (nums.length < 2) { + return nums.length; + } + int done = 0; + for (int i = 1; i < nums.length; i++) { + if (nums[i] != nums[done]) { + nums[++done] = nums[i]; + } + } + return done + 1; + } + +} diff --git a/大厂刷题班/class28/Problem_0034_FindFirstAndLastPositionOfElementInSortedArray.java b/大厂刷题班/class28/Problem_0034_FindFirstAndLastPositionOfElementInSortedArray.java new file mode 100644 index 0000000..92be1c0 --- /dev/null +++ b/大厂刷题班/class28/Problem_0034_FindFirstAndLastPositionOfElementInSortedArray.java @@ -0,0 +1,33 @@ +package class28; + +public class Problem_0034_FindFirstAndLastPositionOfElementInSortedArray { + + public static int[] searchRange(int[] nums, int target) { + if (nums == null || nums.length == 0) { + return new int[] { -1, -1 }; + } + int L = lessMostRight(nums, target) + 1; + if (L == nums.length || nums[L] != target) { + return new int[] { -1, -1 }; + } + return new int[] { L, lessMostRight(nums, target + 1) }; + } + + public static int lessMostRight(int[] arr, int num) { + int L = 0; + int R = arr.length - 1; + int M = 0; + int ans = -1; + while (L <= R) { + M = L + ((R - L) >> 1); + if (arr[M] < num) { + ans = M; + L = M + 1; + } else { + R = M - 1; + } + } + return ans; + } + +} diff --git a/大厂刷题班/class28/Problem_0036_ValidSudoku.java b/大厂刷题班/class28/Problem_0036_ValidSudoku.java new file mode 100644 index 0000000..4a89cf5 --- /dev/null +++ b/大厂刷题班/class28/Problem_0036_ValidSudoku.java @@ -0,0 +1,26 @@ +package class28; + +public class Problem_0036_ValidSudoku { + + public static boolean isValidSudoku(char[][] board) { + boolean[][] row = new boolean[9][10]; + boolean[][] col = new boolean[9][10]; + boolean[][] bucket = new boolean[9][10]; + for (int i = 0; i < 9; i++) { + for (int j = 0; j < 9; j++) { + int bid = 3 * (i / 3) + (j / 3); + if (board[i][j] != '.') { + int num = board[i][j] - '0'; + if (row[i][num] || col[j][num] || bucket[bid][num]) { + return false; + } + row[i][num] = true; + col[j][num] = true; + bucket[bid][num] = true; + } + } + } + return true; + } + +} diff --git a/大厂刷题班/class28/Problem_0037_SudokuSolver.java b/大厂刷题班/class28/Problem_0037_SudokuSolver.java new file mode 100644 index 0000000..7501622 --- /dev/null +++ b/大厂刷题班/class28/Problem_0037_SudokuSolver.java @@ -0,0 +1,61 @@ +package class28; + +public class Problem_0037_SudokuSolver { + + public static void solveSudoku(char[][] board) { + boolean[][] row = new boolean[9][10]; + boolean[][] col = new boolean[9][10]; + boolean[][] bucket = new boolean[9][10]; + initMaps(board, row, col, bucket); + process(board, 0, 0, row, col, bucket); + } + + public static void initMaps(char[][] board, boolean[][] row, boolean[][] col, boolean[][] bucket) { + for (int i = 0; i < 9; i++) { + for (int j = 0; j < 9; j++) { + int bid = 3 * (i / 3) + (j / 3); + if (board[i][j] != '.') { + int num = board[i][j] - '0'; + row[i][num] = true; + col[j][num] = true; + bucket[bid][num] = true; + } + } + } + } + + // 当前来到(i,j)这个位置，如果已经有数字，跳到下一个位置上 + // 如果没有数字，尝试1~9，不能和row、col、bucket冲突 + public static boolean process(char[][] board, int i, int j, boolean[][] row, boolean[][] col, boolean[][] bucket) { + if (i == 9) { + return true; + } + // 当离开(i，j)，应该去哪？(nexti, nextj) + int nexti = j != 8 ? i : i + 1; + int nextj = j != 8 ? j + 1 : 0; + if (board[i][j] != '.') { + return process(board, nexti, nextj, row, col, bucket); + } else { + // 可以尝试1~9 + int bid = 3 * (i / 3) + (j / 3); + for (int num = 1; num <= 9; num++) { // 尝试每一个数字1~9 + if ((!row[i][num]) && (!col[j][num]) && (!bucket[bid][num])) { + // 可以尝试num + row[i][num] = true; + col[j][num] = true; + bucket[bid][num] = true; + board[i][j] = (char) (num + '0'); + if (process(board, nexti, nextj, row, col, bucket)) { + return true; + } + row[i][num] = false; + col[j][num] = false; + bucket[bid][num] = false; + board[i][j] = '.'; + } + } + return false; + } + } + +} diff --git a/大厂刷题班/class28/Problem_0038_CountAndSay.java b/大厂刷题班/class28/Problem_0038_CountAndSay.java new file mode 100644 index 0000000..cd719f3 --- /dev/null +++ b/大厂刷题班/class28/Problem_0038_CountAndSay.java @@ -0,0 +1,33 @@ +package class28; + +public class Problem_0038_CountAndSay { + + public static String countAndSay(int n) { + if (n < 1) { + return ""; + } + if (n == 1) { + return "1"; + } + char[] last = countAndSay(n - 1).toCharArray(); + StringBuilder ans = new StringBuilder(); + int times = 1; + for (int i = 1; i < last.length; i++) { + if (last[i - 1] == last[i]) { + times++; + } else { + ans.append(String.valueOf(times)); + ans.append(String.valueOf(last[i - 1])); + times = 1; + } + } + ans.append(String.valueOf(times)); + ans.append(String.valueOf(last[last.length - 1])); + return ans.toString(); + } + + public static void main(String[] args) { + System.out.println(countAndSay(20)); + } + +} diff --git a/大厂刷题班/class28/Problem_0049_GroupAnagrams.java b/大厂刷题班/class28/Problem_0049_GroupAnagrams.java new file mode 100644 index 0000000..8b325aa --- /dev/null +++ b/大厂刷题班/class28/Problem_0049_GroupAnagrams.java @@ -0,0 +1,52 @@ +package class28; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.HashMap; +import java.util.List; + +public class Problem_0049_GroupAnagrams { + + public static List> groupAnagrams1(String[] strs) { + HashMap> map = new HashMap>(); + for (String str : strs) { + int[] record = new int[26]; + for (char cha : str.toCharArray()) { + record[cha - 'a']++; + } + StringBuilder builder = new StringBuilder(); + for (int value : record) { + builder.append(String.valueOf(value)).append("_"); + } + String key = builder.toString(); + if (!map.containsKey(key)) { + map.put(key, new ArrayList()); + } + map.get(key).add(str); + } + List> res = new ArrayList>(); + for (List list : map.values()) { + res.add(list); + } + return res; + } + + public static List> groupAnagrams2(String[] strs) { + HashMap> map = new HashMap>(); + for (String str : strs) { + char[] chs = str.toCharArray(); + Arrays.sort(chs); + String key = String.valueOf(chs); + if (!map.containsKey(key)) { + map.put(key, new ArrayList()); + } + map.get(key).add(str); + } + List> res = new ArrayList>(); + for (List list : map.values()) { + res.add(list); + } + return res; + } + +} diff --git a/大厂刷题班/class28/说明 b/大厂刷题班/class28/说明 new file mode 100644 index 0000000..ec23675 --- /dev/null +++ b/大厂刷题班/class28/说明 @@ -0,0 +1,33 @@ +leetcode高频题 +leetcode全题目列表 : https://leetcode.com/problemset/all/ +在全题目列表的右侧栏，点击Top Interview Questions +或者直接进入右侧链接 : https://leetcode.com/problemset/all/?listId=wpwgkgt +即可看到leetcode高频题全列表 +本节课解决leetcode高频题列表中的如下题目 : +0008 : 大厂刷题班, 第28节, 本节 +0010 : 大厂刷题班, 第12节第4题 +0011 : 大厂刷题班, 第8节第2题 +0012 : 大厂刷题班, 第28节, 本节 +0013 : 大厂刷题班, 第28节, 本节 +0014 : 大厂刷题班, 第28节, 本节 +0015 : 大厂刷题班, 第25节第2题 +0017 : 大厂刷题班, 第28节, 本节 +0019 : 大厂刷题班, 第28节, 本节 +0020 : 大厂刷题班, 第28节, 本节 +0021 : 算法新手班, 第4节第6题 +0022 : 大厂刷题班, 第28节, 本节 +0023 : 算法新手班, 第6节第1题 +0026 : 大厂刷题班, 第28节, 本节 +0028 : 体系学习班, KMP算法 +0029 : 算法新手班, 第5节第3题 +0034 : 大厂刷题班, 第28节, 本节 +0036 : 大厂刷题班, 第28节, 本节 +0037 : 大厂刷题班, 第28节, 本节 +0038 : 大厂刷题班, 第28节, 本节 +0041 : 大厂刷题班, 第14节第6题 +0042 : 大厂刷题班, 第22节第2题 +0044 : 大厂刷题班, 第12节第4题 +0045 : 大厂刷题班, 第10节第1题 +0046 : 体系学习班, 第17节第4题 +0048 : 体系学习班, 第40节第6题 +0049 : 大厂刷题班, 第28节, 本节 \ No newline at end of file diff --git a/大厂刷题班/class29/Problem_0033_SearchInRotatedSortedArray.java b/大厂刷题班/class29/Problem_0033_SearchInRotatedSortedArray.java new file mode 100644 index 0000000..ba8841a --- /dev/null +++ b/大厂刷题班/class29/Problem_0033_SearchInRotatedSortedArray.java @@ -0,0 +1,67 @@ +package class29; + +public class Problem_0033_SearchInRotatedSortedArray { + + // arr，原本是有序数组，旋转过，而且左部分长度不知道 + // 找num + // num所在的位置返回 + public static int search(int[] arr, int num) { + int L = 0; + int R = arr.length - 1; + int M = 0; + while (L <= R) { + // M = L + ((R - L) >> 1) + M = (L + R) / 2; + if (arr[M] == num) { + return M; + } + // arr[M] != num + // [L] == [M] == [R] != num 无法二分 + if (arr[L] == arr[M] && arr[M] == arr[R]) { + while (L != M && arr[L] == arr[M]) { + L++; + } + // 1) L == M L...M 一路都相等 + // 2) 从L到M终于找到了一个不等的位置 + if (L == M) { // L...M 一路都相等 + L = M + 1; + continue; + } + } + // ... + // arr[M] != num + // [L] [M] [R] 不都一样的情况, 如何二分的逻辑 + if (arr[L] != arr[M]) { + if (arr[M] > arr[L]) { // L...M 一定有序 + if (num >= arr[L] && num < arr[M]) { // 3 [L] == 1 [M] = 5 L...M - 1 + R = M - 1; + } else { // 9 [L] == 2 [M] = 7 M... R + L = M + 1; + } + } else { // [L] > [M] L....M 存在断点 + if (num > arr[M] && num <= arr[R]) { + L = M + 1; + } else { + R = M - 1; + } + } + } else { /// [L] [M] [R] 不都一样， [L] === [M] -> [M]!=[R] + if (arr[M] < arr[R]) { + if (num > arr[M] && num <= arr[R]) { + L = M + 1; + } else { + R = M - 1; + } + } else { + if (num >= arr[L] && num < arr[M]) { + R = M - 1; + } else { + L = M + 1; + } + } + } + } + return -1; + } + +} diff --git a/大厂刷题班/class29/Problem_0050_PowXN.java b/大厂刷题班/class29/Problem_0050_PowXN.java new file mode 100644 index 0000000..77d5288 --- /dev/null +++ b/大厂刷题班/class29/Problem_0050_PowXN.java @@ -0,0 +1,39 @@ +package class29; + +public class Problem_0050_PowXN { + + public static int pow(int a, int n) { + int ans = 1; + int t = a; + while (n != 0) { + if ((n & 1) != 0) { + ans *= t; + } + t *= t; + n >>= 1; + } + return ans; + } + + // x的n次方，n可能是负数 + public static double myPow(double x, int n) { + if (n == 0) { + return 1D; + } + int pow = Math.abs(n == Integer.MIN_VALUE ? n + 1 : n); + double t = x; + double ans = 1D; + while (pow != 0) { + if ((pow & 1) != 0) { + ans *= t; + } + pow >>= 1; + t = t * t; + } + if (n == Integer.MIN_VALUE) { + ans *= x; + } + return n < 0 ? (1D / ans) : ans; + } + +} diff --git a/大厂刷题班/class29/Problem_0056_MergeIntervals.java b/大厂刷题班/class29/Problem_0056_MergeIntervals.java new file mode 100644 index 0000000..d912b49 --- /dev/null +++ b/大厂刷题班/class29/Problem_0056_MergeIntervals.java @@ -0,0 +1,30 @@ +package class29; + +import java.util.Arrays; + +public class Problem_0056_MergeIntervals { + + public static int[][] merge(int[][] intervals) { + if (intervals.length == 0) { + return new int[0][0]; + } + Arrays.sort(intervals, (a, b) -> a[0] - b[0]); + int s = intervals[0][0]; + int e = intervals[0][1]; + int size = 0; + for (int i = 1; i < intervals.length; i++) { + if (intervals[i][0] > e) { + intervals[size][0] = s; + intervals[size++][1] = e; + s = intervals[i][0]; + e = intervals[i][1]; + } else { + e = Math.max(e, intervals[i][1]); + } + } + intervals[size][0] = s; + intervals[size++][1] = e; + return Arrays.copyOf(intervals, size); + } + +} diff --git a/大厂刷题班/class29/Problem_0062_UniquePaths.java b/大厂刷题班/class29/Problem_0062_UniquePaths.java new file mode 100644 index 0000000..c5ffd25 --- /dev/null +++ b/大厂刷题班/class29/Problem_0062_UniquePaths.java @@ -0,0 +1,30 @@ +package class29; + +public class Problem_0062_UniquePaths { + + // m 行 + // n 列 + // 下：m-1 + // 右：n-1 + public static int uniquePaths(int m, int n) { + int right = n - 1; + int all = m + n - 2; + long o1 = 1; + long o2 = 1; + // o1乘进去的个数 一定等于 o2乘进去的个数 + for (int i = right + 1, j = 1; i <= all; i++, j++) { + o1 *= i; + o2 *= j; + long gcd = gcd(o1, o2); + o1 /= gcd; + o2 /= gcd; + } + return (int) o1; + } + + // 调用的时候，请保证初次调用时，m和n都不为0 + public static long gcd(long m, long n) { + return n == 0 ? m : gcd(n, m % n); + } + +} diff --git a/大厂刷题班/class29/Problem_0066_PlusOne.java b/大厂刷题班/class29/Problem_0066_PlusOne.java new file mode 100644 index 0000000..3717e40 --- /dev/null +++ b/大厂刷题班/class29/Problem_0066_PlusOne.java @@ -0,0 +1,19 @@ +package class29; + +public class Problem_0066_PlusOne { + + public static int[] plusOne(int[] digits) { + int n = digits.length; + for (int i = n - 1; i >= 0; i--) { + if (digits[i] < 9) { + digits[i]++; + return digits; + } + digits[i] = 0; + } + int[] ans = new int[n + 1]; + ans[0] = 1; + return ans; + } + +} diff --git a/大厂刷题班/class29/Problem_0069_SqrtX.java b/大厂刷题班/class29/Problem_0069_SqrtX.java new file mode 100644 index 0000000..f58a642 --- /dev/null +++ b/大厂刷题班/class29/Problem_0069_SqrtX.java @@ -0,0 +1,30 @@ +package class29; + +public class Problem_0069_SqrtX { + + // x一定非负，输入可以保证 + public static int mySqrt(int x) { + if (x == 0) { + return 0; + } + if (x < 3) { + return 1; + } + // x >= 3 + long ans = 1; + long L = 1; + long R = x; + long M = 0; + while (L <= R) { + M = (L + R) / 2; + if (M * M <= x) { + ans = M; + L = M + 1; + } else { + R = M - 1; + } + } + return (int) ans; + } + +} diff --git a/大厂刷题班/class29/Problem_0073_SetMatrixZeroes.java b/大厂刷题班/class29/Problem_0073_SetMatrixZeroes.java new file mode 100644 index 0000000..be15049 --- /dev/null +++ b/大厂刷题班/class29/Problem_0073_SetMatrixZeroes.java @@ -0,0 +1,79 @@ +package class29; + +public class Problem_0073_SetMatrixZeroes { + + public static void setZeroes1(int[][] matrix) { + boolean row0Zero = false; + boolean col0Zero = false; + int i = 0; + int j = 0; + for (i = 0; i < matrix[0].length; i++) { + if (matrix[0][i] == 0) { + row0Zero = true; + break; + } + } + for (i = 0; i < matrix.length; i++) { + if (matrix[i][0] == 0) { + col0Zero = true; + break; + } + } + for (i = 1; i < matrix.length; i++) { + for (j = 1; j < matrix[0].length; j++) { + if (matrix[i][j] == 0) { + matrix[i][0] = 0; + matrix[0][j] = 0; + } + } + } + for (i = 1; i < matrix.length; i++) { + for (j = 1; j < matrix[0].length; j++) { + if (matrix[i][0] == 0 || matrix[0][j] == 0) { + matrix[i][j] = 0; + } + } + } + if (row0Zero) { + for (i = 0; i < matrix[0].length; i++) { + matrix[0][i] = 0; + } + } + if (col0Zero) { + for (i = 0; i < matrix.length; i++) { + matrix[i][0] = 0; + } + } + } + + public static void setZeroes2(int[][] matrix) { + boolean col0 = false; + int i = 0; + int j = 0; + for (i = 0; i < matrix.length; i++) { + for (j = 0; j < matrix[0].length; j++) { + if (matrix[i][j] == 0) { + matrix[i][0] = 0; + if (j == 0) { + col0 = true; + } else { + matrix[0][j] = 0; + } + } + } + } + for (i = matrix.length - 1; i >= 0; i--) { + for (j = 1; j < matrix[0].length; j++) { + if (matrix[i][0] == 0 || matrix[0][j] == 0) { + matrix[i][j] = 0; + } + } + } + if (col0) { + for (i = 0; i < matrix.length; i++) { + matrix[i][0] = 0; + } + } + } + +} diff --git a/大厂刷题班/class29/说明 b/大厂刷题班/class29/说明 new file mode 100644 index 0000000..75954f5 --- /dev/null +++ b/大厂刷题班/class29/说明 @@ -0,0 +1,20 @@ +leetcode高频题 +leetcode全题目列表 : https://leetcode.com/problemset/all/ +在全题目列表的右侧栏，点击Top Interview Questions +或者直接进入右侧链接 : https://leetcode.com/problemset/all/?listId=wpwgkgt +即可看到leetcode高频题全列表 +本节课解决leetcode高频题列表中的如下题目 : +0033 : 大厂刷题班, 第29节, 本节 +0050 : 大厂刷题班, 第29节, 本节 +0053 : 体系学习班, 第40节第2题 +0054 : 体系学习班, 第40节第5题 +0055 : 大厂刷题班, 第10节第1题 +0056 : 大厂刷题班, 第29节, 本节 +0062 : 大厂刷题班, 第29节, 本节 +0066 : 大厂刷题班, 第29节, 本节 +0069 : 大厂刷题班, 第29节, 本节 +0070 : 体系学习班, 第26节第2题 +0073 : 大厂刷题班, 第29节, 本节 +0075 : 体系学习班, 第5节第2题, 快排中的荷兰国旗问题 +0076 : 大厂刷题班, 第24节第5题 +0078 : 体系学习班, 第17节题目3, 生成子序列问题和本题一样的 \ No newline at end of file diff --git a/大厂刷题班/class30/Problem_0079_WordSearch.java b/大厂刷题班/class30/Problem_0079_WordSearch.java new file mode 100644 index 0000000..8d3d0da --- /dev/null +++ b/大厂刷题班/class30/Problem_0079_WordSearch.java @@ -0,0 +1,41 @@ +package class30; + +public class Problem_0079_WordSearch { + + public static boolean exist(char[][] board, String word) { + char[] w = word.toCharArray(); + for (int i = 0; i < board.length; i++) { + for (int j = 0; j < board[0].length; j++) { + if (f(board, i, j, w, 0)) { + return true; + } + } + } + return false; + } + + // 目前到达了b[i][j]，word[k....] + // 从b[i][j]出发，能不能搞定word[k....] true false + public static boolean f(char[][] b, int i, int j, char[] w, int k) { + if (k == w.length) { + return true; + } + // word[k.....] 有字符 + // 如果(i,j)越界，返回false + if (i < 0 || i == b.length || j < 0 || j == b[0].length) { + return false; + } + if (b[i][j] != w[k]) { + return false; + } + char tmp = b[i][j]; + b[i][j] = 0; + boolean ans = f(b, i - 1, j, w, k + 1) + || f(b, i + 1, j, w, k + 1) + || f(b, i, j - 1, w, k + 1) + || f(b, i, j + 1, w, k + 1); + b[i][j] = tmp; + return ans; + } + +} diff --git a/大厂刷题班/class30/Problem_0088_MergeSortedArray.java b/大厂刷题班/class30/Problem_0088_MergeSortedArray.java new file mode 100644 index 0000000..4b24fc8 --- /dev/null +++ b/大厂刷题班/class30/Problem_0088_MergeSortedArray.java @@ -0,0 +1,22 @@ +package class30; + +public class Problem_0088_MergeSortedArray { + + public static void merge(int[] nums1, int m, int[] nums2, int n) { + int index = nums1.length; + while (m > 0 && n > 0) { + if (nums1[m - 1] >= nums2[n - 1]) { + nums1[--index] = nums1[--m]; + } else { + nums1[--index] = nums2[--n]; + } + } + while (m > 0) { + nums1[--index] = nums1[--m]; + } + while (n > 0) { + nums1[--index] = nums2[--n]; + } + } + +} diff --git a/大厂刷题班/class30/Problem_0091_DecodeWays.java b/大厂刷题班/class30/Problem_0091_DecodeWays.java new file mode 100644 index 0000000..0db1ac8 --- /dev/null +++ b/大厂刷题班/class30/Problem_0091_DecodeWays.java @@ -0,0 +1,87 @@ +package class30; + +public class Problem_0091_DecodeWays { + + public static int numDecodings1(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + return process(str, 0); + } + + // 潜台词：str[0...index-1]已经转化完了，不用操心了 + // str[index....] 能转出多少有效的，返回方法数 + public static int process(char[] str, int index) { + if (index == str.length) { + return 1; + } + if (str[index] == '0') { + return 0; + } + int ways = process(str, index + 1); + if (index + 1 == str.length) { + return ways; + } + int num = (str[index] - '0') * 10 + str[index + 1] - '0'; + if (num < 27) { + ways += process(str, index + 2); + } + return ways; + } + + public static int numDecodings2(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + int N = str.length; + // dp[i] -> process(str, index)返回值 index 0 ~ N + int[] dp = new int[N + 1]; + dp[N] = 1; + + // dp依次填好 dp[i] dp[i+1] dp[i+2] + for (int i = N - 1; i >= 0; i--) { + if (str[i] != '0') { + dp[i] = dp[i + 1]; + if (i + 1 == str.length) { + continue; + } + int num = (str[i] - '0') * 10 + str[i + 1] - '0'; + if (num <= 26) { + dp[i] += dp[i + 2]; + } + } + } + return dp[0]; + } + + public static int numDecodings(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + int N = str.length; + int[] dp = new int[N + 1]; + dp[N] = 1; + for (int i = N - 1; i >= 0; i--) { + if (str[i] == '0') { + dp[i] = 0; + } else if (str[i] == '1') { + dp[i] = dp[i + 1]; + if (i + 1 < N) { + dp[i] += dp[i + 2]; + } + } else if (str[i] == '2') { + dp[i] = dp[i + 1]; + if (i + 1 < str.length && (str[i + 1] >= '0' && str[i + 1] <= '6')) { + dp[i] += dp[i + 2]; + } + } else { + dp[i] = dp[i + 1]; + } + } + return dp[0]; + } + +} diff --git a/大厂刷题班/class30/Problem_0098_ValidateBinarySearchTree.java b/大厂刷题班/class30/Problem_0098_ValidateBinarySearchTree.java new file mode 100644 index 0000000..d3471d8 --- /dev/null +++ b/大厂刷题班/class30/Problem_0098_ValidateBinarySearchTree.java @@ -0,0 +1,42 @@ +package class30; + +public class Problem_0098_ValidateBinarySearchTree { + + public static class TreeNode { + int val; + TreeNode left; + TreeNode right; + } + + public boolean isValidBST(TreeNode root) { + if (root == null) { + return true; + } + TreeNode cur = root; + TreeNode mostRight = null; + Integer pre = null; + boolean ans = true; + while (cur != null) { + mostRight = cur.left; + if (mostRight != null) { + while (mostRight.right != null && mostRight.right != cur) { + mostRight = mostRight.right; + } + if (mostRight.right == null) { + mostRight.right = cur; + cur = cur.left; + continue; + } else { + mostRight.right = null; + } + } + if (pre != null && pre >= cur.val) { + ans = false; + } + pre = cur.val; + cur = cur.right; + } + return ans; + } + +} diff --git a/大厂刷题班/class30/Problem_0101_SymmetricTree.java b/大厂刷题班/class30/Problem_0101_SymmetricTree.java new file mode 100644 index 0000000..f5cdfe7 --- /dev/null +++ b/大厂刷题班/class30/Problem_0101_SymmetricTree.java @@ -0,0 +1,30 @@ +package class30; + +public class Problem_0101_SymmetricTree { + + public static class TreeNode { + int val; + TreeNode left; + TreeNode right; + } + + public boolean isSymmetric(TreeNode root) { + return isMirror(root, root); + } + + // 一棵树是原始树 head1 + // 另一棵是翻面树 head2 + public static boolean isMirror(TreeNode head1, TreeNode head2) { + if (head1 == null && head2 == null) { + return true; + } + if (head1 != null && head2 != null) { + return head1.val == head2.val + && isMirror(head1.left, head2.right) + && isMirror(head1.right, head2.left); + } + // 一个为空，一个不为空 false + return false; + } + +} diff --git a/大厂刷题班/class30/Problem_0103_BinaryTreeZigzagLevelOrderTraversal.java b/大厂刷题班/class30/Problem_0103_BinaryTreeZigzagLevelOrderTraversal.java new file mode 100644 index 0000000..546fab2 --- /dev/null +++ b/大厂刷题班/class30/Problem_0103_BinaryTreeZigzagLevelOrderTraversal.java @@ -0,0 +1,52 @@ +package class30; + +import java.util.ArrayList; +import java.util.LinkedList; +import java.util.List; + +public class Problem_0103_BinaryTreeZigzagLevelOrderTraversal { + + public static class TreeNode { + int val; + TreeNode left; + TreeNode right; + } + + public static List> zigzagLevelOrder(TreeNode root) { + List> ans = new ArrayList<>(); + if (root == null) { + return ans; + } + LinkedList deque = new LinkedList<>(); + deque.add(root); + int size = 0; + boolean isHead = true; + while (!deque.isEmpty()) { + size = deque.size(); + List curLevel = new ArrayList<>(); + for (int i = 0; i < size; i++) { + TreeNode cur = isHead ? deque.pollFirst() : deque.pollLast(); + curLevel.add(cur.val); + if(isHead) { + if (cur.left != null) { + deque.addLast(cur.left); + } + if (cur.right != null) { + deque.addLast(cur.right); + } + }else { + if (cur.right != null) { + deque.addFirst(cur.right); + } + if (cur.left != null) { + deque.addFirst(cur.left); + } + } + } + ans.add(curLevel); + isHead = !isHead; + } + return ans; + } + +} diff --git a/大厂刷题班/class30/Problem_0108_ConvertSortedArrayToBinarySearchTree.java b/大厂刷题班/class30/Problem_0108_ConvertSortedArrayToBinarySearchTree.java new file mode 100644 index 0000000..0914140 --- /dev/null +++ b/大厂刷题班/class30/Problem_0108_ConvertSortedArrayToBinarySearchTree.java @@ -0,0 +1,33 @@ +package class30; + +public class Problem_0108_ConvertSortedArrayToBinarySearchTree { + + public static class TreeNode { + int val; + TreeNode left; + TreeNode right; + + TreeNode(int val) { + this.val = val; + } + } + + public TreeNode sortedArrayToBST(int[] nums) { + return process(nums, 0, nums.length - 1); + } + + public static TreeNode process(int[] nums, int L, int R) { + if (L > R) { + return null; + } + if (L == R) { + return new TreeNode(nums[L]); + } + int M = (L + R) / 2; + TreeNode head = new TreeNode(nums[M]); + head.left = process(nums, L, M - 1); + head.right = process(nums, M + 1, R); + return head; + } + +} diff --git a/大厂刷题班/class30/Problem_0116_PopulatingNextRightPointersInEachNode.java b/大厂刷题班/class30/Problem_0116_PopulatingNextRightPointersInEachNode.java new file mode 100644 index 0000000..f3d82de --- /dev/null +++ b/大厂刷题班/class30/Problem_0116_PopulatingNextRightPointersInEachNode.java @@ -0,0 +1,77 @@ +package class30; + +public class Problem_0116_PopulatingNextRightPointersInEachNode { + + // 不要提交这个类 + public static class Node { + public int val; + public Node left; + public Node right; + public Node next; + } + + // 提交下面的代码 + public static Node connect(Node root) { + if (root == null) { + return root; + } + MyQueue queue = new MyQueue(); + queue.offer(root); + while (!queue.isEmpty()) { + // 第一个弹出的节点 + Node pre = null; + int size = queue.size; + for (int i = 0; i < size; i++) { + Node cur = queue.poll(); + if (cur.left != null) { + queue.offer(cur.left); + } + if (cur.right != null) { + queue.offer(cur.right); + } + if (pre != null) { + pre.next = cur; + } + pre = cur; + } + } + return root; + } + + public static class MyQueue { + public Node head; + public Node tail; + public int size; + + public MyQueue() { + head = null; + tail = null; + size = 0; + } + + public boolean isEmpty() { + return size == 0; + } + + public void offer(Node cur) { + size++; + if (head == null) { + head = cur; + tail = cur; + } else { + tail.next = cur; + tail = cur; + } + } + + public Node poll() { + size--; + Node ans = head; + head = head.next; + ans.next = null; + return ans; + } + + } + +} diff --git a/大厂刷题班/class30/Problem_0118_PascalTriangle.java b/大厂刷题班/class30/Problem_0118_PascalTriangle.java new file mode 100644 index 0000000..867990f --- /dev/null +++ b/大厂刷题班/class30/Problem_0118_PascalTriangle.java @@ -0,0 +1,23 @@ +package class30; + +import java.util.ArrayList; +import java.util.List; + +public class Problem_0118_PascalTriangle { + + public static List> generate(int numRows) { + List> ans = new ArrayList<>(); + for (int i = 0; i < numRows; i++) { + ans.add(new ArrayList<>()); + ans.get(i).add(1); + } + for (int i = 1; i < numRows; i++) { + for (int j = 1; j < i; j++) { + ans.get(i).add(ans.get(i - 1).get(j - 1) + ans.get(i - 1).get(j)); + } + ans.get(i).add(1); + } + return ans; + } + +} diff --git a/大厂刷题班/class30/Problem_0119_PascalTriangleII.java b/大厂刷题班/class30/Problem_0119_PascalTriangleII.java new file mode 100644 index 0000000..79bfc78 --- /dev/null +++ b/大厂刷题班/class30/Problem_0119_PascalTriangleII.java @@ -0,0 +1,19 @@ +package class30; + +import java.util.ArrayList; +import java.util.List; + +public class Problem_0119_PascalTriangleII { + + public List getRow(int rowIndex) { + List ans = new ArrayList<>(); + for (int i = 0; i <= rowIndex; i++) { + for (int j = i - 1; j > 0; j--) { + ans.set(j, ans.get(j - 1) + ans.get(j)); + } + ans.add(1); + } + return ans; + } + +} diff --git a/大厂刷题班/class30/Problem_0124_BinaryTreeMaximumPathSum.java b/大厂刷题班/class30/Problem_0124_BinaryTreeMaximumPathSum.java new file mode 100644 index 0000000..de6b07c --- /dev/null +++ b/大厂刷题班/class30/Problem_0124_BinaryTreeMaximumPathSum.java @@ -0,0 +1,206 @@ +package class30; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.HashSet; +import java.util.List; + +// follow up : 如果要求返回整个路径怎么做？ +public class Problem_0124_BinaryTreeMaximumPathSum { + + public static class TreeNode { + int val; + TreeNode left; + TreeNode right; + + public TreeNode(int v) { + val = v; + } + + } + + public static int maxPathSum(TreeNode root) { + if (root == null) { + return 0; + } + return process(root).maxPathSum; + } + + // 任何一棵树，必须汇报上来的信息 + public static class Info { + public int maxPathSum; + public int maxPathSumFromHead; + + public Info(int path, int head) { + maxPathSum = path; + maxPathSumFromHead = head; + } + } + + public static Info process(TreeNode x) { + if (x == null) { + return null; + } + Info leftInfo = process(x.left); + Info rightInfo = process(x.right); + // x 1)只有x 2）x往左扎 3）x往右扎 + int maxPathSumFromHead = x.val; + if (leftInfo != null) { + maxPathSumFromHead = Math.max(maxPathSumFromHead, x.val + leftInfo.maxPathSumFromHead); + } + if (rightInfo != null) { + maxPathSumFromHead = Math.max(maxPathSumFromHead, x.val + rightInfo.maxPathSumFromHead); + } + // x整棵树最大路径和 1) 只有x 2)左树整体的最大路径和 3) 右树整体的最大路径和 + int maxPathSum = x.val; + if (leftInfo != null) { + maxPathSum = Math.max(maxPathSum, leftInfo.maxPathSum); + } + if (rightInfo != null) { + maxPathSum = Math.max(maxPathSum, rightInfo.maxPathSum); + } + // 4) x只往左扎 5）x只往右扎 + maxPathSum = Math.max(maxPathSumFromHead, maxPathSum); + // 6）一起扎 + if (leftInfo != null && rightInfo != null && leftInfo.maxPathSumFromHead > 0 + && rightInfo.maxPathSumFromHead > 0) { + maxPathSum = Math.max(maxPathSum, leftInfo.maxPathSumFromHead + rightInfo.maxPathSumFromHead + x.val); + } + return new Info(maxPathSum, maxPathSumFromHead); + } + + // 如果要返回路径的做法 + public static List getMaxSumPath(TreeNode head) { + List ans = new ArrayList<>(); + if (head != null) { + Data data = f(head); + HashMap fmap = new HashMap<>(); + fmap.put(head, head); + fatherMap(head, fmap); + fillPath(fmap, data.from, data.to, ans); + } + return ans; + } + + public static class Data { + public int maxAllSum; + public TreeNode from; + public TreeNode to; + public int maxHeadSum; + public TreeNode end; + + public Data(int a, TreeNode b, TreeNode c, int d, TreeNode e) { + maxAllSum = a; + from = b; + to = c; + maxHeadSum = d; + end = e; + } + } + + public static Data f(TreeNode x) { + if (x == null) { + return null; + } + Data l = f(x.left); + Data r = f(x.right); + int maxHeadSum = x.val; + TreeNode end = x; + if (l != null && l.maxHeadSum > 0 && (r == null || l.maxHeadSum > r.maxHeadSum)) { + maxHeadSum += l.maxHeadSum; + end = l.end; + } + if (r != null && r.maxHeadSum > 0 && (l == null || r.maxHeadSum > l.maxHeadSum)) { + maxHeadSum += r.maxHeadSum; + end = r.end; + } + int maxAllSum = Integer.MIN_VALUE; + TreeNode from = null; + TreeNode to = null; + if (l != null) { + maxAllSum = l.maxAllSum; + from = l.from; + to = l.to; + } + if (r != null && r.maxAllSum > maxAllSum) { + maxAllSum = r.maxAllSum; + from = r.from; + to = r.to; + } + int p3 = x.val + (l != null && l.maxHeadSum > 0 ? l.maxHeadSum : 0) + + (r != null && r.maxHeadSum > 0 ? r.maxHeadSum : 0); + if (p3 > maxAllSum) { + maxAllSum = p3; + from = (l != null && l.maxHeadSum > 0) ? l.end : x; + to = (r != null && r.maxHeadSum > 0) ? r.end : x; + } + return new Data(maxAllSum, from, to, maxHeadSum, end); + } + + public static void fatherMap(TreeNode h, HashMap map) { + if (h.left == null && h.right == null) { + return; + } + if (h.left != null) { + map.put(h.left, h); + fatherMap(h.left, map); + } + if (h.right != null) { + map.put(h.right, h); + fatherMap(h.right, map); + } + } + + public static void fillPath(HashMap fmap, TreeNode a, TreeNode b, List ans) { + if (a == b) { + ans.add(a); + } else { + HashSet ap = new HashSet<>(); + TreeNode cur = a; + while (cur != fmap.get(cur)) { + ap.add(cur); + cur = fmap.get(cur); + } + ap.add(cur); + cur = b; + TreeNode lca = null; + while (lca == null) { + if (ap.contains(cur)) { + lca = cur; + } else { + cur = fmap.get(cur); + } + } + while (a != lca) { + ans.add(a); + a = fmap.get(a); + } + ans.add(lca); + ArrayList right = new ArrayList<>(); + while (b != lca) { + right.add(b); + b = fmap.get(b); + } + for (int i = right.size() - 1; i >= 0; i--) { + ans.add(right.get(i)); + } + } + } + + public static void main(String[] args) { + TreeNode head = new TreeNode(4); + head.left = new TreeNode(-7); + head.right = new TreeNode(-5); + head.left.left = new TreeNode(9); + head.left.right = new TreeNode(9); + head.right.left = new TreeNode(4); + head.right.right = new TreeNode(3); + + List maxPath = getMaxSumPath(head); + + for (TreeNode n : maxPath) { + System.out.println(n.val); + } + } + +} \ No newline at end of file diff --git a/大厂刷题班/class30/Problem_0639_DecodeWaysII.java b/大厂刷题班/class30/Problem_0639_DecodeWaysII.java new file mode 100644 index 0000000..a04422b --- /dev/null +++ b/大厂刷题班/class30/Problem_0639_DecodeWaysII.java @@ -0,0 +1,161 @@ +package class30; + +public class Problem_0639_DecodeWaysII { + + public static int numDecodings0(String str) { + return f(str.toCharArray(), 0); + } + + public static int f(char[] str, int i) { + if (i == str.length) { + return 1; + } + if (str[i] == '0') { + return 0; + } + // str[index]有字符且不是'0' + if (str[i] != '*') { + // str[index] = 1~9 + // i -> 单转 + int p1 = f(str, i + 1); + if (i + 1 == str.length) { + return p1; + } + if (str[i + 1] != '*') { + int num = (str[i] - '0') * 10 + str[i + 1] - '0'; + int p2 = 0; + if (num < 27) { + p2 = f(str, i + 2); + } + return p1 + p2; + } else { // str[i+1] == '*' + // i i+1 -> 一起转 1* 2* 3* 9* + int p2 = 0; + if (str[i] < '3') { + p2 = f(str, i + 2) * (str[i] == '1' ? 9 : 6); + } + return p1 + p2; + } + } else { // str[i] == '*' 1~9 + // i 单转 9种 + int p1 = 9 * f(str, i + 1); + if (i + 1 == str.length) { + return p1; + } + if (str[i + 1] != '*') { + // * 0 10 20 + // * 1 11 21 + // * 2 12 22 + // * 3 13 23 + // * 6 16 26 + // * 7 17 + // * 8 18 + // * 9 19 + int p2 = (str[i + 1] < '7' ? 2 : 1) * f(str, i + 2); + return p1 + p2; + } else { // str[i+1] == * + // ** + // 11~19 9 + // 21 ~26 6 + // 15 + int p2 = 15 * f(str, i + 2); + return p1 + p2; + } + } + } + + public static long mod = 1000000007; + + public static int numDecodings1(String str) { + long[] dp = new long[str.length()]; + return (int) ways1(str.toCharArray(), 0, dp); + } + + public static long ways1(char[] s, int i, long[] dp) { + if (i == s.length) { + return 1; + } + if (s[i] == '0') { + return 0; + } + if (dp[i] != 0) { + return dp[i]; + } + long ans = ways1(s, i + 1, dp) * (s[i] == '*' ? 9 : 1); + if (s[i] == '1' || s[i] == '2' || s[i] == '*') { + if (i + 1 < s.length) { + if (s[i + 1] == '*') { + ans += ways1(s, i + 2, dp) * (s[i] == '*' ? 15 : (s[i] == '1' ? 9 : 6)); + } else { + if (s[i] == '*') { + ans += ways1(s, i + 2, dp) * (s[i + 1] < '7' ? 2 : 1); + } else { + ans += ((s[i] - '0') * 10 + s[i + 1] - '0') < 27 ? ways1(s, i + 2, dp) : 0; + } + } + } + } + ans %= mod; + dp[i] = ans; + return ans; + } + + public static int numDecodings2(String str) { + char[] s = str.toCharArray(); + int n = s.length; + long[] dp = new long[n + 1]; + dp[n] = 1; + for (int i = n - 1; i >= 0; i--) { + if (s[i] != '0') { + dp[i] = dp[i + 1] * (s[i] == '*' ? 9 : 1); + if (s[i] == '1' || s[i] == '2' || s[i] == '*') { + if (i + 1 < n) { + if (s[i + 1] == '*') { + dp[i] += dp[i + 2] * (s[i] == '*' ? 15 : (s[i] == '1' ? 9 : 6)); + } else { + if (s[i] == '*') { + dp[i] += dp[i + 2] * (s[i + 1] < '7' ? 2 : 1); + } else { + dp[i] += ((s[i] - '0') * 10 + s[i + 1] - '0') < 27 ? dp[i + 2] : 0; + } + } + } + } + dp[i] %= mod; + } + } + return (int) dp[0]; + } + + public static int numDecodings3(String str) { + char[] s = str.toCharArray(); + int n = s.length; + long a = 1; + long b = 1; + long c = 0; + for (int i = n - 1; i >= 0; i--) { + if (s[i] != '0') { + c = b * (s[i] == '*' ? 9 : 1); + if (s[i] == '1' || s[i] == '2' || s[i] == '*') { + if (i + 1 < n) { + if (s[i + 1] == '*') { + c += a * (s[i] == '*' ? 15 : (s[i] == '1' ? 9 : 6)); + } else { + if (s[i] == '*') { + c += a * (s[i + 1] < '7' ? 2 : 1); + } else { + c += a * (((s[i] - '0') * 10 + s[i + 1] - '0') < 27 ? 1 : 0); + } + } + } + } + } + c %= mod; + a = b; + b = c; + c = 0; + } + return (int) b; + } + +} diff --git a/大厂刷题班/class30/说明 b/大厂刷题班/class30/说明 new file mode 100644 index 0000000..81b1132 --- /dev/null +++ b/大厂刷题班/class30/说明 @@ -0,0 +1,26 @@ +leetcode高频题 +leetcode全题目列表 : https://leetcode.com/problemset/all/ +在全题目列表的右侧栏，点击Top Interview Questions +或者直接进入右侧链接 : https://leetcode.com/problemset/all/?listId=wpwgkgt +即可看到leetcode高频题全列表 +本节课解决leetcode高频题列表中的如下题目 : +0079 : 大厂刷题班, 第30节, 本节 +0084 : 体系学习班, 第25节第3题 +0088 : 大厂刷题班, 第30节, 本节 +0091 : 体系学习班, 第19节第2题 +0639 : 本题不再高频题列表中, 但本题是0091的难度加强题, 相似度很强, 大厂刷题班, 第30节, 本节 +0094 : 体系学习班, 第30节第1题, Morris遍历 +0098 : 大厂刷题班, 第30节, 本节 +0101 : 大厂刷题班, 第30节, 本节 +0102 : 新手班, 第7节第1题 +0103 : 大厂刷题班, 第30节, 本节 +0104 : 太简单了, 体系学习班, 二叉树的递归套路、Morris遍历都可以做, 跳过 +0105 : 新手班, 第6节第5题 +0108 : 大厂刷题班, 第30节, 本节 +0116 : 大厂刷题班, 第30节, 本节 +0118 : 大厂刷题班, 第30节, 本节 +0119 : 本题不在高频题列表中，但和0118类似, 大厂刷题班, 第30节, 本节 +0121 : 大厂刷题班, 第15节第1题 +0122 : 大厂刷题班, 第15节第2题 +0123 : 大厂刷题班, 第15节第3题 +0124 : 大厂刷题班, 第30节, 本节 \ No newline at end of file diff --git a/大厂刷题班/class31/Problem_0125_ValidPalindrome.java b/大厂刷题班/class31/Problem_0125_ValidPalindrome.java new file mode 100644 index 0000000..e55e04b --- /dev/null +++ b/大厂刷题班/class31/Problem_0125_ValidPalindrome.java @@ -0,0 +1,52 @@ +package class31; + +public class Problem_0125_ValidPalindrome { + + // 忽略空格、忽略大小写 -> 是不是回文 + // 数字不在忽略大小写的范围内 + public static boolean isPalindrome(String s) { + if (s == null || s.length() == 0) { + return true; + } + char[] str = s.toCharArray(); + int L = 0; + int R = str.length - 1; + while (L < R) { + // 英文（大小写） + 数字 + if (validChar(str[L]) && validChar(str[R])) { + if (!equal(str[L], str[R])) { + return false; + } + L++; + R--; + } else { + L += validChar(str[L]) ? 0 : 1; + R -= validChar(str[R]) ? 0 : 1; + } + } + return true; + } + + public static boolean validChar(char c) { + return isLetter(c) || isNumber(c); + } + + public static boolean equal(char c1, char c2) { + if (isNumber(c1) || isNumber(c2)) { + return c1 == c2; + } + // a A 32 + // b B 32 + // c C 32 + return (c1 == c2) || (Math.max(c1, c2) - Math.min(c1, c2) == 32); + } + + public static boolean isLetter(char c) { + return (c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z'); + } + + public static boolean isNumber(char c) { + return (c >= '0' && c <= '9'); + } + +} diff --git a/大厂刷题班/class31/Problem_0127_WordLadder.java b/大厂刷题班/class31/Problem_0127_WordLadder.java new file mode 100644 index 0000000..6a4056f --- /dev/null +++ b/大厂刷题班/class31/Problem_0127_WordLadder.java @@ -0,0 +1,124 @@ +package class31; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.HashSet; +import java.util.LinkedList; +import java.util.List; +import java.util.Queue; + +public class Problem_0127_WordLadder { + + // start，出发的单词 + // to, 目标单位 + // list, 列表 + // to 一定属于list + // start未必 + // 返回变幻的最短路径长度 + public static int ladderLength1(String start, String to, List list) { + list.add(start); + + // key : 列表中的单词，每一个单词都会有记录！ + // value : key这个单词，有哪些邻居！ + HashMap> nexts = getNexts(list); + // abc 出发 abc -> abc 0 + // + // bbc 1 + HashMap distanceMap = new HashMap<>(); + distanceMap.put(start, 1); + HashSet set = new HashSet<>(); + set.add(start); + Queue queue = new LinkedList<>(); + queue.add(start); + while (!queue.isEmpty()) { + String cur = queue.poll(); + Integer distance = distanceMap.get(cur); + for (String next : nexts.get(cur)) { + if (next.equals(to)) { + return distance + 1; + } + if (!set.contains(next)) { + set.add(next); + queue.add(next); + distanceMap.put(next, distance + 1); + } + } + + } + return 0; + } + + public static HashMap> getNexts(List words) { + HashSet dict = new HashSet<>(words); + HashMap> nexts = new HashMap<>(); + for (int i = 0; i < words.size(); i++) { + nexts.put(words.get(i), getNext(words.get(i), dict)); + } + return nexts; + } + + // 应该根据具体数据状况决定用什么来找邻居 + // 1)如果字符串长度比较短，字符串数量比较多，以下方法适合 + // 2)如果字符串长度比较长，字符串数量比较少，以下方法不适合 + public static ArrayList getNext(String word, HashSet dict) { + ArrayList res = new ArrayList(); + char[] chs = word.toCharArray(); + for (int i = 0; i < chs.length; i++) { + for (char cur = 'a'; cur <= 'z'; cur++) { + if (chs[i] != cur) { + char tmp = chs[i]; + chs[i] = cur; + if (dict.contains(String.valueOf(chs))) { + res.add(String.valueOf(chs)); + } + chs[i] = tmp; + } + } + } + return res; + } + + public static int ladderLength2(String beginWord, String endWord, List wordList) { + HashSet dict = new HashSet<>(wordList); + if (!dict.contains(endWord)) { + return 0; + } + HashSet startSet = new HashSet<>(); + HashSet endSet = new HashSet<>(); + HashSet visit = new HashSet<>(); + startSet.add(beginWord); + endSet.add(endWord); + for (int len = 2; !startSet.isEmpty(); len++) { + // startSet是较小的，endSet是较大的 + HashSet nextSet = new HashSet<>(); + for (String w : startSet) { + // w -> a(nextSet) + // a b c + // 0 + // 1 + // 2 + for (int j = 0; j < w.length(); j++) { + char[] ch = w.toCharArray(); + for (char c = 'a'; c <= 'z'; c++) { + if (c != w.charAt(j)) { + ch[j] = c; + String next = String.valueOf(ch); + if (endSet.contains(next)) { + return len; + } + if (dict.contains(next) && !visit.contains(next)) { + nextSet.add(next); + visit.add(next); + } + } + } + } + } + // startSet(小) -> nextSet(某个大小) 和 endSet大小来比 + startSet = (nextSet.size() < endSet.size()) ? nextSet : endSet; + endSet = (startSet == nextSet) ? endSet : nextSet; + } + return 0; + } + +} diff --git a/大厂刷题班/class31/Problem_0130_SurroundedRegions.java b/大厂刷题班/class31/Problem_0130_SurroundedRegions.java new file mode 100644 index 0000000..209ae30 --- /dev/null +++ b/大厂刷题班/class31/Problem_0130_SurroundedRegions.java @@ -0,0 +1,115 @@ +package class31; + +public class Problem_0130_SurroundedRegions { + +// // m -> 二维数组， 不是0就是1 +// // +// public static void infect(int[][] m, int i, int j) { +// if (i < 0 || i == m.length || j < 0 || j == m[0].length || m[i][j] != 1) { +// return; +// } +// // m[i][j] == 1 +// m[i][j] = 2; +// infect(m, i - 1, j); +// infect(m, i + 1, j); +// infect(m, i, j - 1); +// infect(m, i, j + 1); +// } + + public static void solve1(char[][] board) { + boolean[] ans = new boolean[1]; + for (int i = 0; i < board.length; i++) { + for (int j = 0; j < board[0].length; j++) { + if (board[i][j] == 'O') { + ans[0] = true; + can(board, i, j, ans); + board[i][j] = ans[0] ? 'T' : 'F'; + } + } + } + for (int i = 0; i < board.length; i++) { + for (int j = 0; j < board[0].length; j++) { + char can = board[i][j]; + if (can == 'T' || can == 'F') { + board[i][j] = '.'; + change(board, i, j, can); + } + } + } + + } + + public static void can(char[][] board, int i, int j, boolean[] ans) { + if (i < 0 || i == board.length || j < 0 || j == board[0].length) { + ans[0] = false; + return; + } + if (board[i][j] == 'O') { + board[i][j] = '.'; + can(board, i - 1, j, ans); + can(board, i + 1, j, ans); + can(board, i, j - 1, ans); + can(board, i, j + 1, ans); + } + } + + public static void change(char[][] board, int i, int j, char can) { + if (i < 0 || i == board.length || j < 0 || j == board[0].length) { + return; + } + if (board[i][j] == '.') { + board[i][j] = can == 'T' ? 'X' : 'O'; + change(board, i - 1, j, can); + change(board, i + 1, j, can); + change(board, i, j - 1, can); + change(board, i, j + 1, can); + } + } + + // 从边界开始感染的方法，比第一种方法更好 + public static void solve2(char[][] board) { + if (board == null || board.length == 0 || board[0] == null || board[0].length == 0) { + return; + } + int N = board.length; + int M = board[0].length; + for (int j = 0; j < M; j++) { + if (board[0][j] == 'O') { + free(board, 0, j); + } + if (board[N - 1][j] == 'O') { + free(board, N - 1, j); + } + } + for (int i = 1; i < N - 1; i++) { + if (board[i][0] == 'O') { + free(board, i, 0); + } + if (board[i][M - 1] == 'O') { + free(board, i, M - 1); + } + } + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + if (board[i][j] == 'O') { + board[i][j] = 'X'; + } + if (board[i][j] == 'F') { + board[i][j] = 'O'; + } + } + } + } + + public static void free(char[][] board, int i, int j) { + if (i < 0 || i == board.length || j < 0 || j == board[0].length || board[i][j] != 'O') { + return; + } + board[i][j] = 'F'; + free(board, i + 1, j); + free(board, i - 1, j); + free(board, i, j + 1); + free(board, i, j - 1); + } + +} diff --git a/大厂刷题班/class31/Problem_0139_WordBreak.java b/大厂刷题班/class31/Problem_0139_WordBreak.java new file mode 100644 index 0000000..245e1d6 --- /dev/null +++ b/大厂刷题班/class31/Problem_0139_WordBreak.java @@ -0,0 +1,97 @@ +package class31; + +import java.util.List; + +// lintcode也有测试，数据量比leetcode大很多 : https://www.lintcode.com/problem/107/ +public class Problem_0139_WordBreak { + + public static class Node { + public boolean end; + public Node[] nexts; + + public Node() { + end = false; + nexts = new Node[26]; + } + } + + public static boolean wordBreak1(String s, List wordDict) { + Node root = new Node(); + for (String str : wordDict) { + char[] chs = str.toCharArray(); + Node node = root; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = chs[i] - 'a'; + if (node.nexts[index] == null) { + node.nexts[index] = new Node(); + } + node = node.nexts[index]; + } + node.end = true; + } + char[] str = s.toCharArray(); + int N = str.length; + boolean[] dp = new boolean[N + 1]; + dp[N] = true; // dp[i] word[i.....] 能不能被分解 + // dp[N] word[N...] -> "" 能不能够被分解 + // dp[i] ... dp[i+1....] + for (int i = N - 1; i >= 0; i--) { + // i + // word[i....] 能不能够被分解 + // i..i i+1.... + // i..i+1 i+2... + Node cur = root; + for (int end = i; end < N; end++) { + cur = cur.nexts[str[end] - 'a']; + if (cur == null) { + break; + } + // 有路！ + if (cur.end) { + // i...end 真的是一个有效的前缀串 end+1.... 能不能被分解 + dp[i] |= dp[end + 1]; + } + if (dp[i]) { + break; + } + } + } + return dp[0]; + } + + public static int wordBreak2(String s, List wordDict) { + Node root = new Node(); + for (String str : wordDict) { + char[] chs = str.toCharArray(); + Node node = root; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = chs[i] - 'a'; + if (node.nexts[index] == null) { + node.nexts[index] = new Node(); + } + node = node.nexts[index]; + } + node.end = true; + } + char[] str = s.toCharArray(); + int N = str.length; + int[] dp = new int[N + 1]; + dp[N] = 1; + for (int i = N - 1; i >= 0; i--) { + Node cur = root; + for (int end = i; end < N; end++) { + cur = cur.nexts[str[end] - 'a']; + if (cur == null) { + break; + } + if (cur.end) { + dp[i] += dp[end + 1]; + } + } + } + return dp[0]; + } + +} diff --git a/大厂刷题班/class31/Problem_0140_WordBreakII.java b/大厂刷题班/class31/Problem_0140_WordBreakII.java new file mode 100644 index 0000000..3290b88 --- /dev/null +++ b/大厂刷题班/class31/Problem_0140_WordBreakII.java @@ -0,0 +1,104 @@ +package class31; + +import java.util.ArrayList; +import java.util.List; + +public class Problem_0140_WordBreakII { + + public static class Node { + public String path; + public boolean end; + public Node[] nexts; + + public Node() { + path = null; + end = false; + nexts = new Node[26]; + } + } + + public static List wordBreak(String s, List wordDict) { + char[] str = s.toCharArray(); + Node root = gettrie(wordDict); + boolean[] dp = getdp(s, root); + ArrayList path = new ArrayList<>(); + List ans = new ArrayList<>(); + process(str, 0, root, dp, path, ans); + return ans; + } + + // str[index.....] 是要搞定的字符串 + // dp[0...N-1] 0... 1.... 2... N-1... 在dp里 + // root 单词表所有单词生成的前缀树头节点 + // path str[0..index-1]做过决定了，做的决定放在path里 + public static void process(char[] str, int index, Node root, boolean[] dp, ArrayList path, + List ans) { + if (index == str.length) { + StringBuilder builder = new StringBuilder(); + for (int i = 0; i < path.size() - 1; i++) { + builder.append(path.get(i) + " "); + } + builder.append(path.get(path.size() - 1)); + ans.add(builder.toString()); + } else { + Node cur = root; + for (int end = index; end < str.length; end++) { + // str[i..end] （能不能拆出来） + int road = str[end] - 'a'; + if (cur.nexts[road] == null) { + break; + } + cur = cur.nexts[road]; + if (cur.end && dp[end + 1]) { + // [i...end] 前缀串 + // str.subString(i,end+1) [i..end] + path.add(cur.path); + process(str, end + 1, root, dp, path, ans); + path.remove(path.size() - 1); + } + } + } + } + + public static Node gettrie(List wordDict) { + Node root = new Node(); + for (String str : wordDict) { + char[] chs = str.toCharArray(); + Node node = root; + int index = 0; + for (int i = 0; i < chs.length; i++) { + index = chs[i] - 'a'; + if (node.nexts[index] == null) { + node.nexts[index] = new Node(); + } + node = node.nexts[index]; + } + node.path = str; + node.end = true; + } + return root; + } + + public static boolean[] getdp(String s, Node root) { + char[] str = s.toCharArray(); + int N = str.length; + boolean[] dp = new boolean[N + 1]; + dp[N] = true; + for (int i = N - 1; i >= 0; i--) { + Node cur = root; + for (int end = i; end < N; end++) { + int path = str[end] - 'a'; + if (cur.nexts[path] == null) { + break; + } + cur = cur.nexts[path]; + if (cur.end && dp[end + 1]) { + dp[i] = true; + break; + } + } + } + return dp; + } + +} diff --git a/大厂刷题班/class31/Problem_0148_SortList.java b/大厂刷题班/class31/Problem_0148_SortList.java new file mode 100644 index 0000000..8b1bf26 --- /dev/null +++ b/大厂刷题班/class31/Problem_0148_SortList.java @@ -0,0 +1,227 @@ +package class31; + +public class Problem_0148_SortList { + + public static class ListNode { + int val; + ListNode next; + + public ListNode(int v) { + val = v; + } + } + + // 链表的归并排序 + // 时间复杂度O(N*logN), 因为是链表所以空间复杂度O(1) + public static ListNode sortList1(ListNode head) { + int N = 0; + ListNode cur = head; + while (cur != null) { + N++; + cur = cur.next; + } + ListNode h = head; + ListNode teamFirst = head; + ListNode pre = null; + for (int len = 1; len < N; len <<= 1) { + while (teamFirst != null) { + // 左组从哪到哪 ls le + // 右组从哪到哪 rs re + // 左 右 next + ListNode[] hthtn = hthtn(teamFirst, len); + // ls...le rs...re -> merge去 + // 整体的头、整体的尾 + ListNode[] mhmt = merge(hthtn[0], hthtn[1], hthtn[2], hthtn[3]); + if (h == teamFirst) { + h = mhmt[0]; + pre = mhmt[1]; + } else { + pre.next = mhmt[0]; + pre = mhmt[1]; + } + teamFirst = hthtn[4]; + } + teamFirst = h; + pre = null; + } + return h; + } + + public static ListNode[] hthtn(ListNode teamFirst, int len) { + ListNode ls = teamFirst; + ListNode le = teamFirst; + ListNode rs = null; + ListNode re = null; + ListNode next = null; + int pass = 0; + while (teamFirst != null) { + pass++; + if (pass <= len) { + le = teamFirst; + } + if (pass == len + 1) { + rs = teamFirst; + } + if (pass > len) { + re = teamFirst; + } + if (pass == (len << 1)) { + break; + } + teamFirst = teamFirst.next; + } + le.next = null; + if (re != null) { + next = re.next; + re.next = null; + } + return new ListNode[] { ls, le, rs, re, next }; + } + + public static ListNode[] merge(ListNode ls, ListNode le, ListNode rs, ListNode re) { + if (rs == null) { + return new ListNode[] { ls, le }; + } + ListNode head = null; + ListNode pre = null; + ListNode cur = null; + ListNode tail = null; + while (ls != le.next && rs != re.next) { + if (ls.val <= rs.val) { + cur = ls; + ls = ls.next; + } else { + cur = rs; + rs = rs.next; + } + if (pre == null) { + head = cur; + pre = cur; + } else { + pre.next = cur; + pre = cur; + } + } + if (ls != le.next) { + while (ls != le.next) { + pre.next = ls; + pre = ls; + tail = ls; + ls = ls.next; + } + } else { + while (rs != re.next) { + pre.next = rs; + pre = rs; + tail = rs; + rs = rs.next; + } + } + return new ListNode[] { head, tail }; + } + + // 链表的快速排序 + // 时间复杂度O(N*logN), 空间复杂度O(logN) + public static ListNode sortList2(ListNode head) { + int n = 0; + ListNode cur = head; + while (cur != null) { + cur = cur.next; + n++; + } + return process(head, n).head; + } + + public static class HeadAndTail { + public ListNode head; + public ListNode tail; + + public HeadAndTail(ListNode h, ListNode t) { + head = h; + tail = t; + } + } + + public static HeadAndTail process(ListNode head, int n) { + if (n == 0) { + return new HeadAndTail(head, head); + } + int index = (int) (Math.random() * n); + ListNode cur = head; + while (index-- != 0) { + cur = cur.next; + } + Record r = partition(head, cur); + HeadAndTail lht = process(r.lhead, r.lsize); + HeadAndTail rht = process(r.rhead, r.rsize); + if (lht.tail != null) { + lht.tail.next = r.mhead; + } + r.mtail.next = rht.head; + return new HeadAndTail(lht.head != null ? lht.head : r.mhead, rht.tail != null ? rht.tail : r.mtail); + } + + public static class Record { + public ListNode lhead; + public int lsize; + public ListNode rhead; + public int rsize; + public ListNode mhead; + public ListNode mtail; + + public Record(ListNode lh, int ls, ListNode rh, int rs, ListNode mh, ListNode mt) { + lhead = lh; + lsize = ls; + rhead = rh; + rsize = rs; + mhead = mh; + mtail = mt; + } + } + + public static Record partition(ListNode head, ListNode mid) { + ListNode lh = null; + ListNode lt = null; + int ls = 0; + ListNode mh = null; + ListNode mt = null; + ListNode rh = null; + ListNode rt = null; + int rs = 0; + ListNode tmp = null; + while (head != null) { + tmp = head.next; + head.next = null; + if (head.val < mid.val) { + if (lh == null) { + lh = head; + lt = head; + } else { + lt.next = head; + lt = head; + } + ls++; + } else if (head.val > mid.val) { + if (rh == null) { + rh = head; + rt = head; + } else { + rt.next = head; + rt = head; + } + rs++; + } else { + if (mh == null) { + mh = head; + mt = head; + } else { + mt.next = head; + mt = head; + } + } + head = tmp; + } + return new Record(lh, ls, rh, rs, mh, mt); + } + +} diff --git a/大厂刷题班/class31/Problem_0150_EvaluateReversePolishNotation.java b/大厂刷题班/class31/Problem_0150_EvaluateReversePolishNotation.java new file mode 100644 index 0000000..54cf55e --- /dev/null +++ b/大厂刷题班/class31/Problem_0150_EvaluateReversePolishNotation.java @@ -0,0 +1,40 @@ +package class31; + +import java.util.Stack; + +public class Problem_0150_EvaluateReversePolishNotation { + + public static int evalRPN(String[] tokens) { + Stack stack = new Stack<>(); + for (String str : tokens) { + if (str.equals("+") || str.equals("-") || str.equals("*") || str.equals("/")) { + compute(stack, str); + } else { + stack.push(Integer.valueOf(str)); + } + } + return stack.peek(); + } + + public static void compute(Stack stack, String op) { + int num2 = stack.pop(); + int num1 = stack.pop(); + int ans = 0; + switch (op) { + case "+": + ans = num1 + num2; + break; + case "-": + ans = num1 - num2; + break; + case "*": + ans = num1 * num2; + break; + case "/": + ans = num1 / num2; + break; + } + stack.push(ans); + } + +} diff --git a/大厂刷题班/class31/说明 b/大厂刷题班/class31/说明 new file mode 100644 index 0000000..a3b9f3f --- /dev/null +++ b/大厂刷题班/class31/说明 @@ -0,0 +1,21 @@ +leetcode高频题 +leetcode全题目列表 : https://leetcode.com/problemset/all/ +在全题目列表的右侧栏，点击Top Interview Questions +或者直接进入右侧链接 : https://leetcode.com/problemset/all/?listId=wpwgkgt +即可看到leetcode高频题全列表 +本节课解决leetcode高频题列表中的如下题目 : +0125 : 大厂刷题班, 第31节, 本节 +0127 : 大厂刷题班, 第31节, 本节 +0128 : 大厂刷题班, 第12节第3题 +0130 : 大厂刷题班, 第31节, 本节 +0131 : 大厂刷题班, 第11节第2题 +0134 : 体系学习班, 第24节第3题 & 大厂刷题班, 第25节第4题 +0136 : 体系学习班, 第2节第2题 +0138 : 体系学习班, 第9节第4题 +0139 : 大厂刷题班, 第31节, 本节 +0140 : 大厂刷题班, 第31节, 本节 +0141 : 体系学习班, 第10节第1题 +0146 : 大厂刷题班, 第19节第1题 +0148 : 大厂刷题班, 第31节, 本节 +0149 : 大厂刷题班, 第25节第3题 +0150 : 大厂刷题班, 第31节, 本节 \ No newline at end of file diff --git a/大厂刷题班/class32/Problem_0152_MaximumProductSubarray.java b/大厂刷题班/class32/Problem_0152_MaximumProductSubarray.java new file mode 100644 index 0000000..92e5deb --- /dev/null +++ b/大厂刷题班/class32/Problem_0152_MaximumProductSubarray.java @@ -0,0 +1,45 @@ +package class32; + +public class Problem_0152_MaximumProductSubarray { + + + public static double max(double[] arr) { + if(arr == null || arr.length == 0) { + return 0; // 报错！ + } + int n = arr.length; + // 上一步的最大 + double premax = arr[0]; + // 上一步的最小 + double premin = arr[0]; + double ans = arr[0]; + for(int i = 1; i < n; i++) { + double p1 = arr[i]; + double p2 = arr[i] * premax; + double p3 = arr[i] * premin; + double curmax = Math.max(Math.max(p1, p2), p3); + double curmin = Math.min(Math.min(p1, p2), p3); + ans = Math.max(ans, curmax); + premax = curmax; + premin = curmin; + } + return ans; + } + + + + public static int maxProduct(int[] nums) { + int ans = nums[0]; + int min = nums[0]; + int max = nums[0]; + for (int i = 1; i < nums.length; i++) { + int curmin = Math.min(nums[i], Math.min(min * nums[i], max * nums[i])); + int curmax = Math.max(nums[i], Math.max(min * nums[i], max * nums[i])); + min = curmin; + max = curmax; + ans = Math.max(ans, max); + } + return ans; + } + +} diff --git a/大厂刷题班/class32/Problem_0163_MissingRanges.java b/大厂刷题班/class32/Problem_0163_MissingRanges.java new file mode 100644 index 0000000..e6ca7fd --- /dev/null +++ b/大厂刷题班/class32/Problem_0163_MissingRanges.java @@ -0,0 +1,35 @@ +package class32; + +import java.util.ArrayList; +import java.util.List; + +public class Problem_0163_MissingRanges { + + public static List findMissingRanges(int[] nums, int lower, int upper) { + List ans = new ArrayList<>(); + for (int num : nums) { + if (num > lower) { + ans.add(miss(lower, num - 1)); + } + if (num == upper) { + return ans; + } + lower = num + 1; + } + if (lower <= upper) { + ans.add(miss(lower, upper)); + } + return ans; + } + + // 生成"lower->upper"的字符串，如果lower==upper，只用生成"lower" + public static String miss(int lower, int upper) { + String left = String.valueOf(lower); + String right = ""; + if (upper > lower) { + right = "->" + String.valueOf(upper); + } + return left + right; + } + +} diff --git a/大厂刷题班/class32/Problem_0166_FractionToRecurringDecimal.java b/大厂刷题班/class32/Problem_0166_FractionToRecurringDecimal.java new file mode 100644 index 0000000..9dbef7e --- /dev/null +++ b/大厂刷题班/class32/Problem_0166_FractionToRecurringDecimal.java @@ -0,0 +1,46 @@ +package class32; + +import java.util.HashMap; + +public class Problem_0166_FractionToRecurringDecimal { + + public static String fractionToDecimal(int numerator, int denominator) { + if (numerator == 0) { + return "0"; + } + StringBuilder res = new StringBuilder(); + // "+" or "-" + res.append(((numerator > 0) ^ (denominator > 0)) ? "-" : ""); + long num = Math.abs((long) numerator); + long den = Math.abs((long) denominator); + // integral part + res.append(num / den); + num %= den; + if (num == 0) { + return res.toString(); + } + // fractional part + res.append("."); + HashMap map = new HashMap(); + map.put(num, res.length()); + while (num != 0) { + num *= 10; + res.append(num / den); + num %= den; + if (map.containsKey(num)) { + int index = map.get(num); + res.insert(index, "("); + res.append(")"); + break; + } else { + map.put(num, res.length()); + } + } + return res.toString(); + } + + public static void main(String[] args) { + System.out.println(fractionToDecimal(127, 999)); + } + +} diff --git a/大厂刷题班/class32/Problem_0171_ExcelSheetColumnNumber.java b/大厂刷题班/class32/Problem_0171_ExcelSheetColumnNumber.java new file mode 100644 index 0000000..bed8659 --- /dev/null +++ b/大厂刷题班/class32/Problem_0171_ExcelSheetColumnNumber.java @@ -0,0 +1,15 @@ +package class32; + +public class Problem_0171_ExcelSheetColumnNumber { + + // 这道题反过来也要会写 + public static int titleToNumber(String s) { + char[] str = s.toCharArray(); + int ans = 0; + for (int i = 0; i < str.length; i++) { + ans = ans * 26 + (str[i] - 'A') + 1; + } + return ans; + } + +} diff --git a/大厂刷题班/class32/Problem_0172_FactorialTrailingZeroes.java b/大厂刷题班/class32/Problem_0172_FactorialTrailingZeroes.java new file mode 100644 index 0000000..c0ca71e --- /dev/null +++ b/大厂刷题班/class32/Problem_0172_FactorialTrailingZeroes.java @@ -0,0 +1,14 @@ +package class32; + +public class Problem_0172_FactorialTrailingZeroes { + + public static int trailingZeroes(int n) { + int ans = 0; + while (n != 0) { + n /= 5; + ans += n; + } + return ans; + } + +} diff --git a/大厂刷题班/class32/Problem_0189_RotateArray.java b/大厂刷题班/class32/Problem_0189_RotateArray.java new file mode 100644 index 0000000..4eb4480 --- /dev/null +++ b/大厂刷题班/class32/Problem_0189_RotateArray.java @@ -0,0 +1,60 @@ +package class32; + +public class Problem_0189_RotateArray { + + public void rotate1(int[] nums, int k) { + int N = nums.length; + k = k % N; + reverse(nums, 0, N - k - 1); + reverse(nums, N - k, N - 1); + reverse(nums, 0, N - 1); + } + + public static void reverse(int[] nums, int L, int R) { + while (L < R) { + int tmp = nums[L]; + nums[L++] = nums[R]; + nums[R--] = tmp; + } + } + + public static void rotate2(int[] nums, int k) { + int N = nums.length; + k = k % N; + if (k == 0) { + return; + } + int L = 0; + int R = N - 1; + int lpart = N - k; + int rpart = k; + int same = Math.min(lpart, rpart); + int diff = lpart - rpart; + exchange(nums, L, R, same); + while (diff != 0) { + if (diff > 0) { + L += same; + lpart = diff; + } else { + R -= same; + rpart = -diff; + } + same = Math.min(lpart, rpart); + diff = lpart - rpart; + exchange(nums, L, R, same); + } + } + + public static void exchange(int[] nums, int start, int end, int size) { + int i = end - size + 1; + int tmp = 0; + while (size-- != 0) { + tmp = nums[start]; + nums[start] = nums[i]; + nums[i] = tmp; + start++; + i++; + } + } + +} diff --git a/大厂刷题班/class32/Problem_0190_ReverseBits.java b/大厂刷题班/class32/Problem_0190_ReverseBits.java new file mode 100644 index 0000000..5851d87 --- /dev/null +++ b/大厂刷题班/class32/Problem_0190_ReverseBits.java @@ -0,0 +1,48 @@ +package class32; + +public class Problem_0190_ReverseBits { + + // 代码看着很魔幻吧？ + // 给个例子，假设n二进制为： + // 1011 0111 0011 1001 0011 1111 0110 1010 + // 解释一下，第一行，是把n左边16位，和n右边16位交换 + // n = (n >>> 16) | (n << 16); + // 因为 n >>> 16 就是左边16位被移动到了右侧 + // 同时 n << 16 就是右边16位被移动到了左侧 + // 又 | 在了一起，所以，n变成了 + // 0011 1111 0110 1010 1011 0111 0011 1001 + + // 第二行， + // n = ((n & 0xff00ff00) >>> 8) | ((n & 0x00ff00ff) << 8); + // (n & 0xff00ff00) + // 这一句意思是，左侧开始算0~7位，保留；8~15位，全变0；16~23位，保留；24~31位，全变0 + // 0011 1111 0000 0000 1011 0111 0000 0000 + // (n & 0xff00ff00) >>> 8 这句就是上面的值，统一向右移动8位，变成： + // 0000 0000 0011 1111 0000 0000 1011 0111 + // + // + // (n & 0x00ff00ff) + // 这一句意思是，左侧开始算0~7位，全变0；8~15位，保留；16~23位，全变0；24~31位，保留 + // 0000 0000 0110 1010 0000 0000 0011 1001 + // (n & 0x00ff00ff) << 8 这句就是上面的值，统一向左移动8位，变成： + // 0110 1010 0000 0000 0011 1001 0000 0000 + // 那么 ((n & 0xff00ff00) >>> 8) | ((n & 0x00ff00ff) << 8) + // 什么效果？就是n的0~7位和8~15位交换了，16~23位和24~31位交换了 + // 0110 1010 0011 1111 0011 1001 1011 0111 + + // 也就是说，整个过程是n的左16位，和右16位交换 + // n的左16位的内部，左8位和右8位交换；n的右16位的内部，左8位和右8位交换 + // 接下来的一行，其实是，从左边开始算，0~7位内部，左4和右4交换；8~15位，左4和右4交换；... + // 接下来的一行，其实是，从左边开始算，0~3位内部，左2和右2交换；4~7位，左2和右2交换；... + // 最后的一行，其实是，从左边开始算，0~1位内部，左1和右1交换；2~3位，左1和右1交换；... + public static int reverseBits(int n) { + // n的高16位，和n的低16位，交换 + n = (n >>> 16) | (n << 16); + n = ((n & 0xff00ff00) >>> 8) | ((n & 0x00ff00ff) << 8); + n = ((n & 0xf0f0f0f0) >>> 4) | ((n & 0x0f0f0f0f) << 4); + n = ((n & 0xcccccccc) >>> 2) | ((n & 0x33333333) << 2); + n = ((n & 0xaaaaaaaa) >>> 1) | ((n & 0x55555555) << 1); + return n; + } + +} diff --git a/大厂刷题班/class32/Problem_0191_NumberOf1Bits.java b/大厂刷题班/class32/Problem_0191_NumberOf1Bits.java new file mode 100644 index 0000000..f5973f0 --- /dev/null +++ b/大厂刷题班/class32/Problem_0191_NumberOf1Bits.java @@ -0,0 +1,26 @@ +package class32; + +public class Problem_0191_NumberOf1Bits { + + // n的二进制形式，有几个1？ + public static int hammingWeight1(int n) { + int bits = 0; + int rightOne = 0; + while(n != 0) { + bits++; + rightOne = n & (-n); + n ^= rightOne; + } + return bits; + } + + public static int hammingWeight2(int n) { + n = (n & 0x55555555) + ((n >>> 1) & 0x55555555); + n = (n & 0x33333333) + ((n >>> 2) & 0x33333333); + n = (n & 0x0f0f0f0f) + ((n >>> 4) & 0x0f0f0f0f); + n = (n & 0x00ff00ff) + ((n >>> 8) & 0x00ff00ff); + n = (n & 0x0000ffff) + ((n >>> 16) & 0x0000ffff); + return n; + } + +} diff --git a/大厂刷题班/class32/Problem_0202_HappyNumber.java b/大厂刷题班/class32/Problem_0202_HappyNumber.java new file mode 100644 index 0000000..3e54bbc --- /dev/null +++ b/大厂刷题班/class32/Problem_0202_HappyNumber.java @@ -0,0 +1,59 @@ +package class32; + +import java.util.HashSet; +import java.util.TreeSet; + +public class Problem_0202_HappyNumber { + + public static boolean isHappy1(int n) { + HashSet set = new HashSet<>(); + while (n != 1) { + int sum = 0; + while (n != 0) { + int r = n % 10; + sum += r * r; + n /= 10; + } + n = sum; + if (set.contains(n)) { + break; + } + set.add(n); + } + return n == 1; + } + + // 实验代码 + public static TreeSet sum(int n) { + TreeSet set = new TreeSet<>(); + while (!set.contains(n)) { + set.add(n); + int sum = 0; + while (n != 0) { + sum += (n % 10) * (n % 10); + n /= 10; + } + n = sum; + } + return set; + } + + public static boolean isHappy2(int n) { + while (n != 1 && n != 4) { + int sum = 0; + while (n != 0) { + sum += (n % 10) * (n % 10); + n /= 10; + } + n = sum; + } + return n == 1; + } + + public static void main(String[] args) { + for (int i = 1; i < 1000; i++) { + System.out.println(sum(i)); + } + } + +} diff --git a/大厂刷题班/class32/Problem_0204_CountPrimes.java b/大厂刷题班/class32/Problem_0204_CountPrimes.java new file mode 100644 index 0000000..0245051 --- /dev/null +++ b/大厂刷题班/class32/Problem_0204_CountPrimes.java @@ -0,0 +1,30 @@ +package class32; + +public class Problem_0204_CountPrimes { + + public static int countPrimes(int n) { + if (n < 3) { + return 0; + } + // j已经不是素数了，f[j] = true; + boolean[] f = new boolean[n]; + int count = n / 2; // 所有偶数都不要，还剩几个数 + // 跳过了1、2 3、5、7、 + for (int i = 3; i * i < n; i += 2) { + if (f[i]) { + continue; + } + // 3 -> 3 * 3 = 9 3 * 5 = 15 3 * 7 = 21 + // 7 -> 7 * 7 = 49 7 * 9 = 63 + // 13 -> 13 * 13 13 * 15 + for (int j = i * i; j < n; j += 2 * i) { + if (!f[j]) { + --count; + f[j] = true; + } + } + } + return count; + } + +} diff --git a/大厂刷题班/class32/SequenceM.java b/大厂刷题班/class32/SequenceM.java new file mode 100644 index 0000000..daa764f --- /dev/null +++ b/大厂刷题班/class32/SequenceM.java @@ -0,0 +1,162 @@ +package class32; + +import java.util.Arrays; + +// 给定一个数组arr，arr[i] = j，表示第i号试题的难度为j。给定一个非负数M +// 想出一张卷子，对于任何相邻的两道题目，前一题的难度不能超过后一题的难度+M +// 返回所有可能的卷子种数 +public class SequenceM { + + // 纯暴力方法，生成所有排列，一个一个验证 + public static int ways1(int[] arr, int m) { + if (arr == null || arr.length == 0) { + return 0; + } + return process(arr, 0, m); + } + + public static int process(int[] arr, int index, int m) { + if (index == arr.length) { + for (int i = 1; i < index; i++) { + if (arr[i - 1] > arr[i] + m) { + return 0; + } + } + return 1; + } + int ans = 0; + for (int i = index; i < arr.length; i++) { + swap(arr, index, i); + ans += process(arr, index + 1, m); + swap(arr, index, i); + } + return ans; + } + + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + + // 时间复杂度O(N * logN) + // 从左往右的动态规划 + 范围上二分 + public static int ways2(int[] arr, int m) { + if (arr == null || arr.length == 0) { + return 0; + } + Arrays.sort(arr); + int all = 1; + for (int i = 1; i < arr.length; i++) { + all = all * (num(arr, i - 1, arr[i] - m) + 1); + } + return all; + } + + // arr[0..r]上返回>=t的数有几个, 二分的方法 + // 找到 >=t 最左的位置a, 然后返回r - a + 1就是个数 + public static int num(int[] arr, int r, int t) { + int i = 0; + int j = r; + int m = 0; + int a = r + 1; + while (i <= j) { + m = (i + j) / 2; + if (arr[m] >= t) { + a = m; + j = m - 1; + } else { + i = m + 1; + } + } + return r - a + 1; + } + + // 时间复杂度O(N * logV) + // 从左往右的动态规划 + IndexTree + public static int ways3(int[] arr, int m) { + if (arr == null || arr.length == 0) { + return 0; + } + int max = Integer.MIN_VALUE; + int min = Integer.MAX_VALUE; + for (int num : arr) { + max = Math.max(max, num); + min = Math.min(min, num); + } + IndexTree itree = new IndexTree(max - min + 2); + Arrays.sort(arr); + int a = 0; + int b = 0; + int all = 1; + itree.add(arr[0] - min + 1, 1); + for (int i = 1; i < arr.length; i++) { + a = arr[i] - min + 1; + b = i - (a - m - 1 >= 1 ? itree.sum(a - m - 1) : 0); + all = all * (b + 1); + itree.add(a, 1); + } + return all; + } + + // 注意开始下标是1，不是0 + public static class IndexTree { + + private int[] tree; + private int N; + + public IndexTree(int size) { + N = size; + tree = new int[N + 1]; + } + + public int sum(int index) { + int ret = 0; + while (index > 0) { + ret += tree[index]; + index -= index & -index; + } + return ret; + } + + public void add(int index, int d) { + while (index <= N) { + tree[index] += d; + index += index & -index; + } + } + } + + // 为了测试 + public static int[] randomArray(int len, int value) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * (value + 1)); + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int N = 10; + int value = 20; + int testTimes = 1000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + int len = (int) (Math.random() * (N + 1)); + int[] arr = randomArray(len, value); + int m = (int) (Math.random() * (value + 1)); + int ans1 = ways1(arr, m); + int ans2 = ways2(arr, m); + int ans3 = ways3(arr, m); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("出错了!"); + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class32/说明 b/大厂刷题班/class32/说明 new file mode 100644 index 0000000..4f72a2d --- /dev/null +++ b/大厂刷题班/class32/说明 @@ -0,0 +1,25 @@ +leetcode高频题 +leetcode全题目列表 : https://leetcode.com/problemset/all/ +在全题目列表的右侧栏，点击Top Interview Questions +或者直接进入右侧链接 : https://leetcode.com/problemset/all/?listId=wpwgkgt +即可看到leetcode高频题全列表 +本节课解决leetcode高频题列表中的如下题目 : +0152 : 大厂刷题班, 第32节, 本节 +0155 : 体系学习班, 第3节第5题 +0160 : 体系学习班, 第10节第1题 +0162 : 体系学习班, 第1节第6题 +0163 : 大厂刷题班, 第32节, 本节 +0166 : 大厂刷题班, 第32节, 本节 +0169 : 大厂刷题班, 第23节第4题 +0171 : 大厂刷题班, 第32节, 本节 +0172 : 大厂刷题班, 第32节, 本节 +0179 : 体系学习班, 第13节第5题 +0188 : 大厂刷题班, 第15节第4题 +0189 : 大厂刷题班, 第32节, 本节 +0190 : 大厂刷题班, 第32节, 本节 +0191 : 大厂刷题班, 第32节, 本节 +0198 : 大厂刷题班, 第4节第4题 +0200 : 体系学习班, 第15节第2题、第3题 +0202 : 大厂刷题班, 第32节, 本节 +0204 : 大厂刷题班, 第32节, 本节 +补充题 SequenceM : 拼多多, 卷子合法数量问题 \ No newline at end of file diff --git a/大厂刷题班/class33/Problem_0207_CourseSchedule.java b/大厂刷题班/class33/Problem_0207_CourseSchedule.java new file mode 100644 index 0000000..282454a --- /dev/null +++ b/大厂刷题班/class33/Problem_0207_CourseSchedule.java @@ -0,0 +1,113 @@ +package class33; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.LinkedList; +import java.util.Queue; + +public class Problem_0207_CourseSchedule { + + // 一个node，就是一个课程 + // name是课程的编号 + // in是课程的入度 + public static class Course { + public int name; + public int in; + public ArrayList nexts; + + public Course(int n) { + name = n; + in = 0; + nexts = new ArrayList<>(); + } + } + + public static boolean canFinish1(int numCourses, int[][] prerequisites) { + if (prerequisites == null || prerequisites.length == 0) { + return true; + } + // 一个编号 对应 一个课的实例 + HashMap nodes = new HashMap<>(); + for (int[] arr : prerequisites) { + int to = arr[0]; + int from = arr[1]; // from -> to + if (!nodes.containsKey(to)) { + nodes.put(to, new Course(to)); + } + if (!nodes.containsKey(from)) { + nodes.put(from, new Course(from)); + } + Course t = nodes.get(to); + Course f = nodes.get(from); + f.nexts.add(t); + t.in++; + } + int needPrerequisiteNums = nodes.size(); + Queue zeroInQueue = new LinkedList<>(); + for (Course node : nodes.values()) { + if (node.in == 0) { + zeroInQueue.add(node); + } + } + int count = 0; + while (!zeroInQueue.isEmpty()) { + Course cur = zeroInQueue.poll(); + count++; + for (Course next : cur.nexts) { + if (--next.in == 0) { + zeroInQueue.add(next); + } + } + } + return count == needPrerequisiteNums; + } + + // 和方法1算法过程一样 + // 但是写法优化了 + public static boolean canFinish2(int courses, int[][] relation) { + if (relation == null || relation.length == 0) { + return true; + } + // 3 : 0 1 2 + // nexts : 0 {} + // 1 {} + // 2 {} + // 3 {0,1,2} + ArrayList> nexts = new ArrayList<>(); + for (int i = 0; i < courses; i++) { + nexts.add(new ArrayList<>()); + } + // 3 入度 1 in[3] == 1 + int[] in = new int[courses]; + for (int[] arr : relation) { + // arr[1] from arr[0] to + nexts.get(arr[1]).add(arr[0]); + in[arr[0]]++; + } + + // 队列 + int[] zero = new int[courses]; + // 该队列有效范围是[l,r) + // 新来的数，放哪？r位置，r++ + // 出队列的数，从哪拿？l位置，l++ + // l == r 队列无元素 l < r 队列有元素 + int l = 0; + int r = 0; + for (int i = 0; i < courses; i++) { + if (in[i] == 0) { + zero[r++] = i; + } + } + int count = 0; + while (l != r) { + count++; // zero[l] 出队列 l++ + for (int next : nexts.get(zero[l++])) { + if (--in[next] == 0) { + zero[r++] = next; + } + } + } + return count == nexts.size(); + } + +} diff --git a/大厂刷题班/class33/Problem_0210_CourseScheduleII.java b/大厂刷题班/class33/Problem_0210_CourseScheduleII.java new file mode 100644 index 0000000..33017c2 --- /dev/null +++ b/大厂刷题班/class33/Problem_0210_CourseScheduleII.java @@ -0,0 +1,71 @@ +package class33; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.LinkedList; +import java.util.Queue; + +public class Problem_0210_CourseScheduleII { + + public static class Node { + public int name; + public int in; + public ArrayList nexts; + + public Node(int n) { + name = n; + in = 0; + nexts = new ArrayList<>(); + } + } + + public int[] findOrder(int numCourses, int[][] prerequisites) { + int[] ans = new int[numCourses]; + for (int i = 0; i < numCourses; i++) { + ans[i] = i; + } + if (prerequisites == null || prerequisites.length == 0) { + return ans; + } + HashMap nodes = new HashMap<>(); + for (int[] arr : prerequisites) { + int to = arr[0]; + int from = arr[1]; + if (!nodes.containsKey(to)) { + nodes.put(to, new Node(to)); + } + if (!nodes.containsKey(from)) { + nodes.put(from, new Node(from)); + } + Node t = nodes.get(to); + Node f = nodes.get(from); + f.nexts.add(t); + t.in++; + } + int index = 0; + Queue zeroInQueue = new LinkedList<>(); + for (int i = 0; i < numCourses; i++) { + if (!nodes.containsKey(i)) { + ans[index++] = i; + } else { + if (nodes.get(i).in == 0) { + zeroInQueue.add(nodes.get(i)); + } + } + } + int needPrerequisiteNums = nodes.size(); + int count = 0; + while (!zeroInQueue.isEmpty()) { + Node cur = zeroInQueue.poll(); + ans[index++] = cur.name; + count++; + for (Node next : cur.nexts) { + if (--next.in == 0) { + zeroInQueue.add(next); + } + } + } + return count == needPrerequisiteNums ? ans : new int[0]; + } + +} diff --git a/大厂刷题班/class33/Problem_0213_HouseRobberII.java b/大厂刷题班/class33/Problem_0213_HouseRobberII.java new file mode 100644 index 0000000..cc8dcb2 --- /dev/null +++ b/大厂刷题班/class33/Problem_0213_HouseRobberII.java @@ -0,0 +1,50 @@ +package class33; + +public class Problem_0213_HouseRobberII { + + // arr 长度大于等于1 + public static int pickMaxSum(int[] arr) { + int n = arr.length; + // dp[i] : arr[0..i]范围上，随意选择，但是，任何两数不能相邻。得到的最大累加和是多少？ + int[] dp = new int[n]; + dp[0] = arr[0]; + dp[1] = Math.max(arr[0], arr[1]); + for (int i = 2; i < n; i++) { + int p1 = arr[i]; + int p2 = dp[i - 1]; + int p3 = arr[i] + dp[i - 2]; + dp[i] = Math.max(p1, Math.max(p2, p3)); + } + return dp[n - 1]; + } + + public static int rob(int[] nums) { + if (nums == null || nums.length == 0) { + return 0; + } + if (nums.length == 1) { + return nums[0]; + } + if (nums.length == 2) { + return Math.max(nums[0], nums[1]); + } + int pre2 = nums[0]; + int pre1 = Math.max(nums[0], nums[1]); + for (int i = 2; i < nums.length - 1; i++) { + int tmp = Math.max(pre1, nums[i] + pre2); + pre2 = pre1; + pre1 = tmp; + } + int ans1 = pre1; + pre2 = nums[1]; + pre1 = Math.max(nums[1], nums[2]); + for (int i = 3; i < nums.length; i++) { + int tmp = Math.max(pre1, nums[i] + pre2); + pre2 = pre1; + pre1 = tmp; + } + int ans2 = pre1; + return Math.max(ans1, ans2); + } + +} diff --git a/大厂刷题班/class33/Problem_0237_DeleteNodeInLinkedList.java b/大厂刷题班/class33/Problem_0237_DeleteNodeInLinkedList.java new file mode 100644 index 0000000..9aeab6c --- /dev/null +++ b/大厂刷题班/class33/Problem_0237_DeleteNodeInLinkedList.java @@ -0,0 +1,15 @@ +package class33; + +public class Problem_0237_DeleteNodeInLinkedList { + + public static class ListNode { + int val; + ListNode next; + } + + public void deleteNode(ListNode node) { + node.val = node.next.val; + node.next = node.next.next; + } + +} diff --git a/大厂刷题班/class33/Problem_0238_ProductOfArrayExceptSelf.java b/大厂刷题班/class33/Problem_0238_ProductOfArrayExceptSelf.java new file mode 100644 index 0000000..a605d46 --- /dev/null +++ b/大厂刷题班/class33/Problem_0238_ProductOfArrayExceptSelf.java @@ -0,0 +1,24 @@ +package class33; + +public class Problem_0238_ProductOfArrayExceptSelf { + + public int[] productExceptSelf(int[] nums) { + int n = nums.length; + int[] ans = new int[n]; + ans[0] = nums[0]; + for (int i = 1; i < n; i++) { + ans[i] = ans[i - 1] * nums[i]; + } + int right = 1; + for (int i = n - 1; i > 0; i--) { + ans[i] = ans[i - 1] * right; + right *= nums[i]; + } + ans[0] = right; + return ans; + } + + // 扩展 : 如果仅仅是不能用除号，把结果直接填在nums里呢？ + // 解法：数一共几个0；每一个位得到结果就是，a / b，位运算替代 /，之前的课讲过（新手班） + +} diff --git a/大厂刷题班/class33/Problem_0242_ValidAnagram.java b/大厂刷题班/class33/Problem_0242_ValidAnagram.java new file mode 100644 index 0000000..c6f9e49 --- /dev/null +++ b/大厂刷题班/class33/Problem_0242_ValidAnagram.java @@ -0,0 +1,23 @@ +package class33; + +public class Problem_0242_ValidAnagram { + + public static boolean isAnagram(String s, String t) { + if (s.length() != t.length()) { + return false; + } + char[] str1 = s.toCharArray(); + char[] str2 = t.toCharArray(); + int[] count = new int[256]; + for (char cha : str1) { + count[cha]++; + } + for (char cha : str2) { + if (--count[cha] < 0) { + return false; + } + } + return true; + } + +} diff --git a/大厂刷题班/class33/Problem_0251_Flatten2DVector.java b/大厂刷题班/class33/Problem_0251_Flatten2DVector.java new file mode 100644 index 0000000..6dd56e1 --- /dev/null +++ b/大厂刷题班/class33/Problem_0251_Flatten2DVector.java @@ -0,0 +1,53 @@ +package class33; + +public class Problem_0251_Flatten2DVector { + + public static class Vector2D { + private int[][] matrix; + private int row; + private int col; + private boolean curUse; + + public Vector2D(int[][] v) { + matrix = v; + row = 0; + col = -1; + curUse = true; + hasNext(); + } + + public int next() { + int ans = matrix[row][col]; + curUse = true; + hasNext(); + return ans; + } + + public boolean hasNext() { + if (row == matrix.length) { + return false; + } + if (!curUse) { + return true; + } + // (row，col)用过了 + if (col < matrix[row].length - 1) { + col++; + } else { + col = 0; + do { + row++; + } while (row < matrix.length && matrix[row].length == 0); + } + // 新的(row，col) + if (row != matrix.length) { + curUse = false; + return true; + } else { + return false; + } + } + + } + +} diff --git a/大厂刷题班/class33/Problem_0269_AlienDictionary.java b/大厂刷题班/class33/Problem_0269_AlienDictionary.java new file mode 100644 index 0000000..a89e295 --- /dev/null +++ b/大厂刷题班/class33/Problem_0269_AlienDictionary.java @@ -0,0 +1,65 @@ +package class33; + +import java.util.HashMap; +import java.util.HashSet; +import java.util.LinkedList; +import java.util.Queue; + +public class Problem_0269_AlienDictionary { + + public static String alienOrder(String[] words) { + if (words == null || words.length == 0) { + return ""; + } + int N = words.length; + HashMap indegree = new HashMap<>(); + for (int i = 0; i < N; i++) { + for (char c : words[i].toCharArray()) { + indegree.put(c, 0); + } + } + HashMap> graph = new HashMap<>(); + for (int i = 0; i < N - 1; i++) { + char[] cur = words[i].toCharArray(); + char[] nex = words[i + 1].toCharArray(); + int len = Math.min(cur.length, nex.length); + int j = 0; + for (; j < len; j++) { + if (cur[j] != nex[j]) { + if (!graph.containsKey(cur[j])) { + graph.put(cur[j], new HashSet<>()); + } + if (!graph.get(cur[j]).contains(nex[j])) { + graph.get(cur[j]).add(nex[j]); + indegree.put(nex[j], indegree.get(nex[j]) + 1); + } + break; + } + } + if (j < cur.length && j == nex.length) { + return ""; + } + } + StringBuilder ans = new StringBuilder(); + Queue q = new LinkedList<>(); + for (Character key : indegree.keySet()) { + if (indegree.get(key) == 0) { + q.offer(key); + } + } + while (!q.isEmpty()) { + char cur = q.poll(); + ans.append(cur); + if (graph.containsKey(cur)) { + for (char next : graph.get(cur)) { + indegree.put(next, indegree.get(next) - 1); + if (indegree.get(next) == 0) { + q.offer(next); + } + } + } + } + return ans.length() == indegree.size() ? ans.toString() : ""; + } + +} diff --git a/大厂刷题班/class33/Problem_0277_FindTheCelebrity.java b/大厂刷题班/class33/Problem_0277_FindTheCelebrity.java new file mode 100644 index 0000000..a91abc4 --- /dev/null +++ b/大厂刷题班/class33/Problem_0277_FindTheCelebrity.java @@ -0,0 +1,38 @@ +package class33; + +public class Problem_0277_FindTheCelebrity { + + // 提交时不要提交这个函数，因为默认系统会给你这个函数 + // knows方法，自己不认识自己 + public static boolean knows(int x, int i) { + return true; + } + + // 只提交下面的方法 0 ~ n-1 + public int findCelebrity(int n) { + // 谁可能成为明星，谁就是cand + int cand = 0; + for (int i = 0; i < n; ++i) { + if (knows(cand, i)) { + cand = i; + } + } + // cand是什么？唯一可能是明星的人！ + // 下一步就是验证，它到底是不是明星 + // 1) cand是不是不认识所有的人 cand...（右侧cand都不认识） + // 所以，只用验证 ....cand的左侧即可 + for (int i = 0; i < cand; ++i) { + if (knows(cand, i)) { + return -1; + } + } + // 2) 是不是所有的人都认识cand + for (int i = 0; i < n; ++i) { + if (!knows(i, cand)) { + return -1; + } + } + return cand; + } + +} diff --git a/大厂刷题班/class33/Problem_0279_PerfectSquares.java b/大厂刷题班/class33/Problem_0279_PerfectSquares.java new file mode 100644 index 0000000..645680b --- /dev/null +++ b/大厂刷题班/class33/Problem_0279_PerfectSquares.java @@ -0,0 +1,70 @@ +package class33; + +public class Problem_0279_PerfectSquares { + + // 暴力解 + public static int numSquares1(int n) { + int res = n, num = 2; + while (num * num <= n) { + int a = n / (num * num), b = n % (num * num); + res = Math.min(res, a + numSquares1(b)); + num++; + } + return res; + } + + // 1 : 1, 4, 9, 16, 25, 36, ... + // 4 : 7, 15, 23, 28, 31, 39, 47, 55, 60, 63, 71, ... + // 规律解 + // 规律一：个数不超过4 + // 规律二：出现1个的时候，显而易见 + // 规律三：任何数 % 8 == 7，一定是4个 + // 规律四：任何数消去4的因子之后，剩下rest，rest % 8 == 7，一定是4个 + public static int numSquares2(int n) { + int rest = n; + while (rest % 4 == 0) { + rest /= 4; + } + if (rest % 8 == 7) { + return 4; + } + int f = (int) Math.sqrt(n); + if (f * f == n) { + return 1; + } + for (int first = 1; first * first <= n; first++) { + int second = (int) Math.sqrt(n - first * first); + if (first * first + second * second == n) { + return 2; + } + } + return 3; + } + + // 数学解 + // 1）四平方和定理 + // 2）任何数消掉4的因子，结论不变 + public static int numSquares3(int n) { + while (n % 4 == 0) { + n /= 4; + } + if (n % 8 == 7) { + return 4; + } + for (int a = 0; a * a <= n; ++a) { + // a * a + b * b = n + int b = (int) Math.sqrt(n - a * a); + if (a * a + b * b == n) { + return (a > 0 && b > 0) ? 2 : 1; + } + } + return 3; + } + + public static void main(String[] args) { + for (int i = 1; i < 1000; i++) { + System.out.println(i + " , " + numSquares1(i)); + } + } + +} diff --git a/大厂刷题班/class33/Problem_0283_MoveZeroes.java b/大厂刷题班/class33/Problem_0283_MoveZeroes.java new file mode 100644 index 0000000..168e59d --- /dev/null +++ b/大厂刷题班/class33/Problem_0283_MoveZeroes.java @@ -0,0 +1,20 @@ +package class33; + +public class Problem_0283_MoveZeroes { + + public static void moveZeroes(int[] nums) { + int to = 0; + for (int i = 0; i < nums.length; i++) { + if (nums[i] != 0) { + swap(nums, to++, i); + } + } + } + + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + +} diff --git a/大厂刷题班/class33/说明 b/大厂刷题班/class33/说明 new file mode 100644 index 0000000..d91e8cc --- /dev/null +++ b/大厂刷题班/class33/说明 @@ -0,0 +1,32 @@ +leetcode高频题 +leetcode全题目列表 : https://leetcode.com/problemset/all/ +在全题目列表的右侧栏，点击Top Interview Questions +或者直接进入右侧链接 : https://leetcode.com/problemset/all/?listId=wpwgkgt +即可看到leetcode高频题全列表 +本节课解决leetcode高频题列表中的如下题目 : +0206 : 太简单，跳过 +0207 : 大厂刷题班, 第33节, 本节 +0208 : 体系学习班, 第8节第1题 +0210 : 大厂刷题班, 第33节, 本节 +0212 : 大厂刷题班, 第26节第1题 +0213 : 大厂刷题班, 第33节, 本节 +0215 : 体系学习班, 第29节第1题 +0217 : 太简单，跳过 +0218 : 大厂刷题班, 第4节第8题 +0227 : 大厂刷题班, 第8节第1题 +0230 : 太简单了，跳过。二叉树基本遍历、Morris遍历都可以解决 +0234 : 体系学习班, 第9节第2题 +0236 : 体系学习班, 第13节第3题 扩展在 大厂刷题班, 第23节第1题 +0237 : 大厂刷题班, 第33节, 本节 +0238 : 大厂刷题班, 第33节, 本节 +0239 : 体系学习班, 第24节第1题 +0240 : 大厂刷题班, 第17节第1题 +0242 : 大厂刷题班, 第33节, 本节 +0251 : 大厂刷题班, 第33节, 本节 +0253 : 体系学习班, 第14节第3题 +0268 : 大厂刷题班, 第14节第6题 +0269 : 大厂刷题班, 第33节, 本节 +0277 : 大厂刷题班, 第33节, 本节 +0279 : 大厂刷题班, 第33节, 本节 +0283 : 大厂刷题班, 第33节, 本节 +0285 : 太简单了，跳过。用二叉树普通遍历或者Morris遍历都可以解决 \ No newline at end of file diff --git a/大厂刷题班/class34/Problem_0287_FindTheDuplicateNumber.java b/大厂刷题班/class34/Problem_0287_FindTheDuplicateNumber.java new file mode 100644 index 0000000..dcd64b5 --- /dev/null +++ b/大厂刷题班/class34/Problem_0287_FindTheDuplicateNumber.java @@ -0,0 +1,23 @@ +package class34; + +public class Problem_0287_FindTheDuplicateNumber { + + public static int findDuplicate(int[] nums) { + if (nums == null || nums.length < 2) { + return -1; + } + int slow = nums[0]; + int fast = nums[nums[0]]; + while (slow != fast) { + slow = nums[slow]; + fast = nums[nums[fast]]; + } + fast = 0; + while (slow != fast) { + fast = nums[fast]; + slow = nums[slow]; + } + return slow; + } + +} diff --git a/大厂刷题班/class34/Problem_0289_GameOfLife.java b/大厂刷题班/class34/Problem_0289_GameOfLife.java new file mode 100644 index 0000000..c986ef0 --- /dev/null +++ b/大厂刷题班/class34/Problem_0289_GameOfLife.java @@ -0,0 +1,43 @@ +package class34; + +// 有关这个游戏更有意思、更完整的内容： +// https://www.bilibili.com/video/BV1rJ411n7ri +// 也推荐这个up主 +public class Problem_0289_GameOfLife { + + public static void gameOfLife(int[][] board) { + int N = board.length; + int M = board[0].length; + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + int neighbors = neighbors(board, i, j); + if (neighbors == 3 || (board[i][j] == 1 && neighbors == 2)) { + board[i][j] |= 2; + } + } + } + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + board[i][j] >>= 1; + } + } + } + + // b[i][j] 这个位置的数，周围有几个1 + public static int neighbors(int[][] b, int i, int j) { + return f(b, i - 1, j - 1) + + f(b, i - 1, j) + + f(b, i - 1, j + 1) + + f(b, i, j - 1) + + f(b, i, j + 1) + + f(b, i + 1, j - 1) + + f(b, i + 1, j) + + f(b, i + 1, j + 1); + } + + // b[i][j] 上面有1，就返回1，上面不是1，就返回0 + public static int f(int[][] b, int i, int j) { + return (i >= 0 && i < b.length && j >= 0 && j < b[0].length && (b[i][j] & 1) == 1) ? 1 : 0; + } + +} diff --git a/大厂刷题班/class34/Problem_0295_FindMedianFromDataStream.java b/大厂刷题班/class34/Problem_0295_FindMedianFromDataStream.java new file mode 100644 index 0000000..4fa279f --- /dev/null +++ b/大厂刷题班/class34/Problem_0295_FindMedianFromDataStream.java @@ -0,0 +1,46 @@ +package class34; + +import java.util.PriorityQueue; + +public class Problem_0295_FindMedianFromDataStream { + + class MedianFinder { + + private PriorityQueue maxh; + private PriorityQueue minh; + + public MedianFinder() { + maxh = new PriorityQueue<>((a, b) -> b - a); + minh = new PriorityQueue<>((a, b) -> a - b); + } + + public void addNum(int num) { + if (maxh.isEmpty() || maxh.peek() >= num) { + maxh.add(num); + } else { + minh.add(num); + } + balance(); + } + + public double findMedian() { + if (maxh.size() == minh.size()) { + return (double) (maxh.peek() + minh.peek()) / 2; + } else { + return maxh.size() > minh.size() ? maxh.peek() : minh.peek(); + } + } + + private void balance() { + if (Math.abs(maxh.size() - minh.size()) == 2) { + if (maxh.size() > minh.size()) { + minh.add(maxh.poll()); + } else { + maxh.add(minh.poll()); + } + } + } + + } + +} \ No newline at end of file diff --git a/大厂刷题班/class34/Problem_0315_CountOfSmallerNumbersAfterSelf.java b/大厂刷题班/class34/Problem_0315_CountOfSmallerNumbersAfterSelf.java new file mode 100644 index 0000000..7cd8821 --- /dev/null +++ b/大厂刷题班/class34/Problem_0315_CountOfSmallerNumbersAfterSelf.java @@ -0,0 +1,69 @@ +package class34; + +import java.util.ArrayList; +import java.util.List; + +public class Problem_0315_CountOfSmallerNumbersAfterSelf { + + public static class Node { + public int value; + public int index; + + public Node(int v, int i) { + value = v; + index = i; + } + } + + public static List countSmaller(int[] nums) { + List ans = new ArrayList<>(); + if (nums == null) { + return ans; + } + for (int i = 0; i < nums.length; i++) { + ans.add(0); + } + if (nums.length < 2) { + return ans; + } + Node[] arr = new Node[nums.length]; + for (int i = 0; i < arr.length; i++) { + arr[i] = new Node(nums[i], i); + } + process(arr, 0, arr.length - 1, ans); + return ans; + } + + public static void process(Node[] arr, int l, int r, List ans) { + if (l == r) { + return; + } + int mid = l + ((r - l) >> 1); + process(arr, l, mid, ans); + process(arr, mid + 1, r, ans); + merge(arr, l, mid, r, ans); + } + + public static void merge(Node[] arr, int l, int m, int r, List ans) { + Node[] help = new Node[r - l + 1]; + int i = help.length - 1; + int p1 = m; + int p2 = r; + while (p1 >= l && p2 >= m + 1) { + if (arr[p1].value > arr[p2].value) { + ans.set(arr[p1].index, ans.get(arr[p1].index) + p2 - m); + } + help[i--] = arr[p1].value > arr[p2].value ? arr[p1--] : arr[p2--]; + } + while (p1 >= l) { + help[i--] = arr[p1--]; + } + while (p2 >= m + 1) { + help[i--] = arr[p2--]; + } + for (i = 0; i < help.length; i++) { + arr[l + i] = help[i]; + } + } + +} diff --git a/大厂刷题班/class34/Problem_0324_WiggleSortII.java b/大厂刷题班/class34/Problem_0324_WiggleSortII.java new file mode 100644 index 0000000..bd3d929 --- /dev/null +++ b/大厂刷题班/class34/Problem_0324_WiggleSortII.java @@ -0,0 +1,186 @@ +package class34; + +public class Problem_0324_WiggleSortII { + + // 时间复杂度O(N)，额外空间复杂度O(1) + public static void wiggleSort(int[] nums) { + if (nums == null || nums.length < 2) { + return; + } + int N = nums.length; + // 小 中 右 + findIndexNum(nums, 0, nums.length - 1, N / 2); + if ((N & 1) == 0) { + // R L -> L R + shuffle(nums, 0, nums.length - 1); + // R1 L1 R2 L2 R3 L3 R4 L4 + // L4 R4 L3 R3 L2 R2 L1 R1 -> 代码中的方式，可以的！ + // L1 R1 L2 R2 L3 R3 L4 R4 -> 课上的分析，是不行的！不能两两交换！ + reverse(nums, 0, nums.length - 1); + // 做个实验，如果把上一行的code注释掉(reverse过程)，然后跑下面注释掉的for循环代码 + // for循环的代码就是两两交换，会发现对数器报错，说明两两交换是不行的, 必须整体逆序 +// for (int i = 0; i < nums.length; i += 2) { +// swap(nums, i, i + 1); +// } + } else { + shuffle(nums, 1, nums.length - 1); + } + } + + public static int findIndexNum(int[] arr, int L, int R, int index) { + int pivot = 0; + int[] range = null; + while (L < R) { + pivot = arr[L + (int) (Math.random() * (R - L + 1))]; + range = partition(arr, L, R, pivot); + if (index >= range[0] && index <= range[1]) { + return arr[index]; + } else if (index < range[0]) { + R = range[0] - 1; + } else { + L = range[1] + 1; + } + } + return arr[L]; + } + + public static int[] partition(int[] arr, int L, int R, int pivot) { + int less = L - 1; + int more = R + 1; + int cur = L; + while (cur < more) { + if (arr[cur] < pivot) { + swap(arr, ++less, cur++); + } else if (arr[cur] > pivot) { + swap(arr, cur, --more); + } else { + cur++; + } + } + return new int[] { less + 1, more - 1 }; + } + + public static void shuffle(int[] nums, int l, int r) { + while (r - l + 1 > 0) { + int lenAndOne = r - l + 2; + int bloom = 3; + int k = 1; + while (bloom <= lenAndOne / 3) { + bloom *= 3; + k++; + } + int m = (bloom - 1) / 2; + int mid = (l + r) / 2; + rotate(nums, l + m, mid, mid + m); + cycles(nums, l - 1, bloom, k); + l = l + bloom - 1; + } + } + + public static void cycles(int[] nums, int base, int bloom, int k) { + for (int i = 0, trigger = 1; i < k; i++, trigger *= 3) { + int next = (2 * trigger) % bloom; + int cur = next; + int record = nums[next + base]; + int tmp = 0; + nums[next + base] = nums[trigger + base]; + while (cur != trigger) { + next = (2 * cur) % bloom; + tmp = nums[next + base]; + nums[next + base] = record; + cur = next; + record = tmp; + } + } + } + + public static void rotate(int[] arr, int l, int m, int r) { + reverse(arr, l, m); + reverse(arr, m + 1, r); + reverse(arr, l, r); + } + + public static void reverse(int[] arr, int l, int r) { + while (l < r) { + swap(arr, l++, r--); + } + } + + public static void swap(int[] nums, int i, int j) { + int tmp = nums[i]; + nums[i] = nums[j]; + nums[j] = tmp; + } + + // 为了测试，暴力方法 + // 把arr全排列尝试一遍， + // 其中任何一个排列能做到 < > < > ... 返回true; + // 如果没有任何一个排列能做到，返回false; + public static boolean test(int[] arr) { + return process(arr, 0); + } + + // 为了测试 + public static boolean process(int[] arr, int index) { + if (index == arr.length) { + return valid(arr); + } + for (int i = index; i < arr.length; i++) { + swap(arr, index, i); + if (process(arr, index + 1)) { + return true; + } + swap(arr, index, i); + } + return false; + } + + // 为了测试 + public static boolean valid(int[] arr) { + boolean more = true; + for (int i = 1; i < arr.length; i++) { + if ((more && arr[i - 1] >= arr[i]) || (!more && arr[i - 1] <= arr[i])) { + return false; + } + more = !more; + } + return true; + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] ans = new int[n]; + for (int i = 0; i < n; i++) { + ans[i] = (int) (Math.random() * v); + } + return ans; + } + + // 为了测试 + public static int[] copyArray(int[] arr) { + int[] ans = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + ans[i] = arr[i]; + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int N = 10; + int V = 10; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * N) + 1; + int[] arr1 = randomArray(n, V); + int[] arr2 = copyArray(arr1); + wiggleSort(arr1); + if (valid(arr1) != test(arr2)) { + System.out.println("出错了!"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class34/Problem_0326_PowerOfThree.java b/大厂刷题班/class34/Problem_0326_PowerOfThree.java new file mode 100644 index 0000000..48b59c1 --- /dev/null +++ b/大厂刷题班/class34/Problem_0326_PowerOfThree.java @@ -0,0 +1,14 @@ +package class34; + +public class Problem_0326_PowerOfThree { + + // 如果一个数字是3的某次幂，那么这个数一定只含有3这个质数因子 + // 1162261467是int型范围内，最大的3的幂，它是3的19次方 + // 这个1162261467只含有3这个质数因子，如果n也是只含有3这个质数因子，那么 + // 1162261467 % n == 0 + // 反之如果1162261467 % n != 0 说明n一定含有其他因子 + public static boolean isPowerOfThree(int n) { + return (n > 0 && 1162261467 % n == 0); + } + +} diff --git a/大厂刷题班/class34/Problem_0328_OddEvenLinkedList.java b/大厂刷题班/class34/Problem_0328_OddEvenLinkedList.java new file mode 100644 index 0000000..0718c20 --- /dev/null +++ b/大厂刷题班/class34/Problem_0328_OddEvenLinkedList.java @@ -0,0 +1,43 @@ +package class34; + +public class Problem_0328_OddEvenLinkedList { + + // 提交时不要提交这个类 + public static class ListNode { + int val; + ListNode next; + } + + public ListNode oddEvenList(ListNode head) { + ListNode firstOdd = null; + ListNode firstEven = null; + ListNode odd = null; + ListNode even = null; + ListNode next = null; + int count = 1; + while (head != null) { + next = head.next; + head.next = null; + if ((count & 1) == 1) { + firstOdd = firstOdd == null ? head : firstOdd; + if (odd != null) { + odd.next = head; + } + odd = head; + } else { + firstEven = firstEven == null ? head : firstEven; + if (even != null) { + even.next = head; + } + even = head; + } + count++; + head = next; + } + if (odd != null) { + odd.next = firstEven; + } + return firstOdd != null ? firstOdd : firstEven; + } + +} diff --git a/大厂刷题班/class34/Problem_0340_LongestSubstringWithAtMostKDistinctCharacters.java b/大厂刷题班/class34/Problem_0340_LongestSubstringWithAtMostKDistinctCharacters.java new file mode 100644 index 0000000..37124f5 --- /dev/null +++ b/大厂刷题班/class34/Problem_0340_LongestSubstringWithAtMostKDistinctCharacters.java @@ -0,0 +1,29 @@ +package class34; + +public class Problem_0340_LongestSubstringWithAtMostKDistinctCharacters { + + public static int lengthOfLongestSubstringKDistinct(String s, int k) { + if (s == null || s.length() == 0 || k < 1) { + return 0; + } + char[] str = s.toCharArray(); + int N = str.length; + int[] count = new int[256]; + int diff = 0; + int R = 0; + int ans = 0; + for (int i = 0; i < N; i++) { + // R 窗口的右边界 + while (R < N && (diff < k || (diff == k && count[str[R]] > 0))) { + diff += count[str[R]] == 0 ? 1 : 0; + count[str[R++]]++; + } + // R 来到违规的第一个位置 + ans = Math.max(ans, R - i); + diff -= count[str[i]] == 1 ? 1 : 0; + count[str[i]]--; + } + return ans; + } + +} diff --git a/大厂刷题班/class34/Problem_0341_FlattenNestedListIterator.java b/大厂刷题班/class34/Problem_0341_FlattenNestedListIterator.java new file mode 100644 index 0000000..a5bc724 --- /dev/null +++ b/大厂刷题班/class34/Problem_0341_FlattenNestedListIterator.java @@ -0,0 +1,110 @@ +package class34; + +import java.util.Iterator; +import java.util.List; +import java.util.Stack; + +public class Problem_0341_FlattenNestedListIterator { + + // 不要提交这个接口类 + public interface NestedInteger { + + // @return true if this NestedInteger holds a single integer, rather than a + // nested list. + public boolean isInteger(); + + // @return the single integer that this NestedInteger holds, if it holds a + // single integer + // Return null if this NestedInteger holds a nested list + public Integer getInteger(); + + // @return the nested list that this NestedInteger holds, if it holds a nested + // list + // Return null if this NestedInteger holds a single integer + public List getList(); + } + + public class NestedIterator implements Iterator { + + private List list; + private Stack stack; + private boolean used; + + public NestedIterator(List nestedList) { + list = nestedList; + stack = new Stack<>(); + stack.push(-1); + used = true; + hasNext(); + } + + @Override + public Integer next() { + Integer ans = null; + if (!used) { + ans = get(list, stack); + used = true; + hasNext(); + } + return ans; + } + + @Override + public boolean hasNext() { + if (stack.isEmpty()) { + return false; + } + if (!used) { + return true; + } + if (findNext(list, stack)) { + used = false; + } + return !used; + } + + private Integer get(List nestedList, Stack stack) { + int index = stack.pop(); + Integer ans = null; + if (!stack.isEmpty()) { + ans = get(nestedList.get(index).getList(), stack); + } else { + ans = nestedList.get(index).getInteger(); + } + stack.push(index); + return ans; + } + + private boolean findNext(List nestedList, Stack stack) { + int index = stack.pop(); + if (!stack.isEmpty() && findNext(nestedList.get(index).getList(), stack)) { + stack.push(index); + return true; + } + for (int i = index + 1; i < nestedList.size(); i++) { + if (pickFirst(nestedList.get(i), i, stack)) { + return true; + } + } + return false; + } + + private boolean pickFirst(NestedInteger nested, int position, Stack stack) { + if (nested.isInteger()) { + stack.add(position); + return true; + } else { + List actualList = nested.getList(); + for (int i = 0; i < actualList.size(); i++) { + if (pickFirst(actualList.get(i), i, stack)) { + stack.add(position); + return true; + } + } + } + return false; + } + + } + +} diff --git a/大厂刷题班/class34/Problem_0348_DesignTicTacToe.java b/大厂刷题班/class34/Problem_0348_DesignTicTacToe.java new file mode 100644 index 0000000..58f4851 --- /dev/null +++ b/大厂刷题班/class34/Problem_0348_DesignTicTacToe.java @@ -0,0 +1,47 @@ +package class34; + +public class Problem_0348_DesignTicTacToe { + + class TicTacToe { + private int[][] rows; + private int[][] cols; + private int[] leftUp; + private int[] rightUp; + private boolean[][] matrix; + private int N; + + public TicTacToe(int n) { + // rows[a][1] : 1这个人，在a行上，下了几个 + // rows[b][2] : 2这个人，在b行上，下了几个 + rows = new int[n][3]; //0 1 2 + cols = new int[n][3]; + // leftUp[2] = 7 : 2这个人，在左对角线上，下了7个 + leftUp = new int[3]; + // rightUp[1] = 9 : 1这个人，在右对角线上，下了9个 + rightUp = new int[3]; + matrix = new boolean[n][n]; + N = n; + } + + public int move(int row, int col, int player) { + if (matrix[row][col]) { + return 0; + } + matrix[row][col] = true; + rows[row][player]++; + cols[col][player]++; + if (row == col) { + leftUp[player]++; + } + if (row + col == N - 1) { + rightUp[player]++; + } + if (rows[row][player] == N || cols[col][player] == N || leftUp[player] == N || rightUp[player] == N) { + return player; + } + return 0; + } + + } + +} diff --git a/大厂刷题班/class34/Problem_0380_InsertDeleteGetRandom.java b/大厂刷题班/class34/Problem_0380_InsertDeleteGetRandom.java new file mode 100644 index 0000000..2a81aca --- /dev/null +++ b/大厂刷题班/class34/Problem_0380_InsertDeleteGetRandom.java @@ -0,0 +1,52 @@ +package class34; + +import java.util.HashMap; + +// 测试链接 : https://leetcode.com/problems/insert-delete-getrandom-o1/ +public class Problem_0380_InsertDeleteGetRandom { + + public class RandomizedSet { + + private HashMap keyIndexMap; + private HashMap indexKeyMap; + private int size; + + public RandomizedSet() { + keyIndexMap = new HashMap(); + indexKeyMap = new HashMap(); + size = 0; + } + + public boolean insert(int val) { + if (!keyIndexMap.containsKey(val)) { + keyIndexMap.put(val, size); + indexKeyMap.put(size++, val); + return true; + } + return false; + } + + public boolean remove(int val) { + if (keyIndexMap.containsKey(val)) { + int deleteIndex = keyIndexMap.get(val); + int lastIndex = --size; + int lastKey = indexKeyMap.get(lastIndex); + keyIndexMap.put(lastKey, deleteIndex); + indexKeyMap.put(deleteIndex, lastKey); + keyIndexMap.remove(val); + indexKeyMap.remove(lastIndex); + return true; + } + return false; + } + + public int getRandom() { + if (size == 0) { + return -1; + } + int randomIndex = (int) (Math.random() * size); + return indexKeyMap.get(randomIndex); + } + } + +} diff --git a/大厂刷题班/class34/Problem_0384_ShuffleAnArray.java b/大厂刷题班/class34/Problem_0384_ShuffleAnArray.java new file mode 100644 index 0000000..dfb0e0b --- /dev/null +++ b/大厂刷题班/class34/Problem_0384_ShuffleAnArray.java @@ -0,0 +1,34 @@ +package class34; + +public class Problem_0384_ShuffleAnArray { + + class Solution { + private int[] origin; + private int[] shuffle; + private int N; + + public Solution(int[] nums) { + origin = nums; + N = nums.length; + shuffle = new int[N]; + for (int i = 0; i < N; i++) { + shuffle[i] = origin[i]; + } + } + + public int[] reset() { + return origin; + } + + public int[] shuffle() { + for (int i = N - 1; i >= 0; i--) { + int r = (int) (Math.random() * (i + 1)); + int tmp = shuffle[r]; + shuffle[r] = shuffle[i]; + shuffle[i] = tmp; + } + return shuffle; + } + } + +} diff --git a/大厂刷题班/class34/说明 b/大厂刷题班/class34/说明 new file mode 100644 index 0000000..c23f435 --- /dev/null +++ b/大厂刷题班/class34/说明 @@ -0,0 +1,30 @@ +leetcode高频题 +leetcode全题目列表 : https://leetcode.com/problemset/all/ +在全题目列表的右侧栏，点击Top Interview Questions +或者直接进入右侧链接 : https://leetcode.com/problemset/all/?listId=wpwgkgt +即可看到leetcode高频题全列表 +本节课解决leetcode高频题列表中的如下题目 : +0287 : 大厂刷题班, 第34节, 本节 +0289 : 大厂刷题班, 第34节, 本节 +0295 : 大厂刷题班, 第34节, 本节 +0297 : 体系学习班, 第11节第2题 +0300 : 大厂刷题班, 第9节第3题 +0308 : 体系学习班, 第32节第2题 +0309 : 大厂刷题班, 第15节第5题 +0315 : 大厂刷题班, 第34节, 本节 +0322 : 体系学习班, 硬币找零专题 : 第21节第2、3、4题, 第22节第2题, 第24节第4题 +0324 : 大厂刷题班, 第34节, 本节 +0326 : 大厂刷题班, 第34节, 本节 +0328 : 大厂刷题班, 第34节, 本节 +0329 : 大厂刷题班, 第1节第5题 +0334 : 大厂刷题班, 第9节第3题的变形, 问 : 最长递增子序列长度能否超过2而已, 跳过 +0340 : 大厂刷题班, 第34节, 本节 +0341 : 大厂刷题班, 第34节, 本节 +0344 : 太简单了, 跳过 +0348 : 大厂刷题班, 第34节, 本节 +0350 : 太简单了, 跳过 +0371 : 新手班, 第5节第3题 +0378 : 大厂刷题班, 第17节第2题 +0380 : 大厂刷题班, 第34节, 本节 +0384 : 大厂刷题班, 第34节, 本节 +0387 : 太简单了, 跳过 \ No newline at end of file diff --git a/大厂刷题班/class35/Code01_StringKth.java b/大厂刷题班/class35/Code01_StringKth.java new file mode 100644 index 0000000..9ed97dc --- /dev/null +++ b/大厂刷题班/class35/Code01_StringKth.java @@ -0,0 +1,96 @@ +package class35; + +import java.util.ArrayList; +import java.util.List; + +// 给定一个长度len，表示一共有几位 +// 所有字符都是小写(a~z)，可以生成长度为1，长度为2， +// 长度为3...长度为len的所有字符串 +// 如果把所有字符串根据字典序排序，每个字符串都有所在的位置。 +// 给定一个字符串str，给定len，请返回str是总序列中的第几个 +// 比如len = 4，字典序的前几个字符串为: +// a aa aaa aaaa aaab ... aaaz ... azzz b ba baa baaa ... bzzz c ... +// a是这个序列中的第1个，bzzz是这个序列中的第36558个 +public class Code01_StringKth { + // 思路： + // cdb，总共长度为7，请问cdb是第几个？ + // 第一位c : + // 以a开头，剩下长度为(0~6)的所有可能性有几个 + // + + // 以b开头，剩下长度为(0~6)的所有可能性有几个 + // + + // 以c开头，剩下长度为(0)的所有可能性有几个 + // 第二位d : + // + + // 以ca开头的情况下，剩下长度为(0~5)的所有可能性有几个 + // + + // 以cb开头的情况下，剩下长度为(0~5)的所有可能性有几个 + // + + // 以cc开头的情况下，剩下长度为(0~5)的所有可能性有几个 + // + + // 以cd开头的情况下，剩下长度为(0)的所有可能性有几个 + // 第三位b + // + + // 以cda开头的情况下，剩下长度为(0~4)的所有可能性有几个 + // + + // 以cdb开头的情况下，剩下长度为(0)的所有可能性有几个 + public static int kth(String s, int len) { + if (s == null || s.length() == 0 || s.length() > len) { + return -1; + } + char[] num = s.toCharArray(); + int ans = 0; + for (int i = 0, rest = len - 1; i < num.length; i++, rest--) { + ans += (num[i] - 'a') * f(rest) + 1; + } + return ans; + } + + // 不管以什么开头，剩下长度为(0~len)的所有可能性有几个 + public static int f(int len) { + int ans = 1; + for (int i = 1, base = 26; i <= len; i++, base *= 26) { + ans += base; + } + return ans; + } + + // 为了测试 + public static List all(int len) { + List ans = new ArrayList<>(); + for (int i = 1; i <= len; i++) { + char[] path = new char[i]; + process(path, 0, ans); + } + return ans; + } + + // 为了测试 + public static void process(char[] path, int index, List ans) { + if (index == path.length) { + ans.add(String.valueOf(path)); + } else { + for (char c = 'a'; c <= 'z'; c++) { + path[index] = c; + process(path, index + 1, ans); + } + } + } + + public static void main(String[] args) { + int len = 4; + // 暴力方法得到所有字符串 + List ans = all(len); + // 根据字典序排序，所有字符串都在其中 + ans.sort((a, b) -> a.compareTo(b)); + + String test = "bzzz"; + // 根据我们的方法算出test是第几个？ + // 注意我们算出的第几个，是从1开始的 + // 而下标是从0开始的，所以变成index，还需要-1 + int index = kth(test, len) - 1; + // 验证 + System.out.println(ans.get(index)); + } + +} diff --git a/大厂刷题班/class35/Code02_MagicStone.java b/大厂刷题班/class35/Code02_MagicStone.java new file mode 100644 index 0000000..9e77c80 --- /dev/null +++ b/大厂刷题班/class35/Code02_MagicStone.java @@ -0,0 +1,49 @@ +package class35; + +import java.util.Arrays; + +// 来自小红书 +// [0,4,7] ： 0表示这里石头没有颜色，如果变红代价是4，如果变蓝代价是7 +// [1,X,X] ： 1表示这里石头已经是红，而且不能改颜色，所以后两个数X无意义 +// [2,X,X] ： 2表示这里石头已经是蓝，而且不能改颜色，所以后两个数X无意义 +// 颜色只可能是0、1、2，代价一定>=0 +// 给你一批这样的小数组，要求最后必须所有石头都有颜色，且红色和蓝色一样多，返回最小代价 +// 如果怎么都无法做到所有石头都有颜色、且红色和蓝色一样多，返回-1 +public class Code02_MagicStone { + + public static int minCost(int[][] stones) { + int n = stones.length; + if ((n & 1) != 0) { + return -1; + } + Arrays.sort(stones, (a, b) -> a[0] == 0 && b[0] == 0 ? (b[1] - b[2] - a[1] + a[2]) : (a[0] - b[0])); + int zero = 0; + int red = 0; + int blue = 0; + int cost = 0; + for (int i = 0; i < n; i++) { + if (stones[i][0] == 0) { + zero++; + cost += stones[i][1]; + } else if (stones[i][0] == 1) { + red++; + } else { + blue++; + } + } + if (red > (n >> 1) || blue > (n >> 1)) { + return -1; + } + blue = zero - ((n >> 1) - red); + for (int i = 0; i < blue; i++) { + cost += stones[i][2] - stones[i][1]; + } + return cost; + } + + public static void main(String[] args) { + int[][] stones = { { 1, 5, 3 }, { 2, 7, 9 }, { 0, 6, 4 }, { 0, 7, 9 }, { 0, 2, 1 }, { 0, 5, 9 } }; + System.out.println(minCost(stones)); + } + +} diff --git a/大厂刷题班/class35/Code03_WatchMovieMaxTime.java b/大厂刷题班/class35/Code03_WatchMovieMaxTime.java new file mode 100644 index 0000000..137845c --- /dev/null +++ b/大厂刷题班/class35/Code03_WatchMovieMaxTime.java @@ -0,0 +1,136 @@ +package class35; + +import java.util.Arrays; + +// 来自小红书 +// 一场电影开始和结束时间可以用一个小数组来表示["07:30","12:00"] +// 已知有2000场电影开始和结束都在同一天，这一天从00:00开始到23:59结束 +// 一定要选3场完全不冲突的电影来观看，返回最大的观影时间 +// 如果无法选出3场完全不冲突的电影来观看，返回-1 +public class Code03_WatchMovieMaxTime { + + // 暴力方法，枚举前三场所有的可能全排列 + public static int maxEnjoy1(int[][] movies) { + if (movies.length < 3) { + return -1; + } + return process1(movies, 0); + } + + public static int process1(int[][] movies, int index) { + if (index == 3) { + int start = 0; + int watch = 0; + for (int i = 0; i < 3; i++) { + if (start > movies[i][0]) { + return -1; + } + watch += movies[i][1] - movies[i][0]; + start = movies[i][1]; + } + return watch; + } else { + int ans = -1; + for (int i = index; i < movies.length; i++) { + swap(movies, index, i); + ans = Math.max(ans, process1(movies, index + 1)); + swap(movies, index, i); + } + return ans; + } + } + + public static void swap(int[][] movies, int i, int j) { + int[] tmp = movies[i]; + movies[i] = movies[j]; + movies[j] = tmp; + } + + // 优化后的递归解 + public static int maxEnjoy2(int[][] movies) { + Arrays.sort(movies, (a, b) -> a[0] != b[0] ? (a[0] - b[0]) : (a[1] - b[1])); + return process2(movies, 0, 0, 3); + } + + public static int process2(int[][] movies, int index, int time, int rest) { + if (index == movies.length) { + return rest == 0 ? 0 : -1; + } + int p1 = process2(movies, index + 1, time, rest); + int next = movies[index][0] >= time && rest > 0 ? process2(movies, index + 1, movies[index][1], rest - 1) : -1; + int p2 = next != -1 ? (movies[index][1] - movies[index][0] + next) : -1; + return Math.max(p1, p2); + } + + // 记忆化搜索的动态规划 + public static int maxEnjoy3(int[][] movies) { + Arrays.sort(movies, (a, b) -> a[0] != b[0] ? (a[0] - b[0]) : (a[1] - b[1])); + + int max = 0; + for (int[] movie : movies) { + max = Math.max(max, movie[1]); + } + + int[][][] dp = new int[movies.length][max + 1][4]; + for (int i = 0; i < movies.length; i++) { + for (int j = 0; j <= max; j++) { + for (int k = 0; k <= 3; k++) { + dp[i][j][k] = -2;// 用-2代表没算过这个过程 + } + } + } + return process3(movies, 0, 0, 3, dp); + } + + public static int process3(int[][] movies, int index, int time, int rest, int[][][] dp) { + if (index == movies.length) { + return rest == 0 ? 0 : -1; + } + if (dp[index][time][rest] != -2) { + return dp[index][time][rest]; + } + int p1 = process3(movies, index + 1, time, rest, dp); + int next = movies[index][0] >= time && rest > 0 ? process3(movies, index + 1, movies[index][1], rest - 1, dp) + : -1; + int p2 = next != -1 ? (movies[index][1] - movies[index][0] + next) : -1; + int ans = Math.max(p1, p2); + dp[index][time][rest] = ans; + return ans; + } + + // 为了测试 + public static int[][] randomMovies(int len, int time) { + int[][] movies = new int[len][2]; + for (int i = 0; i < len; i++) { + int a = (int) (Math.random() * time); + int b = (int) (Math.random() * time); + movies[i][0] = Math.min(a, b); + movies[i][1] = Math.max(a, b); + } + return movies; + } + + public static void main(String[] args) { + int n = 10; + int t = 20; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * n) + 1; + int[][] movies = randomMovies(len, t); + int ans1 = maxEnjoy1(movies); + int ans2 = maxEnjoy2(movies); + int ans3 = maxEnjoy3(movies); + if (ans1 != ans2 || ans1 != ans3) { + for (int[] m : movies) { + System.out.println(m[0] + " , " + m[1]); + } + System.out.println(ans1); + System.out.println(ans2); + System.out.println("出错了"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class35/Code04_WalkToEnd.java b/大厂刷题班/class35/Code04_WalkToEnd.java new file mode 100644 index 0000000..50ba235 --- /dev/null +++ b/大厂刷题班/class35/Code04_WalkToEnd.java @@ -0,0 +1,53 @@ +package class35; + +import java.util.PriorityQueue; + +// 来自网易 +// map[i][j] == 0，代表(i,j)是海洋，渡过的话代价是2 +// map[i][j] == 1，代表(i,j)是陆地，渡过的话代价是1 +// map[i][j] == 2，代表(i,j)是障碍，无法渡过 +// 每一步上、下、左、右都能走，返回从左上角走到右下角最小代价是多少，如果无法到达返回-1 +public class Code04_WalkToEnd { + + public static int minCost(int[][] map) { + if (map[0][0] == 2) { + return -1; + } + int n = map.length; + int m = map[0].length; + PriorityQueue heap = new PriorityQueue<>((a, b) -> a.cost - b.cost); + boolean[][] visited = new boolean[n][m]; + add(map, 0, 0, 0, heap, visited); + while (!heap.isEmpty()) { + Node cur = heap.poll(); + if (cur.row == n - 1 && cur.col == m - 1) { + return cur.cost; + } + add(map, cur.row - 1, cur.col, cur.cost, heap, visited); + add(map, cur.row + 1, cur.col, cur.cost, heap, visited); + add(map, cur.row, cur.col - 1, cur.cost, heap, visited); + add(map, cur.row, cur.col + 1, cur.cost, heap, visited); + } + return -1; + } + + public static void add(int[][] m, int i, int j, int pre, PriorityQueue heap, boolean[][] visited) { + if (i >= 0 && i < m.length && j >= 0 && j < m[0].length && m[i][j] != 2 && !visited[i][j]) { + heap.add(new Node(i, j, pre + (m[i][j] == 0 ? 2 : 1))); + visited[i][j] = true; + } + } + + public static class Node { + public int row; + public int col; + public int cost; + + public Node(int a, int b, int c) { + row = a; + col = b; + cost = c; + } + } + +} diff --git a/大厂刷题班/class35/Code05_CircleCandy.java b/大厂刷题班/class35/Code05_CircleCandy.java new file mode 100644 index 0000000..e027a13 --- /dev/null +++ b/大厂刷题班/class35/Code05_CircleCandy.java @@ -0,0 +1,58 @@ +package class35; + +// 来自网易 +// 给定一个正数数组arr，表示每个小朋友的得分 +// 任何两个相邻的小朋友，如果得分一样，怎么分糖果无所谓，但如果得分不一样，分数大的一定要比分数少的多拿一些糖果 +// 假设所有的小朋友坐成一个环形，返回在不破坏上一条规则的情况下，需要的最少糖果数 +public class Code05_CircleCandy { + + public static int minCandy(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + if (arr.length == 1) { + return 1; + } + int n = arr.length; + int minIndex = 0; + for (int i = 0; i < n; i++) { + if (arr[i] <= arr[lastIndex(i, n)] && arr[i] <= arr[nextIndex(i, n)]) { + minIndex = i; + break; + } + } + int[] nums = new int[n + 1]; + for (int i = 0; i <= n; i++, minIndex = nextIndex(minIndex, n)) { + nums[i] = arr[minIndex]; + } + int[] left = new int[n + 1]; + left[0] = 1; + for (int i = 1; i <= n; i++) { + left[i] = nums[i] > nums[i - 1] ? (left[i - 1] + 1) : 1; + } + int[] right = new int[n + 1]; + right[n] = 1; + for (int i = n - 1; i >= 0; i--) { + right[i] = nums[i] > nums[i + 1] ? (right[i + 1] + 1) : 1; + } + int ans = 0; + for (int i = 0; i < n; i++) { + ans += Math.max(left[i], right[i]); + } + return ans; + } + + public static int nextIndex(int i, int n) { + return i == n - 1 ? 0 : (i + 1); + } + + public static int lastIndex(int i, int n) { + return i == 0 ? (n - 1) : (i - 1); + } + + public static void main(String[] args) { + int[] arr = { 3, 4, 2, 3, 2 }; + System.out.println(minCandy(arr)); + } + +} diff --git a/大厂刷题班/class35/Problem_0347_TopKFrequentElements.java b/大厂刷题班/class35/Problem_0347_TopKFrequentElements.java new file mode 100644 index 0000000..656e584 --- /dev/null +++ b/大厂刷题班/class35/Problem_0347_TopKFrequentElements.java @@ -0,0 +1,54 @@ +package class35; + +import java.util.Comparator; +import java.util.HashMap; +import java.util.PriorityQueue; + +public class Problem_0347_TopKFrequentElements { + + public static class Node { + public int num; + public int count; + + public Node(int k) { + num = k; + count = 1; + } + } + + public static class CountComparator implements Comparator { + + @Override + public int compare(Node o1, Node o2) { + return o1.count - o2.count; + } + + } + + public static int[] topKFrequent(int[] nums, int k) { + HashMap map = new HashMap<>(); + for (int num : nums) { + if (!map.containsKey(num)) { + map.put(num, new Node(num)); + } else { + map.get(num).count++; + } + } + PriorityQueue heap = new PriorityQueue<>(new CountComparator()); + for (Node node : map.values()) { + if (heap.size() < k || (heap.size() == k && node.count > heap.peek().count)) { + heap.add(node); + } + if (heap.size() > k) { + heap.poll(); + } + } + int[] ans = new int[k]; + int index = 0; + while (!heap.isEmpty()) { + ans[index++] = heap.poll().num; + } + return ans; + } + +} diff --git a/大厂刷题班/class35/Problem_0395_LongestSubstringWithAtLeastKRepeatingCharacters.java b/大厂刷题班/class35/Problem_0395_LongestSubstringWithAtLeastKRepeatingCharacters.java new file mode 100644 index 0000000..9298e38 --- /dev/null +++ b/大厂刷题班/class35/Problem_0395_LongestSubstringWithAtLeastKRepeatingCharacters.java @@ -0,0 +1,110 @@ +package class35; + +public class Problem_0395_LongestSubstringWithAtLeastKRepeatingCharacters { + + public static int longestSubstring1(String s, int k) { + char[] str = s.toCharArray(); + int N = str.length; + int max = 0; + for (int i = 0; i < N; i++) { + int[] count = new int[256]; + int collect = 0; + int satisfy = 0; + for (int j = i; j < N; j++) { + if (count[str[j]] == 0) { + collect++; + } + if (count[str[j]] == k - 1) { + satisfy++; + } + count[str[j]]++; + if (collect == satisfy) { + max = Math.max(max, j - i + 1); + } + } + } + return max; + } + + public static int longestSubstring2(String s, int k) { + char[] str = s.toCharArray(); + int N = str.length; + int max = 0; + for (int require = 1; require <= 26; require++) { + // 3种 + // a~z 出现次数 + int[] count = new int[26]; + // 目前窗口内收集了几种字符了 + int collect = 0; + // 目前窗口内出现次数>=k次的字符，满足了几种 + int satisfy = 0; + // 窗口右边界 + int R = -1; + for (int L = 0; L < N; L++) { // L要尝试每一个窗口的最左位置 + // [L..R] R+1 + while (R + 1 < N && !(collect == require && count[str[R + 1] - 'a'] == 0)) { + R++; + if (count[str[R] - 'a'] == 0) { + collect++; + } + if (count[str[R] - 'a'] == k - 1) { + satisfy++; + } + count[str[R] - 'a']++; + } + // [L...R] + if (satisfy == require) { + max = Math.max(max, R - L + 1); + } + // L++ + if (count[str[L] - 'a'] == 1) { + collect--; + } + if (count[str[L] - 'a'] == k) { + satisfy--; + } + count[str[L] - 'a']--; + } + } + return max; + } + + // 会超时，但是思路的确是正确的 + public static int longestSubstring3(String s, int k) { + return process(s.toCharArray(), 0, s.length() - 1, k); + } + + public static int process(char[] str, int L, int R, int k) { + if (L > R) { + return 0; + } + int[] counts = new int[26]; + for (int i = L; i <= R; i++) { + counts[str[i] - 'a']++; + } + char few = 0; + int min = Integer.MAX_VALUE; + for (int i = 0; i < 26; i++) { + if (counts[i] != 0 && min > counts[i]) { + few = (char) (i + 'a'); + min = counts[i]; + } + } + if (min >= k) { + return R - L + 1; + } + int pre = 0; + int max = Integer.MIN_VALUE; + for (int i = L; i <= R; i++) { + if (str[i] == few) { + max = Math.max(max, process(str, pre, i - 1, k)); + pre = i + 1; + } + } + if (pre != R + 1) { + max = Math.max(max, process(str, pre, R, k)); + } + return max; + } + +} diff --git a/大厂刷题班/class35/Problem_0412_FizzBuzz.java b/大厂刷题班/class35/Problem_0412_FizzBuzz.java new file mode 100644 index 0000000..82268d9 --- /dev/null +++ b/大厂刷题班/class35/Problem_0412_FizzBuzz.java @@ -0,0 +1,24 @@ +package class35; + +import java.util.ArrayList; +import java.util.List; + +public class Problem_0412_FizzBuzz { + + public static List fizzBuzz(int n) { + ArrayList ans = new ArrayList<>(); + for (int i = 1; i <= n; i++) { + if (i % 15 == 0) { + ans.add("FizzBuzz"); + } else if (i % 5 == 0) { + ans.add("Buzz"); + } else if (i % 3 == 0) { + ans.add("Fizz"); + } else { + ans.add(String.valueOf(i)); + } + } + return ans; + } + +} diff --git a/大厂刷题班/class35/Problem_0454_4SumII.java b/大厂刷题班/class35/Problem_0454_4SumII.java new file mode 100644 index 0000000..809fdfa --- /dev/null +++ b/大厂刷题班/class35/Problem_0454_4SumII.java @@ -0,0 +1,32 @@ +package class35; + +import java.util.HashMap; + +public class Problem_0454_4SumII { + + public static int fourSumCount(int[] A, int[] B, int[] C, int[] D) { + HashMap map = new HashMap<>(); + int sum = 0; + for (int i = 0; i < A.length; i++) { + for (int j = 0; j < B.length; j++) { + sum = A[i] + B[j]; + if (!map.containsKey(sum)) { + map.put(sum, 1); + } else { + map.put(sum, map.get(sum) + 1); + } + } + } + int ans = 0; + for (int i = 0; i < C.length; i++) { + for (int j = 0; j < D.length; j++) { + sum = C[i] + D[j]; + if (map.containsKey(-sum)) { + ans += map.get(-sum); + } + } + } + return ans; + } + +} diff --git a/大厂刷题班/class35/Problem_0673_NumberOfLongestIncreasingSubsequence.java b/大厂刷题班/class35/Problem_0673_NumberOfLongestIncreasingSubsequence.java new file mode 100644 index 0000000..72d76b2 --- /dev/null +++ b/大厂刷题班/class35/Problem_0673_NumberOfLongestIncreasingSubsequence.java @@ -0,0 +1,90 @@ +package class35; + +import java.util.ArrayList; +import java.util.TreeMap; + +public class Problem_0673_NumberOfLongestIncreasingSubsequence { + + // 好理解的方法，时间复杂度O(N^2) + public static int findNumberOfLIS1(int[] nums) { + if (nums == null || nums.length == 0) { + return 0; + } + int n = nums.length; + int[] lens = new int[n]; + int[] cnts = new int[n]; + lens[0] = 1; + cnts[0] = 1; + int maxLen = 1; + int allCnt = 1; + for (int i = 1; i < n; i++) { + int preLen = 0; + int preCnt = 1; + for (int j = 0; j < i; j++) { + if (nums[j] >= nums[i] || preLen > lens[j]) { + continue; + } + if (preLen < lens[j]) { + preLen = lens[j]; + preCnt = cnts[j]; + } else { + preCnt += cnts[j]; + } + } + lens[i] = preLen + 1; + cnts[i] = preCnt; + if (maxLen < lens[i]) { + maxLen = lens[i]; + allCnt = cnts[i]; + } else if (maxLen == lens[i]) { + allCnt += cnts[i]; + } + } + return allCnt; + } + + // 优化后的最优解，时间复杂度O(N*logN) + public static int findNumberOfLIS2(int[] nums) { + if (nums == null || nums.length == 0) { + return 0; + } + ArrayList> dp = new ArrayList<>(); + int len = 0; + int cnt = 0; + for (int num : nums) { + // num之前的长度，num到哪个长度len+1 + len = search(dp, num); + // cnt : 最终要去加底下的记录，才是应该填入的value + if (len == 0) { + cnt = 1; + } else { + TreeMap p = dp.get(len - 1); + cnt = p.firstEntry().getValue() - (p.ceilingKey(num) != null ? p.get(p.ceilingKey(num)) : 0); + } + if (len == dp.size()) { + dp.add(new TreeMap()); + dp.get(len).put(num, cnt); + } else { + dp.get(len).put(num, dp.get(len).firstEntry().getValue() + cnt); + } + } + return dp.get(dp.size() - 1).firstEntry().getValue(); + } + + // 二分查找，返回>=num最左的位置 + public static int search(ArrayList> dp, int num) { + int l = 0, r = dp.size() - 1, m = 0; + int ans = dp.size(); + while (l <= r) { + m = (l + r) / 2; + if (dp.get(m).firstKey() >= num) { + ans = m; + r = m - 1; + } else { + l = m + 1; + } + } + return ans; + } + +} diff --git a/大厂刷题班/class35/Problem_0687_LongestUnivaluePath.java b/大厂刷题班/class35/Problem_0687_LongestUnivaluePath.java new file mode 100644 index 0000000..c4466ed --- /dev/null +++ b/大厂刷题班/class35/Problem_0687_LongestUnivaluePath.java @@ -0,0 +1,62 @@ +package class35; + +public class Problem_0687_LongestUnivaluePath { + + public static class TreeNode { + public int val; + public TreeNode left; + public TreeNode right; + + public TreeNode(int v) { + val = v; + } + } + + public static int longestUnivaluePath(TreeNode root) { + if (root == null) { + return 0; + } + return process(root).max - 1; + } + + // 建设以x节点为头的树，返回两个信息 + public static class Info { + // 在一条路径上：要求每个节点通过且只通过一遍 + public int len; // 路径必须从x出发且只能往下走的情况下，路径的最大距离 + public int max; // 路径不要求必须从x出发的情况下，整棵树的合法路径最大距离 + + public Info(int l, int m) { + len = l; + max = m; + } + } + + private static Info process(TreeNode x) { + if (x == null) { + return new Info(0, 0); + } + TreeNode l = x.left; + TreeNode r = x.right; + // 左树上，不要求从左孩子出发，最大路径 + // 左树上，必须从左孩子出发，往下的最大路径 + Info linfo = process(l); + // 右树上，不要求从右孩子出发，最大路径 + // 右树上，必须从右孩子出发，往下的最大路径 + Info rinfo = process(r); + // 必须从x出发的情况下，往下的最大路径 + int len = 1; + if (l != null && l.val == x.val) { + len = linfo.len + 1; + } + if (r != null && r.val == x.val) { + len = Math.max(len, rinfo.len + 1); + } + // 不要求从x出发，最大路径 + int max = Math.max(Math.max(linfo.max, rinfo.max), len); + if (l != null && r != null && l.val == x.val && r.val == x.val) { + max = Math.max(max, linfo.len + rinfo.len + 1); + } + return new Info(len, max); + } + +} diff --git a/大厂刷题班/class35/说明 b/大厂刷题班/class35/说明 new file mode 100644 index 0000000..0c4a832 --- /dev/null +++ b/大厂刷题班/class35/说明 @@ -0,0 +1,21 @@ +leetcode高频题 +leetcode全题目列表 : https://leetcode.com/problemset/all/ +在全题目列表的右侧栏，点击Top Interview Questions +或者直接进入右侧链接 : https://leetcode.com/problemset/all/?listId=wpwgkgt +即可看到leetcode高频题全列表 +本节课解决leetcode高频题列表中的如下题目 : +0347 : 大厂刷题班, 第35节, 本节 +0395 : 大厂刷题班, 第35节, 本节 +0412 : 大厂刷题班, 第35节, 本节 +0454 : 大厂刷题班, 第35节, 本节 +0673 : 大厂刷题班, 第35节, 本节 +0687 : 大厂刷题班, 第35节, 本节 +0772 : 大厂刷题班, 第8节第1题 +至此，Leetcode高频题系列完结 + +本节附加题 +Code01 : 2021年8月大厂真实笔试题 +Code02 : 2021年8月大厂真实笔试题 +Code03 : 2021年8月大厂真实笔试题 +Code04 : 2021年8月大厂真实笔试题 +Code05 : 2021年8月大厂真实笔试题 \ No newline at end of file diff --git a/大厂刷题班/class36/Code01_ReverseInvertString.java b/大厂刷题班/class36/Code01_ReverseInvertString.java new file mode 100644 index 0000000..395b64a --- /dev/null +++ b/大厂刷题班/class36/Code01_ReverseInvertString.java @@ -0,0 +1,108 @@ +package class36; + +// 来自网易 +// 规定：L[1]对应a，L[2]对应b，L[3]对应c，...，L[25]对应y +// S1 = a +// S(i) = S(i-1) + L[i] + reverse(invert(S(i-1))); +// 解释invert操作： +// S1 = a +// S2 = aby +// 假设invert(S(2)) = 甲乙丙 +// a + 甲 = 26, 那么 甲 = 26 - 1 = 25 -> y +// b + 乙 = 26, 那么 乙 = 26 - 2 = 24 -> x +// y + 丙 = 26, 那么 丙 = 26 - 25 = 1 -> a +// 如上就是每一位的计算方式，所以invert(S2) = yxa +// 所以S3 = S2 + L[3] + reverse(invert(S2)) = aby + c + axy = abycaxy +// invert(abycaxy) = yxawyba, 再reverse = abywaxy +// 所以S4 = abycaxy + d + abywaxy = abycaxydabywaxy +// 直到S25结束 +// 给定两个参数n和k，返回Sn的第k位是什么字符，n从1开始，k从1开始 +// 比如n=4，k=2，表示S4的第2个字符是什么，返回b字符 +public class Code01_ReverseInvertString { + + public static int[] lens = null; + + public static void fillLens() { + lens = new int[26]; + lens[1] = 1; + for (int i = 2; i <= 25; i++) { + lens[i] = (lens[i - 1] << 1) + 1; + } + } + + // 求sn中的第k个字符 + // O(n), s <= 25 O(1) + public static char kth(int n, int k) { + if (lens == null) { + fillLens(); + } + if (n == 1) { // 无视k + return 'a'; + } + // sn half + int half = lens[n - 1]; + if (k <= half) { + return kth(n - 1, k); + } else if (k == half + 1) { + return (char) ('a' + n - 1); + } else { + // sn + // 我需要右半区，从左往右的第a个 + // 需要找到，s(n-1)从右往左的第a个 + // 当拿到字符之后，invert一下，就可以返回了！ + return invert(kth(n - 1, ((half + 1) << 1) - k)); + } + } + + public static char invert(char c) { + return (char) (('a' << 1) + 24 - c); + } + + // 为了测试 + public static String generateString(int n) { + String s = "a"; + for (int i = 2; i <= n; i++) { + s = s + (char) ('a' + i - 1) + reverseInvert(s); + } + return s; + } + + // 为了测试 + public static String reverseInvert(String s) { + char[] invert = invert(s).toCharArray(); + for (int l = 0, r = invert.length - 1; l < r; l++, r--) { + char tmp = invert[l]; + invert[l] = invert[r]; + invert[r] = tmp; + } + return String.valueOf(invert); + } + + // 为了测试 + public static String invert(String s) { + char[] str = s.toCharArray(); + for (int i = 0; i < str.length; i++) { + str[i] = invert(str[i]); + } + return String.valueOf(str); + } + + // 为了测试 + public static void main(String[] args) { + int n = 20; + String str = generateString(n); + int len = str.length(); + System.out.println("测试开始"); + for (int i = 1; i <= len; i++) { + if (str.charAt(i - 1) != kth(n, i)) { + System.out.println(i); + System.out.println(str.charAt(i - 1)); + System.out.println(kth(n, i)); + System.out.println("出错了！"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class36/Code02_Ratio01Split.java b/大厂刷题班/class36/Code02_Ratio01Split.java new file mode 100644 index 0000000..e06b928 --- /dev/null +++ b/大厂刷题班/class36/Code02_Ratio01Split.java @@ -0,0 +1,68 @@ +package class36; + +import java.util.HashMap; + +// 来自京东 +// 把一个01字符串切成多个部分，要求每一部分的0和1比例一样，同时要求尽可能多的划分 +// 比如 : 01010101 +// 01 01 01 01 这是一种切法，0和1比例为 1 : 1 +// 0101 0101 也是一种切法，0和1比例为 1 : 1 +// 两种切法都符合要求，但是那么尽可能多的划分为第一种切法，部分数为4 +// 比如 : 00001111 +// 只有一种切法就是00001111整体作为一块，那么尽可能多的划分，部分数为1 +// 给定一个01字符串str，假设长度为N，要求返回一个长度为N的数组ans +// 其中ans[i] = str[0...i]这个前缀串，要求每一部分的0和1比例一样，同时要求尽可能多的划分下，部分数是多少 +// 输入: str = "010100001" +// 输出: ans = [1, 1, 1, 2, 1, 2, 1, 1, 3] +public class Code02_Ratio01Split { + + // 001010010100... + public static int[] split(int[] arr) { + + // key : 分子 + // value : 属于key的分母表, 每一个分母，及其 分子/分母 这个比例，多少个前缀拥有 + HashMap> pre = new HashMap<>(); + int n = arr.length; + int[] ans = new int[n]; + int zero = 0; // 0出现的次数 + int one = 0; // 1出现的次数 + for (int i = 0; i < n; i++) { + if (arr[i] == 0) { + zero++; + } else { + one++; + } + if (zero == 0 || one == 0) { + ans[i] = i + 1; + } else { // 0和1，都有数量 -> 最简分数 + int gcd = gcd(zero, one); + int a = zero / gcd; + int b = one / gcd; + // a / b 比例，之前有多少前缀拥有？ 3+1 4 5+1 6 + if (!pre.containsKey(a)) { + pre.put(a, new HashMap<>()); + } + if (!pre.get(a).containsKey(b)) { + pre.get(a).put(b, 1); + } else { + pre.get(a).put(b, pre.get(a).get(b) + 1); + } + ans[i] = pre.get(a).get(b); + } + } + return ans; + } + + public static int gcd(int m, int n) { + return n == 0 ? m : gcd(n, m % n); + } + + public static void main(String[] args) { + int[] arr = { 0, 1, 0, 1, 0, 1, 1, 0 }; + int[] ans = split(arr); + for (int i = 0; i < ans.length; i++) { + System.out.print(ans[i] + " "); + } + } + +} diff --git a/大厂刷题班/class36/Code03_MatchCount.java b/大厂刷题班/class36/Code03_MatchCount.java new file mode 100644 index 0000000..b7746c0 --- /dev/null +++ b/大厂刷题班/class36/Code03_MatchCount.java @@ -0,0 +1,66 @@ +package class36; + +// 来自美团 +// 给定两个字符串s1和s2 +// 返回在s1中有多少个子串等于s2 +public class Code03_MatchCount { + + public static int sa(String s1, String s2) { + if (s1 == null || s2 == null || s1.length() < s2.length()) { + return 0; + } + char[] str1 = s1.toCharArray(); + char[] str2 = s2.toCharArray(); + return count(str1, str2); + } + + // 改写kmp为这道题需要的功能 + public static int count(char[] str1, char[] str2) { + int x = 0; + int y = 0; + int count = 0; + int[] next = getNextArray(str2); + while (x < str1.length) { + if (str1[x] == str2[y]) { + x++; + y++; + if (y == str2.length) { + count++; + y = next[y]; + } + } else if (next[y] == -1) { + x++; + } else { + y = next[y]; + } + } + return count; + } + + // next数组多求一位 + // 比如：str2 = aaaa + // 那么，next = -1,0,1,2,3 + // 最后一个3表示，终止位置之前的字符串最长前缀和最长后缀的匹配长度 + // 也就是next数组补一位 + public static int[] getNextArray(char[] str2) { + if (str2.length == 1) { + return new int[] { -1, 0 }; + } + int[] next = new int[str2.length + 1]; + next[0] = -1; + next[1] = 0; + int i = 2; + int cn = 0; + while (i < next.length) { + if (str2[i - 1] == str2[cn]) { + next[i++] = ++cn; + } else if (cn > 0) { + cn = next[cn]; + } else { + next[i++] = 0; + } + } + return next; + } + +} diff --git a/大厂刷题班/class36/Code04_ComputeExpressionValue.java b/大厂刷题班/class36/Code04_ComputeExpressionValue.java new file mode 100644 index 0000000..5e8c7dd --- /dev/null +++ b/大厂刷题班/class36/Code04_ComputeExpressionValue.java @@ -0,0 +1,58 @@ +package class36; + +// 来自美团 +// () 分值为2 +// (()) 分值为3 +// ((())) 分值为4 +// 也就是说，每包裹一层，分数就是里面的分值+1 +// ()() 分值为2 * 2 +// (())() 分值为3 * 2 +// 也就是说，每连接一段，分数就是各部分相乘，以下是一个结合起来的例子 +// (()())()(()) -> (2 * 2 + 1) * 2 * 3 -> 30 +// 给定一个括号字符串str，已知str一定是正确的括号结合，不会有违规嵌套 +// 返回分数 +public class Code04_ComputeExpressionValue { + + public static int sores(String s) { + return compute(s.toCharArray(), 0)[0]; + } + + // s[i.....] 遇到 ')' 或者 终止位置 停！ + // 返回值：int[] 长度就是2 + // 0 ：分数是多少 + // 1 : 来到了什么位置停的！ + public static int[] compute(char[] s, int i) { + if (s[i] == ')') { + return new int[] { 1, i }; + } + int ans = 1; + while (i < s.length && s[i] != ')') { + int[] info = compute(s, i + 1); + ans *= info[0] + 1; + i = info[1] + 1; + } + return new int[] { ans, i }; + } + + public static void main(String[] args) { + + String str1 = "(()())()(())"; + System.out.println(sores(str1)); + + // (()()) + (((()))) + ((())()) + // (()()) -> 2 * 2 + 1 -> 5 + // (((()))) -> 5 + // ((())()) -> ((2 + 1) * 2) + 1 -> 7 + // 所以下面的结果应该是175 + String str2 = "(()())(((())))((())())"; + System.out.println(sores(str2)); + + // (()()()) + (()(())) + // (()()()) -> 2 * 2 * 2 + 1 -> 9 + // (()(())) -> 2 * 3 + 1 -> 7 + // 所以下面的结果应该是63 + String str3 = "(()()())(()(()))"; + System.out.println(sores(str3)); + } + +} diff --git a/大厂刷题班/class36/Code05_Query3Problems.java b/大厂刷题班/class36/Code05_Query3Problems.java new file mode 100644 index 0000000..68c68f5 --- /dev/null +++ b/大厂刷题班/class36/Code05_Query3Problems.java @@ -0,0 +1,132 @@ +package class36; + +// 来自美团 +// 给定一个数组arr，长度为N，做出一个结构，可以高效的做如下的查询 +// 1) int querySum(L,R) : 查询arr[L...R]上的累加和 +// 2) int queryAim(L,R) : 查询arr[L...R]上的目标值，目标值定义如下： +// 假设arr[L...R]上的值为[a,b,c,d]，a+b+c+d = s +// 目标值为 : (s-a)^2 + (s-b)^2 + (s-c)^2 + (s-d)^2 +// 3) int queryMax(L,R) : 查询arr[L...R]上的最大值 +// 要求： +// 1) 初始化该结构的时间复杂度不能超过O(N*logN) +// 2) 三个查询的时间复杂度不能超过O(logN) +// 3) 查询时，认为arr的下标从1开始，比如 : +// arr = [ 1, 1, 2, 3 ]; +// querySum(1, 3) -> 4 +// queryAim(2, 4) -> 50 +// queryMax(1, 4) -> 3 +public class Code05_Query3Problems { + + public static class SegmentTree { + private int[] max; + private int[] change; + private boolean[] update; + + public SegmentTree(int N) { + max = new int[N << 2]; + change = new int[N << 2]; + update = new boolean[N << 2]; + for (int i = 0; i < max.length; i++) { + max[i] = Integer.MIN_VALUE; + } + } + + private void pushUp(int rt) { + max[rt] = Math.max(max[rt << 1], max[rt << 1 | 1]); + } + + // ln表示左子树元素结点个数，rn表示右子树结点个数 + private void pushDown(int rt, int ln, int rn) { + if (update[rt]) { + update[rt << 1] = true; + update[rt << 1 | 1] = true; + change[rt << 1] = change[rt]; + change[rt << 1 | 1] = change[rt]; + max[rt << 1] = change[rt]; + max[rt << 1 | 1] = change[rt]; + update[rt] = false; + } + } + + public void update(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + update[rt] = true; + change[rt] = C; + max[rt] = C; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + update(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + update(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + public int query(int L, int R, int l, int r, int rt) { + if (L <= l && r <= R) { + return max[rt]; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + int left = 0; + int right = 0; + if (L <= mid) { + left = query(L, R, l, mid, rt << 1); + } + if (R > mid) { + right = query(L, R, mid + 1, r, rt << 1 | 1); + } + return Math.max(left, right); + } + + } + + public static class Query { + public int[] sum1; + public int[] sum2; + public SegmentTree st; + public int m; + + public Query(int[] arr) { + int n = arr.length; + m = arr.length + 1; + sum1 = new int[m]; + sum2 = new int[m]; + st = new SegmentTree(m); + for (int i = 0; i < n; i++) { + sum1[i + 1] = sum1[i] + arr[i]; + sum2[i + 1] = sum2[i] + arr[i] * arr[i]; + st.update(i + 1, i + 1, arr[i], 1, m, 1); + } + + } + + public int querySum(int L, int R) { + return sum1[R] - sum1[L - 1]; + } + + public int queryAim(int L, int R) { + int sumPower2 = querySum(L, R); + sumPower2 *= sumPower2; + return sum2[R] - sum2[L - 1] + (R - L - 1) * sumPower2; + } + + public int queryMax(int L, int R) { + return st.query(L, R, 1, m, 1); + } + + } + + public static void main(String[] args) { + int[] arr = { 1, 1, 2, 3 }; + Query q = new Query(arr); + System.out.println(q.querySum(1, 3)); + System.out.println(q.queryAim(2, 4)); + System.out.println(q.queryMax(1, 4)); + } + +} diff --git a/大厂刷题班/class36/Code06_NodeWeight.java b/大厂刷题班/class36/Code06_NodeWeight.java new file mode 100644 index 0000000..22c429e --- /dev/null +++ b/大厂刷题班/class36/Code06_NodeWeight.java @@ -0,0 +1,48 @@ +package class36; + +import java.util.HashMap; + +// 来自美团 +// 有一棵树，给定头节点h，和结构数组m，下标0弃而不用 +// 比如h = 1, m = [ [] , [2,3], [4], [5,6], [], [], []] +// 表示1的孩子是2、3; 2的孩子是4; 3的孩子是5、6; 4、5和6是叶节点，都不再有孩子 +// 每一个节点都有颜色，记录在c数组里，比如c[i] = 4, 表示节点i的颜色为4 +// 一开始只有叶节点是有权值的，记录在w数组里， +// 比如，如果一开始就有w[i] = 3, 表示节点i是叶节点、且权值是3 +// 现在规定非叶节点i的权值计算方式： +// 根据i的所有直接孩子来计算，假设i的所有直接孩子，颜色只有a,b,k +// w[i] = Max { +// (颜色为a的所有孩子个数 + 颜色为a的孩子权值之和), +// (颜色为b的所有孩子个数 + 颜色为b的孩子权值之和), +// (颜色为k的所有孩子个数 + 颜色k的孩子权值之和) +// } +// 请计算所有孩子的权值并返回 +public class Code06_NodeWeight { + + // 当前来到h节点， + // h的直接孩子，在哪呢？m[h] = {a,b,c,d,e} + // 每个节点的颜色在哪？比如i号节点，c[i]就是i号节点的颜色 + // 每个节点的权值在哪？比如i号节点，w[i]就是i号节点的权值 + // void : 把w数组填满就是这个函数的目标 + public static void w(int h, int[][] m, int[] w, int[] c) { + if (m[h].length == 0) { // 叶节点 + return; + } + // 有若干个直接孩子 + // 1 7个 + // 3 10个 + HashMap colors = new HashMap(); + // 1 20 + // 3 45 + HashMap weihts = new HashMap(); + for (int child : m[h]) { + w(child, m, w, c); + colors.put(c[child], colors.getOrDefault(c[child], 0) + 1); + weihts.put(c[child], weihts.getOrDefault(c[child], 0) + w[child]); + } + for (int color : colors.keySet()) { + w[h] = Math.max(w[h], colors.get(color) + weihts.get(color)); + } + } + +} diff --git a/大厂刷题班/class36/Code07_PickAddMax.java b/大厂刷题班/class36/Code07_PickAddMax.java new file mode 100644 index 0000000..e0c2b1f --- /dev/null +++ b/大厂刷题班/class36/Code07_PickAddMax.java @@ -0,0 +1,80 @@ +package class36; + +import java.util.Arrays; + +// 来自腾讯 +// 给定一个数组arr，当拿走某个数a的时候，其他所有的数都+a +// 请返回最终所有数都拿走的最大分数 +// 比如: [2,3,1] +// 当拿走3时，获得3分，数组变成[5,4] +// 当拿走5时，获得5分，数组变成[9] +// 当拿走9时，获得9分，数组变成[] +// 这是最大的拿取方式，返回总分17 +public class Code07_PickAddMax { + + // 最优解 + public static int pick(int[] arr) { + Arrays.sort(arr); + int ans = 0; + for (int i = arr.length - 1; i >= 0; i--) { + ans = (ans << 1) + arr[i]; + } + return ans; + } + + // 纯暴力方法，为了测试 + public static int test(int[] arr) { + if (arr.length == 1) { + return arr[0]; + } + int ans = 0; + for (int i = 0; i < arr.length; i++) { + int[] rest = removeAddOthers(arr, i); + ans = Math.max(ans, arr[i] + test(rest)); + } + return ans; + } + + // 为了测试 + public static int[] removeAddOthers(int[] arr, int i) { + int[] rest = new int[arr.length - 1]; + int ri = 0; + for (int j = 0; j < i; j++) { + rest[ri++] = arr[j] + arr[i]; + } + for (int j = i + 1; j < arr.length; j++) { + rest[ri++] = arr[j] + arr[i]; + } + return rest; + } + + // 为了测试 + public static int[] randomArray(int len, int value) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * value) + 1; + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int N = 7; + int V = 10; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * N) + 1; + int[] arr = randomArray(len, V); + int ans1 = pick(arr); + int ans2 = test(arr); + if (ans1 != ans2) { + System.out.println(ans1); + System.out.println(ans2); + System.out.println("出错了!"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class36/Code08_MinBoatEvenNumbers.java b/大厂刷题班/class36/Code08_MinBoatEvenNumbers.java new file mode 100644 index 0000000..2267013 --- /dev/null +++ b/大厂刷题班/class36/Code08_MinBoatEvenNumbers.java @@ -0,0 +1,98 @@ +package class36; + +import java.util.Arrays; + +// 来自腾讯 +// 给定一个正数数组arr，代表每个人的体重。给定一个正数limit代表船的载重，所有船都是同样的载重量 +// 每个人的体重都一定不大于船的载重 +// 要求： +// 1, 可以1个人单独一搜船 +// 2, 一艘船如果坐2人，两个人的体重相加需要是偶数，且总体重不能超过船的载重 +// 3, 一艘船最多坐2人 +// 返回如果想所有人同时坐船，船的最小数量 +public class Code08_MinBoatEvenNumbers { + + public static int minBoat(int[] arr, int limit) { + Arrays.sort(arr); + int odd = 0; + int even = 0; + for (int num : arr) { + if ((num & 1) == 0) { + even++; + } else { + odd++; + } + } + int[] odds = new int[odd]; + int[] evens = new int[even]; + for (int i = arr.length - 1; i >= 0; i--) { + if ((arr[i] & 1) == 0) { + evens[--even] = arr[i]; + } else { + odds[--odd] = arr[i]; + } + } + return min(odds, limit) + min(evens, limit); + } + + public static int min(int[] arr, int limit) { + if (arr == null || arr.length == 0) { + return 0; + } + int N = arr.length; + if (arr[N - 1] > limit) { + return -1; + } + int lessR = -1; + for (int i = N - 1; i >= 0; i--) { + if (arr[i] <= (limit / 2)) { + lessR = i; + break; + } + } + if (lessR == -1) { + return N; + } + int L = lessR; + int R = lessR + 1; + int noUsed = 0; + while (L >= 0) { + int solved = 0; + while (R < N && arr[L] + arr[R] <= limit) { + R++; + solved++; + } + if (solved == 0) { + noUsed++; + L--; + } else { + L = Math.max(-1, L - solved); + } + } + int all = lessR + 1; + int used = all - noUsed; + int moreUnsolved = (N - all) - used; + return used + ((noUsed + 1) >> 1) + moreUnsolved; + } + + // 首尾双指针的解法 + public static int numRescueBoats2(int[] people, int limit) { + Arrays.sort(people); + int ans = 0; + int l = 0; + int r = people.length - 1; + int sum = 0; + while (l <= r) { + sum = l == r ? people[l] : people[l] + people[r]; + if (sum > limit) { + r--; + } else { + l++; + r--; + } + ans++; + } + return ans; + } + +} diff --git a/大厂刷题班/class36/Code09_MaxKLenSequence.java b/大厂刷题班/class36/Code09_MaxKLenSequence.java new file mode 100644 index 0000000..0ba90a1 --- /dev/null +++ b/大厂刷题班/class36/Code09_MaxKLenSequence.java @@ -0,0 +1,87 @@ +package class36; + +import java.util.TreeSet; + +// 来自腾讯 +// 给定一个字符串str，和一个正数k +// 返回长度为k的所有子序列中，字典序最大的子序列 +public class Code09_MaxKLenSequence { + + public static String maxString(String s, int k) { + if (k <= 0 || s.length() < k) { + return ""; + } + char[] str = s.toCharArray(); + int n = str.length; + char[] stack = new char[n]; + int size = 0; + for (int i = 0; i < n; i++) { + while (size > 0 && stack[size - 1] < str[i] && size + n - i > k) { + size--; + } + if (size + n - i == k) { + return String.valueOf(stack, 0, size) + s.substring(i); + } + stack[size++] = str[i]; + } + return String.valueOf(stack, 0, k); + } + + // 为了测试 + public static String test(String str, int k) { + if (k <= 0 || str.length() < k) { + return ""; + } + TreeSet ans = new TreeSet<>(); + process(0, 0, str.toCharArray(), new char[k], ans); + return ans.last(); + } + + // 为了测试 + public static void process(int si, int pi, char[] str, char[] path, TreeSet ans) { + if (si == str.length) { + if (pi == path.length) { + ans.add(String.valueOf(path)); + } + } else { + process(si + 1, pi, str, path, ans); + if (pi < path.length) { + path[pi] = str[si]; + process(si + 1, pi + 1, str, path, ans); + } + } + } + + // 为了测试 + public static String randomString(int len, int range) { + char[] str = new char[len]; + for (int i = 0; i < len; i++) { + str[i] = (char) ((int) (Math.random() * range) + 'a'); + } + return String.valueOf(str); + } + + public static void main(String[] args) { + int n = 12; + int r = 5; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * (n + 1)); + String str = randomString(len, r); + int k = (int) (Math.random() * (str.length() + 1)); + String ans1 = maxString(str, k); + String ans2 = test(str, k); + if (!ans1.equals(ans2)) { + System.out.println("出错了！"); + System.out.println(str); + System.out.println(k); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class36/Code10_StoneGameIV.java b/大厂刷题班/class36/Code10_StoneGameIV.java new file mode 100644 index 0000000..5bb9521 --- /dev/null +++ b/大厂刷题班/class36/Code10_StoneGameIV.java @@ -0,0 +1,63 @@ +package class36; + +// 来自哈喽单车 +// 本题是leetcode原题 : https://leetcode.com/problems/stone-game-iv/ +public class Code10_StoneGameIV { + + // 当前的！先手，会不会赢 + // 打表，不能发现规律 + public static boolean winnerSquareGame1(int n) { + if (n == 0) { + return false; + } + // 当前的先手，会尝试所有的情况，1，4，9，16，25，36.... + for (int i = 1; i * i <= n; i++) { + // 当前的先手，决定拿走 i * i 这个平方数 + // 它的对手会不会赢？ winnerSquareGame1(n - i * i) + if (!winnerSquareGame1(n - i * i)) { + return true; + } + } + return false; + } + + public static boolean winnerSquareGame2(int n) { + int[] dp = new int[n + 1]; + dp[0] = -1; + return process2(n, dp); + } + + public static boolean process2(int n, int[] dp) { + if (dp[n] != 0) { + return dp[n] == 1 ? true : false; + } + boolean ans = false; + for (int i = 1; i * i <= n; i++) { + if (!process2(n - i * i, dp)) { + ans = true; + break; + } + } + dp[n] = ans ? 1 : -1; + return ans; + } + + public static boolean winnerSquareGame3(int n) { + boolean[] dp = new boolean[n + 1]; + for (int i = 1; i <= n; i++) { + for (int j = 1; j * j <= i; j++) { + if (!dp[i - j * j]) { + dp[i] = true; + break; + } + } + } + return dp[n]; + } + + public static void main(String[] args) { + int n = 10000000; + System.out.println(winnerSquareGame3(n)); + } + +} diff --git a/大厂刷题班/class36/Code11_BusRoutes.java b/大厂刷题班/class36/Code11_BusRoutes.java new file mode 100644 index 0000000..2fb7af3 --- /dev/null +++ b/大厂刷题班/class36/Code11_BusRoutes.java @@ -0,0 +1,59 @@ +package class36; + +import java.util.ArrayList; +import java.util.HashMap; + +// 来自三七互娱 +// Leetcode原题 : https://leetcode.com/problems/bus-routes/ +public class Code11_BusRoutes { + + // 0 : [1,3,7,0] + // 1 : [7,9,6,2] + // .... + // 返回：返回换乘几次+1 -> 返回一共坐了多少条线的公交。 + public static int numBusesToDestination(int[][] routes, int source, int target) { + if (source == target) { + return 0; + } + int n = routes.length; + // key : 车站 + // value : list -> 该车站拥有哪些线路！ + HashMap> map = new HashMap<>(); + for (int i = 0; i < n; i++) { + for (int j = 0; j < routes[i].length; j++) { + if (!map.containsKey(routes[i][j])) { + map.put(routes[i][j], new ArrayList<>()); + } + map.get(routes[i][j]).add(i); + } + } + ArrayList queue = new ArrayList<>(); + boolean[] set = new boolean[n]; + for (int route : map.get(source)) { + queue.add(route); + set[route] = true; + } + int len = 1; + while (!queue.isEmpty()) { + ArrayList nextLevel = new ArrayList<>(); + for (int route : queue) { + int[] bus = routes[route]; + for (int station : bus) { + if (station == target) { + return len; + } + for (int nextRoute : map.get(station)) { + if (!set[nextRoute]) { + nextLevel.add(nextRoute); + set[nextRoute] = true; + } + } + } + } + queue = nextLevel; + len++; + } + return -1; + } + +} diff --git a/大厂刷题班/class36/说明 b/大厂刷题班/class36/说明 new file mode 100644 index 0000000..97dd816 --- /dev/null +++ b/大厂刷题班/class36/说明 @@ -0,0 +1,12 @@ +Code01 : 2021年8月大厂真实笔试题 +Code02 : 2021年8月大厂真实笔试题 +Code03 : 2021年8月大厂真实笔试题 +Code04 : 2021年8月大厂真实笔试题 +Code05 : 2021年8月大厂真实笔试题 +Code06 : 2021年8月大厂真实笔试题 +Code07 : 2021年8月大厂真实笔试题 +Code08 : 2021年8月大厂真实笔试题 +Code09 : 2021年8月大厂真实笔试题 +Code10 : 2021年8月大厂真实笔试题 +Code11 : 2021年8月大厂真实笔试题 +Code12 : 2021年8月大厂真实笔试题 \ No newline at end of file diff --git a/大厂刷题班/class37/Code01_ArrangeProject.java b/大厂刷题班/class37/Code01_ArrangeProject.java new file mode 100644 index 0000000..a448960 --- /dev/null +++ b/大厂刷题班/class37/Code01_ArrangeProject.java @@ -0,0 +1,50 @@ +package class37; + +import java.util.ArrayList; +import java.util.LinkedList; +import java.util.Queue; + +// 来自网易 +// 刚入职网易互娱，新人mini项目便如火如荼的开展起来。为了更好的项目协作与管理， +// 小易决定将学到的甘特图知识用于mini项目时间预估。小易先把项目中每一项工作以任务的形式列举出来， +// 每项任务有一个预计花费时间与前置任务表，必须完成了该任务的前置任务才能着手去做该任务。 +// 作为经验PM，小易把任务划分得井井有条，保证没有前置任务或者前置任务全数完成的任务，都可以同时进行。 +// 小易给出了这样一个任务表，请作为程序的你计算需要至少多长时间才能完成所有任务。 +// 输入第一行为一个正整数T，表示数据组数。 +// 对于接下来每组数据，第一行为一个正整数N，表示一共有N项任务。 +// 接下来N行，每行先有两个整数Di和Ki，表示完成第i个任务的预计花费时间为Di天，该任务有Ki个前置任务。 +// 之后为Ki个整数Mj，表示第Mj个任务是第i个任务的前置任务。 +// 数据范围：对于所有数据，满足1<=T<=3, 1<=N, Mj<=100000, 0<=Di<=1000, 0<=sum(Ki)<=N*2。 +public class Code01_ArrangeProject { + + public static int dayCount(ArrayList[] nums, int[] days, int[] headCount) { + Queue head = countHead(headCount); + int maxDay = 0; + int[] countDay = new int[days.length]; + while (!head.isEmpty()) { + int cur = head.poll(); + countDay[cur] += days[cur]; + for (int j = 0; j < nums[cur].size(); j++) { + headCount[nums[cur].get(j)]--; + if (headCount[nums[cur].get(j)] == 0) { + head.offer(nums[cur].get(j)); + } + countDay[nums[cur].get(j)] = Math.max(countDay[nums[cur].get(j)], countDay[cur]); + } + } + for (int i = 0; i < countDay.length; i++) { + maxDay = Math.max(maxDay, countDay[i]); + } + return maxDay; + } + + private static Queue countHead(int[] headCount) { + Queue queue = new LinkedList<>(); + for (int i = 0; i < headCount.length; i++) { + if (headCount[i] == 0) + queue.offer(i); // 没有前驱任务 + } + return queue; + } + +} diff --git a/大厂刷题班/class37/Code02_GameForEveryStepWin.java b/大厂刷题班/class37/Code02_GameForEveryStepWin.java new file mode 100644 index 0000000..077707f --- /dev/null +++ b/大厂刷题班/class37/Code02_GameForEveryStepWin.java @@ -0,0 +1,111 @@ +package class37; + +// 来自字节 +// 扑克牌中的红桃J和梅花Q找不到了，为了利用剩下的牌做游戏，小明设计了新的游戏规则 +// 1) A,2,3,4....10,J,Q,K分别对应1到13这些数字，大小王对应0 +// 2) 游戏人数为2人，轮流从牌堆里摸牌，每次摸到的牌只有“保留”和“使用”两个选项，且当前轮必须做出选择 +// 3) 如果选择“保留”当前牌，那么当前牌的分数加到总分里，并且可以一直持续到游戏结束 +// 4) 如果选择“使用”当前牌，那么当前牌的分数*3，加到总分上去，但是只有当前轮，下一轮，下下轮生效，之后轮效果消失。 +// 5) 每一轮总分大的人获胜 +// 假设小明知道每一轮对手做出选择之后的总分，返回小明在每一轮都赢的情况下，最终的最大分是多少 +// 如果小明怎么都无法保证每一轮都赢，返回-1 +public class Code02_GameForEveryStepWin { + +// public static max(int[] cands, int[] sroces) { +// return f(cands, sroces, 0, 0, 0, 0); +// } + + // 当前来到index位置，牌是cands[index]值 + // 对手第i轮的得分，sroces[i] + // int hold : i之前保留的牌的总分 + // int cur : 当前轮得到的，之前的牌只算上使用的效果，加成是多少 + // int next : 之前的牌，对index的下一轮，使用效果加成是多少 + // 返回值：如果i...最后，不能全赢，返回-1 + // 如果i...最后，能全赢，返回最后一轮的最大值 + + // index -> 26种 + // hold -> (1+2+3+..13) -> 91 -> 91 * 4 - (11 + 12) -> 341 + // cur -> 26 + // next -> 13 + // 26 * 341 * 26 * 13 -> ? * (10 ^ 5) + public static int f(int[] cands, int[] sroces, int index, int hold, int cur, int next) { + if (index == 25) { // 最后一张 + int all = hold + cur + cands[index] * 3; + if (all <= sroces[index]) { + return -1; + } + return all; + } + // 不仅最后一张 + // 保留 + int all1 = hold + cur + cands[index]; + int p1 = -1; + if (all1 > sroces[index]) { + p1 = f(cands, sroces, index + 1, hold + cands[index], next, 0); + } + // 爆发 + int all2 = hold + cur + cands[index] * 3; + int p2 = -1; + if (all2 > sroces[index]) { + p2 = f(cands, sroces, index + 1, hold, next + cands[index] * 3, cands[index] * 3); + } + return Math.max(p1, p2); + } + + // 26 * 341 * 78 * 39 = 2 * (10 ^ 7) + public static int process(int[] cards, int[] scores, int index, int hold, int cur, int next) { + if (index == 25) { + int all = hold + cur + cards[index] * 3; + if (all > scores[index]) { + return all; + } else { + return -1; + } + } else { + int d1 = hold + cur + cards[index]; + int p1 = -1; + if (d1 > scores[index]) { + p1 = process(cards, scores, index + 1, hold + cards[index], next, 0); + } + int d2 = hold + cur + cards[index] * 3; + int p2 = -1; + if (d2 > scores[index]) { + p2 = process(cards, scores, index + 1, hold, next + cards[index] * 3, cards[index] * 3); + } + return Math.max(p1, p2); + } + } + + + + // cur -> 牌点数 -> * 3 之后是效果 + // next -> 牌点数 -> * 3之后是效果 + public static int p(int[] cands, int[] sroces, int index, int hold, int cur, int next) { + if (index == 25) { // 最后一张 + int all = hold + cur * 3 + cands[index] * 3; + if (all <= sroces[index]) { + return -1; + } + return all; + } + // 不仅最后一张 + // 保留 + int all1 = hold + cur * 3 + cands[index]; + int p1 = -1; + if (all1 > sroces[index]) { + p1 = f(cands, sroces, index + 1, hold + cands[index], next, 0); + } + // 爆发 + int all2 = hold + cur * 3 + cands[index] * 3; + int p2 = -1; + if (all2 > sroces[index]) { + p2 = f(cands, sroces, index + 1, hold, next + cands[index], cands[index]); + } + return Math.max(p1, p2); + } + + // 改出动态规划，记忆化搜索！ + + + +} diff --git a/大厂刷题班/class37/Problem_0114_FlattenBinaryTreeToLinkedList.java b/大厂刷题班/class37/Problem_0114_FlattenBinaryTreeToLinkedList.java new file mode 100644 index 0000000..b6ba6ec --- /dev/null +++ b/大厂刷题班/class37/Problem_0114_FlattenBinaryTreeToLinkedList.java @@ -0,0 +1,93 @@ +package class37; + +// 注意，我们课上讲了一个别的题，并不是leetcode 114 +// 我们课上讲的是，把一棵搜索二叉树变成有序链表，怎么做 +// 而leetcode 114是，把一棵树先序遍历的结果串成链表 +// 所以我更新了代码，这个代码是leetcode 114的实现 +// 利用morris遍历 +public class Problem_0114_FlattenBinaryTreeToLinkedList { + + // 这个类不用提交 + public static class TreeNode { + public int val; + public TreeNode left; + public TreeNode right; + + public TreeNode(int value) { + val = value; + } + } + + // 普通解 + public static void flatten1(TreeNode root) { + process(root); + } + + public static class Info { + public TreeNode head; + public TreeNode tail; + + public Info(TreeNode h, TreeNode t) { + head = h; + tail = t; + } + } + + public static Info process(TreeNode head) { + if (head == null) { + return null; + } + Info leftInfo = process(head.left); + Info rightInfo = process(head.right); + head.left = null; + head.right = leftInfo == null ? null : leftInfo.head; + TreeNode tail = leftInfo == null ? head : leftInfo.tail; + tail.right = rightInfo == null ? null : rightInfo.head; + tail = rightInfo == null ? tail : rightInfo.tail; + return new Info(head, tail); + } + + // Morris遍历的解 + public static void flatten2(TreeNode root) { + if (root == null) { + return; + } + TreeNode pre = null; + TreeNode cur = root; + TreeNode mostRight = null; + while (cur != null) { + mostRight = cur.left; + if (mostRight != null) { + while (mostRight.right != null && mostRight.right != cur) { + mostRight = mostRight.right; + } + if (mostRight.right == null) { + mostRight.right = cur; + if (pre != null) { + pre.left = cur; + } + pre = cur; + cur = cur.left; + continue; + } else { + mostRight.right = null; + } + } else { + if (pre != null) { + pre.left = cur; + } + pre = cur; + } + cur = cur.right; + } + cur = root; + TreeNode next = null; + while (cur != null) { + next = cur.left; + cur.left = null; + cur.right = next; + cur = next; + } + } + +} diff --git a/大厂刷题班/class37/Problem_0221_MaximalSquare.java b/大厂刷题班/class37/Problem_0221_MaximalSquare.java new file mode 100644 index 0000000..38bfe06 --- /dev/null +++ b/大厂刷题班/class37/Problem_0221_MaximalSquare.java @@ -0,0 +1,40 @@ +package class37; + +public class Problem_0221_MaximalSquare { + + public static int maximalSquare(char[][] m) { + if (m == null || m.length == 0 || m[0].length == 0) { + return 0; + } + int N = m.length; + int M = m[0].length; + int[][] dp = new int[N + 1][M + 1]; + int max = 0; + for (int i = 0; i < N; i++) { + if (m[i][0] == '1') { + dp[i][0] = 1; + max = 1; + } + } + for (int j = 1; j < M; j++) { + if (m[0][j] == '1') { + dp[0][j] = 1; + max = 1; + } + } + for (int i = 1; i < N; i++) { + for (int j = 1; j < M; j++) { + if (m[i][j] == '1') { + dp[i][j] = Math.min( + Math.min(dp[i - 1][j], + dp[i][j - 1]), + dp[i - 1][j - 1]) + + 1; + max = Math.max(max, dp[i][j]); + } + } + } + return max * max; + } + +} diff --git a/大厂刷题班/class37/Problem_0226_InvertBinaryTree.java b/大厂刷题班/class37/Problem_0226_InvertBinaryTree.java new file mode 100644 index 0000000..d48dd9d --- /dev/null +++ b/大厂刷题班/class37/Problem_0226_InvertBinaryTree.java @@ -0,0 +1,21 @@ +package class37; + +public class Problem_0226_InvertBinaryTree { + + public class TreeNode { + public int val; + public TreeNode left; + public TreeNode right; + } + + public static TreeNode invertTree(TreeNode root) { + if (root == null) { + return null; + } + TreeNode left = root.left; + root.left = invertTree(root.right); + root.right = invertTree(left); + return root; + } + +} diff --git a/大厂刷题班/class37/Problem_0337_HouseRobberIII.java b/大厂刷题班/class37/Problem_0337_HouseRobberIII.java new file mode 100644 index 0000000..88bc718 --- /dev/null +++ b/大厂刷题班/class37/Problem_0337_HouseRobberIII.java @@ -0,0 +1,37 @@ +package class37; + +public class Problem_0337_HouseRobberIII { + + public static class TreeNode { + public int val; + public TreeNode left; + public TreeNode right; + } + + public static int rob(TreeNode root) { + Info info = process(root); + return Math.max(info.no, info.yes); + } + + public static class Info { + public int no; + public int yes; + + public Info(int n, int y) { + no = n; + yes = y; + } + } + + public static Info process(TreeNode x) { + if (x == null) { + return new Info(0, 0); + } + Info leftInfo = process(x.left); + Info rightInfo = process(x.right); + int no = Math.max(leftInfo.no, leftInfo.yes) + Math.max(rightInfo.no, rightInfo.yes); + int yes = x.val + leftInfo.no + rightInfo.no; + return new Info(no, yes); + } + +} diff --git a/大厂刷题班/class37/Problem_0394_DecodeString.java b/大厂刷题班/class37/Problem_0394_DecodeString.java new file mode 100644 index 0000000..0387f14 --- /dev/null +++ b/大厂刷题班/class37/Problem_0394_DecodeString.java @@ -0,0 +1,50 @@ +package class37; + +public class Problem_0394_DecodeString { + + public static String decodeString(String s) { + char[] str = s.toCharArray(); + return process(str, 0).ans; + } + + public static class Info { + public String ans; + public int stop; + + public Info(String a, int e) { + ans = a; + stop = e; + } + } + + // s[i....] 何时停？遇到 ']' 或者遇到 s的终止位置，停止 + // 返回Info + // 0) 串 + // 1) 算到了哪 + public static Info process(char[] s, int i) { + StringBuilder ans = new StringBuilder(); + int count = 0; + while (i < s.length && s[i] != ']') { + if ((s[i] >= 'a' && s[i] <= 'z') || (s[i] >= 'A' && s[i] <= 'Z')) { + ans.append(s[i++]); + } else if (s[i] >= '0' && s[i] <= '9') { + count = count * 10 + s[i++] - '0'; + } else { // str[index] = '[' + Info next = process(s, i + 1); + ans.append(timesString(count, next.ans)); + count = 0; + i = next.stop + 1; + } + } + return new Info(ans.toString(), i); + } + + public static String timesString(int times, String str) { + StringBuilder ans = new StringBuilder(); + for (int i = 0; i < times; i++) { + ans.append(str); + } + return ans.toString(); + } + +} diff --git a/大厂刷题班/class37/Problem_0406_QueueReconstructionByHeight.java b/大厂刷题班/class37/Problem_0406_QueueReconstructionByHeight.java new file mode 100644 index 0000000..793a8cf --- /dev/null +++ b/大厂刷题班/class37/Problem_0406_QueueReconstructionByHeight.java @@ -0,0 +1,264 @@ +package class37; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.Comparator; +import java.util.LinkedList; + +public class Problem_0406_QueueReconstructionByHeight { + + public static int[][] reconstructQueue1(int[][] people) { + int N = people.length; + Unit[] units = new Unit[N]; + for (int i = 0; i < N; i++) { + units[i] = new Unit(people[i][0], people[i][1]); + } + Arrays.sort(units, new UnitComparator()); + ArrayList arrList = new ArrayList<>(); + for (Unit unit : units) { + arrList.add(unit.k, unit); + } + int[][] ans = new int[N][2]; + int index = 0; + for (Unit unit : arrList) { + ans[index][0] = unit.h; + ans[index++][1] = unit.k; + } + return ans; + } + + public static int[][] reconstructQueue2(int[][] people) { + int N = people.length; + Unit[] units = new Unit[N]; + for (int i = 0; i < N; i++) { + units[i] = new Unit(people[i][0], people[i][1]); + } + Arrays.sort(units, new UnitComparator()); + SBTree tree = new SBTree(); + for (int i = 0; i < N; i++) { + tree.insert(units[i].k, i); + } + LinkedList allIndexes = tree.allIndexes(); + int[][] ans = new int[N][2]; + int index = 0; + for (Integer arri : allIndexes) { + ans[index][0] = units[arri].h; + ans[index++][1] = units[arri].k; + } + return ans; + } + + public static class Unit { + public int h; + public int k; + + public Unit(int height, int greater) { + h = height; + k = greater; + } + } + + public static class UnitComparator implements Comparator { + + @Override + public int compare(Unit o1, Unit o2) { + return o1.h != o2.h ? (o2.h - o1.h) : (o1.k - o2.k); + } + + } + + public static class SBTNode { + public int value; + public SBTNode l; + public SBTNode r; + public int size; + + public SBTNode(int arrIndex) { + value = arrIndex; + size = 1; + } + } + + public static class SBTree { + private SBTNode root; + + private SBTNode rightRotate(SBTNode cur) { + SBTNode leftNode = cur.l; + cur.l = leftNode.r; + leftNode.r = cur; + leftNode.size = cur.size; + cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1; + return leftNode; + } + + private SBTNode leftRotate(SBTNode cur) { + SBTNode rightNode = cur.r; + cur.r = rightNode.l; + rightNode.l = cur; + rightNode.size = cur.size; + cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1; + return rightNode; + } + + private SBTNode maintain(SBTNode cur) { + if (cur == null) { + return null; + } + int leftSize = cur.l != null ? cur.l.size : 0; + int leftLeftSize = cur.l != null && cur.l.l != null ? cur.l.l.size : 0; + int leftRightSize = cur.l != null && cur.l.r != null ? cur.l.r.size : 0; + int rightSize = cur.r != null ? cur.r.size : 0; + int rightLeftSize = cur.r != null && cur.r.l != null ? cur.r.l.size : 0; + int rightRightSize = cur.r != null && cur.r.r != null ? cur.r.r.size : 0; + if (leftLeftSize > rightSize) { + cur = rightRotate(cur); + cur.r = maintain(cur.r); + cur = maintain(cur); + } else if (leftRightSize > rightSize) { + cur.l = leftRotate(cur.l); + cur = rightRotate(cur); + cur.l = maintain(cur.l); + cur.r = maintain(cur.r); + cur = maintain(cur); + } else if (rightRightSize > leftSize) { + cur = leftRotate(cur); + cur.l = maintain(cur.l); + cur = maintain(cur); + } else if (rightLeftSize > leftSize) { + cur.r = rightRotate(cur.r); + cur = leftRotate(cur); + cur.l = maintain(cur.l); + cur.r = maintain(cur.r); + cur = maintain(cur); + } + return cur; + } + + private SBTNode insert(SBTNode root, int index, SBTNode cur) { + if (root == null) { + return cur; + } + root.size++; + int leftAndHeadSize = (root.l != null ? root.l.size : 0) + 1; + if (index < leftAndHeadSize) { + root.l = insert(root.l, index, cur); + } else { + root.r = insert(root.r, index - leftAndHeadSize, cur); + } + root = maintain(root); + return root; + } + + private SBTNode get(SBTNode root, int index) { + int leftSize = root.l != null ? root.l.size : 0; + if (index < leftSize) { + return get(root.l, index); + } else if (index == leftSize) { + return root; + } else { + return get(root.r, index - leftSize - 1); + } + } + + private void process(SBTNode head, LinkedList indexes) { + if (head == null) { + return; + } + process(head.l, indexes); + indexes.addLast(head.value); + process(head.r, indexes); + } + + public void insert(int index, int value) { + SBTNode cur = new SBTNode(value); + if (root == null) { + root = cur; + } else { + if (index <= root.size) { + root = insert(root, index, cur); + } + } + } + + public int get(int index) { + SBTNode ans = get(root, index); + return ans.value; + } + + public LinkedList allIndexes() { + LinkedList indexes = new LinkedList<>(); + process(root, indexes); + return indexes; + } + + } + + // 通过以下这个测试， + // 可以很明显的看到LinkedList的插入和get效率不如SBTree + // LinkedList需要找到index所在的位置之后才能插入或者读取，时间复杂度O(N) + // SBTree是平衡搜索二叉树，所以插入或者读取时间复杂度都是O(logN) + public static void main(String[] args) { + // 功能测试 + int test = 10000; + int max = 1000000; + boolean pass = true; + LinkedList list = new LinkedList<>(); + SBTree sbtree = new SBTree(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * (i + 1)); + int randomValue = (int) (Math.random() * (max + 1)); + list.add(randomIndex, randomValue); + sbtree.insert(randomIndex, randomValue); + } + for (int i = 0; i < test; i++) { + if (list.get(i) != sbtree.get(i)) { + pass = false; + break; + } + } + System.out.println("功能测试是否通过 : " + pass); + + // 性能测试 + test = 50000; + list = new LinkedList<>(); + sbtree = new SBTree(); + long start = 0; + long end = 0; + + start = System.currentTimeMillis(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * (i + 1)); + int randomValue = (int) (Math.random() * (max + 1)); + list.add(randomIndex, randomValue); + } + end = System.currentTimeMillis(); + System.out.println("LinkedList插入总时长(毫秒) ： " + (end - start)); + + start = System.currentTimeMillis(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * (i + 1)); + list.get(randomIndex); + } + end = System.currentTimeMillis(); + System.out.println("LinkedList读取总时长(毫秒) : " + (end - start)); + + start = System.currentTimeMillis(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * (i + 1)); + int randomValue = (int) (Math.random() * (max + 1)); + sbtree.insert(randomIndex, randomValue); + } + end = System.currentTimeMillis(); + System.out.println("SBTree插入总时长(毫秒) : " + (end - start)); + + start = System.currentTimeMillis(); + for (int i = 0; i < test; i++) { + int randomIndex = (int) (Math.random() * (i + 1)); + sbtree.get(randomIndex); + } + end = System.currentTimeMillis(); + System.out.println("SBTree读取总时长(毫秒) : " + (end - start)); + + } + +} diff --git a/大厂刷题班/class37/Problem_0437_PathSumIII.java b/大厂刷题班/class37/Problem_0437_PathSumIII.java new file mode 100644 index 0000000..253c27c --- /dev/null +++ b/大厂刷题班/class37/Problem_0437_PathSumIII.java @@ -0,0 +1,44 @@ +package class37; + +import java.util.HashMap; + +public class Problem_0437_PathSumIII { + + public class TreeNode { + public int val; + public TreeNode left; + public TreeNode right; + } + + public static int pathSum(TreeNode root, int sum) { + HashMap preSumMap = new HashMap<>(); + preSumMap.put(0L, 1); + return process(root, sum, 0, preSumMap); + } + + // 返回方法数 + public static int process(TreeNode x, int sum, long preAll, HashMap preSumMap) { + if (x == null) { + return 0; + } + long all = preAll + x.val; + int ans = 0; + if (preSumMap.containsKey(all - sum)) { + ans = preSumMap.get(all - sum); + } + if (!preSumMap.containsKey(all)) { + preSumMap.put(all, 1); + } else { + preSumMap.put(all, preSumMap.get(all) + 1); + } + ans += process(x.left, sum, all, preSumMap); + ans += process(x.right, sum, all, preSumMap); + if (preSumMap.get(all) == 1) { + preSumMap.remove(all); + } else { + preSumMap.put(all, preSumMap.get(all) - 1); + } + return ans; + } + +} diff --git a/大厂刷题班/class37/说明 b/大厂刷题班/class37/说明 new file mode 100644 index 0000000..8ee8a8f --- /dev/null +++ b/大厂刷题班/class37/说明 @@ -0,0 +1,30 @@ +leetcode最受欢迎100题 +leetcode全题目列表 : https://leetcode.com/problemset/all/ +在全题目列表的右侧栏，点击Top 100 Liked Questions +或者直接进入右侧链接 : https://leetcode.com/problemset/all/?listId=79h8rn6 +即可看到leetcode最受欢迎100题全列表 + +大厂刷题班27节~35节已经讲完leetcode高频题系列 +leetcode高频题系列和leetcode最受欢迎100题系列题目有重合 +以下为leetcode最受欢迎100题不和leetcode高频题重合的题号，其他的都重复了，不再讲述 +0032 : 大厂刷题班, 第14节第1题 +0039 : 体系学习班, 硬币找零专题 : 第21节第2、3、4题, 第22节第2题, 第24节第4题。本题就是无限张找零问题，不再重复讲述 +0064 : 体系学习班, 第21节第1题 +0072 : 大厂刷题班, 第5节第3题 +0085 : 体系学习班, 第25节第4题 +0096 : 体系学习班, 第39节第4题, 卡特兰数 +0114 : 大厂刷题班, 第37节, 本节 +0142 : 体系学习班, 第10节第1题 +0221 : 大厂刷题班, 第37节, 本节 +0226 : 大厂刷题班, 第37节, 本节 +0337 : 体系学习班, 第13节, 第4题, 还是这道题的加强版(多叉树) +0338 : 和leetcode第191题重复, 大厂刷题班第30节讲过了 +0394 : 大厂刷题班, 第37节, 本节 +0406 : 大厂刷题班, 第37节, 本节 +0416 : 体系学习班, 第23节第1题, 还是这道题的加强版 +0437 : 大厂刷题班, 第37节, 本节 +剩余题目在下一节 + +本节附加题 +code01 : 2021年8月大厂真实笔试题 +code02 : 2021年8月大厂真实笔试题 \ No newline at end of file diff --git a/大厂刷题班/class38/Code01_FillGapMinStep.java b/大厂刷题班/class38/Code01_FillGapMinStep.java new file mode 100644 index 0000000..361a734 --- /dev/null +++ b/大厂刷题班/class38/Code01_FillGapMinStep.java @@ -0,0 +1,103 @@ +package class38; + +// 来自字节 +// 给定两个数a和b +// 第1轮，把1选择给a或者b +// 第2轮，把2选择给a或者b +// ... +// 第i轮，把i选择给a或者b +// 想让a和b的值一样大，请问至少需要多少轮？ +public class Code01_FillGapMinStep { + + // 暴力方法 + // 不要让a、b过大！ + public static int minStep0(int a, int b) { + if (a == b) { + return 0; + } + int limit = 15; + return process(a, b, 1, limit); + } + + public static int process(int a, int b, int i, int n) { + if (i > n) { + return Integer.MAX_VALUE; + } + if (a + i == b || a == b + i) { + return i; + } + return Math.min(process(a + i, b, i + 1, n), process(a, b + i, i + 1, n)); + } + + public static int minStep1(int a, int b) { + if (a == b) { + return 0; + } + int s = Math.abs(a - b); + int num = 1; + int sum = 0; + for (; !(sum >= s && (sum - s) % 2 == 0); num++) { + sum += num; + } + return num - 1; + } + + public static int minStep2(int a, int b) { + if (a == b) { + return 0; + } + int s = Math.abs(a - b); + // 找到sum >= s, 最小的i + int begin = best(s << 1); + for (; (begin * (begin + 1) / 2 - s) % 2 != 0;) { + begin++; + } + return begin; + } + + public static int best(int s2) { + int L = 0; + int R = 1; + for (; R * (R + 1) < s2;) { + L = R; + R *= 2; + } + int ans = 0; + while (L <= R) { + int M = (L + R) / 2; + if (M * (M + 1) >= s2) { + ans = M; + R = M - 1; + } else { + L = M + 1; + } + } + return ans; + } + + public static void main(String[] args) { + System.out.println("功能测试开始"); + for (int a = 1; a < 100; a++) { + for (int b = 1; b < 100; b++) { + int ans1 = minStep0(a, b); + int ans2 = minStep1(a, b); + int ans3 = minStep2(a, b); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("出错了！"); + System.out.println(a + " , " + b); + break; + } + } + } + System.out.println("功能测试结束"); + + int a = 19019; + int b = 8439284; + int ans2 = minStep1(a, b); + int ans3 = minStep2(a, b); + System.out.println(ans2); + System.out.println(ans3); + + } + +} diff --git a/大厂刷题班/class38/Code02_GreatWall.java b/大厂刷题班/class38/Code02_GreatWall.java new file mode 100644 index 0000000..af9cf35 --- /dev/null +++ b/大厂刷题班/class38/Code02_GreatWall.java @@ -0,0 +1,183 @@ +package class38; + +// 360笔试题 +// 长城守卫军 +// 题目描述： +// 长城上有连成一排的n个烽火台，每个烽火台都有士兵驻守。 +// 第i个烽火台驻守着ai个士兵，相邻峰火台的距离为1。另外，有m位将军， +// 每位将军可以驻守一个峰火台，每个烽火台可以有多个将军驻守， +// 将军可以影响所有距离他驻守的峰火台小于等于x的烽火台。 +// 每个烽火台的基础战斗力为士兵数，另外，每个能影响此烽火台的将军都能使这个烽火台的战斗力提升k。 +// 长城的战斗力为所有烽火台的战斗力的最小值。 +// 请问长城的最大战斗力可以是多少？ +// 输入描述 +// 第一行四个正整数n,m,x,k(1<=x<=n<=10^5,0<=m<=10^5,1<=k<=10^5) +// 第二行n个整数ai(0<=ai<=10^5) +// 输出描述 仅一行，一个整数，表示长城的最大战斗力 +// 样例输入 +// 5 2 1 2 +// 4 4 2 4 4 +// 样例输出 +// 6 +public class Code02_GreatWall { + + public static int maxForce(int[] wall, int m, int x, int k) { + long L = 0; + long R = 0; + for (int num : wall) { + R = Math.max(R, num); + } + R += m * k; + long ans = 0; + while (L <= R) { + long M = (L + R) / 2; + if (can(wall, m, x, k, M)) { + ans = M; + L = M + 1; + } else { + R = M - 1; + } + } + return (int) ans; + } + + public static boolean can(int[] wall, int m, int x, int k, long limit) { + int N = wall.length; + // 注意：下标从1开始 + SegmentTree st = new SegmentTree(wall); + st.build(1, N, 1); + int need = 0; + for (int i = 0; i < N; i++) { + // 注意：下标从1开始 + long cur = st.query(i + 1, i + 1, 1, N, 1); + if (cur < limit) { + int add = (int) ((limit - cur + k - 1) / k); + need += add; + if (need > m) { + return false; + } + st.add(i + 1, Math.min(i + x, N), add * k, 1, N, 1); + } + } + return true; + } + + public static class SegmentTree { + private int MAXN; + private int[] arr; + private int[] sum; + private int[] lazy; + private int[] change; + private boolean[] update; + + public SegmentTree(int[] origin) { + MAXN = origin.length + 1; + arr = new int[MAXN]; + for (int i = 1; i < MAXN; i++) { + arr[i] = origin[i - 1]; + } + sum = new int[MAXN << 2]; + lazy = new int[MAXN << 2]; + change = new int[MAXN << 2]; + update = new boolean[MAXN << 2]; + } + + private void pushUp(int rt) { + sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; + } + + private void pushDown(int rt, int ln, int rn) { + if (update[rt]) { + update[rt << 1] = true; + update[rt << 1 | 1] = true; + change[rt << 1] = change[rt]; + change[rt << 1 | 1] = change[rt]; + lazy[rt << 1] = 0; + lazy[rt << 1 | 1] = 0; + sum[rt << 1] = change[rt] * ln; + sum[rt << 1 | 1] = change[rt] * rn; + update[rt] = false; + } + if (lazy[rt] != 0) { + lazy[rt << 1] += lazy[rt]; + sum[rt << 1] += lazy[rt] * ln; + lazy[rt << 1 | 1] += lazy[rt]; + sum[rt << 1 | 1] += lazy[rt] * rn; + lazy[rt] = 0; + } + } + + public void build(int l, int r, int rt) { + if (l == r) { + sum[rt] = arr[l]; + return; + } + int mid = (l + r) >> 1; + build(l, mid, rt << 1); + build(mid + 1, r, rt << 1 | 1); + pushUp(rt); + } + + public void update(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + update[rt] = true; + change[rt] = C; + sum[rt] = C * (r - l + 1); + lazy[rt] = 0; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + update(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + update(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + public void add(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + sum[rt] += C * (r - l + 1); + lazy[rt] += C; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + add(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + add(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + public long query(int L, int R, int l, int r, int rt) { + if (L <= l && r <= R) { + return sum[rt]; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + long ans = 0; + if (L <= mid) { + ans += query(L, R, l, mid, rt << 1); + } + if (R > mid) { + ans += query(L, R, mid + 1, r, rt << 1 | 1); + } + return ans; + } + + } + + public static void main(String[] args) { + int[] wall = { 3, 1, 1, 1, 3 }; + int m = 2; + int x = 3; + int k = 1; + System.out.println(maxForce(wall, m, x, k)); + } + +} diff --git a/大厂刷题班/class38/Problem_0438_FindAllAnagramsInAString.java b/大厂刷题班/class38/Problem_0438_FindAllAnagramsInAString.java new file mode 100644 index 0000000..3f68453 --- /dev/null +++ b/大厂刷题班/class38/Problem_0438_FindAllAnagramsInAString.java @@ -0,0 +1,58 @@ +package class38; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.List; + +public class Problem_0438_FindAllAnagramsInAString { + + public static List findAnagrams(String s, String p) { + List ans = new ArrayList<>(); + if (s == null || p == null || s.length() < p.length()) { + return ans; + } + char[] str = s.toCharArray(); + int N = str.length; + char[] pst = p.toCharArray(); + int M = pst.length; + HashMap map = new HashMap<>(); + for (char cha : pst) { + if (!map.containsKey(cha)) { + map.put(cha, 1); + } else { + map.put(cha, map.get(cha) + 1); + } + } + int all = M; + for (int end = 0; end < M - 1; end++) { + if (map.containsKey(str[end])) { + int count = map.get(str[end]); + if (count > 0) { + all--; + } + map.put(str[end], count - 1); + } + } + for (int end = M - 1, start = 0; end < N; end++, start++) { + if (map.containsKey(str[end])) { + int count = map.get(str[end]); + if (count > 0) { + all--; + } + map.put(str[end], count - 1); + } + if (all == 0) { + ans.add(start); + } + if (map.containsKey(str[start])) { + int count = map.get(str[start]); + if (count >= 0) { + all++; + } + map.put(str[start], count + 1); + } + } + return ans; + } + +} diff --git a/大厂刷题班/class38/Problem_0448_FindAllNumbersDisappearedInAnArray.java b/大厂刷题班/class38/Problem_0448_FindAllNumbersDisappearedInAnArray.java new file mode 100644 index 0000000..3dcf07e --- /dev/null +++ b/大厂刷题班/class38/Problem_0448_FindAllNumbersDisappearedInAnArray.java @@ -0,0 +1,42 @@ +package class38; + +import java.util.ArrayList; +import java.util.List; + +public class Problem_0448_FindAllNumbersDisappearedInAnArray { + + public static List findDisappearedNumbers(int[] nums) { + List ans = new ArrayList<>(); + if (nums == null || nums.length == 0) { + return ans; + } + int N = nums.length; + for (int i = 0; i < N; i++) { + // 从i位置出发，去玩下标循环怼 + walk(nums, i); + } + for (int i = 0; i < N; i++) { + if (nums[i] != i + 1) { + ans.add(i + 1); + } + } + return ans; + } + + public static void walk(int[] nums, int i) { + while (nums[i] != i + 1) { // 不断从i发货 + int nexti = nums[i] - 1; + if (nums[nexti] == nexti + 1) { + break; + } + swap(nums, i, nexti); + } + } + + public static void swap(int[] nums, int i, int j) { + int tmp = nums[i]; + nums[i] = nums[j]; + nums[j] = tmp; + } + +} diff --git a/大厂刷题班/class38/Problem_0617_MergeTwoBinaryTrees.java b/大厂刷题班/class38/Problem_0617_MergeTwoBinaryTrees.java new file mode 100644 index 0000000..942d269 --- /dev/null +++ b/大厂刷题班/class38/Problem_0617_MergeTwoBinaryTrees.java @@ -0,0 +1,31 @@ +package class38; + +public class Problem_0617_MergeTwoBinaryTrees { + + public static class TreeNode { + public int val; + public TreeNode left; + public TreeNode right; + + public TreeNode(int val) { + this.val = val; + } + } + + // 当前，一棵树的头是t1，另一颗树的头是t2 + // 请返回，整体merge之后的头 + public static TreeNode mergeTrees(TreeNode t1, TreeNode t2) { + if (t1 == null) { + return t2; + } + if (t2 == null) { + return t1; + } + // t1和t2都不是空 + TreeNode merge = new TreeNode(t1.val + t2.val); + merge.left = mergeTrees(t1.left, t2.left); + merge.right = mergeTrees(t1.right, t2.right); + return merge; + } + +} diff --git a/大厂刷题班/class38/Problem_0621_TaskScheduler.java b/大厂刷题班/class38/Problem_0621_TaskScheduler.java new file mode 100644 index 0000000..24b4b59 --- /dev/null +++ b/大厂刷题班/class38/Problem_0621_TaskScheduler.java @@ -0,0 +1,33 @@ +package class38; + +public class Problem_0621_TaskScheduler { + + // ['A', 'B', 'A'] + public static int leastInterval(char[] tasks, int free) { + int[] count = new int[256]; + // 出现最多次的任务，到底是出现了几次 + int maxCount = 0; + for (char task : tasks) { + count[task]++; + maxCount = Math.max(maxCount, count[task]); + } + // 有多少种任务，都出现最多次 + int maxKinds = 0; + for (int task = 0; task < 256; task++) { + if (count[task] == maxCount) { + maxKinds++; + } + } + // maxKinds : 有多少种任务，都出现最多次 + // maxCount : 最多次，是几次？ + // 砍掉最后一组剩余的任务数 + int tasksExceptFinalTeam = tasks.length - maxKinds; + int spaces = (free + 1) * (maxCount - 1); + // 到底几个空格最终会留下！ + int restSpaces = Math.max(0, spaces - tasksExceptFinalTeam); + return tasks.length + restSpaces; + // return Math.max(tasks.length, ((n + 1) * (maxCount - 1) + maxKinds)); + } + + +} diff --git a/大厂刷题班/class38/Problem_0647_PalindromicSubstrings.java b/大厂刷题班/class38/Problem_0647_PalindromicSubstrings.java new file mode 100644 index 0000000..e51180a --- /dev/null +++ b/大厂刷题班/class38/Problem_0647_PalindromicSubstrings.java @@ -0,0 +1,49 @@ +package class38; + +public class Problem_0647_PalindromicSubstrings { + + public static int countSubstrings(String s) { + if (s == null || s.length() == 0) { + return 0; + } + int[] dp = getManacherDP(s); + int ans = 0; + for (int i = 0; i < dp.length; i++) { + ans += dp[i] >> 1; + } + return ans; + } + + public static int[] getManacherDP(String s) { + char[] str = manacherString(s); + int[] pArr = new int[str.length]; + int C = -1; + int R = -1; + for (int i = 0; i < str.length; i++) { + pArr[i] = R > i ? Math.min(pArr[2 * C - i], R - i) : 1; + while (i + pArr[i] < str.length && i - pArr[i] > -1) { + if (str[i + pArr[i]] == str[i - pArr[i]]) + pArr[i]++; + else { + break; + } + } + if (i + pArr[i] > R) { + R = i + pArr[i]; + C = i; + } + } + return pArr; + } + + public static char[] manacherString(String str) { + char[] charArr = str.toCharArray(); + char[] res = new char[str.length() * 2 + 1]; + int index = 0; + for (int i = 0; i != res.length; i++) { + res[i] = (i & 1) == 0 ? '#' : charArr[index++]; + } + return res; + } + +} diff --git a/大厂刷题班/class38/Problem_0739_DailyTemperatures.java b/大厂刷题班/class38/Problem_0739_DailyTemperatures.java new file mode 100644 index 0000000..92b0712 --- /dev/null +++ b/大厂刷题班/class38/Problem_0739_DailyTemperatures.java @@ -0,0 +1,34 @@ +package class38; + +import java.util.ArrayList; +import java.util.List; +import java.util.Stack; + +public class Problem_0739_DailyTemperatures { + + public static int[] dailyTemperatures(int[] arr) { + if (arr == null || arr.length == 0) { + return new int[0]; + } + int N = arr.length; + int[] ans = new int[N]; + Stack> stack = new Stack<>(); + for (int i = 0; i < N; i++) { + while (!stack.isEmpty() && arr[stack.peek().get(0)] < arr[i]) { + List popIs = stack.pop(); + for (Integer popi : popIs) { + ans[popi] = i - popi; + } + } + if (!stack.isEmpty() && arr[stack.peek().get(0)] == arr[i]) { + stack.peek().add(Integer.valueOf(i)); + } else { + ArrayList list = new ArrayList<>(); + list.add(i); + stack.push(list); + } + } + return ans; + } + +} diff --git a/大厂刷题班/class38/Problem_0763_PartitionLabels.java b/大厂刷题班/class38/Problem_0763_PartitionLabels.java new file mode 100644 index 0000000..c72954a --- /dev/null +++ b/大厂刷题班/class38/Problem_0763_PartitionLabels.java @@ -0,0 +1,28 @@ +package class38; + +import java.util.ArrayList; +import java.util.List; + +public class Problem_0763_PartitionLabels { + + public static List partitionLabels(String S) { + char[] str = S.toCharArray(); + int[] far = new int[26]; + for (int i = 0; i < str.length; i++) { + far[str[i] - 'a'] = i; + } + List ans = new ArrayList<>(); + int left = 0; + int right = far[str[0] - 'a']; + for (int i = 1; i < str.length; i++) { + if (i > right) { + ans.add(right - left + 1); + left = i; + } + right = Math.max(right, far[str[i] - 'a']); + } + ans.add(right - left + 1); + return ans; + } + +} diff --git a/大厂刷题班/class38/说明 b/大厂刷题班/class38/说明 new file mode 100644 index 0000000..5fc1f1f --- /dev/null +++ b/大厂刷题班/class38/说明 @@ -0,0 +1,25 @@ +leetcode最受欢迎100题 +leetcode全题目列表 : https://leetcode.com/problemset/all/ +在全题目列表的右侧栏，点击Top 100 Liked Questions +或者直接进入右侧链接 : https://leetcode.com/problemset/all/?listId=79h8rn6 +即可看到leetcode最受欢迎100题全列表 + +大厂刷题班27节~35节已经讲完leetcode高频题系列 +leetcode高频题系列和leetcode最受欢迎100题系列题目有重合 +以下为leetcode最受欢迎100题不和leetcode高频题重合的题号，其他的都重复了，不再讲述 +0438 : 大厂刷题班, 第38节, 本节 +0448 : 大厂刷题班, 第38节, 本节 +0494 : 大厂刷题班, 第1节第7题 +0543 : 体系学习班, 第12节第6题 +0560 : 大厂刷题班, 第37节, Leetcode题目437与本题思路相同, 课上也讲了该题做法 +0581 : 大厂刷题班, 第1节第6题 +0617 : 大厂刷题班, 第38节, 本节 +0621 : 大厂刷题班, 第38节, 本节 +0647 : 大厂刷题班, 第38节, 本节 +0739 : 大厂刷题班, 第38节, 本节 +0763 : 大厂刷题班, 第38节, 本节 +至此，leetcode最受欢迎100题系列完结 + +本节附加题 +code01 : 2021年8月大厂真实笔试题 +code02 : 2021年8月大厂真实笔试题 \ No newline at end of file diff --git a/大厂刷题班/class39/Code01_01AddValue.java b/大厂刷题班/class39/Code01_01AddValue.java new file mode 100644 index 0000000..5609d29 --- /dev/null +++ b/大厂刷题班/class39/Code01_01AddValue.java @@ -0,0 +1,46 @@ +package class39; + +// 来自腾讯 +// 给定一个只由0和1组成的字符串S，假设下标从1开始，规定i位置的字符价值V[i]计算方式如下 : +// 1) i == 1时，V[i] = 1 +// 2) i > 1时，如果S[i] != S[i-1]，V[i] = 1 +// 3) i > 1时，如果S[i] == S[i-1]，V[i] = V[i-1] + 1 +// 你可以随意删除S中的字符，返回整个S的最大价值 +// 字符串长度<=5000 +public class Code01_01AddValue { + + public static int max1(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + int[] arr = new int[str.length]; + for (int i = 0; i < arr.length; i++) { + arr[i] = str[i] == '0' ? 0 : 1; + } + return process1(arr, 0, 0, 0); + } + + // 递归含义 : + // 目前在arr[index...]上做选择, str[index...]的左边，最近的数字是lastNum + // 并且lastNum所带的价值，已经拉高到baseValue + // 返回在str[index...]上做选择，最终获得的最大价值 + // index -> 0 ~ 4999 + // lastNum -> 0 or 1 + // baseValue -> 1 ~ 5000 + // 5000 * 2 * 5000 -> 5 * 10^7(过!) + public static int process1(int[] arr, int index, int lastNum, int baseValue) { + if (index == arr.length) { + return 0; + } + int curValue = lastNum == arr[index] ? (baseValue + 1) : 1; + // 当前index位置的字符保留 + int next1 = process1(arr, index + 1, arr[index], curValue); + // 当前index位置的字符不保留 + int next2 = process1(arr, index + 1, lastNum, baseValue); + return Math.max(curValue + next1, next2); + } + + // 请看体系学习班，动态规划章节，把上面的递归改成动态规划！看完必会 + +} diff --git a/大厂刷题班/class39/Code02_ValidSequence.java b/大厂刷题班/class39/Code02_ValidSequence.java new file mode 100644 index 0000000..37e499e --- /dev/null +++ b/大厂刷题班/class39/Code02_ValidSequence.java @@ -0,0 +1,113 @@ +package class39; + +// 来自腾讯 +// 给定一个长度为n的数组arr，求有多少个子数组满足 : +// 子数组两端的值，是这个子数组的最小值和次小值，最小值和次小值谁在最左和最右无所谓 +// n<=100000（10^5） n*logn O(N) +public class Code02_ValidSequence { + + + public static int nums(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int n = arr.length; + int[] values = new int[n]; + int[] times = new int[n]; + int size = 0; + int ans = 0; + for (int i = 0; i < arr.length; i++) { + while (size != 0 && values[size - 1] > arr[i]) { + size--; + ans += times[size] + cn2(times[size]); + } + if (size != 0 && values[size - 1] == arr[i]) { + times[size - 1]++; + } else { + values[size] = arr[i]; + times[size++] = 1; + } + } + while (size != 0) { + ans += cn2(times[--size]); + } + for (int i = arr.length - 1; i >= 0; i--) { + while (size != 0 && values[size - 1] > arr[i]) { + ans += times[--size]; + } + if (size != 0 && values[size - 1] == arr[i]) { + times[size - 1]++; + } else { + values[size] = arr[i]; + times[size++] = 1; + } + } + return ans; + } + + public static int cn2(int n) { + return (n * (n - 1)) >> 1; + } + + // 为了测试 + // 暴力方法 + public static int test(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int ans = 0; + for (int s = 0; s < arr.length; s++) { + for (int e = s + 1; e < arr.length; e++) { + int max = Math.max(arr[s], arr[e]); + boolean valid = true; + for (int i = s + 1; i < e; i++) { + if (arr[i] < max) { + valid = false; + break; + } + } + ans += valid ? 1 : 0; + } + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * v); + } + return arr; + } + + // 为了测试 + public static void printArray(int[] arr) { + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + } + + // 为了测试 + public static void main(String[] args) { + int n = 30; + int v = 30; + int testTime = 100000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int m = (int) (Math.random() * n); + int[] arr = randomArray(m, v); + int ans1 = nums(arr); + int ans2 = test(arr); + if (ans1 != ans2) { + System.out.println("出错了!"); + printArray(arr); + System.out.println(ans1); + System.out.println(ans2); + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class39/Code03_SequenceKDifferentKinds.java b/大厂刷题班/class39/Code03_SequenceKDifferentKinds.java new file mode 100644 index 0000000..82803e7 --- /dev/null +++ b/大厂刷题班/class39/Code03_SequenceKDifferentKinds.java @@ -0,0 +1,63 @@ +package class39; + +// 来自百度 +// 给定一个字符串str，和一个正数k +// str子序列的字符种数必须是k种，返回有多少子序列满足这个条件 +// 已知str中都是小写字母 +// 原始是取mod +// 本节在尝试上，最难的 +// 搞出桶来，组合公式 +public class Code03_SequenceKDifferentKinds { + + // bu -> {6,7,0,0,6,3} + // 0 1 2 3 4 5 + // a b c d e f + // 在桶数组bu[index....] 一定要凑出rest种来！请问几种方法！ + public static int f(int[] bu, int index, int rest) { + if (index == bu.length) { + return rest == 0 ? 1 : 0; + } + // 最后形成的子序列，一个index代表的字符也没有! + int p1 = f(bu, index + 1, rest); + // 最后形成的子序列，一定要包含index代表的字符，几个呢？(所有可能性都要算上！) + int p2 = 0; + if (rest > 0) { // 剩余的种数，没耗尽，可以包含当前桶的字符 + p2 = (1 << bu[index] - 1) * f(bu, index + 1, rest - 1); + } + return p1 + p2; + } + + public static int nums(String s, int k) { + char[] str = s.toCharArray(); + int[] counts = new int[26]; + for (char c : str) { + counts[c - 97]++; + } + return ways(counts, 0, k); + } + + public static int ways(int[] c, int i, int r) { + if (r == 0) { + return 1; + } + if (i == c.length) { + return 0; + } + // math(n) -> 2 ^ n -1 + return math(c[i]) * ways(c, i + 1, r - 1) + ways(c, i + 1, r); + } + + // n个不同的球 + // 挑出1个的方法数 + 挑出2个的方法数 + ... + 挑出n个的方法数为: + // C(n,1) + C(n,2) + ... + C(n,n) == (2 ^ n) -1 + public static int math(int n) { + return (1 << n) - 1; + } + + public static void main(String[] args) { + String str = "acbbca"; + int k = 3; + System.out.println(nums(str, k)); + } + +} diff --git a/大厂刷题班/class39/Code04_JumpGameOnMatrix.java b/大厂刷题班/class39/Code04_JumpGameOnMatrix.java new file mode 100644 index 0000000..3acbcc4 --- /dev/null +++ b/大厂刷题班/class39/Code04_JumpGameOnMatrix.java @@ -0,0 +1,239 @@ +package class39; + +// 来自京东 +// 给定一个二维数组matrix，matrix[i][j] = k代表: +// 从(i,j)位置可以随意往右跳<=k步，或者从(i,j)位置可以随意往下跳<=k步 +// 如果matrix[i][j] = 0，代表来到(i,j)位置必须停止 +// 返回从matrix左上角到右下角，至少要跳几次 +// 已知matrix中行数n <= 5000, 列数m <= 5000 +// matrix中的值，<= 5000 +// 最弟弟的技巧也过了。最优解 -> dp+枚举优化(线段树，体系学习班) +public class Code04_JumpGameOnMatrix { + + // 暴力方法，仅仅是做对数器 + // 如果无法到达会返回系统最大值 + public static int jump1(int[][] map) { + return process(map, 0, 0); + } + + // 当前来到的位置是(row,col) + // 目标：右下角 + // 当前最大能跳多远，map[row][col]值决定，只能向右、或者向下 + // 返回，到达右下角，最小跳几次？ + // 5000 * 5000 = 25000000 -> 2 * (10 ^ 7) + public static int process(int[][] map, int row, int col) { + if (row == map.length - 1 && col == map[0].length - 1) { + return 0; + } + // 如果没到右下角 + if (map[row][col] == 0) { + return Integer.MAX_VALUE; + } + // 当前位置，可以去很多的位置，next含义： + // 在所有能去的位置里，哪个位置最后到达右下角，跳的次数最少，就是next + int next = Integer.MAX_VALUE; + // 往下能到达的位置，全试一遍 + for (int down = row + 1; down < map.length && (down - row) <= map[row][col]; down++) { + next = Math.min(next, process(map, down, col)); + } + // 往右能到达的位置，全试一遍 + for (int right = col + 1; right < map[0].length && (right - col) <= map[row][col]; right++) { + next = Math.min(next, process(map, row, right)); + } + // 如果所有下一步的位置，没有一个能到右下角，next = 系统最大！ + // 返回系统最大！ + // next != 系统最大 7 + 1 + return next != Integer.MAX_VALUE ? (next + 1) : next; + } + + // 优化方法, 利用线段树做枚举优化 + // 因为线段树，下标从1开始 + // 所以，该方法中所有的下标，请都从1开始，防止乱！ + public static int jump2(int[][] arr) { + int n = arr.length; + int m = arr[0].length; + int[][] map = new int[n + 1][m + 1]; + for (int a = 0, b = 1; a < n; a++, b++) { + for (int c = 0, d = 1; c < m; c++, d++) { + map[b][d] = arr[a][c]; + } + } + SegmentTree[] rowTrees = new SegmentTree[n + 1]; + for (int i = 1; i <= n; i++) { + rowTrees[i] = new SegmentTree(m); + } + SegmentTree[] colTrees = new SegmentTree[m + 1]; + for (int i = 1; i <= m; i++) { + colTrees[i] = new SegmentTree(n); + } + rowTrees[n].update(m, m, 0, 1, m, 1); + colTrees[m].update(n, n, 0, 1, n, 1); + for (int col = m - 1; col >= 1; col--) { + if (map[n][col] != 0) { + int left = col + 1; + int right = Math.min(col + map[n][col], m); + int next = rowTrees[n].query(left, right, 1, m, 1); + if (next != Integer.MAX_VALUE) { + rowTrees[n].update(col, col, next + 1, 1, m, 1); + colTrees[col].update(n, n, next + 1, 1, n, 1); + } + } + } + for (int row = n - 1; row >= 1; row--) { + if (map[row][m] != 0) { + int up = row + 1; + int down = Math.min(row + map[row][m], n); + int next = colTrees[m].query(up, down, 1, n, 1); + if (next != Integer.MAX_VALUE) { + rowTrees[row].update(m, m, next + 1, 1, m, 1); + colTrees[m].update(row, row, next + 1, 1, n, 1); + } + } + } + for (int row = n - 1; row >= 1; row--) { + for (int col = m - 1; col >= 1; col--) { + if (map[row][col] != 0) { + // (row,col) 往右是什么范围呢？[left,right] + int left = col + 1; + int right = Math.min(col + map[row][col], m); + int next1 = rowTrees[row].query(left, right, 1, m, 1); + // (row,col) 往下是什么范围呢？[up,down] + int up = row + 1; + int down = Math.min(row + map[row][col], n); + int next2 = colTrees[col].query(up, down, 1, n, 1); + int next = Math.min(next1, next2); + if (next != Integer.MAX_VALUE) { + rowTrees[row].update(col, col, next + 1, 1, m, 1); + colTrees[col].update(row, row, next + 1, 1, n, 1); + } + } + } + } + return rowTrees[1].query(1, 1, 1, m, 1); + } + + // 区间查询最小值的线段树 + // 注意下标从1开始，不从0开始 + // 比如你传入size = 8 + // 则位置对应为1~8，而不是0~7 + public static class SegmentTree { + private int[] min; + private int[] change; + private boolean[] update; + + public SegmentTree(int size) { + int N = size + 1; + min = new int[N << 2]; + change = new int[N << 2]; + update = new boolean[N << 2]; + update(1, size, Integer.MAX_VALUE, 1, size, 1); + } + + private void pushUp(int rt) { + min[rt] = Math.min(min[rt << 1], min[rt << 1 | 1]); + } + + private void pushDown(int rt, int ln, int rn) { + if (update[rt]) { + update[rt << 1] = true; + update[rt << 1 | 1] = true; + change[rt << 1] = change[rt]; + change[rt << 1 | 1] = change[rt]; + min[rt << 1] = change[rt]; + min[rt << 1 | 1] = change[rt]; + update[rt] = false; + } + } + + // 最后三个参数是固定的, 每次传入相同的值即可: + // l = 1(固定) + // r = size(你设置的线段树大小) + // rt = 1(固定) + public void update(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + update[rt] = true; + change[rt] = C; + min[rt] = C; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + update(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + update(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + // 最后三个参数是固定的, 每次传入相同的值即可: + // l = 1(固定) + // r = size(你设置的线段树大小) + // rt = 1(固定) + public int query(int L, int R, int l, int r, int rt) { + if (L <= l && r <= R) { + return min[rt]; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + int left = Integer.MAX_VALUE; + int right = Integer.MAX_VALUE; + if (L <= mid) { + left = query(L, R, l, mid, rt << 1); + } + if (R > mid) { + right = query(L, R, mid + 1, r, rt << 1 | 1); + } + return Math.min(left, right); + } + + } + + // 为了测试 + public static int[][] randomMatrix(int n, int m, int v) { + int[][] ans = new int[n][m]; + for (int i = 0; i < n; i++) { + for (int j = 0; j < m; j++) { + ans[i][j] = (int) (Math.random() * v); + } + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + // 先展示一下线段树的用法，假设N=100 + // 初始化时，1~100所有位置的值都是系统最大 + System.out.println("线段树展示开始"); + int N = 100; + SegmentTree st = new SegmentTree(N); + // 查询8~19范围上的最小值 + System.out.println(st.query(8, 19, 1, N, 1)); + // 把6~14范围上对应的值都修改成56 + st.update(6, 14, 56, 1, N, 1); + // 查询8~19范围上的最小值 + System.out.println(st.query(8, 19, 1, N, 1)); + // 以上是线段树的用法，你可以随意使用update和query方法 + // 线段树的详解请看体系学习班 + System.out.println("线段树展示结束"); + + // 以下为正式测试 + int len = 10; + int value = 8; + int testTimes = 10000; + System.out.println("对数器测试开始"); + for (int i = 0; i < testTimes; i++) { + int n = (int) (Math.random() * len) + 1; + int m = (int) (Math.random() * len) + 1; + int[][] map = randomMatrix(n, m, value); + int ans1 = jump1(map); + int ans2 = jump2(map); + if (ans1 != ans2) { + System.out.println("出错了!"); + } + } + System.out.println("对数器测试结束"); + } + +} diff --git a/大厂刷题班/class39/Code05_0123Disappear.java b/大厂刷题班/class39/Code05_0123Disappear.java new file mode 100644 index 0000000..c1b64aa --- /dev/null +++ b/大厂刷题班/class39/Code05_0123Disappear.java @@ -0,0 +1,82 @@ +package class39; + +// 真实笔试，忘了哪个公司，但是绝对大厂 +// 一个子序列的消除规则如下: +// 1) 在某一个子序列中，如果'1'的左边有'0'，那么这两个字符->"01"可以消除 +// 2) 在某一个子序列中，如果'3'的左边有'2'，那么这两个字符->"23"可以消除 +// 3) 当这个子序列的某个部分消除之后，认为其他字符会自动贴在一起，可以继续寻找消除的机会 +// 比如，某个子序列"0231"，先消除掉"23"，那么剩下的字符贴在一起变成"01"，继续消除就没有字符了 +// 如果某个子序列通过最优良的方式，可以都消掉，那么这样的子序列叫做“全消子序列” +// 一个只由'0'、'1'、'2'、'3'四种字符组成的字符串str，可以生成很多子序列，返回“全消子序列”的最大长度 +// 字符串str长度 <= 200 +// 体系学习班，代码46节，第2题+第3题 +public class Code05_0123Disappear { + + // str[L...R]上，都能消掉的子序列，最长是多少？ + public static int f(char[] str, int L, int R) { + if (L >= R) { + return 0; + } + if (L == R - 1) { + return (str[L] == '0' && str[R] == '1') || (str[L] == '2' && str[R] == '3') ? 2 : 0; + } + // L...R 有若干个字符 > 2 + // str[L...R]上，都能消掉的子序列，最长是多少？ + // 可能性1，能消掉的子序列完全不考虑str[L]，最长是多少？ + int p1 = f(str, L + 1, R); + if (str[L] == '1' || str[L] == '3') { + return p1; + } + // str[L] =='0' 或者 '2' + // '0' 去找 '1' + // '2' 去找 '3' + char find = str[L] == '0' ? '1' : '3'; + int p2 = 0; + // L() ...... + for (int i = L + 1; i <= R; i++) { + // L(0) ..... i(1) i+1....R + if (str[i] == find) { + p2 = Math.max(p2, f(str, L + 1, i - 1) + 2 + f(str, i + 1, R)); + } + } + return Math.max(p1, p2); + } + + public static int maxDisappear(String str) { + if (str == null || str.length() == 0) { + return 0; + } + return disappear(str.toCharArray(), 0, str.length() - 1); + } + + // s[l..r]范围上，如题目所说的方式，最长的都能消掉的子序列长度 + public static int disappear(char[] s, int l, int r) { + if (l >= r) { + return 0; + } + if (l == r - 1) { + return (s[l] == '0' && s[r] == '1') || (s[l] == '2' && s[r] == '3') ? 2 : 0; + } + int p1 = disappear(s, l + 1, r); + if (s[l] == '1' || s[l] == '3') { + return p1; + } + int p2 = 0; + char find = s[l] == '0' ? '1' : '3'; + for (int i = l + 1; i <= r; i++) { + if (s[i] == find) { + p2 = Math.max(p2, disappear(s, l + 1, i - 1) + 2 + disappear(s, i + 1, r)); + } + } + return Math.max(p1, p2); + } + + public static void main(String[] args) { + String str1 = "010101"; + System.out.println(maxDisappear(str1)); + + String str2 = "021331"; + System.out.println(maxDisappear(str2)); + } + +} diff --git a/大厂刷题班/class40/Code01_SplitTo01.java b/大厂刷题班/class40/Code01_SplitTo01.java new file mode 100644 index 0000000..25713fc --- /dev/null +++ b/大厂刷题班/class40/Code01_SplitTo01.java @@ -0,0 +1,195 @@ +package class40; + +import java.util.ArrayList; +import java.util.HashMap; + +// 腾讯 +// 分裂问题 +// 一个数n，可以分裂成一个数组[n/2, n%2, n/2] +// 这个数组中哪个数不是1或者0，就继续分裂下去 +// 比如 n = 5，一开始分裂成[2, 1, 2] +// [2, 1, 2]这个数组中不是1或者0的数，会继续分裂下去，比如两个2就继续分裂 +// [2, 1, 2] -> [1, 0, 1, 1, 1, 0, 1] +// 那么我们说，5最后分裂成[1, 0, 1, 1, 1, 0, 1] +// 每一个数都可以这么分裂，在最终分裂的数组中，假设下标从1开始 +// 给定三个数n、l、r，返回n的最终分裂数组里[l,r]范围上有几个1 +// n <= 2 ^ 50，n是long类型 +// r - l <= 50000，l和r是int类型 +// 我们的课加个码: +// n是long类型随意多大都行 +// l和r也是long类型随意多大都行，但要保证l<=r +public class Code01_SplitTo01 { + +// public static long nums3(long n, long l, long r) { +// HashMap lenMap = new HashMap<>(); +// len(n, lenMap); +// HashMap onesMap = new HashMap<>(); +// ones(n, onesMap); +// } + + // n = 100 + // n = 100, 最终裂变的数组，长度多少？ + // n = 50, 最终裂变的数组，长度多少？ + // n = 25, 最终裂变的数组，长度多少？ + // .. + // n = 1 ,.最终裂变的数组，长度多少？ + // 请都填写到lenMap中去！ + public static long len(long n, HashMap lenMap) { + if (n == 1 || n == 0) { + lenMap.put(n, 1L); + return 1; + } else { + // n > 1 + long half = len(n / 2, lenMap); + long all = half * 2 + 1; + lenMap.put(n, all); + return all; + } + } + + // n = 100 + // n = 100, 最终裂变的数组中，一共有几个1 + // n = 50, 最终裂变的数组，一共有几个1 + // n = 25, 最终裂变的数组，一共有几个1 + // .. + // n = 1 ,.最终裂变的数组，一共有几个1 + // 请都填写到onesMap中去！ + public static long ones(long num, HashMap onesMap) { + if (num == 1 || num == 0) { + onesMap.put(num, num); + return num; + } + // n > 1 + long half = ones(num / 2, onesMap); + long mid = num % 2 == 1 ? 1 : 0; + long all = half * 2 + mid; + onesMap.put(num, all); + return all; + } + + // + + public static long nums1(long n, long l, long r) { + if (n == 1 || n == 0) { + return n == 1 ? 1 : 0; + } + long half = size(n / 2); + long left = l > half ? 0 : nums1(n / 2, l, Math.min(half, r)); + long mid = (l > half + 1 || r < half + 1) ? 0 : (n & 1); + long right = r > half + 1 ? nums1(n / 2, Math.max(l - half - 1, 1), r - half - 1) : 0; + return left + mid + right; + } + + public static long size(long n) { + if (n == 1 || n == 0) { + return 1; + } else { + long half = size(n / 2); + return (half << 1) + 1; + } + } + + public static long nums2(long n, long l, long r) { + HashMap allMap = new HashMap<>(); + return dp(n, l, r, allMap); + } + + public static long dp(long n, long l, long r, HashMap allMap) { + if (n == 1 || n == 0) { + return n == 1 ? 1 : 0; + } + long half = size(n / 2); + long all = (half << 1) + 1; + long mid = n & 1; + if (l == 1 && r >= all) { + if (allMap.containsKey(n)) { + return allMap.get(n); + } else { + long count = dp(n / 2, 1, half, allMap); + long ans = (count << 1) + mid; + allMap.put(n, ans); + return ans; + } + } else { + mid = (l > half + 1 || r < half + 1) ? 0 : mid; + long left = l > half ? 0 : dp(n / 2, l, Math.min(half, r), allMap); + long right = r > half + 1 ? dp(n / 2, Math.max(l - half - 1, 1), r - half - 1, allMap) : 0; + return left + mid + right; + } + } + + // 为了测试 + // 彻底生成n的最终分裂数组返回 + public static ArrayList test(long n) { + ArrayList arr = new ArrayList<>(); + process(n, arr); + return arr; + } + + public static void process(long n, ArrayList arr) { + if (n == 1 || n == 0) { + arr.add((int) n); + } else { + process(n / 2, arr); + arr.add((int) (n % 2)); + process(n / 2, arr); + } + } + + public static void main(String[] args) { + long num = 671; + ArrayList ans = test(num); + int testTime = 10000; + System.out.println("功能测试开始"); + for (int i = 0; i < testTime; i++) { + int a = (int) (Math.random() * ans.size()) + 1; + int b = (int) (Math.random() * ans.size()) + 1; + int l = Math.min(a, b); + int r = Math.max(a, b); + int ans1 = 0; + for (int j = l - 1; j < r; j++) { + if (ans.get(j) == 1) { + ans1++; + } + } + long ans2 = nums1(num, l, r); + long ans3 = nums2(num, l, r); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("出错了!"); + } + } + System.out.println("功能测试结束"); + System.out.println("=============="); + + System.out.println("性能测试开始"); + num = (2L << 50) + 22517998136L; + long l = 30000L; + long r = 800000200L; + long start; + long end; + start = System.currentTimeMillis(); + System.out.println("nums1结果 : " + nums1(num, l, r)); + end = System.currentTimeMillis(); + System.out.println("nums1花费时间(毫秒) : " + (end - start)); + + start = System.currentTimeMillis(); + System.out.println("nums2结果 : " + nums2(num, l, r)); + end = System.currentTimeMillis(); + System.out.println("nums2花费时间(毫秒) : " + (end - start)); + System.out.println("性能测试结束"); + System.out.println("=============="); + + System.out.println("单独展示nums2方法强悍程度测试开始"); + num = (2L << 55) + 22517998136L; + l = 30000L; + r = 6431000002000L; + start = System.currentTimeMillis(); + System.out.println("nums2结果 : " + nums2(num, l, r)); + end = System.currentTimeMillis(); + System.out.println("nums2花费时间(毫秒) : " + (end - start)); + System.out.println("单独展示nums2方法强悍程度测试结束"); + System.out.println("=============="); + + } + +} diff --git a/大厂刷题班/class40/Code02_Mod3Max.java b/大厂刷题班/class40/Code02_Mod3Max.java new file mode 100644 index 0000000..6bbc985 --- /dev/null +++ b/大厂刷题班/class40/Code02_Mod3Max.java @@ -0,0 +1,247 @@ +package class40; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.TreeSet; + +// 来自去哪儿网 +// 给定一个arr，里面的数字都是0~9 +// 你可以随意使用arr中的数字，哪怕打乱顺序也行 +// 请拼出一个能被3整除的，最大的数字，用str形式返回 +public class Code02_Mod3Max { + + public static String max1(int[] arr) { + Arrays.sort(arr); + for (int l = 0, r = arr.length - 1; l < r; l++, r--) { + int tmp = arr[l]; + arr[l] = arr[r]; + arr[r] = tmp; + } + StringBuilder builder = new StringBuilder(); + TreeSet set = new TreeSet<>((a, b) -> Integer.valueOf(b).compareTo(Integer.valueOf(a))); + process1(arr, 0, builder, set); + return set.isEmpty() ? "" : set.first(); + } + + public static void process1(int[] arr, int index, StringBuilder builder, TreeSet set) { + if (index == arr.length) { + if (builder.length() != 0 && Integer.valueOf(builder.toString()) % 3 == 0) { + set.add(builder.toString()); + } + } else { + process1(arr, index + 1, builder, set); + builder.append(arr[index]); + process1(arr, index + 1, builder, set); + builder.deleteCharAt(builder.length() - 1); + } + } + + public static String max2(int[] arr) { + if (arr == null || arr.length == 0) { + return ""; + } + Arrays.sort(arr); + for (int l = 0, r = arr.length - 1; l < r; l++, r--) { + int tmp = arr[l]; + arr[l] = arr[r]; + arr[r] = tmp; + } + if (arr[0] == 0) { + return "0"; + } + String ans = process2(arr, 0, 0); + String res = ans.replaceAll("^(0+)", ""); + if (!res.equals("")) { + return res; + } + return ans.equals("") ? ans : "0"; + } + + // arr中的数字一定是0~9 + // arr是经过排序的，并且是从大到小排序，比如[9,8,7,7,7,3,1]等 + // 这个递归函数的含义 : + // 在arr[index...一直到最后]上做选择，arr[0...index-1]就当不存在 + // 每个位置的字符可以要、也可以不要，但是！选出来的数字拼完之后的结果，在%3之后，余数一定要是mod！ + // 返回在上面设定的情况下，最大的数是多少？ + // 如果存在这样的数，返回字符串的形式 + // 如果不存在这样的数，返回特殊字符串，比如"$"，代表不可能 + // 这个递归函数可以很轻易的改出动态规划 + public static String process2(int[] arr, int index, int mod) { + if (index == arr.length) { + return mod == 0 ? "" : "$"; + } + String p1 = "$"; + int nextMod = nextMod(mod, arr[index] % 3); + String next = process2(arr, index + 1, nextMod); + if (!next.equals("$")) { + p1 = String.valueOf(arr[index]) + next; + } + String p2 = process2(arr, index + 1, mod); + if (p1.equals("$") && p2.equals("$")) { + return "$"; + } + if (!p1.equals("$") && !p2.equals("$")) { + return smaller(p1, p2) ? p2 : p1; + } + return p1.equals("$") ? p2 : p1; + } + + public static int nextMod(int require, int current) { + if (require == 0) { + if (current == 0) { + return 0; + } else if (current == 1) { + return 2; + } else { + return 1; + } + } else if (require == 1) { + if (current == 0) { + return 1; + } else if (current == 1) { + return 0; + } else { + return 2; + } + } else { // require == 2 + if (current == 0) { + return 2; + } else if (current == 1) { + return 1; + } else { + return 0; + } + } + } + + public static boolean smaller(String p1, String p2) { + if (p1.length() != p2.length()) { + return p1.length() < p2.length(); + } + return p1.compareTo(p2) < 0; + } + + // 贪心的思路解法 : + // 先得到数组的累加和，记为sum + // 1) 如果sum%3==0，说明所有数从大到小拼起来就可以了 + // 2) 如果sum%3==1，说明多了一个余数1， + // 只需要删掉一个最小的数(该数是%3==1的数); + // 如果没有，只需要删掉两个最小的数(这两个数都是%3==2的数); + // 3) 如果sum%3==2，说明多了一个余数2， + // 只需要删掉一个最小的数(该数是%3==2的数); + // 如果没有，只需要删掉两个最小的数(这两个数都是%3==1的数); + // 如果上面都做不到，说明拼不成 + public static String max3(int[] A) { + if (A == null || A.length == 0) { + return ""; + } + int mod = 0; + ArrayList arr = new ArrayList<>(); + for (int num : A) { + arr.add(num); + mod += num; + mod %= 3; + } + if ((mod == 1 || mod == 2) && !remove(arr, mod, 3 - mod)) { + return ""; + } + if (arr.isEmpty()) { + return ""; + } + arr.sort((a, b) -> b - a); + if (arr.get(0) == 0) { + return "0"; + } + StringBuilder builder = new StringBuilder(); + for (int num : arr) { + builder.append(num); + } + return builder.toString(); + } + + // 在arr中，要么删掉最小的一个、且%3之后余数是first的数 + // 如果做不到，删掉最小的两个、且%3之后余数是second的数 + // 如果能做到返回true，不能做到返回false + public static boolean remove(ArrayList arr, int first, int second) { + if (arr.size() == 0) { + return false; + } + arr.sort((a, b) -> compare(a, b, first, second)); + int size = arr.size(); + if (arr.get(size - 1) % 3 == first) { + arr.remove(size - 1); + return true; + } else if (size > 1 && arr.get(size - 1) % 3 == second && arr.get(size - 2) % 3 == second) { + arr.remove(size - 1); + arr.remove(size - 2); + return true; + } else { + return false; + } + } + + // a和b比较: + // 如果余数一样，谁大谁放前面 + // 如果余数不一样，余数是0的放最前面、余数是s的放中间、余数是f的放最后 + public static int compare(int a, int b, int f, int s) { + int ma = a % 3; + int mb = b % 3; + if (ma == mb) { + return b - a; + } else { + if (ma == 0 || mb == 0) { + return ma == 0 ? -1 : 1; + } else { + return ma == s ? -1 : 1; + } + } + } + + // 为了测试 + public static int[] randomArray(int len) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * 10); + } + return arr; + } + + // 为了测试 + public static int[] copyArray(int[] arr) { + int[] ans = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + ans[i] = arr[i]; + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int N = 10; + int testTimes = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + int len = (int) (Math.random() * N); + int[] arr1 = randomArray(len); + int[] arr2 = copyArray(arr1); + int[] arr3 = copyArray(arr1); + String ans1 = max1(arr1); + String ans2 = max2(arr2); + String ans3 = max3(arr3); + if (!ans1.equals(ans2) || !ans1.equals(ans3)) { + System.out.println("出错了！"); + for (int num : arr3) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + break; + } + } + System.out.println("测试结束"); + + } + +} diff --git a/大厂刷题班/class40/Code03_MaxMeetingScore.java b/大厂刷题班/class40/Code03_MaxMeetingScore.java new file mode 100644 index 0000000..467b08d --- /dev/null +++ b/大厂刷题班/class40/Code03_MaxMeetingScore.java @@ -0,0 +1,109 @@ +package class40; + +import java.util.Arrays; +import java.util.PriorityQueue; + +// 给定int[][] meetings，比如 +// { +// {66, 70} 0号会议截止时间66，获得收益70 +// {25, 90} 1号会议截止时间25，获得收益90 +// {50, 30} 2号会议截止时间50，获得收益30 +// } +// 一开始的时间是0，任何会议都持续10的时间，但是一个会议一定要在该会议截止时间之前开始 +// 只有一个会议室，任何会议不能共用会议室，一旦一个会议被正确安排，将获得这个会议的收益 +// 请返回最大的收益 +public class Code03_MaxMeetingScore { + + public static int maxScore1(int[][] meetings) { + Arrays.sort(meetings, (a, b) -> a[0] - b[0]); + int[][] path = new int[meetings.length][]; + int size = 0; + return process1(meetings, 0, path, size); + } + + public static int process1(int[][] meetings, int index, int[][] path, int size) { + if (index == meetings.length) { + int time = 0; + int ans = 0; + for (int i = 0; i < size; i++) { + if (time + 10 <= path[i][0]) { + ans += path[i][1]; + time += 10; + } else { + return 0; + } + } + return ans; + } + int p1 = process1(meetings, index + 1, path, size); + path[size] = meetings[index]; + int p2 = process1(meetings, index + 1, path, size + 1); + // path[size] = null; + return Math.max(p1, p2); + } + + public static int maxScore2(int[][] meetings) { + Arrays.sort(meetings, (a, b) -> a[0] - b[0]); + PriorityQueue heap = new PriorityQueue<>(); + int time = 0; + // 已经把所有会议，按照截止时间，从小到大，排序了！ + // 截止时间一样的，谁排前谁排后，无所谓 + for (int i = 0; i < meetings.length; i++) { + if (time + 10 <= meetings[i][0]) { + heap.add(meetings[i][1]); + time += 10; + } else { + if (!heap.isEmpty() && heap.peek() < meetings[i][1]) { + heap.poll(); + heap.add(meetings[i][1]); + } + } + } + int ans = 0; + while (!heap.isEmpty()) { + ans += heap.poll(); + } + return ans; + } + + public static int[][] randomMeetings(int n, int t, int s) { + int[][] ans = new int[n][2]; + for (int i = 0; i < n; i++) { + ans[i][0] = (int) (Math.random() * t) + 1; + ans[i][1] = (int) (Math.random() * s) + 1; + } + return ans; + } + + public static int[][] copyMeetings(int[][] meetings) { + int n = meetings.length; + int[][] ans = new int[n][2]; + for (int i = 0; i < n; i++) { + ans[i][0] = meetings[i][0]; + ans[i][1] = meetings[i][1]; + } + return ans; + } + + public static void main(String[] args) { + int n = 12; + int t = 100; + int s = 500; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int size = (int) (Math.random() * n) + 1; + int[][] meetings1 = randomMeetings(size, t, s); + int[][] meetings2 = copyMeetings(meetings1); + int ans1 = maxScore1(meetings1); + int ans2 = maxScore2(meetings2); + if (ans1 != ans2) { + System.out.println("出错了!"); + System.out.println(ans1); + System.out.println(ans2); + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class40/Code04_LetASorted.java b/大厂刷题班/class40/Code04_LetASorted.java new file mode 100644 index 0000000..051bd9b --- /dev/null +++ b/大厂刷题班/class40/Code04_LetASorted.java @@ -0,0 +1,54 @@ +package class40; + +// 给定两个数组A和B，长度都是N +// A[i]不可以在A中和其他数交换，只可以选择和B[i]交换(0<=i= lastA && process1(A, B, i + 1, A[i])) { + return true; + } + // 第二种选择 : A[i]和B[i]交换 + if (B[i] >= lastA && process1(A, B, i + 1, B[i])) { + return true; + } + return false; + } + + public static boolean letASorted2(int[] A, int[] B) { + return process2(A, B, 0, true); + } + + // 当前推进到了i位置，对于A和B都是i位置 + // A[i]前一个数字是否来自A ： + // 如果来自A，fromA = true；如果来自B，fromA = false； + // 能否通过题意中的操作，A[i] B[i] 让A整体有序 + // 好处：可变参数成了int + boolean，时间复杂度可以做到O(N) + public static boolean process2(int[] A, int[] B, int i, boolean fromA) { + if (i == A.length) { + return true; + } + if (i == 0 || (A[i] >= (fromA ? A[i - 1] : B[i - 1])) && process2(A, B, i + 1, true)) { + return true; + } + if (i == 0 || (B[i] >= (fromA ? A[i - 1] : B[i - 1])) && process2(A, B, i + 1, false)) { + return true; + } + return false; + } + + // 也可以彻底的贪心：就让A此时的值尽量小！也是可以的。时间复杂度O(N) + +} diff --git a/大厂刷题班/class40/Code05_AllSame.java b/大厂刷题班/class40/Code05_AllSame.java new file mode 100644 index 0000000..ae8359b --- /dev/null +++ b/大厂刷题班/class40/Code05_AllSame.java @@ -0,0 +1,135 @@ +package class40; + +// 来自腾讯 +// 比如arr = {3,1,2,4} +// 下标对应是：0 1 2 3 +// 你最开始选择一个下标进行操作，一旦最开始确定了是哪个下标，以后都只能在这个下标上进行操作 +// 比如你选定1下标，1下标上面的数字是1，你可以选择变化这个数字，比如你让这个数字变成2 +// 那么arr = {3,2,2,4} +// 下标对应是：0 1 2 3 +// 因为你最开始确定了1这个下标，所以你以后都只能对这个下标进行操作， +// 但是，和你此时下标上的数字一样的、且位置连成一片的数字，会跟着一起变 +// 比如你选择让此时下标1的数字2变成3， +// 那么arr = {3,3,3,4} 可以看到下标1和下标2的数字一起变成3，这是规则！一定会一起变 +// 下标对应是：0 1 2 3 +// 接下来，你还是只能对1下标进行操作，那么数字一样的、且位置连成一片的数字(arr[0~2]这个范围)都会一起变 +// 决定变成4 +// 那么arr = {4,4,4,4} +// 下标对应是：0 1 2 3 +// 至此，所有数都成一样的了，你在下标1上做了3个决定(第一次变成2，第二次变成3，第三次变成4)， +// 因为联动规则，arr全刷成一种数字了 +// 给定一个数组arr，最开始选择哪个下标，你随意 +// 你的目的是一定要让arr都成为一种数字，注意联动效果会一直生效 +// 返回最小的变化数 +// arr长度 <= 200, arr中的值 <= 10^6 +public class Code05_AllSame { + + public static int allSame1(int[] arr) { + int ans = Integer.MAX_VALUE; + for (int i = 0; i < arr.length; i++) { + ans = Math.min(ans, process1(arr, i - 1, arr[i], i + 1)); + } + return ans; + } + + // 左边arr[0..left]，如果left == -1，说明没有左边了 + // 右边arr[right...n-1]，如果right == n，说明没有右边了 + // 中间的值是midV，中间的值代表中间一整个部分的值，中间部分有可能是一个数，也可能是一堆数，但是值都为midV + // 返回arr都刷成一样的，最小代价是多少 + // left 可能性 : N + // right 可能性 : N + // midV 可能性 : arr中的最大值！ + public static int process1(int[] arr, int left, int midV, int right) { + for (; left >= 0 && arr[left] == midV;) { + left--; + } + for (; right < arr.length && arr[right] == midV;) { + right++; + } + if (left == -1 && right == arr.length) { + return 0; + } + int p1 = Integer.MAX_VALUE; + if (left >= 0) { + p1 = process1(arr, left - 1, arr[left], right) + 1; + } + int p2 = Integer.MAX_VALUE; + if (right < arr.length) { + p2 = process1(arr, left, arr[right], right + 1) + 1; + } + return Math.min(p1, p2); + } + + public static int allSame2(int[] arr) { + int ans = Integer.MAX_VALUE; + for (int i = 0; i < arr.length; i++) { + ans = Math.min(ans, process2(arr, i - 1, true, i + 1)); + } + return ans; + } + + // 左边arr[0..l]，如果left == -1，说明没有左边了 + // 右边arr[r...n-1]，如果right == n，说明没有右边了 + // 中间的值代表arr[l+1...r-1]这个部分的值已经刷成了一种 + // 中间的值，如果和arr[l+1]一样，isLeft就是true + // 中间的值，如果和arr[r-1]一样，isLeft就是false + // 返回arr都刷成一样的，最小代价是多少 + // left 可能性 : N + // right 可能性 : N + // isLeft 可能性 : true/false,两种 + public static int process2(int[] arr, int l, boolean isLeft, int r) { + int left = l; + for (; left >= 0 && arr[left] == (isLeft ? arr[l + 1] : arr[r - 1]);) { + left--; + } + int right = r; + for (; right < arr.length && arr[right] == (isLeft ? arr[l + 1] : arr[r - 1]);) { + right++; + } + if (left == -1 && right == arr.length) { + return 0; + } + int p1 = Integer.MAX_VALUE; + if (left >= 0) { + p1 = process2(arr, left - 1, true, right) + 1; + } + int p2 = Integer.MAX_VALUE; + if (right < arr.length) { + p2 = process2(arr, left, false, right + 1) + 1; + } + return Math.min(p1, p2); + } + // 如上的递归，请改动态规划，具体参考体系学习班，动态规划大章节！ + + // 为了测试 + public static int[] randomArray(int maxSize, int maxNum) { + int size = 2 + (int) (Math.random() * maxSize); + int[] arr = new int[size]; + for (int i = 0; i < size; i++) { + arr[i] = 1 + (int) (Math.random() * maxSize); + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + System.out.println("测试开始"); + for (int i = 0; i < 10000; i++) { + int[] arr = randomArray(20, 10); + int ans1 = allSame1(arr); + int ans2 = allSame2(arr); + if (ans1 != ans2) { + System.out.println("出错了！！！"); + for (int i1 : arr) { + System.out.print(i1 + " "); + } + System.out.println(); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class41/Code01_MinSwapTimes.java b/大厂刷题班/class41/Code01_MinSwapTimes.java new file mode 100644 index 0000000..70ad912 --- /dev/null +++ b/大厂刷题班/class41/Code01_MinSwapTimes.java @@ -0,0 +1,88 @@ +package class41; + +// 来自小红书 +// 一个无序数组长度为n，所有数字都不一样，并且值都在[0...n-1]范围上 +// 返回让这个无序数组变成有序数组的最小交换次数 +public class Code01_MinSwapTimes { + + // 纯暴力，arr长度大一点都会超时 + // 但是绝对正确 + public static int minSwap1(int[] arr) { + return process1(arr, 0); + } + + // 让arr变有序，最少的交换次数是多少！返回 + // times, 之前已经做了多少次交换 + public static int process1(int[] arr, int times) { + boolean sorted = true; + for (int i = 1; i < arr.length; i++) { + if (arr[i - 1] > arr[i]) { + sorted = false; + break; + } + } + if (sorted) { + return times; + } + // 数组现在是无序的状态！ + if (times >= arr.length - 1) { + return Integer.MAX_VALUE; + } + int ans = Integer.MAX_VALUE; + for (int i = 0; i < arr.length; i++) { + for (int j = i + 1; j < arr.length; j++) { + swap(arr, i, j); + ans = Math.min(ans, process1(arr, times + 1)); + swap(arr, i, j); + } + } + return ans; + } + + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + + // 已知arr中，只有0~n-1这些值，并且都出现1次 + public static int minSwap2(int[] arr) { + int ans = 0; + for (int i = 0; i < arr.length; i++) { + while (i != arr[i]) { + swap(arr, i, arr[i]); + ans++; + } + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int len) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = i; + } + for (int i = 0; i < len; i++) { + swap(arr, i, (int) (Math.random() * len)); + } + return arr; + } + + public static void main(String[] args) { + int n = 6; + int testTime = 2000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * n) + 1; + int[] arr = randomArray(len); + int ans1 = minSwap1(arr); + int ans2 = minSwap2(arr); + if (ans1 != ans2) { + System.out.println("出错了!"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class41/Code02_PoemProblem.java b/大厂刷题班/class41/Code02_PoemProblem.java new file mode 100644 index 0000000..617ca29 --- /dev/null +++ b/大厂刷题班/class41/Code02_PoemProblem.java @@ -0,0 +1,381 @@ +package class41; + +import java.util.Arrays; +import java.util.HashMap; + +// 来自小红书 +// 有四种诗的韵律分别为: AABB、ABAB、ABBA、AAAA +// 比如 : 1 1 3 3就属于AABB型的韵律、6 6 6 6就属于AAAA型的韵律等等 +// 一个数组arr，当然可以生成很多的子序列，如果某个子序列一直以韵律的方式连接起来，我们称这样的子序列是有效的 +// 比如, arr = { 1, 1, 15, 1, 34, 1, 2, 67, 3, 3, 2, 4, 15, 3, 17, 4, 3, 7, 52, 7, 81, 9, 9 } +// arr的一个子序列为{1, 1, 1, 1, 2, 3, 3, 2, 4, 3, 4, 3, 7, 7, 9, 9} +// 其中1, 1, 1, 1是AAAA、2, 3, 3, 2是ABBA、4, 3, 4, 3是ABAB、7, 7, 9, 9是AABB +// 可以看到，整个子序列一直以韵律的方式连接起来，所以这个子序列是有效的 +// 给定一个数组arr, 返回最长的有效子序列长度 +// 题目限制 : arr长度 <= 4000, arr中的值<= 10^9 +// 离散化之后，arr长度 <= 4000, arr中的值<= 4000 +public class Code02_PoemProblem { + + // arr[i.....]符合规则连接的最长子序列长度 +// public static int zuo(int[] arr, int i) { +// if (i + 4 > arr.length) { +// return 0; +// } +// // 最终的，符合规则连接的最长子序列长度，就是不要i位置的字符 +// int p0 = zuo(arr, i + 1); +// // p1使用for循环搞定的！ +// int p1 = 找到arr[i..s]是最短的，且能搞出AABB来的(4个) + zuo(arr, s + 1); +// // p2使用for循环搞定的！ +// int p2 = 找到arr[i..t]是最短的，且能搞出ABAB来的(4个) + zuo(arr, t + 1); +// // p3使用for循环搞定的！ +// int p3 = 找到arr[i..k]是最短的，且能搞出ABBA来的(4个) + zuo(arr, k + 1); +// // p4没用 +// int p4 = 找到arr[i..f]是最短的，且能搞出AAAA来的(4个) + zuo(arr, f + 1); +// return p0~p4的最大值 +// } + + // AABB + // ABAB + // ABBA + // AAAA + public static int maxLen1(int[] arr) { + if (arr == null || arr.length < 4) { + return 0; + } + int[] path = new int[arr.length]; + return process1(arr, 0, path, 0); + } + + public static int process1(int[] arr, int index, int[] path, int size) { + if (index == arr.length) { + if (size % 4 != 0) { + return 0; + } else { + for (int i = 0; i < size; i += 4) { + if (!valid(path, i)) { + return 0; + } + } + return size; + } + } else { + int p1 = process1(arr, index + 1, path, size); + path[size] = arr[index]; + int p2 = process1(arr, index + 1, path, size + 1); + return Math.max(p1, p2); + } + } + + public static boolean valid(int[] p, int i) { + // AABB + // ABAB + // ABBA + // AAAA + return (p[i] == p[i + 1] && p[i + 2] == p[i + 3]) + || (p[i] == p[i + 2] && p[i + 1] == p[i + 3] && p[i] != p[i + 1]) + || (p[i] == p[i + 3] && p[i + 1] == p[i + 2] && p[i] != p[i + 1]); + } + + // 0 : [3,6,9] + // 1 : [2,7,13] + // 2 : [23] + // [ + // [3,6,9] + // ] + public static int maxLen2(int[] arr) { + if (arr == null || arr.length < 4) { + return 0; + } + int n = arr.length; + int[] sorted = Arrays.copyOf(arr, n); + Arrays.sort(sorted); + HashMap vmap = new HashMap<>(); + int index = 0; + vmap.put(sorted[0], index++); + for (int i = 1; i < n; i++) { + if (sorted[i] != sorted[i - 1]) { + vmap.put(sorted[i], index++); + } + } + int[] sizeArr = new int[index]; + for (int i = 0; i < n; i++) { + arr[i] = vmap.get(arr[i]); + sizeArr[arr[i]]++; + } + int[][] imap = new int[index][]; + for (int i = 0; i < index; i++) { + imap[i] = new int[sizeArr[i]]; + } + for (int i = n - 1; i >= 0; i--) { + imap[arr[i]][--sizeArr[arr[i]]] = i; + } + return process2(arr, imap, 0); + } + + // AABB + // ABAB + // ABBA + // AAAA + public static int process2(int[] varr, int[][] imap, int i) { + if (i + 4 > varr.length) { + return 0; + } + int p0 = process2(varr, imap, i + 1); + // AABB + int p1 = 0; + int rightClosedP1A2 = rightClosed(imap, varr[i], i); + if (rightClosedP1A2 != -1) { + for (int next = rightClosedP1A2 + 1; next < varr.length; next++) { + if (varr[i] != varr[next]) { + int rightClosedP1B2 = rightClosed(imap, varr[next], next); + if (rightClosedP1B2 != -1) { + p1 = Math.max(p1, 4 + process2(varr, imap, rightClosedP1B2 + 1)); + } + } + } + } + + // ABAB + int p2 = 0; + for (int p2B1 = i + 1; p2B1 < varr.length; p2B1++) { + if (varr[i] != varr[p2B1]) { + int rightClosedP2A2 = rightClosed(imap, varr[i], p2B1); + if (rightClosedP2A2 != -1) { + int rightClosedP2B2 = rightClosed(imap, varr[p2B1], rightClosedP2A2); + if (rightClosedP2B2 != -1) { + p2 = Math.max(p2, 4 + process2(varr, imap, rightClosedP2B2 + 1)); + } + } + } + } + + // ABBA + int p3 = 0; + for (int p3B1 = i + 1; p3B1 < varr.length; p3B1++) { + if (varr[i] != varr[p3B1]) { + int rightClosedP3B2 = rightClosed(imap, varr[p3B1], p3B1); + if (rightClosedP3B2 != -1) { + int rightClosedP3A2 = rightClosed(imap, varr[i], rightClosedP3B2); + if (rightClosedP3A2 != -1) { + p3 = Math.max(p3, 4 + process2(varr, imap, rightClosedP3A2 + 1)); + } + } + } + } + // AAAA + int p4 = 0; + int rightClosedP4A2 = rightClosed(imap, varr[i], i); + int rightClosedP4A3 = rightClosedP4A2 == -1 ? -1 : rightClosed(imap, varr[i], rightClosedP4A2); + int rightClosedP4A4 = rightClosedP4A3 == -1 ? -1 : rightClosed(imap, varr[i], rightClosedP4A3); + if (rightClosedP4A4 != -1) { + p4 = Math.max(p4, 4 + process2(varr, imap, rightClosedP4A4 + 1)); + } + return Math.max(p0, Math.max(Math.max(p1, p2), Math.max(p3, p4))); + } + + public static int rightClosed(int[][] imap, int v, int i) { + int left = 0; + int right = imap[v].length - 1; + int ans = -1; + while (left <= right) { + int mid = (left + right) / 2; + if (imap[v][mid] <= i) { + left = mid + 1; + } else { + ans = mid; + right = mid - 1; + } + } + return ans == -1 ? -1 : imap[v][ans]; + } + + public static int maxLen3(int[] arr) { + if (arr == null || arr.length < 4) { + return 0; + } + int n = arr.length; + int[] sorted = Arrays.copyOf(arr, n); + Arrays.sort(sorted); + HashMap vmap = new HashMap<>(); + int index = 0; + vmap.put(sorted[0], index++); + for (int i = 1; i < n; i++) { + if (sorted[i] != sorted[i - 1]) { + vmap.put(sorted[i], index++); + } + } + int[] sizeArr = new int[index]; + for (int i = 0; i < n; i++) { + arr[i] = vmap.get(arr[i]); + sizeArr[arr[i]]++; + } + int[][] imap = new int[index][]; + for (int i = 0; i < index; i++) { + imap[i] = new int[sizeArr[i]]; + } + for (int i = n - 1; i >= 0; i--) { + imap[arr[i]][--sizeArr[arr[i]]] = i; + } + int[] dp = new int[n + 1]; + for (int i = n - 4; i >= 0; i--) { + int p0 = dp[i + 1]; + // AABB + int p1 = 0; + int rightClosedP1A2 = rightClosed(imap, arr[i], i); + if (rightClosedP1A2 != -1) { + for (int next = rightClosedP1A2 + 1; next < arr.length; next++) { + if (arr[i] != arr[next]) { + int rightClosedP1B2 = rightClosed(imap, arr[next], next); + if (rightClosedP1B2 != -1) { + p1 = Math.max(p1, 4 + dp[rightClosedP1B2 + 1]); + } + } + } + } + + // ABAB + int p2 = 0; + for (int p2B1 = i + 1; p2B1 < arr.length; p2B1++) { + if (arr[i] != arr[p2B1]) { + int rightClosedP2A2 = rightClosed(imap, arr[i], p2B1); + if (rightClosedP2A2 != -1) { + int rightClosedP2B2 = rightClosed(imap, arr[p2B1], rightClosedP2A2); + if (rightClosedP2B2 != -1) { + p2 = Math.max(p2, 4 + dp[rightClosedP2B2 + 1]); + } + } + } + } + + // ABBA + int p3 = 0; + for (int p3B1 = i + 1; p3B1 < arr.length; p3B1++) { + if (arr[i] != arr[p3B1]) { + int rightClosedP3B2 = rightClosed(imap, arr[p3B1], p3B1); + if (rightClosedP3B2 != -1) { + int rightClosedP3A2 = rightClosed(imap, arr[i], rightClosedP3B2); + if (rightClosedP3A2 != -1) { + p3 = Math.max(p3, 4 + dp[rightClosedP3A2 + 1]); + } + } + } + } + // AAAA + int p4 = 0; + int rightClosedP4A2 = rightClosed(imap, arr[i], i); + int rightClosedP4A3 = rightClosedP4A2 == -1 ? -1 : rightClosed(imap, arr[i], rightClosedP4A2); + int rightClosedP4A4 = rightClosedP4A3 == -1 ? -1 : rightClosed(imap, arr[i], rightClosedP4A3); + if (rightClosedP4A4 != -1) { + p4 = Math.max(p4, 4 + dp[rightClosedP4A4 + 1]); + } + dp[i] = Math.max(p0, Math.max(Math.max(p1, p2), Math.max(p3, p4))); + } + return dp[0]; + } + + // 课堂有同学提出了贪心策略（这题还真是有贪心策略），是正确的 + // AABB + // ABAB + // ABBA + // AAAA + // 先看前三个规则：AABB、ABAB、ABBA + // 首先A、A、B、B的全排列为: + // AABB -> AABB + // ABAB -> ABAB + // ABBA -> ABBA + // BBAA -> 等同于AABB，因为A和B谁在前、谁在后都算是 : AABB的范式 + // BABA -> 等同于ABAB，因为A和B谁在前、谁在后都算是 : ABAB的范式 + // BAAB -> 等同于ABBA，因为A和B谁在前、谁在后都算是 : ABBA的范式 + // 也就是说，AABB、ABAB、ABBA这三个规则，可以这么用： + // 只要有两个不同的数，都出现2次，那么这一共4个数就一定符合韵律规则。 + // 所以： + // 1) 当来到arr中的一个数字num的时候， + // 如果num已经出现了2次了, 只要之前还有一个和num不同的数， + // 也出现了两次，则一定符合了某个规则, 长度直接+4，然后清空所有的统计 + // 2) 当来到arr中的一个数字num的时候, + // 如果num已经出现了4次了(规则四), 长度直接+4，然后清空所有的统计 + // 但是如果我去掉某个规则，该贪心直接报废，比如韵律规则变成: + // AABB、ABAB、AAAA + // 因为少了ABBA, 所以上面的化简不成立了, 得重新分析新规则下的贪心策略 + // 而尝试的方法就更通用(也就是maxLen3)，只是减少一个分支而已 + // 这个贪心费了很多心思，值得点赞！ + public static int maxLen4(int[] arr) { + // 统计某个数(key)，出现的次数(value) + HashMap map = new HashMap<>(); + // tow代表目前有多少数出现了2次 + int two = 0; + // ans代表目前符合韵律链接的子序列增长到了多长 + int ans = 0; + // 当前的num出现了几次 + int numTimes = 0; + for (int num : arr) { + // 对当前的num，做次数统计 + map.put(num, map.getOrDefault(num, 0) + 1); + // 把num出现的次数拿出来 + numTimes = map.get(num); + // 如果num刚刚出现了2次, 那么目前出现了2次的数，的数量，需要增加1个 + two += numTimes == 2 ? 1 : 0; + // 下面的if代表 : + // 如果目前有2个数出现2次了，可以连接了 + // 如果目前有1个数出现4次了，可以连接了 + if (two == 2 || numTimes == 4) { + ans += 4; + map.clear(); + two = 0; + } + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int len, int value) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * value); + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + + // 1111 2332 4343 7799 + int[] test = { 1, 1, 15, 1, 34, 1, 2, 67, 3, 3, 2, 4, 15, 3, 17, 4, 3, 7, 52, 7, 81, 9, 9 }; + System.out.println(maxLen1(test)); + System.out.println(maxLen2(test)); + System.out.println(maxLen3(test)); + System.out.println(maxLen4(test)); + System.out.println("==========="); + + int len = 16; + int value = 10; + int[] arr = randomArray(len, value); + int[] arr1 = Arrays.copyOf(arr, arr.length); + int[] arr2 = Arrays.copyOf(arr, arr.length); + int[] arr3 = Arrays.copyOf(arr, arr.length); + int[] arr4 = Arrays.copyOf(arr, arr.length); + System.out.println(maxLen1(arr1)); + System.out.println(maxLen2(arr2)); + System.out.println(maxLen3(arr3)); + System.out.println(maxLen4(arr4)); + + System.out.println("==========="); + + long start; + long end; + int[] longArr = randomArray(4000, 20); + start = System.currentTimeMillis(); + System.out.println(maxLen3(longArr)); + end = System.currentTimeMillis(); + System.out.println("运行时间(毫秒) : " + (end - start)); + System.out.println("==========="); + + start = System.currentTimeMillis(); + System.out.println(maxLen4(longArr)); + end = System.currentTimeMillis(); + System.out.println("运行时间(毫秒) : " + (end - start)); + System.out.println("==========="); + + } + +} diff --git a/大厂刷题班/class41/Code03_MagicGoToAim.java b/大厂刷题班/class41/Code03_MagicGoToAim.java new file mode 100644 index 0000000..cf6c447 --- /dev/null +++ b/大厂刷题班/class41/Code03_MagicGoToAim.java @@ -0,0 +1,71 @@ +package class41; + +import java.util.PriorityQueue; + +// 来自网易互娱 +// N个结点之间，表世界存在双向通行的道路，里世界存在双向通行的传送门. +// 若走表世界的道路，花费一分钟. +// 若走里世界的传送门，不花费时间，但是接下来一分钟不能走传送门. +// 输入： T为测试用例的组数，对于每组数据: +// 第一行：N M1 M2 N代表结点的个数1到N +// 接下来M1行 每行两个数，u和v，表示表世界u和v之间存在道路 +// 接下来M2行 每行两个数，u和v，表示里世界u和v之间存在传送门 +// 现在处于1号结点，最终要到达N号结点，求最小的到达时间 保证所有输入均有效，不存在环等情况 +public class Code03_MagicGoToAim { + + // 城市编号从0开始，编号对应0~n-1 + // roads[i]是一个数组，表示i能走路达到的城市有哪些，每条路都花费1分钟 + // gates[i]是一个数组，表示i能传送达到的城市有哪些 + // 返回从0到n-1的最少用时 + public static int fast(int n, int[][] roads, int[][] gates) { + int[][] distance = new int[2][n]; + // 因为从0开始走，所以distance[0][0] = 0, distance[1][0] = 0 + // distance[0][i] -> 0 : 前一个城市到达i，是走路的方式, 最小代价，distance[0][i] + // distance[1][i] -> 1 : 前一个城市到达i，是传送的方式, 最小代价，distance[1][i] + for (int i = 1; i < n; i++) { + distance[0][i] = Integer.MAX_VALUE; + distance[1][i] = Integer.MAX_VALUE; + } + // 小根堆，根据距离排序，距离小的点，在上！ + PriorityQueue heap = new PriorityQueue<>((a, b) -> a.cost - b.cost); + heap.add(new Node(0, 0, 0)); + boolean[][] visited = new boolean[2][n]; + while (!heap.isEmpty()) { + Node cur = heap.poll(); + if (visited[cur.preTransfer][cur.city]) { + continue; + } + visited[cur.preTransfer][cur.city] = true; + // 走路的方式 + for (int next : roads[cur.city]) { + if (distance[0][next] > cur.cost + 1) { + distance[0][next] = cur.cost + 1; + heap.add(new Node(0, next, distance[0][next])); + } + } + // 传送的方式 + if (cur.preTransfer == 0) { + for (int next : gates[cur.city]) { + if (distance[1][next] > cur.cost) { + distance[1][next] = cur.cost; + heap.add(new Node(1, next, distance[1][next])); + } + } + } + } + return Math.min(distance[0][n - 1], distance[1][n - 1]); + } + + public static class Node { + public int preTransfer; + public int city; + public int cost; + + public Node(int a, int b, int c) { + preTransfer = a; + city = b; + cost = c; + } + } + +} \ No newline at end of file diff --git a/大厂刷题班/class41/Problem_0031_NextPermutation.java b/大厂刷题班/class41/Problem_0031_NextPermutation.java new file mode 100644 index 0000000..5848f43 --- /dev/null +++ b/大厂刷题班/class41/Problem_0031_NextPermutation.java @@ -0,0 +1,44 @@ +package class41; + +public class Problem_0031_NextPermutation { + + public static void nextPermutation(int[] nums) { + int N = nums.length; + // 从右往左第一次降序的位置 + int firstLess = -1; + for (int i = N - 2; i >= 0; i--) { + if (nums[i] < nums[i + 1]) { + firstLess = i; + break; + } + } + if (firstLess < 0) { + reverse(nums, 0, N - 1); + } else { + int rightClosestMore = -1; + // 找最靠右的、同时比nums[firstLess]大的数，位置在哪 + // 这里其实也可以用二分优化，但是这种优化无关紧要了 + for (int i = N - 1; i > firstLess; i--) { + if (nums[i] > nums[firstLess]) { + rightClosestMore = i; + break; + } + } + swap(nums, firstLess, rightClosestMore); + reverse(nums, firstLess + 1, N - 1); + } + } + + public static void reverse(int[] nums, int L, int R) { + while (L < R) { + swap(nums, L++, R--); + } + } + + public static void swap(int[] nums, int i, int j) { + int tmp = nums[i]; + nums[i] = nums[j]; + nums[j] = tmp; + } + +} diff --git a/大厂刷题班/class42/Problem_0265_PaintHouseII.java b/大厂刷题班/class42/Problem_0265_PaintHouseII.java new file mode 100644 index 0000000..a89f7d9 --- /dev/null +++ b/大厂刷题班/class42/Problem_0265_PaintHouseII.java @@ -0,0 +1,56 @@ +package class42; + +public class Problem_0265_PaintHouseII { + + // costs[i][k] i号房子用k颜色刷的花费 + // 要让0...N-1的房子相邻不同色 + // 返回最小花费 + public static int minCostII(int[][] costs) { + int N = costs.length; + if (N == 0) { + return 0; + } + int K = costs[0].length; + // 之前取得的最小代价、取得最小代价时的颜色 + int preMin1 = 0; + int preEnd1 = -1; + // 之前取得的次小代价、取得次小代价时的颜色 + int preMin2 = 0; + int preEnd2 = -1; + for (int i = 0; i < N; i++) { // i房子 + int curMin1 = Integer.MAX_VALUE; + int curEnd1 = -1; + int curMin2 = Integer.MAX_VALUE; + int curEnd2 = -1; + for (int j = 0; j < K; j++) { // j颜色！ + if (j != preEnd1) { + if (preMin1 + costs[i][j] < curMin1) { + curMin2 = curMin1; + curEnd2 = curEnd1; + curMin1 = preMin1 + costs[i][j]; + curEnd1 = j; + } else if (preMin1 + costs[i][j] < curMin2) { + curMin2 = preMin1 + costs[i][j]; + curEnd2 = j; + } + } else if (j != preEnd2) { + if (preMin2 + costs[i][j] < curMin1) { + curMin2 = curMin1; + curEnd2 = curEnd1; + curMin1 = preMin2 + costs[i][j]; + curEnd1 = j; + } else if (preMin2 + costs[i][j] < curMin2) { + curMin2 = preMin2 + costs[i][j]; + curEnd2 = j; + } + } + } + preMin1 = curMin1; + preEnd1 = curEnd1; + preMin2 = curMin2; + preEnd2 = curEnd2; + } + return preMin1; + } + +} diff --git a/大厂刷题班/class42/Problem_0272_ClosestBinarySearchTreeValueII.java b/大厂刷题班/class42/Problem_0272_ClosestBinarySearchTreeValueII.java new file mode 100644 index 0000000..62d98f2 --- /dev/null +++ b/大厂刷题班/class42/Problem_0272_ClosestBinarySearchTreeValueII.java @@ -0,0 +1,109 @@ +package class42; + +import java.util.LinkedList; +import java.util.List; +import java.util.Stack; + +public class Problem_0272_ClosestBinarySearchTreeValueII { + + public static class TreeNode { + public int val; + public TreeNode left; + public TreeNode right; + + public TreeNode(int val) { + this.val = val; + } + } + + // 这个解法来自讨论区的回答，最优解实现的很易懂且漂亮 + public static List closestKValues(TreeNode root, double target, int k) { + List ret = new LinkedList<>(); + // >=8，最近的节点，而且需要快速找后继的这么一种结构 + Stack moreTops = new Stack<>(); + // <=8，最近的节点，而且需要快速找前驱的这么一种结构 + Stack lessTops = new Stack<>(); + getMoreTops(root, target, moreTops); + getLessTops(root, target, lessTops); + if (!moreTops.isEmpty() && !lessTops.isEmpty() && moreTops.peek().val == lessTops.peek().val) { + getPredecessor(lessTops); + } + while (k-- > 0) { + if (moreTops.isEmpty()) { + ret.add(getPredecessor(lessTops)); + } else if (lessTops.isEmpty()) { + ret.add(getSuccessor(moreTops)); + } else { + double diffs = Math.abs((double) moreTops.peek().val - target); + double diffp = Math.abs((double) lessTops.peek().val - target); + if (diffs < diffp) { + ret.add(getSuccessor(moreTops)); + } else { + ret.add(getPredecessor(lessTops)); + } + } + } + return ret; + } + + // 在root为头的树上 + // 找到>=target，且最接近target的节点 + // 并且找的过程中，只要某个节点x往左走了，就把x放入moreTops里 + public static void getMoreTops(TreeNode root, double target, Stack moreTops) { + while (root != null) { + if (root.val == target) { + moreTops.push(root); + break; + } else if (root.val > target) { + moreTops.push(root); + root = root.left; + } else { + root = root.right; + } + } + } + + // 在root为头的树上 + // 找到<=target，且最接近target的节点 + // 并且找的过程中，只要某个节点x往右走了，就把x放入lessTops里 + public static void getLessTops(TreeNode root, double target, Stack lessTops) { + while (root != null) { + if (root.val == target) { + lessTops.push(root); + break; + } else if (root.val < target) { + lessTops.push(root); + root = root.right; + } else { + root = root.left; + } + } + } + + // 返回moreTops的头部的值 + // 并且调整moreTops : 为了以后能很快的找到返回节点的后继节点 + public static int getSuccessor(Stack moreTops) { + TreeNode cur = moreTops.pop(); + int ret = cur.val; + cur = cur.right; + while (cur != null) { + moreTops.push(cur); + cur = cur.left; + } + return ret; + } + + // 返回lessTops的头部的值 + // 并且调整lessTops : 为了以后能很快的找到返回节点的前驱节点 + public static int getPredecessor(Stack lessTops) { + TreeNode cur = lessTops.pop(); + int ret = cur.val; + cur = cur.left; + while (cur != null) { + lessTops.push(cur); + cur = cur.right; + } + return ret; + } + +} diff --git a/大厂刷题班/class42/Problem_0273_IntegerToEnglishWords.java b/大厂刷题班/class42/Problem_0273_IntegerToEnglishWords.java new file mode 100644 index 0000000..480cab2 --- /dev/null +++ b/大厂刷题班/class42/Problem_0273_IntegerToEnglishWords.java @@ -0,0 +1,78 @@ +package class42; + +public class Problem_0273_IntegerToEnglishWords { + + public static String num1To19(int num) { + if (num < 1 || num > 19) { + return ""; + } + String[] names = { "One ", "Two ", "Three ", "Four ", "Five ", "Six ", "Seven ", "Eight ", "Nine ", "Ten ", + "Eleven ", "Twelve ", "Thirteen ", "Fourteen ", "Fifteen ", "Sixteen ", "Seventeen", "Eighteen ", + "Nineteen " }; + return names[num - 1]; + } + + public static String num1To99(int num) { + if (num < 1 || num > 99) { + return ""; + } + if (num < 20) { + return num1To19(num); + } + int high = num / 10; + String[] tyNames = { "Twenty ", "Thirty ", "Forty ", "Fifty ", "Sixty ", "Seventy ", "Eighty ", "Ninety " }; + return tyNames[high - 2] + num1To19(num % 10); + } + + public static String num1To999(int num) { + if (num < 1 || num > 999) { + return ""; + } + if (num < 100) { + return num1To99(num); + } + int high = num / 100; + return num1To19(high) + "Hundred " + num1To99(num % 100); + } + + public static String numberToWords(int num) { + if (num == 0) { + return "Zero"; + } + String res = ""; + if (num < 0) { + res = "Negative "; + } + if (num == Integer.MIN_VALUE) { + res += "Two Billion "; + num %= -2000000000; + } + num = Math.abs(num); + int high = 1000000000; + int highIndex = 0; + String[] names = { "Billion ", "Million ", "Thousand ", "" }; + while (num != 0) { + int cur = num / high; + num %= high; + if (cur != 0) { + res += num1To999(cur); + res += names[highIndex]; + } + high /= 1000; + highIndex++; + } + return res.trim(); + } + + public static void main(String[] args) { + int test = Integer.MIN_VALUE; + System.out.println(test); + + test = -test; + System.out.println(test); + + int num = -10001; + System.out.println(numberToWords(num)); + } + +} diff --git a/大厂刷题班/class42/Problem_0296_BestMeetingPoint.java b/大厂刷题班/class42/Problem_0296_BestMeetingPoint.java new file mode 100644 index 0000000..8987edf --- /dev/null +++ b/大厂刷题班/class42/Problem_0296_BestMeetingPoint.java @@ -0,0 +1,48 @@ +package class42; + +public class Problem_0296_BestMeetingPoint { + + public static int minTotalDistance(int[][] grid) { + int N = grid.length; + int M = grid[0].length; + int[] iOnes = new int[N]; + int[] jOnes = new int[M]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + if (grid[i][j] == 1) { + iOnes[i]++; + jOnes[j]++; + } + } + } + int total = 0; + int i = 0; + int j = N - 1; + int iRest = 0; + int jRest = 0; + while (i < j) { + if (iOnes[i] + iRest <= iOnes[j] + jRest) { + total += iOnes[i] + iRest; + iRest += iOnes[i++]; + } else { + total += iOnes[j] + jRest; + jRest += iOnes[j--]; + } + } + i = 0; + j = M - 1; + iRest = 0; + jRest = 0; + while (i < j) { + if (jOnes[i] + iRest <= jOnes[j] + jRest) { + total += jOnes[i] + iRest; + iRest += jOnes[i++]; + } else { + total += jOnes[j] + jRest; + jRest += jOnes[j--]; + } + } + return total; + } + +} diff --git a/大厂刷题班/class42/Problem_0335_SelfCrossing.java b/大厂刷题班/class42/Problem_0335_SelfCrossing.java new file mode 100644 index 0000000..cd3d8d8 --- /dev/null +++ b/大厂刷题班/class42/Problem_0335_SelfCrossing.java @@ -0,0 +1,28 @@ +package class42; + +public class Problem_0335_SelfCrossing { + + public static boolean isSelfCrossing(int[] x) { + if (x == null || x.length < 4) { + return false; + } + if ((x.length > 3 && x[2] <= x[0] && x[3] >= x[1]) + || (x.length > 4 + && ((x[3] <= x[1] && x[4] >= x[2]) || (x[3] == x[1] && x[0] + x[4] >= x[2])))) { + return true; + } + for (int i = 5; i < x.length; i++) { + if (x[i - 1] <= x[i - 3] && ((x[i] >= x[i - 2]) + || (x[i - 2] >= x[i - 4] && x[i - 5] + x[i - 1] >= x[i - 3] && x[i - 4] + x[i] >= x[i - 2]))) { + return true; + } + } + return false; + } + + public static void main(String[] args) { + int[] arr = { 2, 2, 3, 2, 2 }; + System.out.println(isSelfCrossing(arr)); + } + +} diff --git a/大厂刷题班/class43/Code01_SumNoPositiveMinCost.java b/大厂刷题班/class43/Code01_SumNoPositiveMinCost.java new file mode 100644 index 0000000..fb039a6 --- /dev/null +++ b/大厂刷题班/class43/Code01_SumNoPositiveMinCost.java @@ -0,0 +1,197 @@ +package class43; + +import java.util.Arrays; + +// 来自微软面试 +// 给定一个正数数组arr长度为n、正数x、正数y +// 你的目标是让arr整体的累加和<=0 +// 你可以对数组中的数num执行以下三种操作中的一种，且每个数最多能执行一次操作 : +// 1）不变 +// 2）可以选择让num变成0，承担x的代价 +// 3）可以选择让num变成-num，承担y的代价 +// 返回你达到目标的最小代价 +// 数据规模 : 面试时面试官没有说数据规模 +public class Code01_SumNoPositiveMinCost { + + // 动态规划 + public static int minOpStep1(int[] arr, int x, int y) { + int sum = 0; + for (int num : arr) { + sum += num; + } + return process1(arr, x, y, 0, sum); + } + + // arr[i...]自由选择，每个位置的数可以执行三种操作中的一种！ + // 执行变0的操作，x操作，代价 -> x + // 执行变相反数的操作，y操作，代价 -> y + // 还剩下sum这么多累加和，需要去搞定！ + // 返回搞定了sum，最低代价是多少？ + public static int process1(int[] arr, int x, int y, int i, int sum) { + if (sum <= 0) { + return 0; + } + // sum > 0 没搞定 + if (i == arr.length) { + return Integer.MAX_VALUE; + } + // 第一选择，什么也不干！ + int p1 = process1(arr, x, y, i + 1, sum); + // 第二选择，执行x的操作，变0 x + 后续 + int p2 = Integer.MAX_VALUE; + int next2 = process1(arr, x, y, i + 1, sum - arr[i]); + if (next2 != Integer.MAX_VALUE) { + p2 = x + next2; + } + // 第三选择，执行y的操作，变相反数 x + 后续 7 -7 -14 + int p3 = Integer.MAX_VALUE; + int next3 = process1(arr, x, y, i + 1, sum - (arr[i] << 1)); + if (next3 != Integer.MAX_VALUE) { + p3 = y + next3; + } + return Math.min(p1, Math.min(p2, p3)); + } + + // 贪心（最优解） + public static int minOpStep2(int[] arr, int x, int y) { + Arrays.sort(arr); // 小 -> 大 + int n = arr.length; + for (int l = 0, r = n - 1; l <= r; l++, r--) { + int tmp = arr[l]; + arr[l] = arr[r]; + arr[r] = tmp; + } + // arr 大 -> 小 + if (x >= y) { // 没有任何必要执行x操作 + int sum = 0; + for (int num : arr) { + sum += num; + } + int cost = 0; + for (int i = 0; i < n && sum > 0; i++) { + sum -= arr[i] << 1; + cost += y; + } + return cost; + } else { + for (int i = n - 2; i >= 0; i--) { + arr[i] += arr[i + 1]; + } + int benefit = 0; + // 注意，可以不二分，用不回退的方式！ + // 执行Y操作的数，有0个的时候 + int left = mostLeft(arr, 0, benefit); + int cost = left * x; + for (int i = 0; i < n - 1; i++) { + // 0..i 这些数，都执行Y + benefit += arr[i] - arr[i + 1]; + left = mostLeft(arr, i + 1, benefit); + cost = Math.min(cost, (i + 1) * y + (left - i - 1) * x); + } + return cost; + } + } + + // arr是后缀和数组， arr[l...]中找到值<=v的最左位置 + public static int mostLeft(int[] arr, int l, int v) { + int r = arr.length - 1; + int m = 0; + int ans = arr.length; + while (l <= r) { + m = (l + r) / 2; + if (arr[m] <= v) { + ans = m; + r = m - 1; + } else { + l = m + 1; + } + } + return ans; + } + + // 不回退 + public static int minOpStep3(int[] arr, int x, int y) { + // 系统排序，小 -> 大 + Arrays.sort(arr); + int n = arr.length; + // 如何变成 大 -> 小 + for (int l = 0, r = n - 1; l <= r; l++, r--) { + int tmp = arr[l]; + arr[l] = arr[r]; + arr[r] = tmp; + } + if (x >= y) { + int sum = 0; + for (int num : arr) { + sum += num; + } + int cost = 0; + for (int i = 0; i < n && sum > 0; i++) { + sum -= arr[i] << 1; + cost += y; + } + return cost; + } else { + // 0个数执行Y + int benefit = 0; + // 全部的数都需要执行x，才能让累加和<=0 + int cost = arr.length * x; + int holdSum = 0; + for (int yRight = 0, holdLeft = n; yRight < holdLeft - 1; yRight++) { + benefit += arr[yRight]; + while (holdLeft - 1 > yRight && holdSum + arr[holdLeft - 1] <= benefit) { + holdSum += arr[holdLeft - 1]; + holdLeft--; + } + // 0...yRight x holdLeft.... + cost = Math.min(cost, (yRight + 1) * y + (holdLeft - yRight - 1) * x); + } + return cost; + } + } + + // 为了测试 + public static int[] randomArray(int len, int v) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * v) + 1; + } + return arr; + } + + // 为了测试 + public static int[] copyArray(int[] arr) { + int[] ans = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + ans[i] = arr[i]; + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int n = 12; + int v = 20; + int c = 10; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * n); + int[] arr = randomArray(len, v); + int[] arr1 = copyArray(arr); + int[] arr2 = copyArray(arr); + int[] arr3 = copyArray(arr); + int x = (int) (Math.random() * c); + int y = (int) (Math.random() * c); + int ans1 = minOpStep1(arr1, x, y); + int ans2 = minOpStep2(arr2, x, y); + int ans3 = minOpStep3(arr3, x, y); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("出错了!"); + } + } + System.out.println("测试结束"); + + } + +} diff --git a/大厂刷题班/class43/Code02_MinCostToYeahArray.java b/大厂刷题班/class43/Code02_MinCostToYeahArray.java new file mode 100644 index 0000000..de9ccdc --- /dev/null +++ b/大厂刷题班/class43/Code02_MinCostToYeahArray.java @@ -0,0 +1,304 @@ +package class43; + +import java.util.Arrays; + +// 来自360笔试 +// 给定一个正数数组arr，长度为n，下标0~n-1 +// arr中的0、n-1位置不需要达标，它们分别是最左、最右的位置 +// 中间位置i需要达标，达标的条件是 : arr[i-1] > arr[i] 或者 arr[i+1] > arr[i]哪个都可以 +// 你每一步可以进行如下操作：对任何位置的数让其-1 +// 你的目的是让arr[1~n-2]都达标，这时arr称之为yeah！数组 +// 返回至少要多少步可以让arr变成yeah！数组 +// 数据规模 : 数组长度 <= 10000，数组中的值<=500 +public class Code02_MinCostToYeahArray { + + public static final int INVALID = Integer.MAX_VALUE; + + // 纯暴力方法，只是为了结果对 + // 时间复杂度极差 + public static int minCost0(int[] arr) { + if (arr == null || arr.length < 3) { + return 0; + } + int n = arr.length; + int min = INVALID; + for (int num : arr) { + min = Math.min(min, num); + } + int base = min - n; + return process0(arr, base, 0); + } + + public static int process0(int[] arr, int base, int index) { + if (index == arr.length) { + for (int i = 1; i < arr.length - 1; i++) { + if (arr[i - 1] <= arr[i] && arr[i] >= arr[i + 1]) { + return INVALID; + } + } + return 0; + } else { + int ans = INVALID; + int tmp = arr[index]; + for (int cost = 0; arr[index] >= base; cost++, arr[index]--) { + int next = process0(arr, base, index + 1); + if (next != INVALID) { + ans = Math.min(ans, cost + next); + } + } + arr[index] = tmp; + return ans; + } + } + + // 递归方法，已经把尝试写出 + public static int minCost1(int[] arr) { + if (arr == null || arr.length < 3) { + return 0; + } + int min = INVALID; + for (int num : arr) { + min = Math.min(min, num); + } + for (int i = 0; i < arr.length; i++) { + arr[i] += arr.length - min; + } + return process1(arr, 1, arr[0], true); + } + + // 当前来到index位置，值arr[index] + // pre : 前一个位置的值，可能减掉了一些，所以不能用arr[index-1] + // preOk : 前一个位置的值，是否被它左边的数变有效了 + // 返回 : 让arr都变有效，最小代价是什么？ + public static int process1(int[] arr, int index, int pre, boolean preOk) { + if (index == arr.length - 1) { // 已经来到最后一个数了 + return preOk || pre < arr[index] ? 0 : INVALID; + } + // 当前index，不是最后一个数！ + int ans = INVALID; + if (preOk) { + for (int cur = arr[index]; cur >= 0; cur--) { + int next = process1(arr, index + 1, cur, cur < pre); + if (next != INVALID) { + ans = Math.min(ans, arr[index] - cur + next); + } + } + } else { + for (int cur = arr[index]; cur > pre; cur--) { + int next = process1(arr, index + 1, cur, false); + if (next != INVALID) { + ans = Math.min(ans, arr[index] - cur + next); + } + } + } + return ans; + } + + // 初改动态规划方法，就是参考minCost1，改出来的版本 + public static int minCost2(int[] arr) { + if (arr == null || arr.length < 3) { + return 0; + } + int min = INVALID; + for (int num : arr) { + min = Math.min(min, num); + } + int n = arr.length; + for (int i = 0; i < n; i++) { + arr[i] += n - min; + } + int[][][] dp = new int[n][2][]; + for (int i = 1; i < n; i++) { + dp[i][0] = new int[arr[i - 1] + 1]; + dp[i][1] = new int[arr[i - 1] + 1]; + Arrays.fill(dp[i][0], INVALID); + Arrays.fill(dp[i][1], INVALID); + } + for (int pre = 0; pre <= arr[n - 2]; pre++) { + dp[n - 1][0][pre] = pre < arr[n - 1] ? 0 : INVALID; + dp[n - 1][1][pre] = 0; + } + for (int index = n - 2; index >= 1; index--) { + for (int pre = 0; pre <= arr[index - 1]; pre++) { + for (int cur = arr[index]; cur > pre; cur--) { + int next = dp[index + 1][0][cur]; + if (next != INVALID) { + dp[index][0][pre] = Math.min(dp[index][0][pre], arr[index] - cur + next); + } + } + for (int cur = arr[index]; cur >= 0; cur--) { + int next = dp[index + 1][cur < pre ? 1 : 0][cur]; + if (next != INVALID) { + dp[index][1][pre] = Math.min(dp[index][1][pre], arr[index] - cur + next); + } + } + } + } + return dp[1][1][arr[0]]; + } + + // minCost2动态规划 + 枚举优化 + // 改出的这个版本，需要一些技巧，但很可惜不是最优解 + // 虽然不是最优解，也足以通过100%的case了， + // 这种技法的练习，非常有意义 + public static int minCost3(int[] arr) { + if (arr == null || arr.length < 3) { + return 0; + } + int min = INVALID; + for (int num : arr) { + min = Math.min(min, num); + } + int n = arr.length; + for (int i = 0; i < n; i++) { + arr[i] += n - min; + } + int[][][] dp = new int[n][2][]; + for (int i = 1; i < n; i++) { + dp[i][0] = new int[arr[i - 1] + 1]; + dp[i][1] = new int[arr[i - 1] + 1]; + } + for (int p = 0; p <= arr[n - 2]; p++) { + dp[n - 1][0][p] = p < arr[n - 1] ? 0 : INVALID; + } + int[][] best = best(dp, n - 1, arr[n - 2]); + for (int i = n - 2; i >= 1; i--) { + for (int p = 0; p <= arr[i - 1]; p++) { + if (arr[i] < p) { + dp[i][1][p] = best[1][arr[i]]; + } else { + dp[i][1][p] = Math.min(best[0][p], p > 0 ? best[1][p - 1] : INVALID); + } + dp[i][0][p] = arr[i] <= p ? INVALID : best[0][p + 1]; + } + best = best(dp, i, arr[i - 1]); + } + return dp[1][1][arr[0]]; + } + + public static int[][] best(int[][][] dp, int i, int v) { + int[][] best = new int[2][v + 1]; + best[0][v] = dp[i][0][v]; + for (int p = v - 1; p >= 0; p--) { + best[0][p] = dp[i][0][p] == INVALID ? INVALID : v - p + dp[i][0][p]; + best[0][p] = Math.min(best[0][p], best[0][p + 1]); + } + best[1][0] = dp[i][1][0] == INVALID ? INVALID : v + dp[i][1][0]; + for (int p = 1; p <= v; p++) { + best[1][p] = dp[i][1][p] == INVALID ? INVALID : v - p + dp[i][1][p]; + best[1][p] = Math.min(best[1][p], best[1][p - 1]); + } + return best; + } + + // 最终的最优解，贪心 + // 时间复杂度O(N) + // 请注意，重点看上面的方法 + // 这个最优解容易理解，但让你学到的东西不是很多 + public static int yeah(int[] arr) { + if (arr == null || arr.length < 3) { + return 0; + } + int n = arr.length; + int[] nums = new int[n + 2]; + nums[0] = Integer.MAX_VALUE; + nums[n + 1] = Integer.MAX_VALUE; + for (int i = 0; i < arr.length; i++) { + nums[i + 1] = arr[i]; + } + int[] leftCost = new int[n + 2]; + int pre = nums[0]; + int change = 0; + for (int i = 1; i <= n; i++) { + change = Math.min(pre - 1, nums[i]); + leftCost[i] = nums[i] - change + leftCost[i - 1]; + pre = change; + } + int[] rightCost = new int[n + 2]; + pre = nums[n + 1]; + for (int i = n; i >= 1; i--) { + change = Math.min(pre - 1, nums[i]); + rightCost[i] = nums[i] - change + rightCost[i + 1]; + pre = change; + } + int ans = Integer.MAX_VALUE; + for (int i = 1; i <= n; i++) { + ans = Math.min(ans, leftCost[i] + rightCost[i + 1]); + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int len, int v) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * v) + 1; + } + return arr; + } + + // 为了测试 + public static int[] copyArray(int[] arr) { + int[] ans = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + ans[i] = arr[i]; + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int len = 7; + int v = 10; + int testTime = 100; + System.out.println("=========="); + System.out.println("功能测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * len) + 1; + int[] arr = randomArray(n, v); + int[] arr0 = copyArray(arr); + int[] arr1 = copyArray(arr); + int[] arr2 = copyArray(arr); + int[] arr3 = copyArray(arr); + int[] arr4 = copyArray(arr); + int ans0 = minCost0(arr0); + int ans1 = minCost1(arr1); + int ans2 = minCost2(arr2); + int ans3 = minCost3(arr3); + int ans4 = yeah(arr4); + if (ans0 != ans1 || ans0 != ans2 || ans0 != ans3 || ans0 != ans4) { + System.out.println("出错了！"); + } + } + System.out.println("功能测试结束"); + System.out.println("=========="); + + System.out.println("性能测试开始"); + + len = 10000; + v = 500; + System.out.println("生成随机数组长度：" + len); + System.out.println("生成随机数组值的范围：[1, " + v + "]"); + int[] arr = randomArray(len, v); + int[] arr3 = copyArray(arr); + int[] arrYeah = copyArray(arr); + long start; + long end; + start = System.currentTimeMillis(); + int ans3 = minCost3(arr3); + end = System.currentTimeMillis(); + System.out.println("minCost3方法:"); + System.out.println("运行结果: " + ans3 + ", 时间(毫秒) : " + (end - start)); + + start = System.currentTimeMillis(); + int ansYeah = yeah(arrYeah); + end = System.currentTimeMillis(); + System.out.println("yeah方法:"); + System.out.println("运行结果: " + ansYeah + ", 时间(毫秒) : " + (end - start)); + + System.out.println("性能测试结束"); + System.out.println("=========="); + + } + +} diff --git a/大厂刷题班/class44/Problem_0248_StrobogrammaticNumberIII.java b/大厂刷题班/class44/Problem_0248_StrobogrammaticNumberIII.java new file mode 100644 index 0000000..c5684db --- /dev/null +++ b/大厂刷题班/class44/Problem_0248_StrobogrammaticNumberIII.java @@ -0,0 +1,216 @@ +package class44; + +public class Problem_0248_StrobogrammaticNumberIII { + + public static int strobogrammaticInRange(String l, String h) { + char[] low = l.toCharArray(); + char[] high = h.toCharArray(); + if (!equalMore(low, high)) { + return 0; + } + int lowLen = low.length; + int highLen = high.length; + if (lowLen == highLen) { + int up1 = up(low, 0, false, 1); + int up2 = up(high, 0, false, 1); + return up1 - up2 + (valid(high) ? 1 : 0); + } + int ans = 0; + // lowLen = 3 hightLen = 7 + // 4 5 6 + for (int i = lowLen + 1; i < highLen; i++) { + ans += all(i); + } + ans += up(low, 0, false, 1); + ans += down(high, 0, false, 1); + return ans; + } + + public static boolean equalMore(char[] low, char[] cur) { + if (low.length != cur.length) { + return low.length < cur.length; + } + for (int i = 0; i < low.length; i++) { + if (low[i] != cur[i]) { + return low[i] < cur[i]; + } + } + return true; + } + + public static boolean valid(char[] str) { + int L = 0; + int R = str.length - 1; + while (L <= R) { + boolean t = L != R; + if (convert(str[L++], t) != str[R--]) { + return false; + } + } + return true; + } + + // left想得到cha字符，right配合应该做什么决定， + // 如果left怎么也得不到cha字符，返回-1；如果能得到，返回right配合应做什么决定 + // 比如，left!=right，即不是同一个位置 + // left想得到0，那么就right就需要是0 + // left想得到1，那么就right就需要是1 + // left想得到6，那么就right就需要是9 + // left想得到8，那么就right就需要是8 + // left想得到9，那么就right就需要是6 + // 除此了这些之外，left不能得到别的了。 + // 比如，left==right，即是同一个位置 + // left想得到0，那么就right就需要是0 + // left想得到1，那么就right就需要是1 + // left想得到8，那么就right就需要是8 + // 除此了这些之外，left不能得到别的了，比如： + // left想得到6，那么就right就需要是9，而left和right是一个位置啊，怎么可能即6又9，返回-1 + // left想得到9，那么就right就需要是6，而left和right是一个位置啊，怎么可能即9又6，返回-1 + public static int convert(char cha, boolean diff) { + switch (cha) { + case '0': + return '0'; + case '1': + return '1'; + case '6': + return diff ? '9' : -1; + case '8': + return '8'; + case '9': + return diff ? '6' : -1; + default: + return -1; + } + } + + // low [左边已经做完决定了 left.....right 右边已经做完决定了] + // 左边已经做完决定的部分，如果大于low的原始，leftMore = true; + // 左边已经做完决定的部分，如果不大于low的原始，那一定是相等，不可能小于，leftMore = false; + // 右边已经做完决定的部分，如果小于low的原始，rightLessEqualMore = 0; + // 右边已经做完决定的部分，如果等于low的原始，rightLessEqualMore = 1; + // 右边已经做完决定的部分，如果大于low的原始，rightLessEqualMore = 2; + // rightLessEqualMore < = > + // 0 1 2 + // 返回 ：没做决定的部分，随意变，几个有效的情况？返回！ + public static int up(char[] low, int left, boolean leftMore, int rightLessEqualMore) { + int N = low.length; + int right = N - 1 - left; + if (left > right) { // 都做完决定了！ + // 如果左边做完决定的部分大于原始 或者 如果左边做完决定的部分等于原始&左边做完决定的部分不小于原始 + // 有效！ + // 否则，无效！ + return leftMore || (!leftMore && rightLessEqualMore != 0) ? 1 : 0; + } + // 如果上面没有return，说明决定没做完，就继续做 + if (leftMore) { // 如果左边做完决定的部分大于原始 + return num(N - (left << 1)); + } else { // 如果左边做完决定的部分等于原始 + int ways = 0; + // 当前left做的决定，大于原始的left + for (char cha = (char) (low[left] + 1); cha <= '9'; cha++) { + if (convert(cha, left != right) != -1) { + ways += up(low, left + 1, true, rightLessEqualMore); + } + } + // 当前left做的决定，等于原始的left + int convert = convert(low[left], left != right); + if (convert != -1) { + if (convert < low[right]) { + ways += up(low, left + 1, false, 0); + } else if (convert == low[right]) { + ways += up(low, left + 1, false, rightLessEqualMore); + } else { + ways += up(low, left + 1, false, 2); + } + } + return ways; + } + } + + // ll < = + // rs < = > + public static int down(char[] high, int left, boolean ll, int rs) { + int N = high.length; + int right = N - 1 - left; + if (left > right) { + return ll || (!ll && rs != 2) ? 1 : 0; + } + if (ll) { + return num(N - (left << 1)); + } else { + int ways = 0; + for (char cha = (N != 1 && left == 0) ? '1' : '0'; cha < high[left]; cha++) { + if (convert(cha, left != right) != -1) { + ways += down(high, left + 1, true, rs); + } + } + int convert = convert(high[left], left != right); + if (convert != -1) { + if (convert < high[right]) { + ways += down(high, left + 1, false, 0); + } else if (convert == high[right]) { + ways += down(high, left + 1, false, rs); + } else { + ways += down(high, left + 1, false, 2); + } + } + return ways; + } + } + + public static int num(int bits) { + if (bits == 1) { + return 3; + } + if (bits == 2) { + return 5; + } + int p2 = 3; + int p1 = 5; + int ans = 0; + for (int i = 3; i <= bits; i++) { + ans = 5 * p2; + p2 = p1; + p1 = ans; + } + return ans; + } + + // 如果是最开始 : + // Y X X X Y + // -> 1 X X X 1 + // -> 8 X X X 8 + // -> 9 X X X 6 + // -> 6 X X X 9 + // 如果不是最开始 : + // Y X X X Y + // -> 0 X X X 0 + // -> 1 X X X 1 + // -> 8 X X X 8 + // -> 9 X X X 6 + // -> 6 X X X 9 + // 所有的len位数，有几个有效的？ + public static int all(int len) { + int ans = (len & 1) == 0 ? 1 : 3; + for (int i = (len & 1) == 0 ? 2 : 3; i < len; i += 2) { + ans *= 5; + } + return ans << 2; + } + + // 我们课上讲的 + public static int all(int len, boolean init) { + if (len == 0) { // init == true，不可能调用all(0) + return 1; + } + if (len == 1) { + return 3; + } + if (init) { + return all(len - 2, false) << 2; + } else { + return all(len - 2, false) * 5; + } + } + +} diff --git a/大厂刷题班/class44/Problem_0317_ShortestDistanceFromAllBuildings.java b/大厂刷题班/class44/Problem_0317_ShortestDistanceFromAllBuildings.java new file mode 100644 index 0000000..6a6648b --- /dev/null +++ b/大厂刷题班/class44/Problem_0317_ShortestDistanceFromAllBuildings.java @@ -0,0 +1,237 @@ +package class44; + +import java.util.HashMap; +import java.util.LinkedList; +import java.util.Queue; + +public class Problem_0317_ShortestDistanceFromAllBuildings { + + // 如果grid中0比较少，用这个方法比较好 + public static int shortestDistance1(int[][] grid) { + int ans = Integer.MAX_VALUE; + int N = grid.length; + int M = grid[0].length; + int buildings = 0; + Position[][] positions = new Position[N][M]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + if (grid[i][j] == 1) { + buildings++; + } + positions[i][j] = new Position(i, j, grid[i][j]); + } + } + if (buildings == 0) { + return 0; + } + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + ans = Math.min(ans, bfs(positions, buildings, i, j)); + } + } + return ans == Integer.MAX_VALUE ? -1 : ans; + } + + public static int bfs(Position[][] positions, int buildings, int i, int j) { + if (positions[i][j].v != 0) { + return Integer.MAX_VALUE; + } + HashMap levels = new HashMap<>(); + Queue queue = new LinkedList<>(); + Position from = positions[i][j]; + levels.put(from, 0); + queue.add(from); + int ans = 0; + int solved = 0; + while (!queue.isEmpty() && solved != buildings) { + Position cur = queue.poll(); + int level = levels.get(cur); + if (cur.v == 1) { + ans += level; + solved++; + } else { + add(queue, levels, positions, cur.r - 1, cur.c, level + 1); + add(queue, levels, positions, cur.r + 1, cur.c, level + 1); + add(queue, levels, positions, cur.r, cur.c - 1, level + 1); + add(queue, levels, positions, cur.r, cur.c + 1, level + 1); + } + } + return solved == buildings ? ans : Integer.MAX_VALUE; + } + + public static class Position { + public int r; + public int c; + public int v; + + public Position(int row, int col, int value) { + r = row; + c = col; + v = value; + } + } + + public static void add(Queue q, HashMap l, Position[][] p, int i, int j, int level) { + if (i >= 0 && i < p.length && j >= 0 && j < p[0].length && p[i][j].v != 2 && !l.containsKey(p[i][j])) { + l.put(p[i][j], level); + q.add(p[i][j]); + } + } + + // 如果grid中1比较少，用这个方法比较好 + public static int shortestDistance2(int[][] grid) { + int N = grid.length; + int M = grid[0].length; + int ones = 0; + int zeros = 0; + Info[][] infos = new Info[N][M]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + if (grid[i][j] == 1) { + infos[i][j] = new Info(i, j, 1, ones++); + } else if (grid[i][j] == 0) { + infos[i][j] = new Info(i, j, 0, zeros++); + } else { + infos[i][j] = new Info(i, j, 2, Integer.MAX_VALUE); + } + } + } + if (ones == 0) { + return 0; + } + int[][] distance = new int[ones][zeros]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + if (infos[i][j].v == 1) { + bfs(infos, i, j, distance); + } + } + } + int ans = Integer.MAX_VALUE; + for (int i = 0; i < zeros; i++) { + int sum = 0; + for (int j = 0; j < ones; j++) { + if (distance[j][i] == 0) { + sum = Integer.MAX_VALUE; + break; + } else { + sum += distance[j][i]; + } + } + ans = Math.min(ans, sum); + } + return ans == Integer.MAX_VALUE ? -1 : ans; + } + + public static class Info { + public int r; + public int c; + public int v; + public int t; + + public Info(int row, int col, int value, int th) { + r = row; + c = col; + v = value; + t = th; + } + } + + public static void bfs(Info[][] infos, int i, int j, int[][] distance) { + HashMap levels = new HashMap<>(); + Queue queue = new LinkedList<>(); + Info from = infos[i][j]; + add(queue, levels, infos, from.r - 1, from.c, 1); + add(queue, levels, infos, from.r + 1, from.c, 1); + add(queue, levels, infos, from.r, from.c - 1, 1); + add(queue, levels, infos, from.r, from.c + 1, 1); + while (!queue.isEmpty()) { + Info cur = queue.poll(); + int level = levels.get(cur); + distance[from.t][cur.t] = level; + add(queue, levels, infos, cur.r - 1, cur.c, level + 1); + add(queue, levels, infos, cur.r + 1, cur.c, level + 1); + add(queue, levels, infos, cur.r, cur.c - 1, level + 1); + add(queue, levels, infos, cur.r, cur.c + 1, level + 1); + } + } + + public static void add(Queue q, HashMap l, Info[][] infos, int i, int j, int level) { + if (i >= 0 && i < infos.length && j >= 0 && j < infos[0].length && infos[i][j].v == 0 + && !l.containsKey(infos[i][j])) { + l.put(infos[i][j], level); + q.add(infos[i][j]); + } + } + + // 方法三的大流程和方法二完全一样，从每一个1出发，而不从0出发 + // 运行时间快主要是因为常数优化，以下是优化点： + // 1) 宽度优先遍历时，一次解决一层，不是一个一个遍历： + // int size = que.size(); + // level++; + // for (int k = 0; k < size; k++) { ... } + // 2) pass的值每次减1何用？只有之前所有的1都到达的0，才有必要继续尝试的意思 + // 也就是说，如果某个1，自我封闭，之前的1根本到不了现在这个1附近的0，就没必要继续尝试了 + // if (nextr >= 0 && nextr < grid.length + // && nextc >= 0 && nextc < grid[0].length + // && grid[nextr][nextc] == pass) + // 3) int[] trans = { 0, 1, 0, -1, 0 }; 的作用是迅速算出上、下、左、右 + // 4) 如果某个1在计算时，它周围已经没有pass值了，可以提前宣告1之间是不连通的 + // step = bfs(grid, dist, i, j, pass--, trans); + // if (step == Integer.MAX_VALUE) { + // return -1; + // } + // 5) 最要的优化，每个1到某个0的距离是逐渐叠加的，每个1给所有的0叠一次（宽度优先遍历） + // dist[nextr][nextc] += level; + public static int shortestDistance3(int[][] grid) { + int[][] dist = new int[grid.length][grid[0].length]; + int pass = 0; + int step = Integer.MAX_VALUE; + int[] trans = { 0, 1, 0, -1, 0 }; + for (int i = 0; i < grid.length; i++) { + for (int j = 0; j < grid[0].length; j++) { + if (grid[i][j] == 1) { + step = bfs(grid, dist, i, j, pass--, trans); + if (step == Integer.MAX_VALUE) { + return -1; + } + } + } + } + return step == Integer.MAX_VALUE ? -1 : step; + } + + // 原始矩阵是grid，但是所有的路(0)，被改了 + // 改成了啥？改成认为，pass才是路！原始矩阵中的1和2呢？不变！ + // dist，距离压缩表，之前的bfs，也就是之前每个1，走到某个0，总距离和都在dist里 + // row,col 宽度优先遍历的，出发点！ + // trans -> 炫技的，上下左右 + // 返回值代表，进行完这一遍bfs，压缩距离表中(dist)，最小值是谁？ + // 如果突然发现，无法联通！返回系统最大！ + public static int bfs(int[][] grid, int[][] dist, int row, int col, int pass, int[] trans) { + Queue que = new LinkedList(); + que.offer(new int[] { row, col }); + int level = 0; + int ans = Integer.MAX_VALUE; + while (!que.isEmpty()) { + int size = que.size(); + level++; + for (int k = 0; k < size; k++) { + int[] node = que.poll(); + for (int i = 1; i < trans.length; i++) { // 上下左右 + int nextr = node[0] + trans[i - 1]; + int nextc = node[1] + trans[i]; + if (nextr >= 0 && nextr < grid.length && nextc >= 0 && nextc < grid[0].length + && grid[nextr][nextc] == pass) { + que.offer(new int[] { nextr, nextc }); + dist[nextr][nextc] += level; + ans = Math.min(ans, dist[nextr][nextc]); + grid[nextr][nextc]--; + } + } + } + } + return ans; + } + +} diff --git a/大厂刷题班/class44/Problem_0992_SubarraysWithKDifferentIntegers.java b/大厂刷题班/class44/Problem_0992_SubarraysWithKDifferentIntegers.java new file mode 100644 index 0000000..97c111c --- /dev/null +++ b/大厂刷题班/class44/Problem_0992_SubarraysWithKDifferentIntegers.java @@ -0,0 +1,71 @@ +package class44; + +import java.util.HashMap; + +public class Problem_0992_SubarraysWithKDifferentIntegers { + + // nums 数组，题目规定，nums中的数字，不会超过nums的长度 + // [ ]长度为5，0~5 + public static int subarraysWithKDistinct1(int[] nums, int k) { + int n = nums.length; + // k-1种数的窗口词频统计 + int[] lessCounts = new int[n + 1]; + // k种数的窗口词频统计 + int[] equalCounts = new int[n + 1]; + int lessLeft = 0; + int equalLeft = 0; + int lessKinds = 0; + int equalKinds = 0; + int ans = 0; + for (int r = 0; r < n; r++) { + // 当前刚来到r位置！ + if (lessCounts[nums[r]] == 0) { + lessKinds++; + } + if (equalCounts[nums[r]] == 0) { + equalKinds++; + } + lessCounts[nums[r]]++; + equalCounts[nums[r]]++; + while (lessKinds == k) { + if (lessCounts[nums[lessLeft]] == 1) { + lessKinds--; + } + lessCounts[nums[lessLeft++]]--; + } + while (equalKinds > k) { + if (equalCounts[nums[equalLeft]] == 1) { + equalKinds--; + } + equalCounts[nums[equalLeft++]]--; + } + ans += lessLeft - equalLeft; + } + return ans; + } + + public static int subarraysWithKDistinct2(int[] arr, int k) { + return numsMostK(arr, k) - numsMostK(arr, k - 1); + } + + public static int numsMostK(int[] arr, int k) { + int i = 0, res = 0; + HashMap count = new HashMap<>(); + for (int j = 0; j < arr.length; ++j) { + if (count.getOrDefault(arr[j], 0) == 0) { + k--; + } + count.put(arr[j], count.getOrDefault(arr[j], 0) + 1); + while (k < 0) { + count.put(arr[i], count.get(arr[i]) - 1); + if (count.get(arr[i]) == 0) { + k++; + } + i++; + } + res += j - i + 1; + } + return res; + } + +} diff --git a/大厂刷题班/class45/Code01_SplitBuildingBlock.java b/大厂刷题班/class45/Code01_SplitBuildingBlock.java new file mode 100644 index 0000000..b63b36c --- /dev/null +++ b/大厂刷题班/class45/Code01_SplitBuildingBlock.java @@ -0,0 +1,87 @@ +package class45; + +import java.util.Arrays; + +// 来自京东笔试 +// 小明手中有n块积木，并且小明知道每块积木的重量。现在小明希望将这些积木堆起来 +// 要求是任意一块积木如果想堆在另一块积木上面，那么要求： +// 1) 上面的积木重量不能小于下面的积木重量 +// 2) 上面积木的重量减去下面积木的重量不能超过x +// 3) 每堆中最下面的积木没有重量要求 +// 现在小明有一个机会，除了这n块积木，还可以获得k块任意重量的积木。 +// 小明希望将积木堆在一起，同时希望积木堆的数量越少越好，你能帮他找到最好的方案么？ +// 输入描述: +// 第一行三个整数n,k,x，1<=n<=200000，0<=x,k<=1000000000 +// 第二行n个整数，表示积木的重量，任意整数范围都在[1,1000000000] +// 样例输出： +// 13 1 38 +// 20 20 80 70 70 70 420 5 1 5 1 60 90 +// 1 1 5 5 20 20 60 70 70 70 80 90 420 -> 只有1块魔法积木，x = 38 +// 输出：2 +// 解释： +// 两堆分别是 +// 1 1 5 5 20 20 (50) 60 70 70 70 80 90 +// 420 +// 其中x是一个任意重量的积木，夹在20和60之间可以让积木继续往上搭 +public class Code01_SplitBuildingBlock { + + // 这是启发解 + // arr是从小到大排序的，x是限制，固定参数 + // 当前来到i位置，积木重量arr[i] + // 潜台词 : 当前i位置的积木在一个堆里，堆的开头在哪？之前已经决定了 + // i i+1 该在一起 or 该用魔法积木弥合 or 该分家 + // 返回值：arr[i....]最少能分几个堆？ + public static int zuo(int[] arr, int x, int i, int r) { + if (i == arr.length - 1) { + return 1; + } + // i没到最后一个数 + if (arr[i + 1] - arr[i] <= x) { // 一定贴在一起 + return zuo(arr, x, i + 1, r); + } else { // 弥合！分家 + // 分家 + int p1 = 1 + zuo(arr, x, i + 1, r); + // 弥合 + int p2 = Integer.MAX_VALUE; + int need = (arr[i + 1] - arr[i] - 1) / x; + if (r >= need) { + p2 = zuo(arr, x, i + 1, r - need); + } + return Math.min(p1, p2); + } + } + + // 这是最优解 + // arr里装着所有积木的重量 + // k是魔法积木的数量，每一块魔法积木都能变成任何重量 + // x差值，后 - 前 <= x + public static int minSplit(int[] arr, int k, int x) { + Arrays.sort(arr); + int n = arr.length; + int[] needs = new int[n]; + int size = 0; + int splits = 1; + for (int i = 1; i < n; i++) { + if (arr[i] - arr[i - 1] > x) { + needs[size++] = arr[i] - arr[i - 1]; + splits++; + } + } + if (splits == 1 || x == 0 || k == 0) { + return splits; + } + // 试图去利用魔法积木，弥合堆！ + Arrays.sort(needs, 0, size); + for (int i = 0; i < size; i++) { + int need = (needs[i] - 1) / x; + if (k >= need) { + splits--; + k -= need; + } else { + break; + } + } + return splits; + } + +} diff --git a/大厂刷题班/class45/Problem_0291_WordPatternII.java b/大厂刷题班/class45/Problem_0291_WordPatternII.java new file mode 100644 index 0000000..a961a6d --- /dev/null +++ b/大厂刷题班/class45/Problem_0291_WordPatternII.java @@ -0,0 +1,62 @@ +package class45; + +import java.util.HashSet; + +public class Problem_0291_WordPatternII { + + public static boolean wordPatternMatch(String pattern, String str) { + return match(str, pattern, 0, 0, new String[26], new HashSet<>()); + } + + // 题目有限制，str和pattern其中的字符，一定是a~z小写 + // p[a] -> "abc" + // p[b] -> "fbf" + // 需要指代的表最多26长度 + // String[] map -> new String[26] + // p[a] -> "abc" map[0] -> "abc" + // p[b] -> "fbf" map[1] -> "fbf"; + // p[z] -> "kfk" map[25] -> "kfk" + // HashSet set -> map中指代了哪些字符串 + // str[si.......] 是不是符合 p[pi......]？符合返回true，不符合返回false + // 之前的决定！由map和set，告诉我！不能冲突！ + public static boolean match(String s, String p, int si, int pi, String[] map, HashSet set) { + if (pi == p.length() && si == s.length()) { + return true; + } + // str和pattern，并没有都结束！ + if (pi == p.length() || si == s.length()) { + return false; + } + // str和pattern，都没结束！ + + char ch = p.charAt(pi); + String cur = map[ch - 'a']; + if (cur != null) { // 当前p[pi]已经指定过了！ + return si + cur.length() <= s.length() // 不能越界！ + && cur.equals(s.substring(si, si + cur.length())) + && match(s, p, si + cur.length(), pi + 1, map, set); + } + // p[pi]没指定！ + int end = s.length(); + // 剪枝！重要的剪枝！ + for (int i = p.length() - 1; i > pi; i--) { + end -= map[p.charAt(i) - 'a'] == null ? 1 : map[p.charAt(i) - 'a'].length(); + } + for (int i = si; i < end; i++) { + // 从si出发的所有前缀串，全试 + cur = s.substring(si, i + 1); + // 但是，只有这个前缀串，之前没占过别的坑！才能去尝试 + if (!set.contains(cur)) { + set.add(cur); + map[ch - 'a'] = cur; + if (match(s, p, i + 1, pi + 1, map, set)) { + return true; + } + map[ch - 'a'] = null; + set.remove(cur); + } + } + return false; + } + +} diff --git a/大厂刷题班/class45/Problem_0403_FrogJump.java b/大厂刷题班/class45/Problem_0403_FrogJump.java new file mode 100644 index 0000000..08540e5 --- /dev/null +++ b/大厂刷题班/class45/Problem_0403_FrogJump.java @@ -0,0 +1,40 @@ +package class45; + +import java.util.HashMap; +import java.util.HashSet; + +public class Problem_0403_FrogJump { + + public static boolean canCross(int[] stones) { + HashSet set = new HashSet<>(); + for (int num : stones) { + set.add(num); + } + HashMap> dp = new HashMap<>(); + return jump(1, 1, stones[stones.length - 1], set, dp); + } + + public static boolean jump(int cur, int pre, int end, HashSet set, + HashMap> dp) { + if (cur == end) { + return true; + } + if (!set.contains(cur)) { + return false; + } + if (dp.containsKey(cur) && dp.get(cur).containsKey(pre)) { + return dp.get(cur).get(pre); + } + boolean ans = (pre > 1 && jump(cur + pre - 1, pre - 1, end, set, dp)) + || jump(cur + pre, pre, end, set, dp) + || jump(cur + pre + 1, pre + 1, end, set, dp); + if (!dp.containsKey(cur)) { + dp.put(cur, new HashMap<>()); + } + if (!dp.get(cur).containsKey(pre)) { + dp.get(cur).put(pre, ans); + } + return ans; + } + +} diff --git a/大厂刷题班/class45/Problem_2035_PartitionArrayIntoTwoArraysToMinimizeSumDifference.java b/大厂刷题班/class45/Problem_2035_PartitionArrayIntoTwoArraysToMinimizeSumDifference.java new file mode 100644 index 0000000..3592e53 --- /dev/null +++ b/大厂刷题班/class45/Problem_2035_PartitionArrayIntoTwoArraysToMinimizeSumDifference.java @@ -0,0 +1,58 @@ +package class45; + +import java.util.HashMap; +import java.util.TreeSet; + +public class Problem_2035_PartitionArrayIntoTwoArraysToMinimizeSumDifference { + + public static int minimumDifference(int[] arr) { + int size = arr.length; + int half = size >> 1; + HashMap> lmap = new HashMap<>(); + process(arr, 0, half, 0, 0, lmap); + HashMap> rmap = new HashMap<>(); + process(arr, half, size, 0, 0, rmap); + int sum = 0; + for (int num : arr) { + sum += num; + } + int ans = Integer.MAX_VALUE; + for (int leftNum : lmap.keySet()) { + for (int leftSum : lmap.get(leftNum)) { + Integer rightSum = rmap.get(half - leftNum).floor((sum >> 1) - leftSum); + if (rightSum != null) { + int pickSum = leftSum + rightSum; + int restSum = sum - pickSum; + ans = Math.min(ans, restSum - pickSum); + } + } + } + return ans; + } + + + // arr -> 8 0 1 2 3 4 5 6 7 + // process(arr, 0, 4) [0,4) + // process(arr, 4, 8) [4,8) + // arr[index....end-1]这个范围上，去做选择 + // pick挑了几个数！ + // sum挑的这些数，累加和是多少！ + // map记录结果 + // HashMap> map + // key -> 挑了几个数，比如挑了3个数，但是形成累加和可能多个！ + // value -> 有序表，都记下来！ + // 整个过程，纯暴力！2^15 -> 3万多，纯暴力跑完，依然很快！ + public static void process(int[] arr, int index, int end, int pick, int sum, + HashMap> map) { + if (index == end) { + if (!map.containsKey(pick)) { + map.put(pick, new TreeSet<>()); + } + map.get(pick).add(sum); + } else { + process(arr, index + 1, end, pick, sum, map); + process(arr, index + 1, end, pick + 1, sum + arr[index], map); + } + } + +} diff --git a/大厂刷题班/class46/Problem_0363_MaxSumOfRectangleNoLargerThanK.java b/大厂刷题班/class46/Problem_0363_MaxSumOfRectangleNoLargerThanK.java new file mode 100644 index 0000000..2694c13 --- /dev/null +++ b/大厂刷题班/class46/Problem_0363_MaxSumOfRectangleNoLargerThanK.java @@ -0,0 +1,77 @@ +package class46; + +import java.util.TreeSet; + +public class Problem_0363_MaxSumOfRectangleNoLargerThanK { + + public static int nearK(int[] arr, int k) { + if (arr == null || arr.length == 0) { + return Integer.MIN_VALUE; + } + TreeSet set = new TreeSet<>(); + set.add(0); + int ans = Integer.MIN_VALUE; + int sum = 0; + for (int i = 0; i < arr.length; i++) { + // 讨论子数组必须以i位置结尾，最接近k的累加和是多少？ + sum += arr[i]; + // 找之前哪个前缀和 >= sum - k 且最接近 + // 有序表中，ceiling(x) 返回>=x且最接近的！ + // 有序表中，floor(x) 返回<=x且最接近的！ + Integer find = set.ceiling(sum - k); + if(find != null) { + int curAns = sum - find; + ans = Math.max(ans, curAns); + } + set.add(sum); + } + return ans; + } + + public static int maxSumSubmatrix(int[][] matrix, int k) { + if (matrix == null || matrix[0] == null) { + return 0; + } + if (matrix.length > matrix[0].length) { + matrix = rotate(matrix); + } + int row = matrix.length; + int col = matrix[0].length; + int res = Integer.MIN_VALUE; + TreeSet sumSet = new TreeSet<>(); + for (int s = 0; s < row; s++) { + int[] colSum = new int[col]; + for (int e = s; e < row; e++) { + // s ~ e 这些行 选的子矩阵必须包含、且只包含s行~e行的数据 + // 0 ~ 0 0 ~ 1 0 ~ 2 。。。 + // 1 ~ 2 1 ~ 2 1 ~ 3 。。。 + sumSet.add(0); + int rowSum = 0; + for (int c = 0; c < col; c++) { + colSum[c] += matrix[e][c]; + rowSum += colSum[c]; + Integer it = sumSet.ceiling(rowSum - k); + if (it != null) { + res = Math.max(res, rowSum - it); + } + sumSet.add(rowSum); + } + sumSet.clear(); + } + } + return res; + } + + public static int[][] rotate(int[][] matrix) { + int N = matrix.length; + int M = matrix[0].length; + int[][] r = new int[M][N]; + for (int i = 0; i < N; i++) { + for (int j = 0; j < M; j++) { + r[j][i] = matrix[i][j]; + } + } + return r; + } + +} diff --git a/大厂刷题班/class46/Problem_0391_PerfectRectangle.java b/大厂刷题班/class46/Problem_0391_PerfectRectangle.java new file mode 100644 index 0000000..b041a47 --- /dev/null +++ b/大厂刷题班/class46/Problem_0391_PerfectRectangle.java @@ -0,0 +1,59 @@ +package class46; + +import java.util.HashMap; + +public class Problem_0391_PerfectRectangle { + + public static boolean isRectangleCover(int[][] matrix) { + if (matrix.length == 0 || matrix[0].length == 0) { + return false; + } + int l = Integer.MAX_VALUE; + int r = Integer.MIN_VALUE; + int d = Integer.MAX_VALUE; + int u = Integer.MIN_VALUE; + HashMap> map = new HashMap<>(); + int area = 0; + for (int[] rect : matrix) { + add(map, rect[0], rect[1]); + add(map, rect[0], rect[3]); + add(map, rect[2], rect[1]); + add(map, rect[2], rect[3]); + area += (rect[2] - rect[0]) * (rect[3] - rect[1]); + l = Math.min(rect[0], l); + d = Math.min(rect[1], d); + r = Math.max(rect[2], r); + u = Math.max(rect[3], u); + } + return checkPoints(map, l, d, r, u) && area == (r - l) * (u - d); + } + + public static void add(HashMap> map, int row, int col) { + if (!map.containsKey(row)) { + map.put(row, new HashMap<>()); + } + map.get(row).put(col, map.get(row).getOrDefault(col, 0) + 1); + } + + public static boolean checkPoints(HashMap> map, int l, int d, int r, int u) { + if (map.get(l).getOrDefault(d, 0) != 1 + || map.get(l).getOrDefault(u, 0) != 1 + || map.get(r).getOrDefault(d, 0) != 1 + || map.get(r).getOrDefault(u, 0) != 1) { + return false; + } + map.get(l).remove(d); + map.get(l).remove(u); + map.get(r).remove(d); + map.get(r).remove(u); + for (int key : map.keySet()) { + for (int value : map.get(key).values()) { + if ((value & 1) != 0) { + return false; + } + } + } + return true; + } + +} diff --git a/大厂刷题班/class46/Problem_0411_MinimumUniqueWordAbbreviation.java b/大厂刷题班/class46/Problem_0411_MinimumUniqueWordAbbreviation.java new file mode 100644 index 0000000..29d840f --- /dev/null +++ b/大厂刷题班/class46/Problem_0411_MinimumUniqueWordAbbreviation.java @@ -0,0 +1,174 @@ +package class46; + +public class Problem_0411_MinimumUniqueWordAbbreviation { + + // 区分出来之后，缩写的长度，最短是多少？ + public static int min = Integer.MAX_VALUE; + + // 取得缩写的长度最短的时候，决定是什么(fix) + public static int best = 0; + + public static int abbrLen(int fix, int len) { + int ans = 0; + int cnt = 0; + for (int i = 0; i < len; i++) { + if ((fix & (1 << i)) != 0) { + ans++; + if (cnt != 0) { + ans += (cnt > 9 ? 2 : 1) - cnt; + } + cnt = 0; + } else { + cnt++; + } + } + if (cnt != 0) { + ans += (cnt > 9 ? 2 : 1) - cnt; + } + return ans; + } + + // 原始的字典，被改了 + // target : abc 字典中的词 : bbb -> 101 -> int -> + // fix -> int -> 根本不用值，用状态 -> 每一位保留还是不保留的决定 + public static boolean canFix(int[] words, int fix) { + for (int word : words) { + if ((fix & word) == 0) { + return false; + } + } + return true; + } + + // 利用位运算加速 + public static String minAbbreviation1(String target, String[] dictionary) { + min = Integer.MAX_VALUE; + best = 0; + char[] t = target.toCharArray(); + int len = t.length; + int siz = 0; + for (String word : dictionary) { + if (word.length() == len) { + siz++; + } + } + int[] words = new int[siz]; + int index = 0; + for (String word : dictionary) { + if (word.length() == len) { + char[] w = word.toCharArray(); + int status = 0; + for (int j = 0; j < len; j++) { + if (t[j] != w[j]) { + status |= 1 << j; + } + } + words[index++] = status; + } + } + dfs1(words, len, 0, 0); + StringBuilder builder = new StringBuilder(); + int count = 0; + for (int i = 0; i < len; i++) { + if ((best & (1 << i)) != 0) { + if (count > 0) { + builder.append(count); + } + builder.append(t[i]); + count = 0; + } else { + count++; + } + } + if (count > 0) { + builder.append(count); + } + return builder.toString(); + } + + // 所有字典中的单词现在都变成了int，放在words里 + // 0....len-1 位去决定保留还是不保留！当前来到index位 + // 之前做出的决定! + public static void dfs1(int[] words, int len, int fix, int index) { + if (!canFix(words, fix)) { + if (index < len) { + dfs1(words, len, fix, index + 1); + dfs1(words, len, fix | (1 << index), index + 1); + } + } else { + // 决定是fix，一共的长度是len，求出缩写是多长？ + int ans = abbrLen(fix, len); + if (ans < min) { + min = ans; + best = fix; + } + } + } + + // 进一步设计剪枝，注意diff的用法 + public static String minAbbreviation2(String target, String[] dictionary) { + min = Integer.MAX_VALUE; + best = 0; + char[] t = target.toCharArray(); + int len = t.length; + int siz = 0; + for (String word : dictionary) { + if (word.length() == len) { + siz++; + } + } + int[] words = new int[siz]; + int index = 0; + // 用来剪枝 + int diff = 0; + for (String word : dictionary) { + if (word.length() == len) { + char[] w = word.toCharArray(); + int status = 0; + for (int j = 0; j < len; j++) { + if (t[j] != w[j]) { + status |= 1 << j; + } + } + words[index++] = status; + diff |= status; + } + } + dfs2(words, len, diff, 0, 0); + StringBuilder builder = new StringBuilder(); + int count = 0; + for (int i = 0; i < len; i++) { + if ((best & (1 << i)) != 0) { + if (count > 0) { + builder.append(count); + } + builder.append(t[i]); + count = 0; + } else { + count++; + } + } + if (count > 0) { + builder.append(count); + } + return builder.toString(); + } + + public static void dfs2(int[] words, int len, int diff, int fix, int index) { + if (!canFix(words, fix)) { + if (index < len) { + dfs2(words, len, diff, fix, index + 1); + if ((diff & (1 << index)) != 0) { + dfs2(words, len, diff, fix | (1 << index), index + 1); + } + } + } else { + int ans = abbrLen(fix, len); + if (ans < min) { + min = ans; + best = fix; + } + } + } + +} diff --git a/大厂刷题班/class46/Problem_0425_WordSquares.java b/大厂刷题班/class46/Problem_0425_WordSquares.java new file mode 100644 index 0000000..24f2418 --- /dev/null +++ b/大厂刷题班/class46/Problem_0425_WordSquares.java @@ -0,0 +1,87 @@ +package class46; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.LinkedList; +import java.util.List; + +// 注意！课上介绍题目设定的时候，有一点点小错 +// 题目描述如下： +// 给定n个字符串，并且每个字符串长度一定是n，请组成单词方阵，比如： +// 给定4个字符串，长度都是4，["ball","area","lead","lady"] +// 可以组成如下的方阵： +// b a l l +// a r e a +// l e a d +// l a d y +// 什么叫单词方阵？如上的方阵可以看到， +// 第1行和第1列都是"ball"，第2行和第2列都是"area"，第3行和第3列都是"lead"，第4行和第4列都是"lady" +// 所以如果有N个单词，单词方阵是指: +// 一个N*N的二维矩阵，并且i行和i列都是某个单词，不要求全部N个单词都在这个方阵里。 +// 请返回所有可能的单词方阵。 +// 示例： +// 输入: words = ["abat","baba","atan","atal"] +// 输出: [["baba","abat","baba","atal"],["baba","abat","baba","atan"]] +// 解释： +// 可以看到输出里，有两个链表，代表两个单词方阵 +// 第一个如下： +// b a b a +// a b a t +// b a b a +// a t a l +// 这个方阵里没有atan，因为不要求全部单词都在方阵里 +// 第二个如下： +// b a b a +// a b a t +// b a b a +// a t a n +// 这个方阵里没有atal，因为不要求全部单词都在方阵里 +// 课上说的是：一个N*N的二维矩阵，并且i行和i列都是某个单词，要求全部N个单词都在这个方阵里 +// 原题说的是：一个N*N的二维矩阵，并且i行和i列都是某个单词，不要求全部N个单词都在这个方阵里 +// 讲的过程没错，但是介绍题意的时候，这里失误了 +public class Problem_0425_WordSquares { + + public static List> wordSquares(String[] words) { + int n = words[0].length(); + // 所有单词，所有前缀字符串，都会成为key！ + HashMap> map = new HashMap<>(); + for (String word : words) { + for (int end = 0; end <= n; end++) { + String prefix = word.substring(0, end); + if (!map.containsKey(prefix)) { + map.put(prefix, new ArrayList<>()); + } + map.get(prefix).add(word); + } + } + List> ans = new ArrayList<>(); + process(0, n, map, new LinkedList<>(), ans); + return ans; + } + + // i, 当前填到第i号单词，从0开始，填到n-1 + // map, 前缀所拥有的单词 + // path, 之前填过的单词, 0...i-1填过的 + // ans, 收集答案 + public static void process(int i, int n, HashMap> map, LinkedList path, + List> ans) { + if (i == n) { + ans.add(new ArrayList<>(path)); + } else { + // 把限制求出来，前缀的限制！ + StringBuilder builder = new StringBuilder(); + for (String pre : path) { + builder.append(pre.charAt(i)); + } + String prefix = builder.toString(); + if (map.containsKey(prefix)) { + for (String next : map.get(prefix)) { + path.addLast(next); + process(i + 1, n, map, path, ans); + path.pollLast(); + } + } + } + } + +} diff --git a/大厂刷题班/class47/Code01_DynamicSegmentTree.java b/大厂刷题班/class47/Code01_DynamicSegmentTree.java new file mode 100644 index 0000000..6a75a39 --- /dev/null +++ b/大厂刷题班/class47/Code01_DynamicSegmentTree.java @@ -0,0 +1,135 @@ +package class47; + +// 只支持单点增加 + 范围查询的动态开点线段树（累加和） +public class Code01_DynamicSegmentTree { + + public static class Node { + public int sum; + public Node left; + public Node right; + } + + // arr[0] -> 1 + // 线段树，从1开始下标! + public static class DynamicSegmentTree { + public Node root; + public int size; + + public DynamicSegmentTree(int max) { + root = new Node(); + size = max; + } + + // 下标i这个位置的数，增加v + public void add(int i, int v) { + add(root, 1, size, i, v); + } + + // c-> cur 当前节点！表达的范围 l~r + // i位置的数，增加v + // 潜台词！i一定在l~r范围上！ + private void add(Node c, int l, int r, int i, int v) { + if (l == r) { + c.sum += v; + } else { // l~r 还可以划分 + int mid = (l + r) / 2; + if (i <= mid) { // l ~ mid + if (c.left == null) { + c.left = new Node(); + } + add(c.left, l, mid, i, v); + } else { // mid + 1 ~ r + if (c.right == null) { + c.right = new Node(); + } + add(c.right, mid + 1, r, i, v); + } + c.sum = (c.left != null ? c.left.sum : 0) + (c.right != null ? c.right.sum : 0); + } + } + + // s~e范围的累加和，告诉我！ + public int query(int s, int e) { + return query(root, 1, size, s, e); + } + + // 当前节点c，表达的范围l~r + // 收到了一个任务，s~e这个任务！ + // s~e这个任务，影响了多少l~r范围的数，把答案返回！ + private int query(Node c, int l, int r, int s, int e) { + if (c == null) { + return 0; + } + if (s <= l && r <= e) { // 3~6 1~100任务 + return c.sum; + } + // 有影响，但又不是全影响 + // l ~ r + // l~mid mid+1~r + int mid = (l + r) / 2; + // 1~100 + // 1~50 51 ~ 100 + // 任务 s~e 53~76 + if (e <= mid) { + return query(c.left, l, mid, s, e); + } else if (s > mid) { + return query(c.right, mid + 1, r, s, e); + } else { + return query(c.left, l, mid, s, e) + query(c.right, mid + 1, r, s, e); + } + } + + } + + public static class Right { + public int[] arr; + + public Right(int size) { + arr = new int[size + 1]; + } + + public void add(int i, int v) { + arr[i] += v; + } + + public int query(int s, int e) { + int sum = 0; + for (int i = s; i <= e; i++) { + sum += arr[i]; + } + return sum; + } + + } + + public static void main(String[] args) { + int size = 10000; + int testTime = 50000; + int value = 500; + DynamicSegmentTree dst = new DynamicSegmentTree(size); + Right right = new Right(size); + System.out.println("测试开始"); + for (int k = 0; k < testTime; k++) { + if (Math.random() < 0.5) { + int i = (int) (Math.random() * size) + 1; + int v = (int) (Math.random() * value); + dst.add(i, v); + right.add(i, v); + } else { + int a = (int) (Math.random() * size) + 1; + int b = (int) (Math.random() * size) + 1; + int s = Math.min(a, b); + int e = Math.max(a, b); + int ans1 = dst.query(s, e); + int ans2 = right.query(s, e); + if (ans1 != ans2) { + System.out.println("出错了!"); + System.out.println(ans1); + System.out.println(ans2); + } + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class47/Code02_DynamicSegmentTree.java b/大厂刷题班/class47/Code02_DynamicSegmentTree.java new file mode 100644 index 0000000..1dca034 --- /dev/null +++ b/大厂刷题班/class47/Code02_DynamicSegmentTree.java @@ -0,0 +1,198 @@ +package class47; + +// 同时支持范围增加 + 范围修改 + 范围查询的动态开点线段树（累加和） +// 真的用到！才去建立 +// 懒更新，及其所有的东西，和普通线段树，没有任何区别！ +public class Code02_DynamicSegmentTree { + + public static class Node { + public int sum; + public int lazy; + public int change; + public boolean update; + public Node left; + public Node right; + } + + public static class DynamicSegmentTree { + public Node root; + public int size; + + public DynamicSegmentTree(int max) { + root = new Node(); + size = max; + } + + private void pushUp(Node c) { + c.sum = c.left.sum + c.right.sum; + } + + private void pushDown(Node p, int ln, int rn) { + if (p.left == null) { + p.left = new Node(); + } + if (p.right == null) { + p.right = new Node(); + } + if (p.update) { + p.left.update = true; + p.right.update = true; + p.left.change = p.change; + p.right.change = p.change; + p.left.lazy = 0; + p.right.lazy = 0; + p.left.sum = p.change * ln; + p.right.sum = p.change * rn; + p.update = false; + } + if (p.lazy != 0) { + p.left.lazy += p.lazy; + p.right.lazy += p.lazy; + p.left.sum += p.lazy * ln; + p.right.sum += p.lazy * rn; + p.lazy = 0; + } + } + + public void update(int s, int e, int v) { + update(root, 1, size, s, e, v); + } + + private void update(Node c, int l, int r, int s, int e, int v) { + if (s <= l && r <= e) { + c.update = true; + c.change = v; + c.sum = v * (r - l + 1); + c.lazy = 0; + } else { + int mid = (l + r) >> 1; + pushDown(c, mid - l + 1, r - mid); + if (s <= mid) { + update(c.left, l, mid, s, e, v); + } + if (e > mid) { + update(c.right, mid + 1, r, s, e, v); + } + pushUp(c); + } + } + + public void add(int s, int e, int v) { + add(root, 1, size, s, e, v); + } + + private void add(Node c, int l, int r, int s, int e, int v) { + if (s <= l && r <= e) { + c.sum += v * (r - l + 1); + c.lazy += v; + } else { + int mid = (l + r) >> 1; + pushDown(c, mid - l + 1, r - mid); + if (s <= mid) { + add(c.left, l, mid, s, e, v); + } + if (e > mid) { + add(c.right, mid + 1, r, s, e, v); + } + pushUp(c); + } + } + + public int query(int s, int e) { + return query(root, 1, size, s, e); + } + + private int query(Node c, int l, int r, int s, int e) { + if (s <= l && r <= e) { + return c.sum; + } + int mid = (l + r) >> 1; + pushDown(c, mid - l + 1, r - mid); + int ans = 0; + if (s <= mid) { + ans += query(c.left, l, mid, s, e); + } + if (e > mid) { + ans += query(c.right, mid + 1, r, s, e); + } + return ans; + } + + } + + public static class Right { + public int[] arr; + + public Right(int size) { + arr = new int[size + 1]; + } + + public void add(int s, int e, int v) { + for (int i = s; i <= e; i++) { + arr[i] += v; + } + } + + public void update(int s, int e, int v) { + for (int i = s; i <= e; i++) { + arr[i] = v; + } + } + + public int query(int s, int e) { + int sum = 0; + for (int i = s; i <= e; i++) { + sum += arr[i]; + } + return sum; + } + + } + + public static void main(String[] args) { + int n = 1000; + int value = 50; + int createTimes = 5000; + int operateTimes = 5000; + System.out.println("测试开始"); + for (int i = 0; i < createTimes; i++) { + int size = (int) (Math.random() * n) + 1; + DynamicSegmentTree dst = new DynamicSegmentTree(size); + Right right = new Right(size); + for (int k = 0; k < operateTimes; k++) { + double choose = Math.random(); + if (choose < 0.333) { + int a = (int) (Math.random() * size) + 1; + int b = (int) (Math.random() * size) + 1; + int s = Math.min(a, b); + int e = Math.max(a, b); + int v = (int) (Math.random() * value); + dst.update(s, e, v); + right.update(s, e, v); + } else if (choose < 0.666) { + int a = (int) (Math.random() * size) + 1; + int b = (int) (Math.random() * size) + 1; + int s = Math.min(a, b); + int e = Math.max(a, b); + int v = (int) (Math.random() * value); + dst.add(s, e, v); + right.add(s, e, v); + } else { + int a = (int) (Math.random() * size) + 1; + int b = (int) (Math.random() * size) + 1; + int s = Math.min(a, b); + int e = Math.max(a, b); + int ans1 = dst.query(s, e); + int ans2 = right.query(s, e); + if (ans1 != ans2) { + System.out.println("出错了!"); + System.out.println(ans1); + System.out.println(ans2); + } + } + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class47/Problem_0315_CountOfSmallerNumbersAfterSelf.java b/大厂刷题班/class47/Problem_0315_CountOfSmallerNumbersAfterSelf.java new file mode 100644 index 0000000..7f01d20 --- /dev/null +++ b/大厂刷题班/class47/Problem_0315_CountOfSmallerNumbersAfterSelf.java @@ -0,0 +1,94 @@ +package class47; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.List; + +// 利用只支持单点增加 + 范围查询的动态开点线段树（累加和），解决leetcode 315 +public class Problem_0315_CountOfSmallerNumbersAfterSelf { + + public static class Node { + public int sum; + public Node left; + public Node right; + } + + public static class DynamicSegmentTree { + public Node root; + public int size; + + public DynamicSegmentTree(int max) { + root = new Node(); + size = max; + } + + public void add(int i, int v) { + add(root, 1, size, i, v); + } + + private void add(Node c, int l, int r, int i, int v) { + if (l == r) { + c.sum += v; + } else { + int mid = (l + r) / 2; + if (i <= mid) { + if (c.left == null) { + c.left = new Node(); + } + add(c.left, l, mid, i, v); + } else { + if (c.right == null) { + c.right = new Node(); + } + add(c.right, mid + 1, r, i, v); + } + c.sum = (c.left != null ? c.left.sum : 0) + (c.right != null ? c.right.sum : 0); + } + } + + public int query(int s, int e) { + return query(root, 1, size, s, e); + } + + private int query(Node c, int l, int r, int s, int e) { + if (c == null) { + return 0; + } + if (s <= l && r <= e) { + return c.sum; + } + int mid = (l + r) / 2; + if (e <= mid) { + return query(c.left, l, mid, s, e); + } else if (s > mid) { + return query(c.right, mid + 1, r, s, e); + } else { + return query(c.left, l, mid, s, e) + query(c.right, mid + 1, r, s, e); + } + } + + } + + public static List countSmaller(int[] nums) { + List ans = new ArrayList<>(); + if (nums == null || nums.length == 0) { + return ans; + } + int n = nums.length; + for (int i = 0; i < n; i++) { + ans.add(0); + } + int[][] arr = new int[n][]; + for (int i = 0; i < n; i++) { + arr[i] = new int[] { nums[i], i }; + } + Arrays.sort(arr, (a, b) -> (a[0] - b[0])); + DynamicSegmentTree dst = new DynamicSegmentTree(n); + for (int[] num : arr) { + ans.set(num[1], dst.query(num[1] + 1, n)); + dst.add(num[1] + 1, 1); + } + return ans; + } + +} diff --git a/大厂刷题班/class47/Problem_0358_RearrangeStringKDistanceApart.java b/大厂刷题班/class47/Problem_0358_RearrangeStringKDistanceApart.java new file mode 100644 index 0000000..60319c7 --- /dev/null +++ b/大厂刷题班/class47/Problem_0358_RearrangeStringKDistanceApart.java @@ -0,0 +1,64 @@ +package class47; + +import java.util.ArrayList; +import java.util.Arrays; + +// 本题的解法思路与leetcode 621题 TaskScheduler 问题是一样的 +public class Problem_0358_RearrangeStringKDistanceApart { + + public String rearrangeString(String s, int k) { + if (s == null || s.length() < k) { + return s; + } + char[] str = s.toCharArray(); + int[][] cnts = new int[256][2]; + for (int i = 0; i < 256; i++) { + cnts[i] = new int[] { i, 0 }; + } + int maxCount = 0; + for (char task : str) { + cnts[task][1]++; + maxCount = Math.max(maxCount, cnts[task][1]); + } + int maxKinds = 0; + for (int task = 0; task < 256; task++) { + if (cnts[task][1] == maxCount) { + maxKinds++; + } + } + int N = str.length; + if (!isValid(N, k, maxCount, maxKinds)) { + return ""; + } + ArrayList ans = new ArrayList<>(); + for (int i = 0; i < maxCount; i++) { + ans.add(new StringBuilder()); + } + Arrays.sort(cnts, (a, b) -> (b[1] - a[1])); + int i = 0; + for (; i < 256 && cnts[i][1] == maxCount; i++) { + for (int j = 0; j < maxCount; j++) { + ans.get(j).append((char) cnts[i][0]); + } + } + int out = 0; + for (; i < 256; i++) { + for (int j = 0; j < cnts[i][1]; j++) { + ans.get(out).append((char) cnts[i][0]); + out = out == ans.size() - 2 ? 0 : out + 1; + } + } + StringBuilder builder = new StringBuilder(); + for (StringBuilder b : ans) { + builder.append(b.toString()); + } + return builder.toString(); + } + + public static boolean isValid(int N, int k, int maxCount, int maxKinds) { + int restTasks = N - maxKinds; + int spaces = k * (maxCount - 1); + return spaces - restTasks <= 0; + } + +} diff --git a/大厂刷题班/class47/Problem_0428_SerializeAndDeserializeNaryTree.java b/大厂刷题班/class47/Problem_0428_SerializeAndDeserializeNaryTree.java new file mode 100644 index 0000000..6b9c1df --- /dev/null +++ b/大厂刷题班/class47/Problem_0428_SerializeAndDeserializeNaryTree.java @@ -0,0 +1,103 @@ +package class47; + +import java.util.ArrayList; +import java.util.LinkedList; +import java.util.List; +import java.util.Queue; + +public class Problem_0428_SerializeAndDeserializeNaryTree { + + // 不要提交这个类 + public static class Node { + public int val; + public List children; + + public Node() { + children = new ArrayList<>(); + } + + public Node(int _val) { + val = _val; + children = new ArrayList<>(); + } + + public Node(int _val, List _children) { + val = _val; + children = _children; + } + }; + + // 提交下面这个类 + public static class Codec { + + public static String serialize(Node root) { + if (root == null) { // 空树！直接返回# + return "#"; + } + StringBuilder builder = new StringBuilder(); + serial(builder, root); + return builder.toString(); + } + + // 当前来到head + private static void serial(StringBuilder builder, Node head) { + builder.append(head.val + ","); + if (!head.children.isEmpty()) { + builder.append("[,"); + for (Node next : head.children) { + serial(builder, next); + } + builder.append("],"); + } + } + + public static Node deserialize(String data) { + if (data.equals("#")) { + return null; + } + String[] splits = data.split(","); + Queue queue = new LinkedList<>(); + for (String str : splits) { + queue.offer(str); + } + return deserial(queue); + } + + private static Node deserial(Queue queue) { + Node cur = new Node(Integer.valueOf(queue.poll())); + cur.children = new ArrayList<>(); + if (!queue.isEmpty() && queue.peek().equals("[")) { + queue.poll(); + while (!queue.peek().equals("]")) { + Node child = deserial(queue); + cur.children.add(child); + } + queue.poll(); + } + return cur; + } + + } + + public static void main(String[] args) { + // 如果想跑以下的code，请把Codec类描述和内部所有方法改成static的 + Node a = new Node(1); + Node b = new Node(2); + Node c = new Node(3); + Node d = new Node(4); + Node e = new Node(5); + Node f = new Node(6); + Node g = new Node(7); + a.children.add(b); + a.children.add(c); + a.children.add(d); + b.children.add(e); + b.children.add(f); + e.children.add(g); + String content = Codec.serialize(a); + System.out.println(content); + Node head = Codec.deserialize(content); + System.out.println(content.equals(Codec.serialize(head))); + } + +} diff --git a/大厂刷题班/class47/Problem_0465_OptimalAccountBalancing.java b/大厂刷题班/class47/Problem_0465_OptimalAccountBalancing.java new file mode 100644 index 0000000..d1fe945 --- /dev/null +++ b/大厂刷题班/class47/Problem_0465_OptimalAccountBalancing.java @@ -0,0 +1,149 @@ +package class47; + +import java.util.Arrays; +import java.util.HashMap; + +// 需要证明： +// 一个集合中，假设整体的累加和为K， +// 不管该集合用了什么样的0集合划分方案，当一个新的数到来时： +// 1) 如果该数是-K，那么任何0集合的划分方案中，因为新数字的加入，0集合的数量都会+1 +// 2) 如果该数不是-K，那么任何0集合的划分方案中，0集合的数量都会不变 +public class Problem_0465_OptimalAccountBalancing { + + // 用位信息替代集合结构的暴力尝试 + public static int minTransfers1(int[][] transactions) { + // 不管转账有几笔，最终每个人收到的钱，如果是0，不进入debt数组 + // 只有最终收到的钱，不等于0的人，进入debt数组 + // 收到的钱，4，说明该给出去！ + // 收到的钱，-4，说明该要回来！ + // debt数组的累加和，必为0！ + int[] debt = debts(transactions); + int N = debt.length; + return N - process1(debt, (1 << N) - 1, 0, N); + } + + // set -> int -> 不使用值 -> 只使用状态！ + // 001110 0号人，不在集合里；1、2、3号人在集合里，4、5号人不在集合里！ + // sum -> set这个集合累加和是多少？sum被set决定的！ + // debt数组，收到的钱的数组(固定) + // N, debt的长度(固定) + // 返回值含义 : set这个集合中，最多能划分出多少个小集合累加和是0，返回累加和是0的小集合最多的数量 + public static int process1(int[] debt, int set, int sum, int N) { + // set中只有一个人的时候！ + // debt中，没有0的，所以每个人一定都需要转账！ + if ((set & (set - 1)) == 0) { + return 0; + } + int value = 0; + int max = 0; + // 尝试，每一个人都最后考虑 + // 0,如果最后考虑 + // 1,如果最后考虑 + // 2,如果最后考虑 + // .... + // n-1，最后考虑 + // 011001 + // 0（在） + // 1（不能考虑！因为不在集合里！） + for (int i = 0; i < N; i++) { + value = debt[i]; + if ((set & (1 << i)) != 0) { // i号人真的在集合里，才能考虑！ + // 011001 + // 3号人在 + // 3号人之前，010001（考虑0号人和4号人剩下的情况） + // process ( set ^ (1 << i) , sum - value ) + max = Math.max(max, process1(debt, set ^ (1 << i), sum - value, N)); + } + } + return sum == 0 ? max + 1 : max; + } + + // 上面的尝试过程 + 记忆化搜索 + // 最优解 + public static int minTransfers2(int[][] transactions) { + int[] debt = debts(transactions); + int N = debt.length; + int sum = 0; + for (int num : debt) { + sum += num; + } + int[] dp = new int[1 << N]; + Arrays.fill(dp, -1); + return N - process2(debt, (1 << N) - 1, sum, N, dp); + } + + public static int process2(int[] debt, int set, int sum, int N, int[] dp) { + if (dp[set] != -1) { + return dp[set]; + } + int ans = 0; + if ((set & (set - 1)) != 0) { + int value = 0; + int max = 0; + for (int i = 0; i < N; i++) { + value = debt[i]; + if ((set & (1 << i)) != 0) { + max = Math.max(max, process2(debt, set ^ (1 << i), sum - value, N, dp)); + } + } + ans = sum == 0 ? max + 1 : max; + } + dp[set] = ans; + return ans; + } + + public static int[] debts(int[][] transactions) { + HashMap map = new HashMap<>(); + for (int[] tran : transactions) { + map.put(tran[0], map.getOrDefault(tran[0], 0) + tran[2]); + map.put(tran[1], map.getOrDefault(tran[1], 0) - tran[2]); + } + int N = 0; + for (int value : map.values()) { + if (value != 0) { + N++; + } + } + int[] debt = new int[N]; + int index = 0; + for (int value : map.values()) { + if (value != 0) { + debt[index++] = value; + } + } + return debt; + } + + // 为了测试 + public static int[][] randomTrans(int s, int n, int m) { + int[][] trans = new int[s][3]; + for (int i = 0; i < s; i++) { + trans[i][0] = (int) (Math.random() * n); + trans[i][1] = (int) (Math.random() * n); + trans[i][2] = (int) (Math.random() * m) + 1; + } + return trans; + } + + // 为了测试 + public static void main(String[] args) { + int s = 8; + int n = 8; + int m = 10; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int[][] trans = randomTrans(s, n, m); + int ans1 = minTransfers1(trans); + int ans2 = minTransfers2(trans); + if (ans1 != ans2) { + System.out.println("Oops!"); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/大厂刷题班/class47/Problem_0475_Heaters.java b/大厂刷题班/class47/Problem_0475_Heaters.java new file mode 100644 index 0000000..cd0f7b0 --- /dev/null +++ b/大厂刷题班/class47/Problem_0475_Heaters.java @@ -0,0 +1,113 @@ +package class47; + +import java.util.Arrays; + +public class Problem_0475_Heaters { + + // 比如地点是7, 9, 14 + // 供暖点的位置: 1 3 4 5 13 15 17 + // 先看地点7 + // 由1供暖，半径是6 + // 由3供暖，半径是4 + // 由4供暖，半径是3 + // 由5供暖，半径是2 + // 由13供暖，半径是6 + // 由此可知，地点7应该由供暖点5来供暖，半径是2 + // 再看地点9 + // 供暖点不回退 + // 由5供暖，半径是4 + // 由13供暖，半径是4 + // 由15供暖，半径是6 + // 由此可知，地点9应该由供暖点13来供暖，半径是4 + // 为什么是13而不是5？因为接下来的地点都会更靠右，所以半径一样的时候，就应该选更右的供暖点 + // 再看地点14 + // 供暖点不回退 + // 由13供暖，半径是1 + // 由15供暖，半径是1 + // 由17供暖，半径是3 + // 由此可知，地点14应该由供暖点15来供暖，半径是1 + // 以此类推 + public static int findRadius(int[] houses, int[] heaters) { + Arrays.sort(houses); + Arrays.sort(heaters); + int ans = 0; + // 时间复杂度O(N) + // i是地点，j是供暖点 + for (int i = 0, j = 0; i < houses.length; i++) { + while (!best(houses, heaters, i, j)) { + j++; + } + ans = Math.max(ans, Math.abs(heaters[j] - houses[i])); + } + return ans; + } + + // 这个函数含义： + // 当前的地点houses[i]由heaters[j]来供暖是最优的吗？ + // 当前的地点houses[i]由heaters[j]来供暖，产生的半径是a + // 当前的地点houses[i]由heaters[j + 1]来供暖，产生的半径是b + // 如果a < b, 说明是最优，供暖不应该跳下一个位置 + // 如果a >= b, 说明不是最优，应该跳下一个位置 + public static boolean best(int[] houses, int[] heaters, int i, int j) { + return j == heaters.length - 1 + || Math.abs(heaters[j] - houses[i]) < Math.abs(heaters[j + 1] - houses[i]); + } + + // 下面这个方法不对，你能找出原因嘛？^_^ + public static int findRadius2(int[] houses, int[] heaters) { + Arrays.sort(houses); + Arrays.sort(heaters); + int ans = 0; + // 时间复杂度O(N) + // i是地点，j是供暖点 + for (int i = 0, j = 0; i < houses.length; i++) { + while (!best2(houses, heaters, i, j)) { + j++; + } + ans = Math.max(ans, Math.abs(heaters[j] - houses[i])); + } + return ans; + } + + public static boolean best2(int[] houses, int[] heaters, int i, int j) { + return j == heaters.length - 1 || Math.abs(heaters[j] - houses[i]) <= Math.abs(heaters[j + 1] - houses[i]); + } + + // 为了测试 + public static int[] randomArray(int len, int v) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * v) + 1; + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int len = 5; + int v = 10; + int testTime = 10000; + for (int i = 0; i < testTime; i++) { + int[] a = randomArray(len, v); + int[] b = randomArray(len, v); + int ans1 = findRadius(a, b); + int ans2 = findRadius2(a, b); + if (ans1 != ans2) { + System.out.println("A : "); + for (int num : a) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println("B : "); + for (int num : b) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + } + +} diff --git a/大厂刷题班/class48/Code01_MinKthPairMinusABS.java b/大厂刷题班/class48/Code01_MinKthPairMinusABS.java new file mode 100644 index 0000000..14868c8 --- /dev/null +++ b/大厂刷题班/class48/Code01_MinKthPairMinusABS.java @@ -0,0 +1,122 @@ +package class48; + +import java.util.Arrays; + +// 来自学员问题 +// 比如{ 5, 3, 1, 4 } +// 全部数字对是：(5,3)、(5,1)、(5,4)、(3,1)、(3,4)、(1,4) +// 数字对的差值绝对值： 2、4、1、2、1、3 +// 差值绝对值排序后：1、1、2、2、3、4 +// 给定一个数组arr，和一个正数k +// 返回arr中所有数字对差值的绝对值，第k小的是多少 +// arr = { 5, 3, 1, 4 }, k = 4 +// 返回2 +public class Code01_MinKthPairMinusABS { + + // 暴力解，生成所有数字对差值绝对值，排序，拿出第k个，k从1开始 + public static int kthABS1(int[] arr, int k) { + int n = arr.length; + int m = ((n - 1) * n) >> 1; + if (m == 0 || k < 1 || k > m) { + return -1; + } + int[] abs = new int[m]; + int size = 0; + for (int i = 0; i < n; i++) { + for (int j = i + 1; j < n; j++) { + abs[size++] = Math.abs(arr[i] - arr[j]); + } + } + Arrays.sort(abs); + return abs[k - 1]; + } + + // 二分 + 不回退 + public static int kthABS2(int[] arr, int k) { + int n = arr.length; + if (n < 2 || k < 1 || k > ((n * (n - 1)) >> 1)) { + return -1; + } + Arrays.sort(arr); + // 0 ~ 大-小 二分 + // l ~ r + int left = 0; + int right = arr[n - 1] - arr[0]; + int mid = 0; + int rightest = -1; + while (left <= right) { + mid = (left + right) / 2; + // 数字对差值的绝对值<=mid的数字对个数，是不是 < k个的！ + if (valid(arr, mid, k)) { + rightest = mid; + left = mid + 1; + } else { + right = mid - 1; + } + } + return rightest + 1; + } + + // 假设arr中的所有数字对，差值绝对值<=limit的个数为x + // 如果 x < k，达标，返回true + // 如果 x >= k，不达标，返回false + public static boolean valid(int[] arr, int limit, int k) { + int x = 0; + for (int l = 0, r = 1; l < arr.length; r = Math.max(r, ++l)) { + while (r < arr.length && arr[r] - arr[l] <= limit) { + r++; + } + x += r - l - 1; + } + return x < k; + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] ans = new int[n]; + for (int i = 0; i < ans.length; i++) { + ans[i] = (int) (Math.random() * v); + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int size = 100; + int value = 100; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * size); + int k = (int) (Math.random() * (n * (n - 1) / 2)) + 1; + int[] arr = randomArray(n, value); + int ans1 = kthABS1(arr, k); + int ans2 = kthABS2(arr, k); + if (ans1 != ans2) { + System.out.print("arr : "); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println("k : " + k); + System.out.println("ans1 : " + ans1); + System.out.println("ans2 : " + ans2); + System.out.println("出错了！"); + break; + } + } + System.out.println("测试结束"); + + long start; + long end; + int n = 500000; + int v = 50000000; + int[] arr = randomArray(n, v); + int k = (int) (Math.random() * (n * (n - 1) / 2)) + 1; + start = System.currentTimeMillis(); + kthABS2(arr, k); + end = System.currentTimeMillis(); + System.out.println("大数据量运行时间（毫秒）：" + (end - start)); + } + +} diff --git a/大厂刷题班/class48/Problem_0472_ConcatenatedWords.java b/大厂刷题班/class48/Problem_0472_ConcatenatedWords.java new file mode 100644 index 0000000..533b537 --- /dev/null +++ b/大厂刷题班/class48/Problem_0472_ConcatenatedWords.java @@ -0,0 +1,124 @@ +package class48; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.List; + +public class Problem_0472_ConcatenatedWords { + + public static class TrieNode { + public boolean end; + public TrieNode[] nexts; + + public TrieNode() { + end = false; + nexts = new TrieNode[26]; + } + } + + public static void insert(TrieNode root, char[] s) { + int path = 0; + for (char c : s) { + path = c - 'a'; + if (root.nexts[path] == null) { + root.nexts[path] = new TrieNode(); + } + root = root.nexts[path]; + } + root.end = true; + } + + // 方法1：前缀树优化 + public static List findAllConcatenatedWordsInADict1(String[] words) { + List ans = new ArrayList<>(); + if (words == null || words.length < 3) { + return ans; + } + // 字符串数量 >= 3个 + Arrays.sort(words, (str1, str2) -> str1.length() - str2.length()); + TrieNode root = new TrieNode(); + for (String str : words) { + char[] s = str.toCharArray(); // "" 题目要求 + if (s.length > 0 && split1(s, root, 0)) { + ans.add(str); + } else { + insert(root, s); + } + } + return ans; + } + + // 字符串s[i....]能不能被分解？ + // 之前的元件，全在前缀树上，r就是前缀树头节点 + public static boolean split1(char[] s, TrieNode r, int i) { + boolean ans = false; + if (i == s.length) { // 没字符了！ + ans = true; + } else { // 还有字符 + TrieNode c = r; + // s[i.....] + // s[i..end]作前缀，看看是不是一个元件！f(end+1)... + for (int end = i; end < s.length; end++) { + int path = s[end] - 'a'; + if (c.nexts[path] == null) { + break; + } + c = c.nexts[path]; + if (c.end && split1(s, r, end + 1)) { + ans = true; + break; + } + } + } + return ans; + } + + // 提前准备好动态规划表 + public static int[] dp = new int[1000]; + + // 方法二：前缀树优化 + 动态规划优化 + public static List findAllConcatenatedWordsInADict2(String[] words) { + List ans = new ArrayList<>(); + if (words == null || words.length < 3) { + return ans; + } + Arrays.sort(words, (str1, str2) -> str1.length() - str2.length()); + TrieNode root = new TrieNode(); + for (String str : words) { + char[] s = str.toCharArray(); + Arrays.fill(dp, 0, s.length + 1, 0); + if (s.length > 0 && split2(s, root, 0, dp)) { + ans.add(str); + } else { + insert(root, s); + } + } + return ans; + } + + public static boolean split2(char[] s, TrieNode r, int i, int[] dp) { + if (dp[i] != 0) { + return dp[i] == 1; + } + boolean ans = false; + if (i == s.length) { + ans = true; + } else { + TrieNode c = r; + for (int end = i; end < s.length; end++) { + int path = s[end] - 'a'; + if (c.nexts[path] == null) { + break; + } + c = c.nexts[path]; + if (c.end && split2(s, r, end + 1, dp)) { + ans = true; + break; + } + } + } + dp[i] = ans ? 1 : -1; + return ans; + } + +} diff --git a/大厂刷题班/class48/Problem_0483_SmallestGoodBase.java b/大厂刷题班/class48/Problem_0483_SmallestGoodBase.java new file mode 100644 index 0000000..b11f38d --- /dev/null +++ b/大厂刷题班/class48/Problem_0483_SmallestGoodBase.java @@ -0,0 +1,33 @@ +package class48; + +public class Problem_0483_SmallestGoodBase { + + // ""4651" -> 4651 + public static String smallestGoodBase(String n) { + long num = Long.valueOf(n); + // n这个数，需要从m位开始试，固定位数，一定要有m位！ + for (int m = (int) (Math.log(num + 1) / Math.log(2)); m > 2; m--) { + // num开m次方 + long l = (long) (Math.pow(num, 1.0 / m)); + long r = (long) (Math.pow(num, 1.0 / (m - 1))) + 1L; + while (l <= r) { + long k = l + ((r - l) >> 1); + long sum = 0L; + long base = 1L; + for (int i = 0; i < m && sum <= num; i++) { + sum += base; + base *= k; + } + if (sum < num) { + l = k + 1; + } else if (sum > num) { + r = k - 1; + } else { + return String.valueOf(k); + } + } + } + return String.valueOf(num - 1); + } + +} diff --git a/大厂刷题班/class48/Problem_0499_TheMazeIII.java b/大厂刷题班/class48/Problem_0499_TheMazeIII.java new file mode 100644 index 0000000..38bd2ec --- /dev/null +++ b/大厂刷题班/class48/Problem_0499_TheMazeIII.java @@ -0,0 +1,87 @@ +package class48; + +public class Problem_0499_TheMazeIII { + + // 节点：来到了哪？(r,c)这个位置 + // 从哪个方向来的！d -> 0 1 2 3 4 + // 之前做了什么决定让你来到这个位置。 + public static class Node { + public int r; + public int c; + public int d; + public String p; + + public Node(int row, int col, int dir, String path) { + r = row; + c = col; + d = dir; + p = path; + } + + } + + public static String findShortestWay(int[][] maze, int[] ball, int[] hole) { + int n = maze.length; + int m = maze[0].length; + Node[] q1 = new Node[n * m], q2 = new Node[n * m]; + int s1 = 0, s2 = 0; + boolean[][][] visited = new boolean[maze.length][maze[0].length][4]; + s1 = spread(maze, n, m, new Node(ball[0], ball[1], 4, ""), visited, q1, s1); + while (s1 != 0) { + for (int i = 0; i < s1; i++) { + Node cur = q1[i]; + if (hole[0] == cur.r && hole[1] == cur.c) { + return cur.p; + } + s2 = spread(maze, n, m, cur, visited, q2, s2); + } + Node[] tmp = q1; + q1 = q2; + q2 = tmp; + s1 = s2; + s2 = 0; + } + return "impossible"; + } + + public static int[][] to = { { 1, 0 }, { 0, -1 }, { 0, 1 }, { -1, 0 }, { 0, 0 } }; + + public static String[] re = { "d", "l", "r", "u" }; + + // maze迷宫，走的格子 + // n 行数 + // m 列数 + // 当前来到的节点，cur -> (r,c) 方向 路径（决定） + // v [行][列][方向] 一个格子，其实在宽度有限遍历时，是4个点！ + // q 下一层的队列 + // s 下一层队列填到了哪，size + // 当前点cur，该分裂分裂，该继续走继续走，所产生的一下层的点，进入q，s++ + // 返回值：q增长到了哪？返回size -> s + public static int spread(int[][] maze, int n, int m, + Node cur, boolean[][][] v, Node[] q, int s) { + int d = cur.d; + int r = cur.r + to[d][0]; + int c = cur.c + to[d][1]; + // 分裂去！ + if (d == 4 || r < 0 || r == n || c < 0 || c == m || maze[r][c] != 0) { + for (int i = 0; i < 4; i++) { + if (i != d) { + r = cur.r + to[i][0]; + c = cur.c + to[i][1]; + if (r >= 0 && r < n && c >= 0 && c < m && maze[r][c] == 0 && !v[r][c][i]) { + v[r][c][i] = true; + Node next = new Node(r, c, i, cur.p + re[i]); + q[s++] = next; + } + } + } + } else { // 不分裂！继续走！ + if (!v[r][c][d]) { + v[r][c][d] = true; + q[s++] = new Node(r, c, d, cur.p); + } + } + return s; + } + +} diff --git a/大厂刷题班/class49/Problem_0377_CombinationSumIV.java b/大厂刷题班/class49/Problem_0377_CombinationSumIV.java new file mode 100644 index 0000000..13e7ffa --- /dev/null +++ b/大厂刷题班/class49/Problem_0377_CombinationSumIV.java @@ -0,0 +1,64 @@ +package class49; + +import java.util.Arrays; + +public class Problem_0377_CombinationSumIV { + + // 当前剩下的值是rest， + // nums中所有的值，都可能作为分解rest的，第一块！全试一试 + // nums, 无重复，都是正 + // rest, + public static int ways(int rest, int[] nums) { + if (rest < 0) { + return 0; + } + if (rest == 0) { + return 1; + } + int ways = 0; + for (int num : nums) { + ways += ways(rest - num, nums); + } + return ways; + } + + public static int[] dp = new int[1001]; + + public static int combinationSum41(int[] nums, int target) { + Arrays.fill(dp, 0, target + 1, -1); + return process1(nums, target); + } + + public static int process1(int[] nums, int rest) { + if (rest < 0) { + return 0; + } + if (dp[rest] != -1) { + return dp[rest]; + } + int ans = 0; + if (rest == 0) { + ans = 1; + } else { + for (int num : nums) { + ans += process1(nums, rest - num); + } + } + dp[rest] = ans; + return ans; + } + + // 剪枝 + 严格位置依赖的动态规划 + public static int combinationSum42(int[] nums, int target) { + Arrays.sort(nums); + int[] dp = new int[target + 1]; + dp[0] = 1; + for (int rest = 1; rest <= target; rest++) { + for (int i = 0; i < nums.length && nums[i] <= rest; i++) { + dp[rest] += dp[rest - nums[i]]; + } + } + return dp[target]; + } + +} diff --git a/大厂刷题班/class49/Problem_0440_KthSmallestInLexicographicalOrder.java b/大厂刷题班/class49/Problem_0440_KthSmallestInLexicographicalOrder.java new file mode 100644 index 0000000..9514700 --- /dev/null +++ b/大厂刷题班/class49/Problem_0440_KthSmallestInLexicographicalOrder.java @@ -0,0 +1,88 @@ +package class49; + +// 这道题在leetcode上，所有题解都只能做到O( (logN) 平方)的解 +// 我们课上讲的是O(logN)的解 +// 打败所有题解 +public class Problem_0440_KthSmallestInLexicographicalOrder { + + public static int[] offset = { 0, 1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000 }; + + public static int[] number = { 0, 1, 11, 111, 1111, 11111, 111111, 1111111, 11111111, 111111111, 1111111111 }; + + public static int findKthNumber(int n, int k) { + // 数字num，有几位，len位 + // 65237, 5位，len = 5 + int len = len(n); + // 65237, 开头数字，6，first + int first = n / offset[len]; + // 65237，左边有几个？ + int left = (first - 1) * number[len]; + int pick = 0; + int already = 0; + if (k <= left) { + // k / a 向上取整-> (k + a - 1) / a + pick = (k + number[len] - 1) / number[len]; + already = (pick - 1) * number[len]; + return kth((pick + 1) * offset[len] - 1, len, k - already); + } + int mid = number[len - 1] + (n % offset[len]) + 1; + if (k - left <= mid) { + return kth(n, len, k - left); + } + k -= left + mid; + len--; + pick = (k + number[len] - 1) / number[len] + first; + already = (pick - first - 1) * number[len]; + return kth((pick + 1) * offset[len] - 1, len, k - already); + } + + public static int len(int n) { + int len = 0; + while (n != 0) { + n /= 10; + len++; + } + return len; + } + + public static int kth(int max, int len, int kth) { + // 中间范围还管不管的着！ + // 有任何一步，中间位置没命中，左或者右命中了，那以后就都管不着了！ + // 但是开始时，肯定是管的着的！ + boolean closeToMax = true; + int ans = max / offset[len]; + while (--kth > 0) { + max %= offset[len--]; + int pick = 0; + if (!closeToMax) { + pick = (kth - 1) / number[len]; + ans = ans * 10 + pick; + kth -= pick * number[len]; + } else { + int first = max / offset[len]; + int left = first * number[len]; + if (kth <= left) { + closeToMax = false; + pick = (kth - 1) / number[len]; + ans = ans * 10 + pick; + kth -= pick * number[len]; + continue; + } + kth -= left; + int mid = number[len - 1] + (max % offset[len]) + 1; + if (kth <= mid) { + ans = ans * 10 + first; + continue; + } + closeToMax = false; + kth -= mid; + len--; + pick = (kth + number[len] - 1) / number[len] + first; + ans = ans * 10 + pick; + kth -= (pick - first - 1) * number[len]; + } + } + return ans; + } + +} diff --git a/大厂刷题班/class49/Problem_0446_ArithmeticSlicesIISubsequence.java b/大厂刷题班/class49/Problem_0446_ArithmeticSlicesIISubsequence.java new file mode 100644 index 0000000..c34f2f3 --- /dev/null +++ b/大厂刷题班/class49/Problem_0446_ArithmeticSlicesIISubsequence.java @@ -0,0 +1,30 @@ +package class49; + +import java.util.ArrayList; +import java.util.HashMap; + +public class Problem_0446_ArithmeticSlicesIISubsequence { + + // 时间复杂度是O(N^2)，最优解的时间复杂度 + public static int numberOfArithmeticSlices(int[] arr) { + int N = arr.length; + int ans = 0; + ArrayList> maps = new ArrayList<>(); + for (int i = 0; i < N; i++) { + maps.add(new HashMap<>()); + // ....j...i（结尾） + for (int j = i - 1; j >= 0; j--) { + long diff = (long) arr[i] - (long) arr[j]; + if (diff <= Integer.MIN_VALUE || diff > Integer.MAX_VALUE) { + continue; + } + int dif = (int) diff; + int count = maps.get(j).getOrDefault(dif, 0); + ans += count; + maps.get(i).put(dif, maps.get(i).getOrDefault(dif, 0) + count + 1); + } + } + return ans; + } + +} diff --git a/大厂刷题班/class49/Problem_0489_RobotRoomCleaner.java b/大厂刷题班/class49/Problem_0489_RobotRoomCleaner.java new file mode 100644 index 0000000..a52a3d2 --- /dev/null +++ b/大厂刷题班/class49/Problem_0489_RobotRoomCleaner.java @@ -0,0 +1,57 @@ +package class49; + +import java.util.HashSet; + +public class Problem_0489_RobotRoomCleaner { + + // 不要提交这个接口的内容 + interface Robot { + public boolean move(); + + public void turnLeft(); + + public void turnRight(); + + public void clean(); + } + + // 提交下面的内容 + public static void cleanRoom(Robot robot) { + clean(robot, 0, 0, 0, new HashSet<>()); + } + + private static final int[][] ds = { { -1, 0 }, { 0, 1 }, { 1, 0 }, { 0, -1 } }; + + // 机器人robot， + // 当前来到的位置(x,y)，且之前没来过 + // 机器人脸冲什么方向d，0 1 2 3 + // visited里记录了机器人走过哪些位置 + // 函数的功能：不要重复走visited里面的位置，把剩下的位置，都打扫干净！ + // 而且要回去！ + public static void clean(Robot robot, int x, int y, int d, HashSet visited) { + robot.clean(); + visited.add(x + "_" + y); + for (int i = 0; i < 4; i++) { + // d = 0 : 0 1 2 3 + // d = 1 : 1 2 3 0 + // d = 2 : 2 3 0 1 + // d = 3 : 3 0 1 2 + // 下一步的方向！ + int nd = (i + d) % 4; + // 当下一步的方向定了！下一步的位置在哪？(nx, ny) + int nx = ds[nd][0] + x; + int ny = ds[nd][1] + y; + if (!visited.contains(nx + "_" + ny) && robot.move()) { + clean(robot, nx, ny, nd, visited); + } + robot.turnRight(); + } + // 负责回去：之前的位置，怎么到你的！你要回去，而且方向和到你之前，要一致！ + robot.turnRight(); + robot.turnRight(); + robot.move(); + robot.turnRight(); + robot.turnRight(); + } + +} diff --git a/大厂刷题班/class49/Problem_0527_WordAbbreviation.java b/大厂刷题班/class49/Problem_0527_WordAbbreviation.java new file mode 100644 index 0000000..97440d6 --- /dev/null +++ b/大厂刷题班/class49/Problem_0527_WordAbbreviation.java @@ -0,0 +1,48 @@ +package class49; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.List; + +public class Problem_0527_WordAbbreviation { + + public static List wordsAbbreviation(List words) { + int len = words.size(); + List res = new ArrayList<>(); + HashMap> map = new HashMap<>(); + for (int i = 0; i < len; i++) { + res.add(makeAbbr(words.get(i), 1)); + List list = map.getOrDefault(res.get(i), new ArrayList<>()); + list.add(i); + map.put(res.get(i), list); + } + int[] prefix = new int[len]; + for (int i = 0; i < len; i++) { + if (map.get(res.get(i)).size() > 1) { + List indexes = map.get(res.get(i)); + map.remove(res.get(i)); + for (int j : indexes) { + prefix[j]++; + res.set(j, makeAbbr(words.get(j), prefix[j])); + List list = map.getOrDefault(res.get(j), new ArrayList<>()); + list.add(j); + map.put(res.get(j), list); + } + i--; + } + } + return res; + } + + public static String makeAbbr(String s, int k) { + if (k >= s.length() - 2) { + return s; + } + StringBuilder builder = new StringBuilder(); + builder.append(s.substring(0, k)); + builder.append(s.length() - 1 - k); + builder.append(s.charAt(s.length() - 1)); + return builder.toString(); + } + +} diff --git a/大厂刷题班/class49/Problem_0548_SplitArrayEithEqualSum.java b/大厂刷题班/class49/Problem_0548_SplitArrayEithEqualSum.java new file mode 100644 index 0000000..3009e59 --- /dev/null +++ b/大厂刷题班/class49/Problem_0548_SplitArrayEithEqualSum.java @@ -0,0 +1,45 @@ +package class49; + +public class Problem_0548_SplitArrayEithEqualSum { + + public static boolean splitArray(int[] nums) { + if (nums.length < 7) { + return false; + } + int[] sumLeftToRight = new int[nums.length]; + int[] sumRightToLeft = new int[nums.length]; + int s = 0; + for (int i = 0; i < nums.length; i++) { + sumLeftToRight[i] = s; + s += nums[i]; + } + s = 0; + for (int i = nums.length - 1; i >= 0; i--) { + sumRightToLeft[i] = s; + s += nums[i]; + } + for (int i = 1; i < nums.length - 5; i++) { + for (int j = nums.length - 2; j > i + 3; j--) { + if (sumLeftToRight[i] == sumRightToLeft[j] && find(sumLeftToRight, sumRightToLeft, i, j)) { + return true; + } + } + } + return false; + } + + public static boolean find(int[] sumLeftToRight, int[] sumRightToLeft, int l, int r) { + int s = sumLeftToRight[l]; + int prefix = sumLeftToRight[l + 1]; + int suffix = sumRightToLeft[r - 1]; + for (int i = l + 2; i < r - 1; i++) { + int s1 = sumLeftToRight[i] - prefix; + int s2 = sumRightToLeft[i] - suffix; + if (s1 == s2 && s1 == s) { + return true; + } + } + return false; + } + +} diff --git a/大厂刷题班/class49/Problem_0564_FindTheClosestPalindrome.java b/大厂刷题班/class49/Problem_0564_FindTheClosestPalindrome.java new file mode 100644 index 0000000..b8ff906 --- /dev/null +++ b/大厂刷题班/class49/Problem_0564_FindTheClosestPalindrome.java @@ -0,0 +1,72 @@ +package class49; + +public class Problem_0564_FindTheClosestPalindrome { + + public static String nearestPalindromic(String n) { + Long num = Long.valueOf(n); + Long raw = getRawPalindrome(n); + Long big = raw > num ? raw : getBigPalindrome(raw); + Long small = raw < num ? raw : getSmallPalindrome(raw); + return String.valueOf(big - num >= num - small ? small : big); + } + + public static Long getRawPalindrome(String n) { + char[] chs = n.toCharArray(); + int len = chs.length; + for (int i = 0; i < len / 2; i++) { + chs[len - 1 - i] = chs[i]; + } + return Long.valueOf(String.valueOf(chs)); + } + + public static Long getBigPalindrome(Long raw) { + char[] chs = String.valueOf(raw).toCharArray(); + char[] res = new char[chs.length + 1]; + res[0] = '0'; + for (int i = 0; i < chs.length; i++) { + res[i + 1] = chs[i]; + } + int size = chs.length; + for (int j = (size - 1) / 2 + 1; j >= 0; j--) { + if (++res[j] > '9') { + res[j] = '0'; + } else { + break; + } + } + int offset = res[0] == '1' ? 1 : 0; + size = res.length; + for (int i = size - 1; i >= (size + offset) / 2; i--) { + res[i] = res[size - i - offset]; + } + return Long.valueOf(String.valueOf(res)); + } + + public static Long getSmallPalindrome(Long raw) { + char[] chs = String.valueOf(raw).toCharArray(); + char[] res = new char[chs.length]; + int size = res.length; + for (int i = 0; i < size; i++) { + res[i] = chs[i]; + } + for (int j = (size - 1) / 2; j >= 0; j--) { + if (--res[j] < '0') { + res[j] = '9'; + } else { + break; + } + } + if (res[0] == '0') { + res = new char[size - 1]; + for (int i = 0; i < res.length; i++) { + res[i] = '9'; + } + return size == 1 ? 0 : Long.parseLong(String.valueOf(res)); + } + for (int k = 0; k < size / 2; k++) { + res[size - 1 - k] = res[k]; + } + return Long.valueOf(String.valueOf(res)); + } + +} diff --git a/大厂刷题班/class50/Problem_0568_MaximumVacationDays.java b/大厂刷题班/class50/Problem_0568_MaximumVacationDays.java new file mode 100644 index 0000000..0821911 --- /dev/null +++ b/大厂刷题班/class50/Problem_0568_MaximumVacationDays.java @@ -0,0 +1,50 @@ +package class50; + +public class Problem_0568_MaximumVacationDays { + + public static int maxVacationDays(int[][] fly, int[][] day) { + int n = fly.length; + int k = day[0].length; + // pas[i] = {a, b, c} + // 从a、b、c能飞到i + int[][] pass = new int[n][]; + for (int i = 0; i < n; i++) { + int s = 0; + for (int j = 0; j < n; j++) { + if (fly[j][i] != 0) { + s++; + } + } + pass[i] = new int[s]; + for (int j = n - 1; j >= 0; j--) { + if (fly[j][i] != 0) { + pass[i][--s] = j; + } + } + } + // dp[i][j] -> 第i周必须在j这座城，0~i-1周（随意），最大休假天数 + int[][] dp = new int[k][n]; + // 飞的时机，是周一早上飞，认为对时间没有影响，直接到某个城，然后过一周 + dp[0][0] = day[0][0]; + for (int j = 1; j < n; j++) { + dp[0][j] = fly[0][j] != 0 ? day[j][0] : -1; + } + for (int i = 1; i < k; i++) { // 第i周 + for (int j = 0; j < n; j++) { // 在j号城过！ + // 第i周，要怎么到j号城 + // 下面max的初始值，我第i-1周，就在j号城，选择不动地方，进入第i周 + int max = dp[i - 1][j]; + for (int p : pass[j]) { // 枚举什么？能到j号城的城市p + max = Math.max(max, dp[i - 1][p]); + } + dp[i][j] = max != -1 ? max + day[j][i] : -1; + } + } + int ans = 0; + for (int i = 0; i < n; i++) { + ans = Math.max(ans, dp[k - 1][i]); + } + return ans; + } + +} diff --git a/大厂刷题班/class50/Problem_0587_ErectTheFence.java b/大厂刷题班/class50/Problem_0587_ErectTheFence.java new file mode 100644 index 0000000..8b7aabb --- /dev/null +++ b/大厂刷题班/class50/Problem_0587_ErectTheFence.java @@ -0,0 +1,55 @@ +package class50; + +import java.util.Arrays; + +public class Problem_0587_ErectTheFence { + + public static int[][] outerTrees(int[][] points) { + int n = points.length; + int s = 0; + int[][] stack = new int[n << 1][]; + // x小的排前面，x一样的，y小的排前面 + Arrays.sort(points, (a, b) -> a[0] != b[0] ? a[0] - b[0] : a[1] - b[1]); + for (int i = 0; i < n; i++) { + while (s > 1 && cross(stack[s - 2], stack[s - 1], points[i]) > 0) { + s--; + } + stack[s++] = points[i]; + } + for (int i = n - 2; i >= 0; i--) { + while (s > 1 && cross(stack[s - 2], stack[s - 1], points[i]) > 0) { + s--; + } + stack[s++] = points[i]; + } + // 去重返回 + Arrays.sort(stack, 0, s, (a, b) -> b[0] == a[0] ? b[1] - a[1] : b[0] - a[0]); + n = 1; + for (int i = 1; i < s; i++) { + // 如果i点，x和y，与i-1点，x和y都一样 + // i点与i-1点，在同一个位置，此时，i点不保留 + if (stack[i][0] != stack[i - 1][0] || stack[i][1] != stack[i - 1][1]) { + stack[n++] = stack[i]; + } + } + return Arrays.copyOf(stack, n); + } + + // 叉乘的实现 + // 假设有a、b、c三个点，并且给出每个点的(x,y)位置 + // 从a到c的向量，在从a到b的向量的哪一侧？ + // 如果a到c的向量，在从a到b的向量右侧，返回正数 + // 如果a到c的向量，在从a到b的向量左侧，返回负数 + // 如果a到c的向量，和从a到b的向量重合，返回0 + public static int cross(int[] a, int[] b, int[] c) { + return (b[1] - a[1]) * (c[0] - b[0]) - (b[0] - a[0]) * (c[1] - b[1]); + } + + public static void main(String[] args) { + int[] a = { 4, 4 }; + int[] b = { 1, 1 }; + int[] c = { 1, 5 }; + System.out.println(cross(a, b, c)); + } + +} diff --git a/大厂刷题班/class50/Problem_0588_DesignInMemoryFileSystem.java b/大厂刷题班/class50/Problem_0588_DesignInMemoryFileSystem.java new file mode 100644 index 0000000..e44ea18 --- /dev/null +++ b/大厂刷题班/class50/Problem_0588_DesignInMemoryFileSystem.java @@ -0,0 +1,107 @@ +package class50; + +import java.util.ArrayList; +import java.util.List; +import java.util.TreeMap; + +public class Problem_0588_DesignInMemoryFileSystem { + + class FileSystem { + + public class Node { + // 文件名、目录名 + public String name; + // content == null 意味着这个节点是目录 + // content != null 意味着这个节点是文件 + public StringBuilder content; + public TreeMap nexts; + + // 构造目录 + public Node(String n) { + name = n; + content = null; + nexts = new TreeMap<>(); + } + + // 构造文件，c是文件内容 + public Node(String n, String c) { + name = n; + content = new StringBuilder(c); + nexts = new TreeMap<>(); + } + + } + + public Node head; + + public FileSystem() { + head = new Node(""); + } + + public List ls(String path) { + List ans = new ArrayList<>(); + Node cur = head; + String[] parts = path.split("/"); + int n = parts.length; + for (int i = 1; i < n; i++) { + if (!cur.nexts.containsKey(parts[i])) { + return ans; + } + cur = cur.nexts.get(parts[i]); + } + // cur结束了！来到path最后的节点，该返回了 + // ls a/b/c cur 来到c目录 + // 如果c是目录，那么就要返回c下面所有的东西！ + // 如果c是文件，那么就值返回c + if (cur.content == null) { + ans.addAll(cur.nexts.keySet()); + } else { + ans.add(cur.name); + } + return ans; + } + + public void mkdir(String path) { + Node cur = head; + String[] parts = path.split("/"); + int n = parts.length; + for (int i = 1; i < n; i++) { + if (!cur.nexts.containsKey(parts[i])) { + cur.nexts.put(parts[i], new Node(parts[i])); + } + cur = cur.nexts.get(parts[i]); + } + } + + public void addContentToFile(String path, String content) { + Node cur = head; + String[] parts = path.split("/"); + int n = parts.length; + for (int i = 1; i < n - 1; i++) { + if (!cur.nexts.containsKey(parts[i])) { + cur.nexts.put(parts[i], new Node(parts[i])); + } + cur = cur.nexts.get(parts[i]); + } + // 来到的是倒数第二的节点了！注意for！ + if (!cur.nexts.containsKey(parts[n - 1])) { + cur.nexts.put(parts[n - 1], new Node(parts[n - 1], "")); + } + cur.nexts.get(parts[n - 1]).content.append(content); + } + + public String readContentFromFile(String path) { + Node cur = head; + String[] parts = path.split("/"); + int n = parts.length; + for (int i = 1; i < n; i++) { + if (!cur.nexts.containsKey(parts[i])) { + cur.nexts.put(parts[i], new Node(parts[i])); + } + cur = cur.nexts.get(parts[i]); + } + return cur.content.toString(); + } + } + +} diff --git a/大厂刷题班/class50/Problem_0600_NonnegativeIntegersWithoutConsecutiveOnes.java b/大厂刷题班/class50/Problem_0600_NonnegativeIntegersWithoutConsecutiveOnes.java new file mode 100644 index 0000000..48f461d --- /dev/null +++ b/大厂刷题班/class50/Problem_0600_NonnegativeIntegersWithoutConsecutiveOnes.java @@ -0,0 +1,80 @@ +package class50; + +public class Problem_0600_NonnegativeIntegersWithoutConsecutiveOnes { + +// // f(0, false, 5,n) +// // 6 5 ... 0 -1 n +// // 0 ? 停 +// +// // 5 4 3 2 1 0 -1 +// // n : 1 0 1 1 1 0 +// // 0 1 i +// // pre : 第i+1位做的决定，0还是1 +// // 在 第i+1位做的决定 是pre的情况下，从index位开始，往后都做决定！ +// // 但是，不能有相邻的1，请问有多少决定！返回！ +// // alreadyLess : 之前的决定，是不是已经导致你到index的时候，已经比n小了！ +// // pre -> 0 1 +// // alreadyLess -> true false +// // index -> n的位数，（logN） +// // 2 * 2 * logN +// // dp[2][] +// // int alreadyLess 0 1 +// public int f(int pre, boolean alreadyLess, int index, int n) { +// if (index == -1) { +// return 1; +// } +// if (pre == 1) { +// // 只能做0的决定，然后去往i-1位置 +// boolean curLessOrMore = alreadyLess | ((n & 1 << index) != 0); +// return f(0, curLessOrMore, index - 1, n); +// } else { // pre == 0的决定 +// // 可能性1，继续做0的决定 +// boolean curLessOrMore = alreadyLess | ((n & 1 << index) != 0); +// int p1 = f(0, curLessOrMore, index - 1, n); +// // 可能性2，做1的决定 +// // 1)pre == 0的决定, 2) +// int p2 = 0; +// if (alreadyLess || (n & 1 << index) != 0) { +// p2 = f(1, alreadyLess, index - 1, n); +// } +// return p1 + p2; +// } +// } + + public static int findIntegers(int n) { + int i = 31; + for (; i >= 0; i--) { + if ((n & (1 << i)) != 0) { + break; + } + } + // for循环出来之后，i表示，n最高位的1，在哪？ + // 从这个位置，往右边低位上走！ + int[][][] dp = new int[2][2][i + 1]; + return f(0, 0, i, n, dp); + } + + + public static int f(int pre, int alreadyLess, int index, int num, int[][][] dp) { + if (index == -1) { + return 1; + } + if (dp[pre][alreadyLess][index] != 0) { + return dp[pre][alreadyLess][index]; + } + int ans = 0; + if (pre == 1) { + ans = f(0, Math.max(alreadyLess, (num & (1 << index)) != 0 ? 1 : 0), index - 1, num, dp); + } else { + if ((num & (1 << index)) == 0 && alreadyLess == 0) { + ans = f(0, alreadyLess, index - 1, num, dp); + } else { + ans = f(1, alreadyLess, index - 1, num, dp) + + f(0, Math.max(alreadyLess, (num & (1 << index)) != 0 ? 1 : 0), index - 1, num, dp); + } + } + dp[pre][alreadyLess][index] = ans; + return ans; + } + +} diff --git a/大厂刷题班/class51/LCP_0003_Robot.java b/大厂刷题班/class51/LCP_0003_Robot.java new file mode 100644 index 0000000..14451b3 --- /dev/null +++ b/大厂刷题班/class51/LCP_0003_Robot.java @@ -0,0 +1,144 @@ +package class51; + +import java.util.Arrays; +import java.util.HashSet; + +// leetcode题目 : https://leetcode-cn.com/problems/programmable-robot/ +public class LCP_0003_Robot { + + public static boolean robot1(String command, int[][] obstacles, int x, int y) { + int X = 0; + int Y = 0; + HashSet set = new HashSet<>(); + set.add(0); + for (char c : command.toCharArray()) { + X += c == 'R' ? 1 : 0; + Y += c == 'U' ? 1 : 0; + set.add((X << 10) | Y); + } + // 不考虑任何额外的点，机器人能不能到达，(x，y) + if (!meet1(x, y, X, Y, set)) { + return false; + } + for (int[] ob : obstacles) { // ob[0] ob[1] + if (ob[0] <= x && ob[1] <= y && meet1(ob[0], ob[1], X, Y, set)) { + return false; + } + } + return true; + } + + // 一轮以内，X，往右一共有几个单位 + // Y, 往上一共有几个单位 + // set, 一轮以内的所有可能性 + // (x,y)要去的点 + // 机器人从(0,0)位置，能不能走到(x,y) + public static boolean meet1(int x, int y, int X, int Y, HashSet set) { + if (X == 0) { // Y != 0 往上肯定走了！ + return x == 0; + } + if (Y == 0) { + return y == 0; + } + // 至少几轮？ + int atLeast = Math.min(x / X, y / Y); + // 经历过最少轮数后，x剩多少？ + int rx = x - atLeast * X; + // 经历过最少轮数后，y剩多少？ + int ry = y - atLeast * Y; + return set.contains((rx << 10) | ry); + } + + // 此处为一轮以内，x和y最大能移动的步数，对应的2的几次方 + // 比如本题，x和y最大能移动1000步，就对应2的10次方 + // 如果换一个数据量，x和y最大能移动5000步，就对应2的13次方 + // 只需要根据数据量修改这一个变量，剩下的代码不需要调整 + public static final int bit = 10; + // 如果，x和y最大能移动的步数，对应2的bit次方 + // 那么一个坐标(x,y)，所有的可能性就是：(2 ^ bit) ^ 2 = 2 ^ (bit * 2) + // 也就是，(1 << (bit << 1))个状态，记为bits + public static int bits = (1 << (bit << 1)); + // 为了表示下bits个状态，需要几个整数？ + // 32位只需要一个整数，所以bits个状态，需要bits / 32 个整数 + // 即整型长度需要 : bits >> 5 + public static int[] set = new int[bits >> 5]; + + public static boolean robot2(String command, int[][] obstacles, int x, int y) { + Arrays.fill(set, 0); + set[0] = 1; + int X = 0; + int Y = 0; + for (char c : command.toCharArray()) { + X += c == 'R' ? 1 : 0; + Y += c == 'U' ? 1 : 0; + add((X << 10) | Y); + } + if (!meet2(x, y, X, Y)) { + return false; + } + for (int[] ob : obstacles) { + if (ob[0] <= x && ob[1] <= y && meet2(ob[0], ob[1], X, Y)) { + return false; + } + } + return true; + } + + public static boolean meet2(int x, int y, int X, int Y) { + if (X == 0) { + return x == 0; + } + if (Y == 0) { + return y == 0; + } + int atLeast = Math.min(x / X, y / Y); + int rx = x - atLeast * X; + int ry = y - atLeast * Y; + return contains((rx << 10) | ry); + } + + public static void add(int status) { + set[status >> 5] |= 1 << (status & 31); + } + + public static boolean contains(int status) { + return (status < bits) && (set[status >> 5] & (1 << (status & 31))) != 0; + } + + // int num -> 32位的状态 + // 请打印这32位状态 + public static void printBinary(int num) { + for (int i = 31; i >= 0; i--) { + System.out.print((num & (1 << i)) != 0 ? "1" : "0"); + } + System.out.println(); + } + + public static void main(String[] args) { + int x = 7; + printBinary(x); + + int y = 4; + + printBinary(y); + + // x_y 组合！ + int c = (x << 10) | y; + printBinary(c); + + System.out.println(c); + + // 0 ~ 1048575 任何一个数，bit来表示的！ +// int[] set = new int[32768]; +// set[0] = int 32 位 0~31这些数出现过没出现过 +// set[1] = int 32 位 32~63这些数出现过没出现过 + + // 0 ~ 1048575 +// int[] set = new int[32768]; +// int num = 738473; // 32 bit int +//// set[ 734873 / 32 ] // 734873 % 32 +// boolean exist = (set[num / 32] & (1 << (num % 32))) != 0; + + } + +} diff --git a/大厂刷题班/class51/Problem_0630_CourseScheduleIII.java b/大厂刷题班/class51/Problem_0630_CourseScheduleIII.java new file mode 100644 index 0000000..fb7e70d --- /dev/null +++ b/大厂刷题班/class51/Problem_0630_CourseScheduleIII.java @@ -0,0 +1,33 @@ +package class51; + +import java.util.Arrays; +import java.util.PriorityQueue; + +public class Problem_0630_CourseScheduleIII { + + public static int scheduleCourse(int[][] courses) { + // courses[i] = {花费，截止} + Arrays.sort(courses, (a, b) -> a[1] - b[1]); + // 花费时间的大根堆 + PriorityQueue heap = new PriorityQueue<>((a, b) -> b - a); + // 时间点 + int time = 0; + for (int[] c : courses) { + // + if (time + c[0] <= c[1]) { // 当前时间 + 花费 <= 截止时间的 + heap.add(c[0]); + time += c[0]; + } else { // 当前时间 + 花费 > 截止时间的, 只有淘汰掉某课，当前的课才能进来！ + if (!heap.isEmpty() && heap.peek() > c[0]) { +// time -= heap.poll(); +// heap.add(c[0]); +// time += c[0]; + heap.add(c[0]); + time += c[0] - heap.poll(); + } + } + } + return heap.size(); + } + +} diff --git a/大厂刷题班/class51/Problem_0642_DesignSearchAutocompleteSystem.java b/大厂刷题班/class51/Problem_0642_DesignSearchAutocompleteSystem.java new file mode 100644 index 0000000..6eec56f --- /dev/null +++ b/大厂刷题班/class51/Problem_0642_DesignSearchAutocompleteSystem.java @@ -0,0 +1,134 @@ +package class51; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.List; +import java.util.TreeSet; + +public class Problem_0642_DesignSearchAutocompleteSystem { + + class AutocompleteSystem { + + public class TrieNode { + public TrieNode father; + public String path; + public TrieNode[] nexts; + + public TrieNode(TrieNode f, String p) { + father = f; + path = p; + nexts = new TrieNode[27]; + } + } + + public class WordCount implements Comparable { + public String word; + public int count; + + public WordCount(String w, int c) { + word = w; + count = c; + } + + public int compareTo(WordCount o) { + return count != o.count ? (o.count - count) : word.compareTo(o.word); + } + } + + // 题目的要求，只输出排名前3的列表 + public final int top = 3; + public final TrieNode root = new TrieNode(null, ""); + // 某个前缀树节点，上面的有序表，不在这个节点内部 + // 外挂 + public HashMap> nodeRankMap = new HashMap<>(); + + // 字符串 "abc" 7次 -> ("abc", 7) + public HashMap wordCountMap = new HashMap<>(); + + + public String path; + public TrieNode cur; + + // ' ' -> 0 + // 'a' -> 1 + // 'b' -> 2 + // ... + // 'z' -> 26 + // '`' a b .. z + private int f(char c) { + return c == ' ' ? 0 : (c - '`'); + } + + public AutocompleteSystem(String[] sentences, int[] times) { + path = ""; + cur = root; + for (int i = 0; i < sentences.length; i++) { + String word = sentences[i]; + int count = times[i]; + WordCount wc = new WordCount(word, count - 1); + wordCountMap.put(word, wc); + for (char c : word.toCharArray()) { + input(c); + } + input('#'); + } + } + + // 之前已经有一些历史了！ + // 当前键入 c 字符 + // 请顺着之前的历史，根据c字符是什么，继续 + // path : 之前键入的字符串整体 + // cur : 当前滑到了前缀树的哪个节点 + + public List input(char c) { + List ans = new ArrayList<>(); + if (c != '#') { + path += c; + int i = f(c); + if (cur.nexts[i] == null) { + cur.nexts[i] = new TrieNode(cur, path); + } + cur = cur.nexts[i]; + if (!nodeRankMap.containsKey(cur)) { + nodeRankMap.put(cur, new TreeSet<>()); + } + int k = 0; + // for循环本身就是根据排序后的顺序来遍历！ + for (WordCount wc : nodeRankMap.get(cur)) { + if (k == top) { + break; + } + ans.add(wc.word); + k++; + } + } + // c = # path = "abcde" + // # + // # + // # + // a b .. # + if (c == '#' && !path.equals("")) { + // 真的有一个，有效字符串需要加入！path + if (!wordCountMap.containsKey(path)) { + wordCountMap.put(path, new WordCount(path, 0)); + } + // 有序表内部的小对象，该小对象参与排序的指标数据改变 + // 但是有序表并不会自动刷新 + // 所以，删掉老的，加新的！ + WordCount oldOne = wordCountMap.get(path); + WordCount newOne = new WordCount(path, oldOne.count + 1); + while (cur != root) { + nodeRankMap.get(cur).remove(oldOne); + nodeRankMap.get(cur).add(newOne); + cur = cur.father; + } + wordCountMap.put(path, newOne); + path = ""; + // cur 回到头了 + } + return ans; + } + + } + +} diff --git a/大厂刷题班/class51/Problem_0875_KokoEatingBananas.java b/大厂刷题班/class51/Problem_0875_KokoEatingBananas.java new file mode 100644 index 0000000..9a8f4a4 --- /dev/null +++ b/大厂刷题班/class51/Problem_0875_KokoEatingBananas.java @@ -0,0 +1,34 @@ +package class51; + +public class Problem_0875_KokoEatingBananas { + + public static int minEatingSpeed(int[] piles, int h) { + int L = 1; + int R = 0; + for (int pile : piles) { + R = Math.max(R, pile); + } + int ans = 0; + int M = 0; + while (L <= R) { + M = L + ((R - L) >> 1); + if (hours(piles, M) <= h) { + ans = M; + R = M - 1; + } else { + L = M + 1; + } + } + return ans; + } + + public static long hours(int[] piles, int speed) { + long ans = 0; + int offset = speed - 1; + for (int pile : piles) { + ans += (pile + offset) / speed; + } + return ans; + } + +} diff --git a/大厂刷题班/class51/Problem_1035_UncrossedLines.java b/大厂刷题班/class51/Problem_1035_UncrossedLines.java new file mode 100644 index 0000000..04d1c4d --- /dev/null +++ b/大厂刷题班/class51/Problem_1035_UncrossedLines.java @@ -0,0 +1,86 @@ +package class51; + +import java.util.HashMap; + +public class Problem_1035_UncrossedLines { + + // 针对这个题的题意，做的动态规划 + public static int maxUncrossedLines1(int[] A, int[] B) { + if (A == null || A.length == 0 || B == null || B.length == 0) { + return 0; + } + int N = A.length; + int M = B.length; + // dp[i][j]代表: A[0...i]对应B[0...j]最多能划几条线 + int[][] dp = new int[N][M]; + if (A[0] == B[0]) { + dp[0][0] = 1; + } + for (int j = 1; j < M; j++) { + dp[0][j] = A[0] == B[j] ? 1 : dp[0][j - 1]; + } + for (int i = 1; i < N; i++) { + dp[i][0] = A[i] == B[0] ? 1 : dp[i - 1][0]; + } + // 某个值(key)，上次在A中出现的位置(value) + HashMap AvalueLastIndex = new HashMap<>(); + AvalueLastIndex.put(A[0], 0); + // 某个值(key)，上次在B中出现的位置(value) + HashMap BvalueLastIndex = new HashMap<>(); + for (int i = 1; i < N; i++) { + AvalueLastIndex.put(A[i], i); + BvalueLastIndex.put(B[0], 0); + for (int j = 1; j < M; j++) { + BvalueLastIndex.put(B[j], j); + // 可能性1，就是不让A[i]去划线 + int p1 = dp[i - 1][j]; + // 可能性2，就是不让B[j]去划线 + int p2 = dp[i][j - 1]; + // 可能性3，就是要让A[i]去划线，那么如果A[i]==5，它跟谁划线？ + // 贪心的点：一定是在B[0...j]中，尽量靠右侧的5 + int p3 = 0; + if (BvalueLastIndex.containsKey(A[i])) { + int last = BvalueLastIndex.get(A[i]); + p3 = (last > 0 ? dp[i - 1][last - 1] : 0) + 1; + } + // 可能性4，就是要让B[j]去划线，那么如果B[j]==7，它跟谁划线？ + // 贪心的点：一定是在A[0...i]中，尽量靠右侧的7 + int p4 = 0; + if (AvalueLastIndex.containsKey(B[j])) { + int last = AvalueLastIndex.get(B[j]); + p4 = (last > 0 ? dp[last - 1][j - 1] : 0) + 1; + } + dp[i][j] = Math.max(Math.max(p1, p2), Math.max(p3, p4)); + } + BvalueLastIndex.clear(); + } + return dp[N - 1][M - 1]; + } + + // 但是其实这个题，不就是求两个数组的最长公共子序列吗？ + public static int maxUncrossedLines2(int[] A, int[] B) { + if (A == null || A.length == 0 || B == null || B.length == 0) { + return 0; + } + int N = A.length; + int M = B.length; + int[][] dp = new int[N][M]; + dp[0][0] = A[0] == B[0] ? 1 : 0; + for (int j = 1; j < M; j++) { + dp[0][j] = A[0] == B[j] ? 1 : dp[0][j - 1]; + } + for (int i = 1; i < N; i++) { + dp[i][0] = A[i] == B[0] ? 1 : dp[i - 1][0]; + } + for (int i = 1; i < N; i++) { + for (int j = 1; j < M; j++) { + int p1 = dp[i - 1][j]; + int p2 = dp[i][j - 1]; + int p3 = A[i] == B[j] ? (1 + dp[i - 1][j - 1]) : 0; + dp[i][j] = Math.max(p1, Math.max(p2, p3)); + } + } + return dp[N - 1][M - 1]; + } + +} diff --git a/大厂刷题班/class52/Problem_0656_CoinPath.java b/大厂刷题班/class52/Problem_0656_CoinPath.java new file mode 100644 index 0000000..04439c5 --- /dev/null +++ b/大厂刷题班/class52/Problem_0656_CoinPath.java @@ -0,0 +1,67 @@ +package class52; + +import java.util.Arrays; +import java.util.LinkedList; +import java.util.List; + +public class Problem_0656_CoinPath { + + // arr 0 -> n-1 + // arr[i] = -1 死了！ + public static int minCost(int[] arr, int jump) { + if (arr == null || arr.length == 0) { + return 0; + } + int n = arr.length; + if (arr[0] == -1 || arr[n - 1] == -1) { + return -1; + } + // dp[i] : 从0位置开始出发，到达i位置的最小代价 + int[] dp = new int[n]; + dp[0] = arr[0]; + for (int i = 1; i < n; i++) { + dp[i] = Integer.MAX_VALUE; + if (arr[i] != -1) { + for (int pre = Math.max(0, i - jump); pre < i; pre++) { + if (dp[pre] != -1) { + dp[i] = Math.min(dp[i], dp[pre] + arr[i]); + } + } + } + dp[i] = dp[i] == Integer.MAX_VALUE ? -1 : dp[i]; + } + return dp[n - 1]; + } + + public static List cheapestJump(int[] arr, int jump) { + int n = arr.length; + int[] best = new int[n]; + int[] last = new int[n]; + int[] size = new int[n]; + Arrays.fill(best, Integer.MAX_VALUE); + Arrays.fill(last, -1); + best[0] = 0; + for (int i = 0; i < n; i++) { + if (arr[i] != -1) { + for (int j = Math.max(0, i - jump); j < i; j++) { + if (arr[j] != -1) { + int cur = best[j] + arr[i]; + // 1) 代价低换方案！ + // 2) 代价一样，但是点更多，换方案！ + if (cur < best[i] || (cur == best[i] && size[i] - 1 < size[j])) { + best[i] = cur; + last[i] = j; + size[i] = size[j] + 1; + } + } + } + } + } + List path = new LinkedList<>(); + for (int cur = n - 1; cur >= 0; cur = last[cur]) { + path.add(0, cur + 1); + } + return path.get(0) != 1 ? new LinkedList<>() : path; + } + +} diff --git a/大厂刷题班/class52/Problem_0683_KEmptySlots.java b/大厂刷题班/class52/Problem_0683_KEmptySlots.java new file mode 100644 index 0000000..7d9b8d7 --- /dev/null +++ b/大厂刷题班/class52/Problem_0683_KEmptySlots.java @@ -0,0 +1,75 @@ +package class52; + +public class Problem_0683_KEmptySlots { + + public static int kEmptySlots1(int[] bulbs, int k) { + int n = bulbs.length; + int[] days = new int[n]; + for (int i = 0; i < n; i++) { + days[bulbs[i] - 1] = i + 1; + } + int ans = Integer.MAX_VALUE; + if (k == 0) { + for (int i = 1; i < n; i++) { + ans = Math.min(ans, Math.max(days[i - 1], days[i])); + } + } else { + int[] minq = new int[n]; + int l = 0; + int r = -1; + for (int i = 1; i < n && i < k; i++) { + while (l <= r && days[minq[r]] >= days[i]) { + r--; + } + minq[++r] = i; + } + for (int i = 1, j = k; j < n - 1; i++, j++) { + while (l <= r && days[minq[r]] >= days[j]) { + r--; + } + minq[++r] = j; + int cur = Math.max(days[i - 1], days[j + 1]); + if (days[minq[l]] > cur) { + ans = Math.min(ans, cur); + } + if (i == minq[l]) { + l++; + } + } + } + return (ans == Integer.MAX_VALUE) ? -1 : ans; + } + + public static int kEmptySlots2(int[] bulbs, int k) { + int n = bulbs.length; + int[] days = new int[n]; + for (int i = 0; i < n; i++) { + days[bulbs[i] - 1] = i + 1; + } + int ans = Integer.MAX_VALUE; + for (int left = 0, mid = 1, right = k + 1; right < n; mid++) { + // 验证指针mid + // mid 永远不和left撞上的！ + // 1) mid在left和right中间验证的时候，没通过！ + // 2) mid是撞上right的时候 + if (days[mid] <= Math.max(days[left], days[right])) { +// if(mid == right) { // left...right 达标的！ +// int cur = Math.max(days[left], days[right]); +// ans = Math.min(ans, cur); +// left = mid; +// right = mid + k + 1; +// } else { // 验证不通过！ +// left = mid; +// right = mid + k + 1; +// } + if (mid == right) { // 收答案！ + ans = Math.min(ans, Math.max(days[left], days[right])); + } + left = mid; + right = mid + k + 1; + } + } + return (ans == Integer.MAX_VALUE) ? -1 : ans; + } + +} diff --git a/大厂刷题班/class52/Problem_1488_AvoidFloodInTheCity.java b/大厂刷题班/class52/Problem_1488_AvoidFloodInTheCity.java new file mode 100644 index 0000000..ac6ec55 --- /dev/null +++ b/大厂刷题班/class52/Problem_1488_AvoidFloodInTheCity.java @@ -0,0 +1,81 @@ +package class52; + +import java.util.HashMap; +import java.util.HashSet; +import java.util.LinkedList; +import java.util.PriorityQueue; + +public class Problem_1488_AvoidFloodInTheCity { + + // rains[i] = j 第i天轮到j号湖泊下雨 + // 规定，下雨日，干啥 : -1 + // 不下雨日，如果没有湖泊可抽 : 1 + public static int[] avoidFlood(int[] rains) { + int n = rains.length; + int[] ans = new int[n]; + int[] invalid = new int[0]; + // key : 某个湖 + // value : 这个湖在哪些位置降雨 + // 4 : {3,7,19,21} + // 1 : { 13 } + // 2 : {4, 56} + HashMap> map = new HashMap<>(); + for (int i = 0; i < n; i++) { + if (rains[i] != 0) { // 第i天要下雨，rains[i] + // 3天 9号 + // 9号 { 3 } + // 9号 {1, 3} + if (!map.containsKey(rains[i])) { + map.put(rains[i], new LinkedList<>()); + } + map.get(rains[i]).addLast(i); + } + } + // 没抽干的湖泊表 + // 某个湖如果满了，加入到set里 + // 某个湖被抽干了，从set中移除 + HashSet set = new HashSet<>(); + // 这个堆的堆顶表示最先处理的湖是哪个 + PriorityQueue heap = new PriorityQueue<>(); + for (int i = 0; i < n; i++) { // 0 1 2 3 ... + if (rains[i] != 0) { + if (set.contains(rains[i])) { + return invalid; + } + // 放入到没抽干的表里 + set.add(rains[i]); + map.get(rains[i]).pollFirst(); + if (!map.get(rains[i]).isEmpty()) { + heap.add(new Work(rains[i], map.get(rains[i]).peekFirst())); + } + // 题目规定 + ans[i] = -1; + } else { // 今天干活！ + if (heap.isEmpty()) { + ans[i] = 1; + } else { + Work cur = heap.poll(); + set.remove(cur.lake); + ans[i] = cur.lake; + } + } + } + return ans; + } + + public static class Work implements Comparable { + public int lake; + public int nextRain; + + public Work(int l, int p) { + lake = l; + nextRain = p; + } + + @Override + public int compareTo(Work o) { + return nextRain - o.nextRain; + } + } + +} diff --git a/算法周更班/class_2021_11_4_week/Code01_RetainTree.java b/算法周更班/class_2021_11_4_week/Code01_RetainTree.java new file mode 100644 index 0000000..f787c7b --- /dev/null +++ b/算法周更班/class_2021_11_4_week/Code01_RetainTree.java @@ -0,0 +1,89 @@ +package class_2021_11_4_week; + +import java.util.ArrayList; +import java.util.List; + +public class Code01_RetainTree { + + public static class Node { + // 值 + public int value; + // 是否保留 + public boolean retain; + // 下级节点 + public List nexts; + + public Node(int v, boolean r) { + value = v; + retain = r; + nexts = new ArrayList<>(); + } + } + + // 给定一棵树的头节点head + // 请按照题意，保留节点，没有保留的节点删掉 + // 树调整完之后，返回头节点 + public static Node retain(Node x) { + if (x.nexts.isEmpty()) { + return x.retain ? x : null; + } + // x下层有节点 + List newNexts = new ArrayList<>(); + for (Node next : x.nexts) { + Node newNext = retain(next); + if (newNext != null) { + newNexts.add(newNext); + } + } + // x.nexts 老的链表，下级节点 + // newNexts 新的链表，只有保留的在里面 + // + if (!newNexts.isEmpty() || x.retain) { + x.nexts = newNexts; + return x; + } + return null; + } + + // 先序打印 + public static void preOrderPrint(Node head) { + System.out.println(head.value); + for (Node next : head.nexts) { + preOrderPrint(next); + } + } + + public static void main(String[] args) { + Node n1 = new Node(1, false); + Node n2 = new Node(2, true); + Node n3 = new Node(3, false); + Node n4 = new Node(4, false); + Node n5 = new Node(5, false); + Node n6 = new Node(6, true); + Node n7 = new Node(7, true); + Node n8 = new Node(8, false); + Node n9 = new Node(9, false); + Node n10 = new Node(10, false); + Node n11 = new Node(11, false); + Node n12 = new Node(12, false); + Node n13 = new Node(13, true); + + n1.nexts.add(n2); + n1.nexts.add(n3); + n2.nexts.add(n4); + n2.nexts.add(n5); + n3.nexts.add(n6); + n3.nexts.add(n7); + n6.nexts.add(n8); + n6.nexts.add(n9); + n6.nexts.add(n10); + n7.nexts.add(n11); + n7.nexts.add(n12); + n9.nexts.add(n13); + + Node head = retain(n1); + preOrderPrint(head); + + } + +} diff --git a/算法周更班/class_2021_11_4_week/Code02_GuessNumberHigherOrLowerII.java b/算法周更班/class_2021_11_4_week/Code02_GuessNumberHigherOrLowerII.java new file mode 100644 index 0000000..acc1216 --- /dev/null +++ b/算法周更班/class_2021_11_4_week/Code02_GuessNumberHigherOrLowerII.java @@ -0,0 +1,107 @@ +package class_2021_11_4_week; + +// 测试链接 : https://leetcode.com/problems/guess-number-higher-or-lower-ii/ +public class Code02_GuessNumberHigherOrLowerII { + + // 正确的数字，在1~n之间 + // 每次猜错，花费就是你猜的数字 + // 返回：永远最倒霉的情况下，也能赢的胜利，所需要的最少钱数 + public static int minGold(int n) { + return zuo(1, n); + } + + // 目前锁定了，正确的数字，在L~R范围上，除了这个范围都不可能了！ + // 返回，永远最倒霉的情况下，猜中这个数字，所需要的最少钱数 + public static int zuo(int L, int R) { + if (L == R) { + return 0; + } + if (L == R - 1) { + return L; + } + int min = Math.min(L + zuo(L + 1, R), R + zuo(L, R - 1)); + for (int i = L + 1; i < R; i++) { + min = Math.min(min, i + Math.max(zuo(L, i - 1), zuo(i + 1, R))); + } + return min; + } + + public static int minGold2(int n) { + + // L -> 1~n + // R -> 1~n + int[][] dp = new int[n + 1][n + 1]; + // 因为初始化都是0，所以dp的对角线，不用填了 + for (int i = 1; i < n; i++) { + dp[i][i + 1] = i; + } + for (int L = n - 2; L >= 1; L--) { + for (int R = L + 2; R <= n; R++) { + int min = Math.min(L + dp[L + 1][R], R + dp[L][R - 1]); + for (int i = L + 1; i < R; i++) { + min = Math.min(min, i + Math.max(dp[L][i - 1], dp[i + 1][R])); + } + dp[L][R] = min; + } + } + return dp[1][n]; + } + + // 暴力递归 + public static int getMoneyAmount1(int n) { + return process1(1, n); + } + + // 假设现在在L ~ R的范围上, 猜数字 + // 返回：确保获胜的最小现金数，不管答案是哪个数字 + // 注意：所谓的“确保获胜”，以及“不管答案是哪个数字”，意味着你每次永远面临猜错的最差情况！ + public static int process1(int L, int R) { + // 说明L~R范围，只剩下一个数字了，那不用猜了，获胜了 + if (L == R) { + return 0; + } + // 说明L~R范围，只剩下两个数字了 + // 比如： 5 6 + // 假设永远会遇到最差情况， + // 那么当然猜5，因为最差情况下，也只需要耗费5的代价，然后就知道了答案是6 + // 不能猜6，因为最差情况下，需要耗费6的代价，然后才知道答案是5 + // 所以当然选代价低的！请深刻理解：每次永远面临猜错的最差情况！ + if (L == R - 1) { + return L; + } + // 如果说明L~R范围，不仅仅两个数字 + // 比如：5 6 7 8 9 + // 首先尝试5，如果最差情况出现，代价为：5 + 6~9范围上的尝试 + // 最后尝试9，如果最差情况出现，代价为：9 + 5~8范围上的尝试 + int ans = Math.min(L + process1(L + 1, R), R + process1(L, R - 1)); + // 进而尝试6，如果最差情况出现，代价为：6 + Max { 5~5范围上的尝试 ，7~9范围上的尝试} + // 这是因为猜了6，会告诉你，猜高了还是猜低了，所以左右两侧的待定范围，一定会只走一侧 + // 又因为永远会遇到最差情况，所以，一定会走最难受的那一侧，所以是 Max { 5~5范围上的尝试 ，7~9范围上的尝试} + // 进而尝试7，如果最差情况出现，代价为：7 + Max { 5~6范围上的尝试 ，8~9范围上的尝试} + // 进而尝试8，如果最差情况出现，代价为：8 + Max { 5~7范围上的尝试 ，9~9范围上的尝试} + // 所有尝试中，取代价最小值 + for (int M = L + 1; M < R; M++) { + ans = Math.min(ans, M + Math.max(process1(L, M - 1), process1(M + 1, R))); + } + return ans; + } + + // 上面的暴力递归改动态规划 + // 提交到leetcode上可以直接通过 + public static int getMoneyAmount2(int n) { + int[][] dp = new int[n + 1][n + 1]; + for (int i = 1; i < n; i++) { + dp[i][i + 1] = i; + } + for (int L = n - 2; L >= 1; L--) { + for (int R = L + 2; R <= n; R++) { + dp[L][R] = Math.min(L + dp[L + 1][R], R + dp[L][R - 1]); + for (int M = L + 1; M < R; M++) { + dp[L][R] = Math.min(dp[L][R], M + Math.max(dp[L][M - 1], dp[M + 1][R])); + } + } + } + return dp[1][n]; + } + +} diff --git a/算法周更班/class_2021_11_4_week/Code03_StartToEndBinaryOneTarget.java b/算法周更班/class_2021_11_4_week/Code03_StartToEndBinaryOneTarget.java new file mode 100644 index 0000000..fb477d1 --- /dev/null +++ b/算法周更班/class_2021_11_4_week/Code03_StartToEndBinaryOneTarget.java @@ -0,0 +1,337 @@ +package class_2021_11_4_week; + +import java.util.Arrays; + +// 来自真实面试，同学给我的问题 +// 限制：0 <= start <= end，0 <= target <= 64 +// [start,end]范围上的数字，有多少数字二进制中1的个数等于target +public class Code03_StartToEndBinaryOneTarget { + + // 暴力方法 + // 为了验证 + public static long nums1(long start, long end, int target) { + if (start < 0 || end < 0 || start > end || target < 0) { + return -1; + } + long ans = 0; + for (long i = start; i <= end; i++) { + if (bitOne(i) == target) { + ans++; + } + } + return ans; + } + + public static int bitOne(long num) { + int bits = 0; + for (int i = 63; i >= 0; i--) { + if ((num & (1L << i)) != 0) { + bits++; + } + } + return bits; + } + + // 正式方法 + // 递归展示其思路，但是不做动态规划的优化 + public static long nums2(long start, long end, int target) { + if (start < 0 || end < 0 || start > end || target < 0) { + return -1; + } + if (start == 0 && end == 0) { + return target == 0 ? 1 : 0; + } + // 寻找end这数，最高位的1在哪？ + int ehigh = 62; + while ((end & (1L << ehigh)) == 0) { + ehigh--; + } + if (start == 0) { + return process2(ehigh, 0, target, end); + } else { // 170 ~ 3657 0 ~ 169 0 ~ 3657 + start--; + int shigh = 62; + while (shigh >= 0 && (start & (1L << shigh)) == 0) { + shigh--; + } + return process2(ehigh, 0, target, end) - process2(shigh, 0, target, start); + } + } + // 11 10 9 8 7 6 5 4 3 2 1 0 + // num : 0 1 1 0 0 1 1 0 0 1 0 1 0 + // 1..... + // 0..... + + // 如果num最高位的1在i位，也就是说num[i...0]才有意义，比i再高的位置都是0 + // 那么从第i位开始做决定，依次往低位进行决定 + // index : 当前来到哪一位做决定，index在i~0之间，从高到低 + // less : 之间做的决定是不是已经比num小了 + // rest : 还剩几个1需要出现 + // long process(index, less, rest, num)含义 : + // [i...index+1]上面已经做过决定了，接下来要在[index...0]上面做决定 + // 在[i...index+1]上面的决定是不是比num[i...index+1]小， + // 如果是，less = 1 + // 如果还没有，一定说明之前的决定==num[i...index+1]，此时less = 0 + // 在[index...0]上面做决定的过程中一定要出现rest个1 + // 返回[index...0]上，有多少合法的决定 + // 这个方法可以改动态规划，因为：index范围62~0, less范围0或者1，rest范围0~target + // 自己改动态规划吧 + // 只有index、less、rest这三个有效可变参数、num是固定参数 + // 所以可以改成三维动态规划 + // + // num [h....index+1] 决定完了，需要保证之前做的决定，不能比num大， + // 1) 之前做的决定 已经小于 num所对应的前缀状态了， + // 2) 之前做的决定 等于 num所对应的前缀状态了， + // index..... 去做决定吧！ + // less == 1 之前做的决定 已经小于 num所对应的前缀状态了 + // less == 0 之前做的决定 等于 num所对应的前缀状态 + // 剩余几个1，需要出现！ + // [index.....] + // index -> 62~0 63种变化 + // less -> 0 1 2种变化 + // rest -> 0 64种变化 + // 63 * 2 * 64 + public static long process2(int index, int less, int rest, long num) { + if (rest > index + 1) { + return 0; + } + // rest <= index + 1 + if (rest == 0) { + return 1L; + } + // 0 < rest <= index + 1 + // 还有1需要去消耗 + // 位数也够用 + if (less == 1) { // less == 1 之前做的决定 已经小于 num所对应的前缀状态了 + if (rest == index + 1) { + return 1; + } else { + // 后面剩余的位数 > 需要消耗掉1的数量的！ + // 某些位置填1，某些位置填0 + // index 0 1 + // index 0 + // process2(index - 1, 1, rest, num ); + // index 1 + // process2(index - 1, 1, rest - 1, num); + return process2(index - 1, 1, rest - 1, num) + process2(index - 1, 1, rest, num); + } + } else { // less == 0, 之前做的决定 等于 num所对应的前缀状态的 + if (rest == index + 1) { // 后面剩余的位数，必须都填1，才能消耗完 + // index + // 1 1 1 1 1 1 1 1 1 + // num + // 1 + // index 1 + // + // num 111111111 1 + return (num & (1L << index)) == 0 ? 0 : process2(index - 1, 0, rest - 1, num); + } else { + // less == 0, 之前做的决定 等于 num所对应的前缀状态的 + // 后面剩余的位数 > 需要消耗掉1的数量的！ + // 某些位置填1，某些位置填0 + if ((num & (1L << index)) == 0) { + return process2(index - 1, 0, rest, num); + } else { // num 当前位置 1 + // index 1 + // index 0 + return process2(index - 1, 0, rest - 1, num) + process2(index - 1, 1, rest, num); + } + } + } + } + + // 最优解方法 + // 方法二的思路 + 动态规划 + public static long nums3(long start, long end, int target) { + if (start < 0 || end < 0 || start > end || target < 0) { + return -1; + } + if (start == 0 && end == 0) { + return target == 0 ? 1 : 0; + } + int ehigh = 62; + while ((end & (1L << ehigh)) == 0) { + ehigh--; + } + long[][][] dpe = new long[ehigh + 1][2][target + 1]; + for (int i = 0; i <= ehigh; i++) { + Arrays.fill(dpe[i][0], -1); + Arrays.fill(dpe[i][1], -1); + } + long anse = process3(ehigh, 0, target, end, dpe); + if (start == 0) { + return anse; + } else { + start--; + int shigh = 62; + while (shigh >= 0 && (start & (1L << shigh)) == 0) { + shigh--; + } + long[][][] dps = new long[shigh + 1][2][target + 1]; + for (int i = 0; i <= shigh; i++) { + Arrays.fill(dps[i][0], -1); + Arrays.fill(dps[i][1], -1); + } + long anss = process3(shigh, 0, target, start, dps); + return anse - anss; + } + } + + // dp 傻缓存 + // dp 63 * 2 * 64 + // 一定能装下所有的解！ + // index+less+rest + // dp[index][less][rest] 请直接拿数据 + // dp[index][less][rest] == -1 表示 没算过，去算！ + // dp[index][less][rest] != -1 表示 算过！结果就是dp[index][less][rest] + public static long process3(int index, int less, int rest, long num, long[][][] dp) { + if (rest > index + 1) { + return 0; + } + if (rest == 0) { + return 1L; + } + if (dp[index][less][rest] != -1) { + return dp[index][less][rest]; + } + // 没算过！ + long ans = 0; + if (less == 1) { + if (rest == index + 1) { + ans = 1; + } else { + ans = process3(index - 1, 1, rest - 1, num, dp) + process3(index - 1, 1, rest, num, dp); + } + } else { + if (rest == index + 1) { + ans = (num & (1L << index)) == 0 ? 0 : process3(index - 1, 0, rest - 1, num, dp); + } else { + if ((num & (1L << index)) == 0) { + ans = process3(index - 1, 0, rest, num, dp); + } else { + ans = process3(index - 1, 0, rest - 1, num, dp) + process3(index - 1, 1, rest, num, dp); + } + } + } + dp[index][less][rest] = ans; + return ans; + } + + // 利用排列组合的方法 + public static long nums4(long start, long end, int target) { + if (start < 0 || end < 0 || start > end || target < 0) { + return -1; + } + long anse = process4(63, target, end); + if (start == 0) { + return anse; + } else { + long anss = process4(63, target, start - 1); + return anse - anss; + } + } + + public static long process4(int index, int rest, long num) { + if (rest > index + 1) { + return 0; + } + if (rest == 0) { + return 1; + } + if ((num & (1L << index)) == 0) { + return process4(index - 1, rest, num); + } else { + return c(index, rest) + process4(index - 1, rest - 1, num); + } + } + + // 求C(N,A)的解 + // N! / (A! * (N - A)!) + // 即 : (A+1 * A+2 * ... * N) / (2 * 3 * 4 * (N-A)) + // 为了不溢出，每一步求一个最大公约数，然后消掉 + public static long c(long n, long a) { + if (n < a) { + return 0L; + } + long up = 1L; + long down = 1L; + for (long i = a + 1, j = 2; i <= n || j <= n - a;) { + if (i <= n) { + up *= i++; + } + if (j <= n - a) { + down *= j++; + } + long gcd = gcd(up, down); + up /= gcd; + down /= gcd; + } + return up / down; + } + + // 求m和n的最大公约数 + public static long gcd(long m, long n) { + return n == 0 ? m : gcd(n, m % n); + } + + public static void main(String[] args) { + long range = 600L; + System.out.println("功能测试开始"); + for (long start = 0L; start < range; start++) { + for (long end = start; end < range; end++) { + int target = (int) (Math.random() * 10); + long ans1 = nums1(start, end, target); + long ans2 = nums2(start, end, target); + long ans3 = nums3(start, end, target); + long ans4 = nums4(start, end, target); + if (ans1 != ans2 || ans1 != ans3 || ans1 != ans4) { + System.out.println("出错了！"); + } + } + } + System.out.println("功能测试结束"); + + long start = 33281731L; + long end = 204356810L; + int target = 17; + long startTime; + long endTime; + long ans1; + long ans2; + long ans3; + long ans4; + + System.out.println("大范围性能测试，开始"); + startTime = System.currentTimeMillis(); + ans1 = nums1(start, end, target); + endTime = System.currentTimeMillis(); + System.out.println("方法一答案：" + ans1 + ", 运行时间(毫秒) : " + (endTime - startTime)); + startTime = System.currentTimeMillis(); + ans2 = nums2(start, end, target); + endTime = System.currentTimeMillis(); + System.out.println("方法二答案：" + ans2 + ", 运行时间(毫秒) : " + (endTime - startTime)); + startTime = System.currentTimeMillis(); + ans3 = nums3(start, end, target); + endTime = System.currentTimeMillis(); + System.out.println("方法三答案：" + ans3 + ", 运行时间(毫秒) : " + (endTime - startTime)); + ans4 = nums4(start, end, target); + endTime = System.currentTimeMillis(); + System.out.println("方法四答案：" + ans4 + ", 运行时间(毫秒) : " + (endTime - startTime)); + System.out.println("大范围性能测试，结束"); + + System.out.println("超大范围性能测试，开始"); + start = 88193819381L; + end = 92371283713182371L; + target = 30; + startTime = System.currentTimeMillis(); + ans3 = nums3(start, end, target); + endTime = System.currentTimeMillis(); + System.out.println("方法三答案：" + ans3 + ", 运行时间(毫秒) : " + (endTime - startTime)); + startTime = System.currentTimeMillis(); + ans4 = nums4(start, end, target); + endTime = System.currentTimeMillis(); + System.out.println("方法四答案：" + ans4 + ", 运行时间(毫秒) : " + (endTime - startTime)); + System.out.println("超大范围性能测试，结束"); + } + +} diff --git a/算法周更班/class_2021_12_1_week/Code01_XtoYMinDistance.java b/算法周更班/class_2021_12_1_week/Code01_XtoYMinDistance.java new file mode 100644 index 0000000..d0aca05 --- /dev/null +++ b/算法周更班/class_2021_12_1_week/Code01_XtoYMinDistance.java @@ -0,0 +1,171 @@ +package class_2021_12_1_week; + +import java.util.PriorityQueue; + +// 来自真实面试，同学给我的问题 +public class Code01_XtoYMinDistance { + + // 暴力方法 + // dfs尝试所有情况 + // 没有优化，就是纯暴力 + public static int minDistance1(int n, int[][] roads, int x, int y) { + // 第一步生成邻接矩阵 + int[][] map = new int[n + 1][n + 1]; + for (int i = 0; i <= n; i++) { + for (int j = 0; j <= n; j++) { + map[i][j] = Integer.MAX_VALUE; + } + } + for (int[] road : roads) { + map[road[0]][road[1]] = Math.min(map[road[0]][road[1]], road[2]); + map[road[1]][road[0]] = Math.min(map[road[1]][road[0]], road[2]); + } + boolean[] visited = new boolean[n + 1]; + return process(x, y, n, map, visited); + } + + // 当前来到的城市是cur，最终目的地是aim，一共有1~n这些城市 + // 所有城市之间的距离都在map里 + // 之前已经走过了哪些城市都记录在了visited里面，请不要重复经过 + // 返回从cur到aim所有可能的路里，最小距离是多少 + public static int process(int cur, int aim, int n, int[][] map, boolean[] visited) { + if (visited[cur]) { + return Integer.MAX_VALUE; + } + if (cur == aim) { + return 0; + } + visited[cur] = true; + int ans = Integer.MAX_VALUE; + for (int next = 1; next <= n; next++) { + if (next != cur && map[cur][next] != Integer.MAX_VALUE) { + int rest = process(next, aim, n, map, visited); + if (rest != Integer.MAX_VALUE) { + ans = Math.min(ans, map[cur][next] + rest); + } + } + } + visited[cur] = false; + return ans; + } + + // Dijkstra的解 + // n是城市数量 + // 城市编号：1 ~ n 0弃而不用 + public static int minDistance2(int n, int[][] roads, int x, int y) { + // 第一步生成邻接矩阵 + int[][] map = new int[n + 1][n + 1]; + for (int i = 0; i <= n; i++) { + for (int j = 0; j <= n; j++) { + map[i][j] = Integer.MAX_VALUE; + } + } + // 建立路！ + for (int[] road : roads) { + map[road[0]][road[1]] = Math.min(map[road[0]][road[1]], road[2]); + map[road[1]][road[0]] = Math.min(map[road[1]][road[0]], road[2]); + } + // computed[i] = true，表示从源出发点到i这个城市，已经计算出最短距离了 + // computed[i] = false，表示从源出发点到i这个城市，还没有计算出最短距离 + boolean[] computed = new boolean[n + 1]; + // 距离小根堆 + PriorityQueue heap = new PriorityQueue<>((a, b) -> (a.pathSum - b.pathSum)); + heap.add(new Node(x, 0)); + while (!heap.isEmpty()) { + // x -> ... -> 当前的城市, 有距离 + Node cur = heap.poll(); + if (computed[cur.city]) { + continue; + } + // 没算过 + // 开始算！ + if (cur.city == y) { + return cur.pathSum; + } + computed[cur.city] = true; + for (int next = 1; next <= n; next++) { + if (next != cur.city && map[cur.city][next] != Integer.MAX_VALUE && !computed[next]) { + heap.add(new Node(next, cur.pathSum + map[cur.city][next])); + } + } + } + return Integer.MAX_VALUE; + } + + // 当前来到的Node，注意这不是城市的意思，这是就是一个普通的封装类 + // Node封装了，当前来到的城市是什么，以及，从源出发点到这个城市的路径和是多少？ + public static class Node { + // 当前来到的城市编号 + public int city; + // 从源出发点到这个城市的路径和 + public int pathSum; + + public Node(int c, int p) { + city = c; + pathSum = p; + } + } + + // 为了测试 + // 城市1~n + // 随机生成m条道路 + // 每一条路的距离，在1~v之间 + public static int[][] randomRoads(int n, int m, int v) { + int[][] roads = new int[m][3]; + for (int i = 0; i < m; i++) { + int from = (int) (Math.random() * n) + 1; + int to = (int) (Math.random() * n) + 1; + int distance = (int) (Math.random() * v) + 1; + roads[i] = new int[] { from, to, distance }; + } + return roads; + } + + // 为了测试 + public static void main(String[] args) { + // 城市数量n，下标从1开始，不从0开始 + int n = 4; + // 边的数量m，m的值不能大于n * (n-1) / 2 + int m = 4; + // 所的路有m条 + // [a,b,c]表示a和b之间有路，距离为3，根据题意，本题中的边都是无向边 + // 假设有两条路 + // [1,3,7]，这条路是从1到3，距离是7 + // [1,3,4]，这条路是从1到3，距离是4 + // 那么应该忽略[1,3,7]，因为[1,3,4]比它好 + int[][] roads = new int[m][3]; + roads[0] = new int[] { 1, 2, 4 }; + roads[1] = new int[] { 1, 3, 1 }; + roads[2] = new int[] { 1, 4, 1 }; + roads[3] = new int[] { 2, 3, 1 }; + // 求从x到y的最短距离是多少，x和y应该在[1,n]之间 + int x = 2; + int y = 4; + + // 暴力方法的解 + System.out.println(minDistance1(n, roads, x, y)); + + // Dijkstra的解 + System.out.println(minDistance2(n, roads, x, y)); + + // 下面开始随机验证 + int cityMaxSize = 12; + int pathMax = 30; + int testTimes = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + n = (int) (Math.random() * cityMaxSize) + 1; + m = (int) (Math.random() * n * (n - 1) / 2) + 1; + roads = randomRoads(n, m, pathMax); + x = (int) (Math.random() * n) + 1; + y = (int) (Math.random() * n) + 1; + int ans1 = minDistance1(n, roads, x, y); + int ans2 = minDistance2(n, roads, x, y); + if (ans1 != ans2) { + System.out.println("出错了！"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2021_12_1_week/Code02_4KeysKeyboard.java b/算法周更班/class_2021_12_1_week/Code02_4KeysKeyboard.java new file mode 100644 index 0000000..7b37442 --- /dev/null +++ b/算法周更班/class_2021_12_1_week/Code02_4KeysKeyboard.java @@ -0,0 +1,38 @@ +package class_2021_12_1_week; + +// 测试链接 : https://leetcode.com/problems/4-keys-keyboard/ +public class Code02_4KeysKeyboard { + + // 可以证明： + // 来到i的时候，包括i在内最多有连续4次粘贴行为 + // 不可能更多，如果有连续5次粘贴，一定就不再是最优解 + // 假设开始时，A的数量为S，看如下的变化过程，我们称这是行为一： + // 开始 全选 复制(粘贴板S个A) 粘贴 粘贴 粘贴 粘贴 粘贴 + // S S S 2*S 3*S 4*S 5*S 6*S + // 但是，注意看如下的行为二： + // 开始 全选 复制(粘贴板S个A) 粘贴 全选 复制(粘贴板2S个A) 粘贴 粘贴 + // S S S 2*S 2*S 2*S 4*S 6*S + // 行为一，经历8步，最后是6*S个A + // 行为二，经历8步，最后是6*S个A + // 但是行为二在粘贴板上有2S个A，而行为一在粘贴板上有S个A + // 所以行为一没有行为二优 + // 以此说明：来到i的时候，包括i在内最多有连续4次粘贴行为 + // 那么就尝试：连续1次、连续2次、连续3次、连续4次粘贴行为即可 + public static int maxA(int n) { + // dp[0] 1步以内的最优解 + // dp[1] 2步以内的最优解 + // dp[2] 3步以内的最优解 + // dp[i] i+1步以内的最优解 + int[] dp = new int[n]; + for (int i = 0; i < 6 && i < n; i++) { + dp[i] = i + 1; + } + for (int i = 6; i < n; i++) { + dp[i] = Math.max( + Math.max(dp[i - 3] * 2, dp[i - 4] * 3), + Math.max(dp[i - 5] * 4, dp[i - 6] * 5)); + } + return dp[n - 1]; + } + +} diff --git a/算法周更班/class_2021_12_1_week/Code03_RedundantConnectionII.java b/算法周更班/class_2021_12_1_week/Code03_RedundantConnectionII.java new file mode 100644 index 0000000..97ea609 --- /dev/null +++ b/算法周更班/class_2021_12_1_week/Code03_RedundantConnectionII.java @@ -0,0 +1,102 @@ +package class_2021_12_1_week; + +// 来自微软，真正考的时候阉割了难度 +// 测试链接 : https://leetcode.com/problems/redundant-connection-ii/ +public class Code03_RedundantConnectionII { + + // [ + // [1 , 5] 1 > 5 + // [7 , 3] 7 > 3 + + + // ] + // 边的数量 = 点的数量！ + public static int[] findRedundantDirectedConnection(int[][] edges) { + // N是点的数量 + // 点的编号，1~N，没有0 + int N = edges.length; + // 并查集！N个点，去初始化，每个点各自是一个集合 + UnionFind uf = new UnionFind(N); + // pre[i] = 0 来到i节点是第一次 + // pre[i] = 6 之前来过i，是从6来的！ + int[] pre = new int[N + 1]; + + // 如果，没有入度为2的点， + // first second 都维持是null + // 如果，有入度为2的点，那么也只可能有一个 + // 比如入度为2的点，是5 + // first = [3,5] + // second = [12,5] + int[] first = null; + int[] second = null; + // 有没有环！非常不单纯！含义复杂！ + int[] circle = null; + for (int i = 0; i < N; i++) { // 遍历每条边！ + int from = edges[i][0]; + int to = edges[i][1]; + if (pre[to] != 0) { // 不止一次来过to！ + first = new int[] { pre[to], to }; + second = edges[i]; + } else { // 第一次到达to， + pre[to] = from; + if (uf.same(from, to)) { + circle = edges[i]; + } else { + uf.union(from, to); + } + } + } + // 重点解析！这是啥？？？ + // first != null + // 有入度为2的点！ + return first != null ? (circle != null ? first : second) : circle; + } + + public static class UnionFind { + private int[] f; + private int[] s; + private int[] h; + + public UnionFind(int N) { + f = new int[N + 1]; + s = new int[N + 1]; + h = new int[N + 1]; + for (int i = 0; i <= N; i++) { + f[i] = i; + s[i] = 1; + } + } + + private int find(int i) { + int hi = 0; + while (i != f[i]) { + h[hi++] = i; + i = f[i]; + } + while (hi > 0) { + f[h[--hi]] = i; + } + return i; + } + + public boolean same(int i, int j) { + return find(i) == find(j); + } + + public void union(int i, int j) { + int fi = find(i); + int fj = find(j); + if (fi != fj) { + if (s[fi] >= s[fj]) { + f[fj] = fi; + s[fi] = s[fi] + s[fj]; + } else { + f[fi] = fj; + s[fj] = s[fi] + s[fj]; + } + } + } + + } + +} diff --git a/算法周更班/class_2021_12_2_week/Code01_FindAllPeopleWithSecret.java b/算法周更班/class_2021_12_2_week/Code01_FindAllPeopleWithSecret.java new file mode 100644 index 0000000..e59d154 --- /dev/null +++ b/算法周更班/class_2021_12_2_week/Code01_FindAllPeopleWithSecret.java @@ -0,0 +1,102 @@ +package class_2021_12_2_week; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.List; + +// 链接测试 : https://leetcode.com/problems/find-all-people-with-secret/ +public class Code01_FindAllPeopleWithSecret { + + public List findAllPeople(int n, int[][] meetings, int firstPerson) { + // 0~n-1号专家，各自建立小集合 + // (0, firstPerson)合在一起，作为知道秘密的集合 + UnionFind uf = new UnionFind(n, firstPerson); + int m = meetings.length; + Arrays.sort(meetings, (a, b) -> a[2] - b[2]); + // [1,7,1] [2,4,2] [3,6,2] + // 1,7 2,4 3,6 + int[] help = new int[m << 1]; + help[0] = meetings[0][0]; + help[1] = meetings[0][1]; + int size = 2; + for (int i = 1; i < m; i++) { + // i 2 + if (meetings[i][2] != meetings[i - 1][2]) { + share(help, size, uf); + help[0] = meetings[i][0]; + help[1] = meetings[i][1]; + size = 2; + } else { + help[size++] = meetings[i][0]; + help[size++] = meetings[i][1]; + } + } + share(help, size, uf); + List ans = new ArrayList<>(); + for (int i = 0; i < n; i++) { + if (uf.know(i)) { + ans.add(i); + } + } + return ans; + } + + public static void share(int[] help, int size, UnionFind uf) { + for (int i = 0; i < size; i += 2) { + uf.union(help[i], help[i + 1]); + } + for (int i = 0; i < size; i++) { + if (!uf.know(help[i])) { + uf.isolate(help[i]); + } + } + } + + public static class UnionFind { + public int[] father; + public boolean[] sect; + public int[] help; + + public UnionFind(int n, int first) { + father = new int[n]; + sect = new boolean[n]; + help = new int[n]; + for (int i = 1; i < n; i++) { + father[i] = i; + } + father[first] = 0; + sect[0] = true; + } + + private int find(int i) { + int hi = 0; + while (i != father[i]) { + help[hi++] = i; + i = father[i]; + } + for (hi--; hi >= 0; hi--) { + father[help[hi]] = i; + } + return i; + } + + public void union(int i, int j) { + int fatheri = find(i); + int fatherj = find(j); + if (fatheri != fatherj) { + father[fatherj] = fatheri; + sect[fatheri] |= sect[fatherj]; + } + } + + public boolean know(int i) { + return sect[find(i)]; + } + + public void isolate(int i) { + father[i] = i; + } + + } + +} diff --git a/算法周更班/class_2021_12_2_week/Code02_AwayFromBlackHole.java b/算法周更班/class_2021_12_2_week/Code02_AwayFromBlackHole.java new file mode 100644 index 0000000..709dcd8 --- /dev/null +++ b/算法周更班/class_2021_12_2_week/Code02_AwayFromBlackHole.java @@ -0,0 +1,122 @@ +package class_2021_12_2_week; + +// 来自美团 +// 所有黑洞的中心点记录在holes数组里 +// 比如[[3,5] [6,9]]表示，第一个黑洞在(3,5)，第二个黑洞在(6,9) +// 并且所有黑洞的中心点都在左下角(0,0)，右上角(x,y)的区域里 +// 飞船一旦开始进入黑洞，就会被吸进黑洞里 +// 返回： +// 如果统一所有黑洞的半径，最大半径是多少，依然能保证飞船从(0,0)能到达(x,y) +// 1000 1000*1000 10^6 * 二分 +public class Code02_AwayFromBlackHole { + + public static int maxRadius(int[][] holes, int x, int y) { + int L = 1; + int R = Math.max(x, y); + int ans = 0; + while (L <= R) { + int M = (L + R) / 2; + if (ok(holes, M, x, y)) { + ans = M; + L = M + 1; + } else { + R = M - 1; + } + } + return ans; + } + + public static boolean ok(int[][] holes, int r, int x, int y) { + int n = holes.length; + UnionFind uf = new UnionFind(holes, n, r); + for (int i = 0; i < n; i++) { + for (int j = i; j < n; j++) { + if (touch(holes[i][0], holes[i][1], holes[j][0], holes[j][1], r)) { + uf.union(i, j); + } + if (uf.block(i, x, y)) { + return false; + } + } + } + return true; + } + + public static boolean touch(int x1, int y1, int x2, int y2, int r) { + return (r << 1) >= Math.sqrt((Math.pow(Math.abs(x1 - x2), 2) + Math.pow(Math.abs(y1 - y2), 2))); + } + + public static class UnionFind { + public int[] father; + public int[] size; + public int[] xmin; + public int[] xmax; + public int[] ymin; + public int[] ymax; + public int[] help; + + public UnionFind(int[][] holes, int n, int r) { + father = new int[n]; + size = new int[n]; + xmin = new int[n]; + xmax = new int[n]; + ymin = new int[n]; + ymax = new int[n]; + help = new int[n]; + for (int i = 0; i < n; i++) { + father[i] = i; + size[i] = 1; + xmin[i] = holes[i][0] - r; + xmax[i] = holes[i][0] + r; + ymin[i] = holes[i][1] - r; + ymax[i] = holes[i][1] + r; + } + } + + private int find(int i) { + int hi = 0; + while (i != father[i]) { + help[hi++] = i; + i = father[i]; + } + for (hi--; hi >= 0; hi--) { + father[help[hi]] = i; + } + return i; + } + + public void union(int i, int j) { + int fatheri = find(i); + int fatherj = find(j); + if (fatheri != fatherj) { + int sizei = size[fatheri]; + int sizej = size[fatherj]; + int big = sizei >= sizej ? fatheri : fatherj; + int small = big == fatheri ? fatherj : fatheri; + father[small] = big; + size[big] = sizei + sizej; + xmin[big] = Math.min(xmin[big], xmin[small]); + xmax[big] = Math.max(xmax[big], xmax[small]); + ymin[big] = Math.min(ymin[big], ymin[small]); + ymax[big] = Math.max(ymax[big], ymax[small]); + } + } + + public boolean block(int i, int x, int y) { + i = find(i); + return (xmin[i] <= 0 && xmax[i] >= x) + || (ymin[i] <= 0 && ymax[i] >= y) + || (xmin[i] <= 0 && ymin[i] <= 0) + || (xmax[i] >= x && ymax[i] >= y); + } + + } + + public static void main(String[] args) { + int[][] holes = { { 1, 2 }, { 4, 4 }, { 3, 0 }, { 5, 2 } }; + int x = 4; + int y = 6; + System.out.println(maxRadius(holes, x, y)); + } + +} diff --git a/算法周更班/class_2021_12_2_week/Code03_MagicSum.java b/算法周更班/class_2021_12_2_week/Code03_MagicSum.java new file mode 100644 index 0000000..5105972 --- /dev/null +++ b/算法周更班/class_2021_12_2_week/Code03_MagicSum.java @@ -0,0 +1,301 @@ +package class_2021_12_2_week; + +import java.util.Arrays; + +// 来自真实面试，同学给我的问题 +// arr数组长度为n, magic数组长度为m +// 含义： +// arr = { 3, 1, 4, 5, 7 } +// 如果完全不改变arr中的值，那么收益就是累加和 = 3 + 1 + 4 + 5 + 7 = 20 +// magics = { +// {0,2,5} 表示arr[0~2]中的任何一个值，都能改成5 +// {3,4,3} 表示arr[3~4]中的任何一个值，都能改成3 +// {1,3,7} 表示arr[1~3]中的任何一个值，都能改成7 +// } +// 就是说，magics中的任何一组数据{a,b,c}，都表示一种操作， +// 你可以选择arr[a~b]中任何一个数字，变成c。 +// 并且每一种操作，都可以执行任意次 +// 其中 0 <= a <= b < n +// 那么经过若干次的魔法操作，你当然可能得到arr的更大的累加和 +// 返回arr尽可能大的累加和 +// n <= 10^7 +// m <= 10^6 +// arr中的值和c的范围 <= 10^12 +public class Code03_MagicSum { + + // 暴力解，写出来为了验证正式方法而已 + public static int maxSum1(int[] arr, int[][] magics) { + int[] help = Arrays.copyOf(arr, arr.length); + for (int[] m : magics) { + int l = m[0]; + int r = m[1]; + int c = m[2]; + for (int i = l; i <= r; i++) { + help[i] = Math.max(help[i], c); + } + } + int sum = 0; + for (int num : help) { + sum += num; + } + return sum; + } + + // O(N) + O(M * logM) + O(M * logN) + O(N * logN) + public static int maxSum2(int[] arr, int[][] magics) { + int n = arr.length; + // 线段树里的下标，从1开始，不从0开始！ + SegmentTree2 st = new SegmentTree2(n); + Arrays.sort(magics, (a, b) -> (a[2] - b[2])); + for (int[] magic : magics) { + st.update(magic[0] + 1, magic[1] + 1, magic[2], 1, n, 1); + } + int ans = 0; + for (int i = 0; i < n; i++) { + ans += Math.max(st.query(i + 1, i + 1, 1, n, 1), arr[i]); + } + return ans; + } + + // 这是一棵普通的线段树 + // 区间上维持最大值的线段树 + // 支持区间值更新 + // 支持区间最大值查询 + public static class SegmentTree2 { + private int[] max; + private int[] change; + private boolean[] update; + + public SegmentTree2(int size) { + int N = size + 1; + max = new int[N << 2]; + change = new int[N << 2]; + update = new boolean[N << 2]; + } + + private void pushUp(int rt) { + max[rt] = Math.max(max[rt << 1], max[rt << 1 | 1]); + } + + private void pushDown(int rt, int ln, int rn) { + if (update[rt]) { + update[rt << 1] = true; + update[rt << 1 | 1] = true; + change[rt << 1] = change[rt]; + change[rt << 1 | 1] = change[rt]; + max[rt << 1] = change[rt]; + max[rt << 1 | 1] = change[rt]; + update[rt] = false; + } + } + + public void update(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + update[rt] = true; + change[rt] = C; + max[rt] = C; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + update(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + update(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + public int query(int L, int R, int l, int r, int rt) { + if (L <= l && r <= R) { + return max[rt]; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + int left = 0; + int right = 0; + if (L <= mid) { + left = query(L, R, l, mid, rt << 1); + } + if (R > mid) { + right = query(L, R, mid + 1, r, rt << 1 | 1); + } + return Math.max(left, right); + } + + } + + // O(N) + O(M * logM) + O(M * logN) + O(N) + public static int maxSum3(int[] arr, int[][] magics) { + int n = arr.length; + SegmentTree3 st = new SegmentTree3(n); + Arrays.sort(magics, (a, b) -> (a[2] - b[2])); + for (int[] magic : magics) { + st.update(magic[0] + 1, magic[1] + 1, magic[2], 1, n, 1); + } + int ans = 0; + int[] query = st.buildSingleQuery(n); + for (int i = 0; i < n; i++) { + ans += Math.max(query[i], arr[i]); + } + return ans; + } + + // 为方法三特别定制的线段树 + // 区间上维持最大值的线段树 + // 支持区间值更新 + // 为本道题定制了一个方法： + // 假设全是单点查询，请统一返回所有单点的结果(一个结果数组，里面有所有单点记录) + public static class SegmentTree3 { + private int[] max; + private int[] change; + private boolean[] update; + + public SegmentTree3(int size) { + int N = size + 1; + max = new int[N << 2]; + change = new int[N << 2]; + update = new boolean[N << 2]; + } + + private void pushUp(int rt) { + max[rt] = Math.max(max[rt << 1], max[rt << 1 | 1]); + } + + private void pushDown(int rt, int ln, int rn) { + if (update[rt]) { + update[rt << 1] = true; + update[rt << 1 | 1] = true; + change[rt << 1] = change[rt]; + change[rt << 1 | 1] = change[rt]; + max[rt << 1] = change[rt]; + max[rt << 1 | 1] = change[rt]; + update[rt] = false; + } + } + + public void update(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + update[rt] = true; + change[rt] = C; + max[rt] = C; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + update(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + update(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + public int index = 0; + + public int[] buildSingleQuery(int n) { + int[] ans = new int[n + 1]; + process(ans, 1, n, 1); + return ans; + } + + private void process(int[] ans, int l, int r, int rt) { + if (l == r) { + ans[index++] = max[rt]; + } else { + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + process(ans, l, mid, rt << 1); + process(ans, mid + 1, r, rt << 1 | 1); + } + } + + } + + // 为了测试 + public static int[] generateRandomArray(int n, int value) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * value) + 1; + } + return arr; + } + + // 为了测试 + public static int[][] generateRandomMagics(int n, int m, int value) { + int[][] magics = new int[m][3]; + for (int[] magic : magics) { + int a = (int) (Math.random() * n); + int b = (int) (Math.random() * n); + int c = (int) (Math.random() * value) + 1; + magic[0] = Math.min(a, b); + magic[1] = Math.max(a, b); + magic[2] = c; + } + return magics; + } + + // 为了测试 + public static void main(String[] args) { + int n = 30; + int m = 15; + int v = 1000; + int testTimes = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + int N = (int) (Math.random() * n) + 1; + int M = (int) (Math.random() * m) + 1; + int[] arr = generateRandomArray(N, v); + int[][] magics = generateRandomMagics(N, M, v); + int ans1 = maxSum1(arr, magics); + int ans2 = maxSum2(arr, magics); + int ans3 = maxSum3(arr, magics); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + for (int[] magic : magics) { + System.out.println("[ " + magic[0] + " , " + magic[1] + " , " + magic[2] + " ] "); + } + System.out.println("出错了！"); + break; + } + } + System.out.println("测试结束"); + + System.out.println("性能测试开始"); + System.out.println("n的数据量将达到10^7"); + System.out.println("m的数据量将达到10^6"); + System.out.println("为了防止溢出，每个值的范围控制在10以内"); + + n = 10000000; + m = 1000000; + v = 10; + + int[] arr = generateRandomArray(n, v); + int[][] magics = generateRandomMagics(n, m, v); + + long start; + long end; + + start = System.currentTimeMillis(); + int ans2 = maxSum2(arr, magics); + end = System.currentTimeMillis(); + System.out.println("方法二的结果 : " + ans2 + ", 方法二的运行时间: " + (end - start) + " 毫秒"); + + start = System.currentTimeMillis(); + int ans3 = maxSum3(arr, magics); + end = System.currentTimeMillis(); + System.out.println("方法三的结果 : " + ans3 + ", 方法三的运行时间: " + (end - start) + " 毫秒"); + + System.out.println("性能测试结束"); + + } + +} diff --git a/算法周更班/class_2021_12_2_week/Code04_LowestCommonAncestorOfABinaryTreeIV.java b/算法周更班/class_2021_12_2_week/Code04_LowestCommonAncestorOfABinaryTreeIV.java new file mode 100644 index 0000000..a1d654a --- /dev/null +++ b/算法周更班/class_2021_12_2_week/Code04_LowestCommonAncestorOfABinaryTreeIV.java @@ -0,0 +1,57 @@ +package class_2021_12_2_week; + +import java.util.HashSet; + +// 测试链接 : https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree-iv/ +public class Code04_LowestCommonAncestorOfABinaryTreeIV { + + public static class TreeNode { + public int val; + public TreeNode left; + public TreeNode right; + } + + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode[] nodes) { + HashSet set = new HashSet<>(); + for (TreeNode node : nodes) { + set.add(node.val); + } + return process(root, set, set.size()).find; + } + + public static class Info { + // 找没找到最低公共祖先 + // 没找到，find = null + // 找到了最低公共祖先，find是最低公共祖先 + public TreeNode find; + // 我这颗子树上，删掉了几个节点！ + public int removes; + + public Info(TreeNode f, int r) { + find = f; + removes = r; + } + } + + public static Info process(TreeNode x, HashSet set, int all) { + if (x == null) { + return new Info(null, 0); + } + Info left = process(x.left, set, all); + if (left.find != null) { + return left; + } + Info right = process(x.right, set, all); + if (right.find != null) { + return right; + } + int cur = set.contains(x.val) ? 1 : 0; + set.remove(x.val); + if (left.removes + right.removes + cur == all) { + return new Info(x, all); + } else { + return new Info(null, left.removes + right.removes + cur); + } + } + +} diff --git a/算法周更班/class_2021_12_2_week/Code05_Colors.java b/算法周更班/class_2021_12_2_week/Code05_Colors.java new file mode 100644 index 0000000..20be746 --- /dev/null +++ b/算法周更班/class_2021_12_2_week/Code05_Colors.java @@ -0,0 +1,325 @@ +package class_2021_12_2_week; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.HashSet; +import java.util.LinkedList; +import java.util.List; +import java.util.Queue; + +// 来自美团 +// 给定一棵多叉树的头节点head +// 每个节点的颜色只会是0、1、2、3中的一种 +// 任何两个节点之间的都有路径 +// 如果节点a和节点b的路径上，包含全部的颜色，这条路径算达标路径 +// (a -> ... -> b)和(b -> ... -> a)算两条路径 +// 求多叉树上达标的路径一共有多少？ +// 点的数量 <= 10^5 +public class Code05_Colors { + + public static class Node { + public int color; + public List nexts; + + public Node(int c) { + color = c; + nexts = new ArrayList<>(); + } + } + + // 暴力方法 + // 为了验证 + public static int colors1(Node head) { + if (head == null) { + return 0; + } + HashMap map = new HashMap<>(); + parentMap(head, null, map); + List allNodes = new ArrayList<>(); + for (Node cur : map.keySet()) { + allNodes.add(cur); + } + int ans = 0; + for (int i = 0; i < allNodes.size(); i++) { + for (int j = i + 1; j < allNodes.size(); j++) { + if (ok(allNodes.get(i), allNodes.get(j), map)) { + ans++; + } + } + } + return ans << 1; + } + + public static void parentMap(Node cur, Node pre, HashMap map) { + if (cur != null) { + map.put(cur, pre); + for (Node next : cur.nexts) { + parentMap(next, cur, map); + } + } + } + + public static boolean ok(Node a, Node b, HashMap map) { + HashSet aPath = new HashSet<>(); + Node cur = a; + while (cur != null) { + aPath.add(cur); + cur = map.get(cur); + } + Node lowest = b; + while (!aPath.contains(lowest)) { + lowest = map.get(lowest); + } + int colors = 1 << lowest.color; + cur = a; + while (cur != lowest) { + colors |= (1 << cur.color); + cur = map.get(cur); + } + cur = b; + while (cur != lowest) { + colors |= (1 << cur.color); + cur = map.get(cur); + } + return colors == 15; + } + + // 正式方法 + public static long colors2(Node head) { + if (head == null) { + return 0; + } + return process2(head).all; + } + + public static class Info { + // 我这棵子树，总共合法的路径有多少？ + public long all; + // 课上没有强调！但是请务必注意！ + // 一定要从头节点出发的情况下！ + // 一定要从头节点出发的情况下！ + // 一定要从头节点出发的情况下！ + // 走出来每种状态路径的条数 + public long[] colors; + + public Info() { + all = 0; + colors = new long[16]; + } + } + + public static Info process2(Node h) { + Info ans = new Info(); + // 头节点拥有的颜色 + // 2 0100 0 0001 3 1000 + int hs = 1 << h.color; + ans.colors[hs] = 1; + if (!h.nexts.isEmpty()) { + int n = h.nexts.size(); + // 0(不用) 1 2 3 4 + Info[] infos = new Info[n + 1]; + for (int i = 1; i <= n; i++) { + infos[i] = process2(h.nexts.get(i - 1)); + ans.all += infos[i].all; + } + long[][] lefts = new long[n + 2][16]; + for (int i = 1; i <= n; i++) { + for (int status = 1; status < 16; status++) { + lefts[i][status] = lefts[i - 1][status] + infos[i].colors[status]; + } + } + long[][] rights = new long[n + 2][16]; + for (int i = n; i >= 1; i--) { + for (int status = 1; status < 16; status++) { + rights[i][status] = rights[i + 1][status] + infos[i].colors[status]; + } + } + for (int status = 1; status < 16; status++) { + // x : 0010 子：0001 10个 + // 0011 + 10个 + ans.colors[status | hs] += rights[1][status]; + } + // 头节点出发，全颜色搞定，100个，200 + ans.all += ans.colors[15] << 1; + for (int from = 1; from <= n; from++) { + for (int fromStatus = 1; fromStatus < 16; fromStatus++) { + for (int toStatus = 1; toStatus < 16; toStatus++) { + if ((fromStatus | toStatus | hs) == 15) { + ans.all += infos[from].colors[fromStatus] + * (lefts[from - 1][toStatus] + rights[from + 1][toStatus]); + } + } + } + } + } + return ans; + } + + // 最后的优化版本 + // 和方法二没有本质区别 + // 优化的点：每个状态需要和哪些状态结合，都放在辅助数组consider里 + public static long colors3(Node head) { + if (head == null) { + return 0; + } + return process3(head).all; + } + + public static int[][] consider = { {}, // 0 + { 14, 15 }, // 1 -> 0001 + { 13, 15 }, // 2 -> 0010 + { 12, 13, 14, 15 }, // 3 -> 0011 + { 11, 15 }, // 4 -> 0100 + { 10, 11, 14, 15 }, // 5 -> 0101 + { 9, 11, 13, 15 }, // 6 -> 0110 + { 8, 9, 10, 11, 12, 13, 14, 15 }, // 7 -> 0111 + { 7, 15 }, // 8 -> 1000 + { 6, 7, 14, 15 }, // 9 -> 1001 + { 5, 7, 13, 15 }, // 10 -> 1010 + { 4, 5, 6, 7, 12, 13, 14, 15 }, // 11 -> 1011 + { 3, 7, 11, 15 }, // 12 -> 1100 + { 2, 3, 6, 7, 10, 11, 14, 15 }, // 13 -> 1101 + { 1, 3, 5, 7, 9, 11, 13, 15 }, // 14 -> 1110 + { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 } // 15 -> 1111 + }; + + public static Info process3(Node h) { + Info ans = new Info(); + int hs = 1 << h.color; + ans.colors[hs] = 1; + if (!h.nexts.isEmpty()) { + int n = h.nexts.size(); + Info[] infos = new Info[n + 1]; + for (int i = 1; i <= n; i++) { + infos[i] = process3(h.nexts.get(i - 1)); + ans.all += infos[i].all; + } + long[][] lefts = new long[n + 2][16]; + for (int i = 1; i <= n; i++) { + for (int status = 1; status < 16; status++) { + lefts[i][status] = lefts[i - 1][status] + infos[i].colors[status]; + } + } + long[][] rights = new long[n + 2][16]; + for (int i = n; i >= 1; i--) { + for (int status = 1; status < 16; status++) { + rights[i][status] = rights[i + 1][status] + infos[i].colors[status]; + } + } + for (int status = 1; status < 16; status++) { + ans.colors[status | hs] += rights[1][status]; + } + ans.all += ans.colors[15] << 1; + for (int from = 1; from <= n; from++) { + for (int fromStatus = 1; fromStatus < 16; fromStatus++) { + for (int toStatus : consider[fromStatus | hs]) { + ans.all += infos[from].colors[fromStatus] + * (lefts[from - 1][toStatus] + rights[from + 1][toStatus]); + } + } + } + } + return ans; + } + + // 为了测试 + public static Node randomTree(int len, int childs) { + Node head = new Node((int) (Math.random() * 4)); + generate(head, len - 1, childs); + return head; + } + + // 为了测试 + public static void generate(Node pre, int restLen, int childs) { + if (restLen == 0) { + return; + } + int size = (int) (Math.random() * childs); + for (int i = 0; i < size; i++) { + Node next = new Node((int) (Math.random() * 4)); + generate(next, restLen - 1, childs); + pre.nexts.add(next); + } + } + + // 为了测试 + public static void printTree(Node head) { + System.out.print(head.color + " "); + if (!head.nexts.isEmpty()) { + System.out.print("( "); + for (Node next : head.nexts) { + printTree(next); + System.out.print(" , "); + } + System.out.print(") "); + } + } + + // 为了测试 + // 生成高度为9的满5叉树，每个节点的颜色在0~3上随机 + // 这棵树的节点个数已经达到5 * 10^5的规模 + public static Node randomTree() { + Queue curq = new LinkedList<>(); + Queue nexq = new LinkedList<>(); + Node head = new Node((int) (Math.random() * 4)); + curq.add(head); + for (int len = 1; len < 9; len++) { + while (!curq.isEmpty()) { + Node cur = curq.poll(); + for (int i = 0; i < 5; i++) { + Node next = new Node((int) (Math.random() * 4)); + cur.nexts.add(next); + nexq.add(next); + } + } + Queue tmp = nexq; + nexq = curq; + curq = tmp; + } + return head; + } + + // 为了测试 + public static void main(String[] args) { + int len = 6; + int childs = 6; + int testTime = 3000; + System.out.println("功能测试开始"); + for (int i = 0; i < testTime; i++) { + Node head = randomTree(len, childs); + int ans1 = colors1(head); + long ans2 = colors2(head); + long ans3 = colors3(head); + if (ans1 != ans2 || ans2 != ans3) { + System.out.println("出错了"); + printTree(head); + System.out.println(); + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + break; + } + } + System.out.println("功能测试结束"); + + System.out.println("性能测试开始"); + + Node h = randomTree(); + System.out.println("节点数量达到 5*(10^5) 规模"); + long start; + long end; + + start = System.currentTimeMillis(); + long ans2 = colors2(h); + end = System.currentTimeMillis(); + System.out.println("方法二答案 : " + ans2 + ", 方法二运行时间 : " + (end - start) + " 毫秒"); + + start = System.currentTimeMillis(); + long ans3 = colors3(h); + end = System.currentTimeMillis(); + System.out.println("方法三答案 : " + ans3 + ", 方法三运行时间 : " + (end - start) + " 毫秒"); + + System.out.println("性能测试结束"); + } + +} diff --git a/算法周更班/class_2021_12_3_week/Code01_RightMoveInBinaryTree.java b/算法周更班/class_2021_12_3_week/Code01_RightMoveInBinaryTree.java new file mode 100644 index 0000000..b493b9f --- /dev/null +++ b/算法周更班/class_2021_12_3_week/Code01_RightMoveInBinaryTree.java @@ -0,0 +1,74 @@ +package class_2021_12_3_week; + +// 测试链接 : https://www.nowcoder.com/test/33701596/summary +// 本题目为第1题 +// 牛客网判题过程不好，卡常数了 +// 课上会说什么是"卡常数" +// 卡常数怎么反馈？ +public class Code01_RightMoveInBinaryTree { + + // 这个类不需要提交 + public static class TreeNode { + int val = 0; + TreeNode left = null; + TreeNode right = null; + + public TreeNode(int val) { + this.val = val; + } + } + + // 提交下面的代码 + + public static TreeNode[] queue = new TreeNode[300000]; + + public static int[] ends = new int[50]; + + public static TreeNode cyclicShiftTree(TreeNode root, int k) { + int l = 0; + int r = 0; + queue[r++] = root; + int level = 0; + while (l != r) { + ends[level] = r; + while (l < ends[level]) { + TreeNode cur = queue[l++]; + if (cur != null) { + queue[r++] = cur.left; + queue[r++] = cur.right; + } + } + level++; + } + for (int i = level - 1; i > 0; i--) { + + // 当前层 : curLeft....curRight + // 3(null) 4(a) 5(null) 6(b) + // 下一层 ：downLeft....downRight + // 7 8 9 10 + // downIndex : 下一层需要根据，k和下一层的长度，来右移。右移之后，从哪个位置开始，分配节点给当前层第一个不空的节点 + int downLeft = ends[i - 1]; + int downRight = ends[i] - 1; + int downRightSize = k % (downRight - downLeft + 1); + int downIndex = downRightSize == 0 ? downLeft : (downRight - downRightSize + 1); + int curLeft = i - 2 >= 0 ? ends[i - 2] : 0; + int curRight = ends[i - 1] - 1; + for (int j = curLeft; j <= curRight; j++) { + if (queue[j] != null) { + queue[j].left = queue[downIndex]; + downIndex = nextIndex(downIndex, downLeft, downRight); + queue[j].right = queue[downIndex]; + downIndex = nextIndex(downIndex, downLeft, downRight); + } + } + } + return root; + } + + // l......r i -> next index + // 4.....9 i = 7 8 9 4 + public static int nextIndex(int i, int l, int r) { + return i == r ? l : i + 1; + } + +} diff --git a/算法周更班/class_2021_12_3_week/Code02_BinaryNegate.java b/算法周更班/class_2021_12_3_week/Code02_BinaryNegate.java new file mode 100644 index 0000000..bf2e23e --- /dev/null +++ b/算法周更班/class_2021_12_3_week/Code02_BinaryNegate.java @@ -0,0 +1,25 @@ +package class_2021_12_3_week; + +//测试链接 : https://www.nowcoder.com/test/33701596/summary +//本题目为第2题 +public class Code02_BinaryNegate { + + public static String maxLexicographical(String num) { + char[] arr = num.toCharArray(); + int i = 0; + while (i < arr.length) { + if (arr[i] == '0') { + break; + } + i++; + } + while(i < arr.length) { + if(arr[i] == '1') { + break; + } + arr[i++] = '1'; + } + return String.valueOf(arr); + } + +} diff --git a/算法周更班/class_2021_12_3_week/Code03_OneCountsInKSystem.java b/算法周更班/class_2021_12_3_week/Code03_OneCountsInKSystem.java new file mode 100644 index 0000000..b7b6d2d --- /dev/null +++ b/算法周更班/class_2021_12_3_week/Code03_OneCountsInKSystem.java @@ -0,0 +1,58 @@ +package class_2021_12_3_week; + +// 测试链接 : https://www.nowcoder.com/test/33701596/summary +// 本题目为第3题 +// 核心方法，在大厂刷题班19节，第3题 +public class Code03_OneCountsInKSystem { + + public static long minM(int n, int k) { + int len = bits(n, k); + long l = 1; + long r = power(k, len + 1); + long ans = r; + while (l <= r) { + long m = l + ((r - l) >> 1); + if (ones(m, k) >= n) { + ans = m; + r = m - 1; + } else { + l = m + 1; + } + } + return ans; + } + + public static int bits(long num, int k) { + int len = 0; + while (num != 0) { + len++; + num /= k; + } + return len; + } + + public static long power(long base, int power) { + long ans = 1; + while (power != 0) { + if ((power & 1) != 0) { + ans *= base; + } + base *= base; + power >>= 1; + } + return ans; + } + + public static long ones(long num, int k) { + int len = bits(num, k); + if (len <= 1) { + return len; + } + long offset = power(k, len - 1); + long first = num / offset; + long curOne = first == 1 ? (num % offset) + 1 : offset; + long restOne = first * (len - 1) * (offset / k); + return curOne + restOne + ones(num % offset, k); + } + +} diff --git a/算法周更班/class_2021_12_3_week/Code04_CutOffTreesForGolfEvent.java b/算法周更班/class_2021_12_3_week/Code04_CutOffTreesForGolfEvent.java new file mode 100644 index 0000000..87cd3a5 --- /dev/null +++ b/算法周更班/class_2021_12_3_week/Code04_CutOffTreesForGolfEvent.java @@ -0,0 +1,79 @@ +package class_2021_12_3_week; + +import java.util.ArrayList; +import java.util.LinkedList; +import java.util.List; + +// 测试链接 : https://leetcode.com/problems/cut-off-trees-for-golf-event/ +public class Code04_CutOffTreesForGolfEvent { + + public static int cutOffTree(List> forest) { + int n = forest.size(); + int m = forest.get(0).size(); + // [ [3,5,2], [1,9,4] , [2,6,10] ] + // 低 中 高 + ArrayList cells = new ArrayList<>(); + for (int i = 0; i < n; i++) { + for (int j = 0; j < m; j++) { + int val = forest.get(i).get(j); + if (val > 1) { + cells.add(new int[] { i, j, val }); + } + } + } + cells.sort((a, b) -> a[2] - b[2]); + int ans = 0, lastR = 0, lastC = 0; + for (int[] cell : cells) { + int step = bestWalk(forest, lastR, lastC, cell[0], cell[1]); + if (step == -1) { + return -1; + } + ans += step; + lastR = cell[0]; + lastC = cell[1]; + forest.get(lastR).set(lastC, 1); + } + return ans; + } + + public static int[] next = { -1, 0, 1, 0, -1 }; + + // 0 1 2 3 4 + // i + // 行 + next[i-1] + // 列 + next[i] + // i == 1 -> 上 + // i == 2 -> 右 + // i == 3 -> 下 + // i == 4 -> 左 + public static int bestWalk(List> forest, int sr, int sc, int tr, int tc) { + int n = forest.size(); + int m = forest.get(0).size(); + boolean[][] seen = new boolean[n][m]; + LinkedList deque = new LinkedList<>(); + deque.offerFirst(new int[] { 0, sr, sc }); + while (!deque.isEmpty()) { + int[] cur = deque.pollFirst(); + int step = cur[0], r = cur[1], c = cur[2]; + if (r == tr && c == tc) { + return step; + } + seen[r][c] = true; + for (int i = 1; i < 5; i++) { // (r,c) 上下左右，全试试！ + int nr = r + next[i - 1]; + int nc = c + next[i]; + if (nr >= 0 && nr < n && nc >= 0 && nc < m && !seen[nr][nc] && forest.get(nr).get(nc) > 0) { + int[] move = { step + 1, nr, nc }; + // 更近的话 + if ((i == 1 && r > tr) || (i == 2 && c < tc) || (i == 3 && r < tr) || (i == 4 && c > tc)) { + deque.offerFirst(move); + } else { // 更远的话，放到尾部！ + deque.offerLast(move); + } + } + } + } + return -1; + } + +} diff --git a/算法周更班/class_2021_12_3_week/Code05_MinContinuousFragment.java b/算法周更班/class_2021_12_3_week/Code05_MinContinuousFragment.java new file mode 100644 index 0000000..002800a --- /dev/null +++ b/算法周更班/class_2021_12_3_week/Code05_MinContinuousFragment.java @@ -0,0 +1,162 @@ +package class_2021_12_3_week; + +// 来自CMU入学申请考试 +// 给定一个长度为 N 的字符串 S，由字符'a'和'b'组成，空隙由 '?' 表示 +// 你的任务是用a字符或b字符替换每个间隙， +// 替换完成后想让连续出现同一种字符的最长子串尽可能短 +// 例如，S = "aa??bbb"， +// 如果将"??"替换为"aa" ，即"aaaabbb"，则由相等字符组成的最长子串长度为4 +// 如果将"??"替换为"ba" ，即"aababbb"，则由相等字符组成的最长子串长度为3 +// 那么方案二是更好的结果，返回3 +// S的长度 <= 10^6 +public class Code05_MinContinuousFragment { + + // 暴力方法 + // 为了验证 + public static int minContinuous1(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + return process1(str, 0); + } + + public static int process1(char[] str, int index) { + if (index == str.length) { + return maxLen(str); + } else { + if (str[index] != '?') { + return process1(str, index + 1); + } else { + str[index] = 'a'; + int p1 = process1(str, index + 1); + str[index] = 'b'; + int p2 = process1(str, index + 1); + str[index] = '?'; + return Math.min(p1, p2); + } + } + } + + // 正式方法 + // 时间复杂度O(N) + public static int minContinuous2(String s) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + int N = str.length; + int L = 0; + int R = -1; + for (int i = 0; i < N; i++) { + if (str[i] != '?') { + set(str, L, R); + L = i + 1; + R = i; + } else { + R++; + } + } + set(str, L, R); + // 下面的for循环，是单独处理，条件5） + for (int i = 1; i < N; i++) { + if (str[i] == '?') { + // baaaa?bbbbbbbba + for (L = i - 1; L >= 0 && str[L] == str[i - 1]; L--) + ; + for (R = i + 1; R < N && str[R] == str[i + 1]; R++) + ; + L = i - L - 1; + R = R - i - 1; + if (L <= R) { + str[i] = str[i - 1]; + } else { + str[i] = str[i + 1]; + } + } + } + return maxLen(str); + } + + // L...R 都是？ + // 如果这一坨问号，满足1）2）3）4）中的一种，就填好 + // 如果满足5），就不填！a?b + public static void set(char[] str, int L, int R) { + int N = str.length; + if (L > R) { + return; + } + if (L == 0 && R == N - 1) { + for (int i = 0; i < N; i++) { + str[i] = (i & 1) == 0 ? 'a' : 'b'; + } + } else if (L == 0) { + for (int i = R; i >= 0; i--) { + str[i] = str[i + 1] == 'a' ? 'b' : 'a'; + } + } else if (R == N - 1) { + for (int i = L; i < str.length; i++) { + str[i] = str[i - 1] == 'a' ? 'b' : 'a'; + } + } else { + if (str[L - 1] == str[R + 1] || L != R) { + for (; L <= R; L++, R--) { + str[L] = str[L - 1] == 'a' ? 'b' : 'a'; + str[R] = str[R + 1] == 'a' ? 'b' : 'a'; + } + } + } + } + + public static int maxLen(char[] str) { + int ans = 1; + int cur = 1; + for (int i = 1; i < str.length; i++) { + if (str[i] != str[i - 1]) { + ans = Math.max(ans, cur); + cur = 1; + } else { + cur++; + } + } + ans = Math.max(ans, cur); + return ans; + } + + public static char[] arr = { 'a', 'b', '?' }; + + public static String randomString(int len) { + int N = (int) (Math.random() * (len + 1)); + char[] str = new char[N]; + for (int i = 0; i < N; i++) { + str[i] = arr[(int) (Math.random() * 3)]; + } + return String.valueOf(str); + } + + public static void main(String[] args) { + int len = 35; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + String s = randomString(len); + int ans1 = minContinuous1(s); + int ans2 = minContinuous2(s); + if (ans1 != ans2) { + System.out.println(s); + System.out.println(ans1); + System.out.println(ans2); + } + } + System.out.println("测试结束"); + + len = 10000000; + String s = randomString(len); + long start = System.currentTimeMillis(); + minContinuous2(s); + long end = System.currentTimeMillis(); + System.out.println("运行时间（毫秒）：" + (end - start)); + + } + +} diff --git a/算法周更班/class_2021_12_4_week/Code01_FiveNodesListNumbers.java b/算法周更班/class_2021_12_4_week/Code01_FiveNodesListNumbers.java new file mode 100644 index 0000000..82ed17e --- /dev/null +++ b/算法周更班/class_2021_12_4_week/Code01_FiveNodesListNumbers.java @@ -0,0 +1,58 @@ +package class_2021_12_4_week; + +import java.util.HashSet; + +// 来自美团 +// 给定一个无向图 +// 从任何一个点x出发，比如有一条路径: x -> a -> b -> c -> y +// 这条路径上有5个点并且5个点都不一样的话，我们说(x,a,b,c,y)是一条合法路径 +// 这条合法路径的代表，就是x,a,b,c,y所组成的集合，我们叫做代表集合 +// 如果从b到y，还有一条路径叫(b,a,c,x,y)，那么(x,a,b,c,y)和(b,a,c,x,y)是同一个代表集合 +// 返回这个无向图中所有合法路径的代表集合数量 +// 题目给定点的数量n <= 15，边的数量m <= 60 +// 所有的点编号都是从0~n-1的 +public class Code01_FiveNodesListNumbers { + + // graph[i] = { a, b, c} 代表：点i直接相邻的节点有a,b,c + // graph[j] = { d } 代表：点j直接相邻的节点有d + // 所以二维数组graph可以表示无向图 + // 0 : + // 1 : + // 2 : + // n-1 : + public static int validPathSets(int[][] graph) { + int n = graph.length; + // 任何一个合法路径的集合，都被弄成了整数形式 + // 0010010011 -> int + // 甲 ： 0011010011 + // 乙 ： 0011010011 + // 丙 ： 0011010011 + HashSet set = new HashSet<>(); + // 下面的过程：从每个点出发，0、1、2、3、。。。 + // 从x点出发，往外最多迈5步，所产生的所有路径，都要！ + for (int from = 0; from < n; from++) { + dfs(0, 0, from, graph, set); + } + return set.size(); + } + + // int status -> 已经走过了哪些点的集合 -> 00001101 + // int len -> 已经往外几步了！ + // int cur -> 当前来到的是几号点！ + // int[][] graph -> 图 + // HashSet set -> 收集所有合法路径的点集合！ + public static void dfs(int status, int len, int cur, int[][] graph, HashSet set) { + if ((status & (1 << cur)) == 0) { // 之前走过的点，不包括cur，迈上去！ + len++; + status |= 1 << cur; + if (len == 5) { + set.add(status); + } else { + for (int next : graph[cur]) { + dfs(status, len, next, graph, set); + } + } + } + } + +} diff --git a/算法周更班/class_2021_12_4_week/Code02_MergeArea.java b/算法周更班/class_2021_12_4_week/Code02_MergeArea.java new file mode 100644 index 0000000..ea97aa0 --- /dev/null +++ b/算法周更班/class_2021_12_4_week/Code02_MergeArea.java @@ -0,0 +1,143 @@ +package class_2021_12_4_week; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.List; + +// 来自北京北明数科信息技术有限公司 +// area表示的是地区全路径,最多可能有6级,用分隔符连接,分隔符是 spliter, +// 分隔符是逗号 +// 例如： +// area = 中国,四川,成都 或者 中国,浙江,杭州 或者 中国,浙江,义乌 +// spliter = , +// count表示门店数 +// class AreaResource { +// String area; +// String spliter; +// long count; +// } + +// area = "中国,四川,成都" +// spliter = "," +// count = 10 + +// 现在需要把 List 进行字符串转换，供下游处理，需要做到同级的地域能合并 +// 比如 +// area为 中国,四川,成都 有10个门店 +// 中国,浙江,杭州 有25个门店 +// 中国,浙江,义乌 有22个门店 +// 中国,四川,成都 有25个门店 +// spliter为逗号 "," 最终转化成JSON的形式，并且同级的地域需要被合并，最终生成的JSON字符串如下所示 +// +// 返回: { +// "中国": +// {"四川":{"成都":35]}, +// "浙江":{"义乌":22,"杭州":25}}} +// 请实现下面的方法 public String mergeCount(List areas) +public class Code02_MergeArea { + + // 系统给你的原始类 + public static class AreaResource { + public String area; + public String spliter; + public long count; + + public AreaResource(String a, String s, long c) { + area = a; + spliter = s; + count = c; + } + } + + // 要实现的方法 + public static String mergeCount(List areas) { + Area all = new Area("", 0); + for (AreaResource r : areas) { + // 中国,四川,成都 , 10个门店 + String[] path = r.area.split(r.spliter); + // 中国 四川 成都 + long count = r.count; + f(path, 0, all, count); + } + return all.toString(); + } + + // [中国，四川，成都] + // 0 1 2 + public static void f(String[] path, int index, Area pre, long count) { + if (index == path.length) { + pre.count += count; + } else { + // 前一个节点 pre 中国 + // cur = 四川 + String cur = path[index]; + if (!pre.next.containsKey(cur)) { + pre.next.put(cur, new Area(cur, 0)); + } + f(path, index + 1, pre.next.get(cur), count); + } + } + + // 自己定义的！前缀树节点 + public static class Area { + // 地区名称：中国 + // 四川 + // 成都 + public String name; + // 中国 key : 省名 value: 下级节点 + public HashMap next; + public long count; + + public Area(String n, long c) { + name = n; + count = c; + next = new HashMap<>(); + } + + public String toString() { + StringBuilder ans = new StringBuilder(); + // : "" + // str = "中国": + // str = "成都":100 + if (!name.equals("")) { + ans.append("\"" + name + "\"" + ":"); + } + if (next.isEmpty()) { + ans.append(count); + } else { + // "中国":{ 四川如何如何,河南如何如何,江苏如何如何} + ans.append("{"); + for (Area child : next.values()) { + ans.append(child.toString() + ","); + } + ans.setCharAt(ans.length() - 1, '}'); + } + return ans.toString(); + } + } + + public static void main(String[] args) { + AreaResource a1 = new AreaResource("中国,四川,成都", ",", 10); + AreaResource a2 = new AreaResource("中国,浙江,杭州", ",", 50); + AreaResource a3 = new AreaResource("中国,浙江,杭州", ",", 25); + AreaResource a4 = new AreaResource("中国,浙江,义务", ",", 22); + AreaResource a5 = new AreaResource("中国,四川,成都", ",", 15); + AreaResource a6 = new AreaResource("中国,四川,攀枝花", ",", 12); + AreaResource a7 = new AreaResource("中国,浙江,宁波", ",", 16); + + List areas = new ArrayList<>(); + areas.add(a1); + areas.add(a2); + areas.add(a3); + areas.add(a4); + areas.add(a5); + areas.add(a6); + areas.add(a7); + + String ans = mergeCount(areas); + + System.out.println(ans); + + } + +} diff --git a/算法周更班/class_2021_12_4_week/Code03_HowManyObtuseAngles.java b/算法周更班/class_2021_12_4_week/Code03_HowManyObtuseAngles.java new file mode 100644 index 0000000..1cae781 --- /dev/null +++ b/算法周更班/class_2021_12_4_week/Code03_HowManyObtuseAngles.java @@ -0,0 +1,42 @@ +package class_2021_12_4_week; + +import java.util.Arrays; + +// 来自hulu +// 有一个以原点为圆心，半径为1的圆 +// 在这个圆的圆周上，有一些点 +// 因为所有的点都在圆周上，所以每个点可以有很简练的表达 +// 比如：用0来表示一个圆周上的点，这个点就在(1,0)位置 +// 比如：用6000来表示一个点，这个点是(1,0)点沿着圆周逆时针转60.00度之后所在的位置 +// 比如：用18034来表示一个点，这个点是(1,0)点沿着圆周逆时针转180.34度之后所在的位置 +// 这样一来，所有的点都可以用[0, 36000)范围上的数字来表示 +// 那么任意三个点都可以组成一个三角形，返回能组成钝角三角形的数量 +public class Code03_HowManyObtuseAngles { + + public static long obtuseAngles(int[] arr) { + // n长度的排序，O(N * logN) + // O(N) + int n = arr.length; + int m = n << 1; + int[] enlarge = new int[m]; + Arrays.sort(arr); + for (int i = 0; i < n; i++) { + enlarge[i] = arr[i]; + enlarge[i + n] = arr[i] + 36000; + } + long ans = 0; + // 这里不用二分查找(太慢)，能做一个不回退的优化 + for (int L = 0, R = 0; L < n; L++) { + while (R < m && enlarge[R] - enlarge[L] < 18000) { + R++; + } + ans += num(R - L - 1); + } + return ans; + } + + public static long num(long nodes) { + return nodes < 2 ? 0 : ((nodes - 1) * nodes) >> 1; + } + +} diff --git a/算法周更班/class_2021_12_4_week/Code04_MaximumNumberOfVisiblePoints.java b/算法周更班/class_2021_12_4_week/Code04_MaximumNumberOfVisiblePoints.java new file mode 100644 index 0000000..9e05247 --- /dev/null +++ b/算法周更班/class_2021_12_4_week/Code04_MaximumNumberOfVisiblePoints.java @@ -0,0 +1,43 @@ +package class_2021_12_4_week; + +import java.util.Arrays; +import java.util.List; + +// 整个二维平面算是一张地图，给定[x,y]，表示你站在x行y列 +// 你可以选择面朝的任何方向 +// 给定一个正数值angle，表示你视野的角度为 +// 这个角度内你可以看无穷远，这个角度外你看不到任何东西 +// 给定一批点的二维坐标，返回你在朝向最好的情况下，最多能看到几个点 +// 测试链接 : https://leetcode.com/problems/maximum-number-of-visible-points/ +public class Code04_MaximumNumberOfVisiblePoints { + + public static int visiblePoints(List> points, int angle, List location) { + int n = points.size(); + int a = location.get(0); + int b = location.get(1); + int zero = 0; + double[] arr = new double[n << 1]; + int m = 0; + for (int i = 0; i < n; i++) { + int x = points.get(i).get(0) - a; + int y = points.get(i).get(1) - b; + if (x == 0 && y == 0) { + zero++; + } else { + arr[m] = Math.toDegrees(Math.atan2(y, x)); + arr[m + 1] = arr[m] + 360; + m += 2; + } + } + Arrays.sort(arr, 0, m); + int max = 0; + for (int L = 0, R = 0; L < n; L++) { + while (R < m && arr[R] - arr[L] <= angle) { + R++; + } + max = Math.max(max, R - L); + } + return max + zero; + } + +} diff --git a/算法周更班/class_2021_12_4_week/Code05_SplitApples.java b/算法周更班/class_2021_12_4_week/Code05_SplitApples.java new file mode 100644 index 0000000..f8d6e42 --- /dev/null +++ b/算法周更班/class_2021_12_4_week/Code05_SplitApples.java @@ -0,0 +1,123 @@ +package class_2021_12_4_week; + +//有m个同样的苹果，认为苹果之间无差别 +//有n个同样的盘子，认为盘子之间也无差别 +//还有，比如5个苹果如果放进3个盘子， +//那么1、3、1和1、1、3和3、1、1的放置方法，也认为是一种方法 +//如上的设定下，返回有多少种放置方法 +//测试链接 : https://www.nowcoder.com/practice/bfd8234bb5e84be0b493656e390bdebf +//提交以下的code，提交时请把类名改成"Main" +import java.util.Arrays; +import java.util.Scanner; + +public class Code05_SplitApples { + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + while (sc.hasNext()) { + int m = sc.nextInt(); + int n = sc.nextInt(); + int ways = ways3(m, n); + System.out.println(ways); + } + sc.close(); + } + + // 思路来自于分裂数问题 + // 体系学习班代码第22节，题目3，split number问题 + public static int ways1(int apples, int plates) { + return process1(1, apples, plates); + } + + // pre : 上一个盘子分到的苹果数量，当前的盘子分到的数量不能小于pre + // apples : 剩余的苹果数量 + // plates : 剩余的盘子数量 + // 在盘子够用的情况下，把苹果分完，有几种方法 + public static int process1(int pre, int apples, int plates) { + if (apples == 0) { + return 1; + } + // apples != 0 + if (plates == 0) { + return 0; + } + // apples != 0 && plates != 0 + if (pre > apples) { + return 0; + } + // apples != 0 && plates != 0 && pre <= apples + int way = 0; + // 之前的盘子分了3个苹果，现在还剩下8个苹果 + // 当前的盘子，可以装几个苹果：3、4、5、6、7、8 + for (int cur = pre; cur <= apples; cur++) { + way += process1(cur, apples - cur, plates - 1); + } + return way; + } + + // 新的尝试，最优解 + // 苹果有apples个，盘子有plates个 + // 返回有几种摆法 + // 如果苹果数为0，有1种摆法：什么也不摆 + // 如果苹果数不为0，但是盘子数为0，有0种摆法（做不到） + // 如果苹果数不为0，盘子数也不为0，进行如下的情况讨论： + // 假设苹果数为apples，盘子数为plates + // 可能性 1) apples < plates + // 这种情况下，一定有多余的盘子，这些盘子完全没用，所以砍掉 + // 后续是f(apples, apples) + // 可能性 2) apples >= plates + // 在可能性2)下，讨论摆法，有如下两种选择 + // 选择a) 不是所有的盘子都使用 + // 选择b) 就是所有的盘子都使用 + // 对于选择a)，既然不是所有盘子都使用，那么后续就是f(apples, plates - 1) + // 意思是：既然不是所有盘子都使用，那盘子减少一个，然后继续讨论吧！ + // 对于选择b)，既然就是所有的盘子都使用，那么先把所有盘子都摆上1个苹果。 + // 剩余苹果数 = apples - plates + // 然后继续讨论，剩下的这些苹果，怎么摆进plates个盘子里， + // 所以后续是f(apples - plates, plates) + public static int ways2(int apples, int plates) { + if (apples == 0) { + return 1; + } + if (plates == 0) { + return 0; + } + if (plates > apples) { + return ways2(apples, apples); + } else { // apples >= plates; + return ways2(apples, plates - 1) + ways2(apples - plates, plates); + } + } + + // 上面最优解尝试的记忆化搜索版本 + public static int[][] dp = null; + + public static int ways3(int apples, int plates) { + if (dp == null) { + dp = new int[11][11]; + for (int i = 0; i <= 10; i++) { + Arrays.fill(dp[i], -1); + } + } + return process3(apples, plates, dp); + } + + public static int process3(int apples, int plates, int[][] dp) { + if (dp[apples][plates] != -1) { + return dp[apples][plates]; + } + int ans = 0; + if (apples == 0) { + ans = 1; + } else if (plates == 0) { + ans = 0; + } else if (plates > apples) { + ans = process3(apples, apples, dp); + } else { + ans = process3(apples, plates - 1, dp) + process3(apples - plates, plates, dp); + } + dp[apples][plates] = ans; + return ans; + } + +} \ No newline at end of file diff --git a/算法周更班/class_2021_12_5_week/Code01_LoudAndRich.java b/算法周更班/class_2021_12_5_week/Code01_LoudAndRich.java new file mode 100644 index 0000000..36fccd4 --- /dev/null +++ b/算法周更班/class_2021_12_5_week/Code01_LoudAndRich.java @@ -0,0 +1,68 @@ +package class_2021_12_5_week; + +import java.util.ArrayList; + +// 测试链接 : https://leetcode.com/problems/loud-and-rich/ +public class Code01_LoudAndRich { + + // richer[i] = {a, b} a比b更有钱 a -> b + // quiet[i] = k, i这个人安静值是k + public static int[] loudAndRich(int[][] richer, int[] quiet) { + int N = quiet.length; + // a -> b + // a -> c + // b -> c + // a : b c + // b : c + // nexts[0] = {5,7,3} + // 0 : 5 7 3 + // 5最没钱的， + // nexts[5] = { } + ArrayList> nexts = new ArrayList<>(); + for (int i = 0; i < N; i++) { + // 0 : {} + // 1 : {} + // n-1 : {} + nexts.add(new ArrayList<>()); + } + // 入度 + // 0 : 0 + // 1 : 2 + int[] degree = new int[N]; + for (int[] r : richer) { + // [a,b] a -> b + nexts.get(r[0]).add(r[1]); + degree[r[1]]++; + } + // 所有入度为0的点，入队列 + int[] zeroQueue = new int[N]; + int l = 0; + int r = 0; + for (int i = 0; i < N; i++) { + if (degree[i] == 0) { + zeroQueue[r++] = i; + } + } + // ans[i] = j : 比i有钱的所有人里，j最安静 + int[] ans = new int[N]; + for (int i = 0; i < N; i++) { + ans[i] = i; + } + while (l < r) { // 如果队列不空 + // 弹出一个入度为0的点 + int cur = zeroQueue[l++]; + // 1) 消除当前cur的影响！ + for (int next : nexts.get(cur)) { + // cur : 比cur有钱，最安静的！ans[cur] + if (quiet[ans[next]] > quiet[ans[cur]]) { + ans[next] = ans[cur]; + } + if (--degree[next] == 0) { + zeroQueue[r++] = next; + } + } + } + return ans; + } + +} diff --git a/算法周更班/class_2021_12_5_week/Code02_DoAllJobs.java b/算法周更班/class_2021_12_5_week/Code02_DoAllJobs.java new file mode 100644 index 0000000..1081e2b --- /dev/null +++ b/算法周更班/class_2021_12_5_week/Code02_DoAllJobs.java @@ -0,0 +1,129 @@ +package class_2021_12_5_week; + +import java.util.Arrays; +import java.util.PriorityQueue; + +// 来自hulu +// 有n个人，m个任务，任务之间有依赖记录在int[][] depends里 +// 比如: depends[i] = [a, b]，表示a任务依赖b任务的完成 +// 其中 0 <= a < m，0 <= b < m +// 1个人1天可以完成1个任务，每个人都会选当前能做任务里，标号最小的任务 +// 一个任务所依赖的任务都完成了，该任务才能开始做 +// 返回n个人做完m个任务，需要几天 +public class Code02_DoAllJobs { + + public static int days(int n, int m, int[][] depends) { + if (n < 1) { + return -1; + } + if (m <= 0) { + return 0; + } + // nexts[0] = {1,4} + int[][] nexts = nexts(depends, m); + int[] indegree = indegree(nexts, m); + // 工人队列！ + PriorityQueue workers = new PriorityQueue<>(); + for (int i = 0; i < n; i++) { + workers.add(0); + } + // zeroIn ： 放着工作，放着可以开始做的工作，不能做的任务，不在其中 + // 小根堆：标号小的任务，一定要先做！ + PriorityQueue zeroIn = new PriorityQueue<>(); + for (int i = 0; i < m; i++) { + if (indegree[i] == 0) { + zeroIn.add(i); + } + } + // start[i] ：i之前必须完成的任务，占了几天，导致i任务只能从那天开始！ + int[] start = new int[m]; + // 完成所有任务的最大天数 + int finishAll = 0; + int done = 0; + while (!zeroIn.isEmpty()) { // 有任务可做 + // 当前可以做的任务中，标号最小的任务 + int job = zeroIn.poll(); + // 当前可用的工人里，最早醒的！ + int wake = workers.poll(); + // job 何时完成呢？ + // (工人醒来，开工时间)最晚的！+1 + int finish = Math.max(start[job], wake) + 1; + finishAll = Math.max(finishAll, finish); + done++; + // 消除影响 + for (int next : nexts[job]) { + start[next] = Math.max(start[next], finish); + if (--indegree[next] == 0) { + zeroIn.add(next); + } + } + workers.add(finish); + } + return done == m ? finishAll : -1; + } + + public static int[][] nexts(int[][] depends, int m) { + Arrays.sort(depends, (a, b) -> a[1] - b[1]); + int n = depends.length; + int[][] nexts = new int[m][0]; + if (n == 0) { + return nexts; + } + int size = 1; + for (int i = 1; i < n; i++) { + if (depends[i - 1][1] != depends[i][1]) { + int from = depends[i - 1][1]; + nexts[from] = new int[size]; + for (int k = 0, j = i - size; k < size; k++, j++) { + nexts[from][k] = depends[j][0]; + } + size = 1; + } else { + size++; + } + } + int from = depends[n - 1][1]; + nexts[from] = new int[size]; + for (int k = 0, j = n - size; k < size; k++, j++) { + nexts[from][k] = depends[j][0]; + } + return nexts; + } + + public static int[] indegree(int[][] nexts, int m) { + int[] indegree = new int[m]; + for (int i = 0; i < m; i++) { + for (int j = 0; j < nexts[i].length; j++) { + indegree[nexts[i][j]]++; + } + } + return indegree; + } + + public static void main(String[] args) { + // 2 -> 5 -> 6 + // | + // v + // 1 -> 4 -> 7 + // ^ + // | + // 0 -> 3 + // 两个人 + // {1，2} 工人队列 + // 0 : 干0号工作 ，1 + // 0 : 干1号工作 ，1 + // 1 : 干2号工作，2 + int[][] d = { + { 3, 0 }, + { 4, 1 }, + { 5, 2 }, + { 4, 3 }, + { 6, 5 }, + { 7, 4 }, + { 7, 6 } + }; + System.out.println(days(3, 8, d)); + System.out.println(days(2, 8, d)); + } + +} diff --git a/算法周更班/class_2021_12_5_week/Code03_WaysToBuildWall.java b/算法周更班/class_2021_12_5_week/Code03_WaysToBuildWall.java new file mode 100644 index 0000000..5cdf26b --- /dev/null +++ b/算法周更班/class_2021_12_5_week/Code03_WaysToBuildWall.java @@ -0,0 +1,64 @@ +package class_2021_12_5_week; + +// 来自hulu +// 你只有1*1、1*2、1*3、1*4，四种规格的砖块 +// 你想铺满n行m列的区域，规则如下： +// 1）不管那种规格的砖，都只能横着摆 +// 比如1*3这种规格的砖，3长度是水平方向，1长度是竖直方向 +// 2）会有很多方法铺满整个区域，整块区域哪怕有一点点不一样，就算不同的方法 +// 3）区域内部(不算区域整体的4条边界)，不能有任何砖块的边界线，是从上一直贯穿到下的直线 +// 返回符合三条规则下，铺满n行m列的区域，有多少种不同的摆放方法 +public class Code03_WaysToBuildWall { + + public static long[] r = { 0, 1, 2, 4, 8 }; + + public static long ways(int n, int m) { + if (n <= 0 || m <= 1) { + return 1; + } + // len[i] = 一共有1行的情况下，列的长度为i的时候有几种摆法(所有，不分合法和非法) + long[] len = new long[m + 1]; + for (int i = 1; i <= Math.min(m, 4); i++) { + len[i] = r[i]; + } + for (int i = 5; i <= m; i++) { + len[i] = len[i - 1] + len[i - 2] + len[i - 3] + len[i - 4]; + } + // any[i] = 一共有n行的情况下，列的长度为i的时候有几种摆法(所有，不分合法和非法) + long[] any = new long[m + 1]; + for (int i = 1; i <= m; i++) { + // n * i的区域：总共的摆法！不区分合法、不合法 + any[i] = power(len[i], n); + } + // solid[i] = 一共有n行的情况下，列的长度为i的时候有几种合法的摆法 + long[] solid = new long[m + 1]; + solid[1] = 1; + for (int i = 2; i <= m; i++) { + long invalid = 0; + // N * i + // 1) （N * 1 合法） * （N * (i-1) 总共） + // 2) （N * 2 合法） * （N * (i-2) 总共） + // 3) （N * 3 合法） * （N * (i-3) 总共） + // + // j) （N * j 合法） * （N * (i-j) 总共） + for (int j = 1; j < i; j++) { + invalid += solid[j] * any[i - j]; + } + solid[i] = any[i] - invalid; + } + return solid[m]; + } + + public static long power(long base, int power) { + long ans = 1; + while (power != 0) { + if ((power & 1) != 0) { + ans *= base; + } + base *= base; + power >>= 1; + } + return ans; + } + +} diff --git a/算法周更班/class_2022_01_1_week/Code01_ABDisappear.java b/算法周更班/class_2022_01_1_week/Code01_ABDisappear.java new file mode 100644 index 0000000..c842d9d --- /dev/null +++ b/算法周更班/class_2022_01_1_week/Code01_ABDisappear.java @@ -0,0 +1,131 @@ +package class_2022_01_1_week; + +// 来自阿里 +// 给定一个只由'a'和'b'组成的字符串str， +// str中"ab"和"ba"子串都可以消除，消除之后剩下字符会重新靠在一起，继续出现可以消除的子串... +// 你的任务是决定一种消除的顺序，最后让str消除到尽可能的短 +// 返回尽可能的短的剩余字符串 +public class Code01_ABDisappear { + + // 暴力尝试，尝试所有的可能性 + // 所有情况都枚举，所有先后消除的顺序都暴力尝试 + public static String disappear1(String str) { + String ans = str; + for (int i = 1; i < str.length(); i++) { + // i == 1 0 1 是不是a和b都全，2... + // i == 2 1 2 是不是a和b都全，03.... + // i == 3 2 3 是不是a和b都全，014.... + // (i-1 i)扣掉！ + boolean hasA = str.charAt(i - 1) == 'a' || str.charAt(i) == 'a'; + boolean hasB = str.charAt(i - 1) == 'b' || str.charAt(i) == 'b'; + if (hasA && hasB) { + String curAns = disappear1(str.substring(0, i - 1) + str.substring(i + 1)); + if (curAns.length() < ans.length()) { + ans = curAns; + } + } + } + return ans; + } + + // 好一点的方法尝试 + // 遇到第一个能消除的就消除， + // 剩下的字符串继续：遇到第一个能消除的就消除 + // 剩下的字符串继续：遇到第一个能消除的就消除 + // ... + public static String disappear2(String str) { + String ans = str; + for (int i = 1; i < str.length(); i++) { + boolean hasA = str.charAt(i - 1) == 'a' || str.charAt(i) == 'a'; + boolean hasB = str.charAt(i - 1) == 'b' || str.charAt(i) == 'b'; + if (hasA && hasB) { + return disappear2(str.substring(0, i - 1) + str.substring(i + 1)); + } + } + return ans; + } + + // 时间复杂度O(N) + // 利用栈 + public static String disappear3(String s) { + char[] str = s.toCharArray(); + int n = str.length; + // 用数组结构，自己实现栈 + int[] stack = new int[n]; + int size = 0; + for (int i = 0; i < n; i++) { + boolean hasA = size != 0 && str[stack[size - 1]] == 'a'; + boolean hasB = size != 0 && str[stack[size - 1]] == 'b'; + hasA |= str[i] == 'a'; + hasB |= str[i] == 'b'; + if (hasA && hasB) { + size--; + } else { + stack[size++] = i; + } + } + StringBuilder builder = new StringBuilder(); + for (int i = 0; i < size; i++) { + builder.append(str[stack[i]]); + } + return builder.toString(); + } + + // 贪心，时间复杂度O(N) + // A的个数如果比B的个数多，那么一定只剩下A + // B的个数如果比A的个数多，那么一定只剩下B + public static String disappear4(String s) { + int count = 0; + for (char cha : s.toCharArray()) { + count += cha == 'a' ? 1 : -1; + } + StringBuilder builder = new StringBuilder(); + char rest = count > 0 ? 'a' : 'b'; + count = Math.abs(count); + for (int i = 0; i < count; i++) { + builder.append(rest); + } + return builder.toString(); + } + + // 为了测试 + public static String randomString(int len, int varible) { + char[] str = new char[len]; + for (int i = 0; i < str.length; i++) { + str[i] = (char) ((int) (Math.random() * varible) + 'a'); + } + return String.valueOf(str); + } + + // 为了测试 + // 对数器，三种截然不同的方法进行验证 + public static void main(String[] args) { + // 字符串最大长度，可以随意改变 + // 不过长度太大的话，方法一会超时 + int n = 14; + // 字符的种类，可以随意改变 + int v = 2; + // 测试次数 + int testTime = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * n) + 1; + String test = randomString(len, v); + String ans1 = disappear1(test); + String ans2 = disappear2(test); + String ans3 = disappear3(test); + String ans4 = disappear4(test); + if (!ans1.equals(ans2) || !ans1.equals(ans3) || !ans1.equals(ans4)) { + System.out.println(test); + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + System.out.println(ans4); + System.out.println("出错了!"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_01_1_week/Code02_CatAndMouse.java b/算法周更班/class_2022_01_1_week/Code02_CatAndMouse.java new file mode 100644 index 0000000..2803f1a --- /dev/null +++ b/算法周更班/class_2022_01_1_week/Code02_CatAndMouse.java @@ -0,0 +1,233 @@ +package class_2022_01_1_week; + +import java.util.ArrayList; +import java.util.Arrays; + +// 测试链接 : https://leetcode.com/problems/cat-and-mouse/ +public class Code02_CatAndMouse { + + // 不贪心，就递归 + 记忆化搜索 + public static int catMouseGame1(int[][] graph) { + int n = graph.length; + // 9 : 2 * 8 * 9 再大，平局 + int limit = ((n * (n - 1)) << 1) + 1; + int[][][] dp = new int[n][n][limit]; + for (int i = 0; i < n; i++) { + for (int j = 0; j < n; j++) { + Arrays.fill(dp[i][j], -1); + } + } + return process(graph, limit, 2, 1, 1, dp); + } + + // 贪心 + 递归 + 记忆化搜索 + // 但是不对！ + // 再次强调！不要去证明！ + public static int catMouseGame2(int[][] graph) { + int n = graph.length; + // 这里！ + // int limit = (n << 1) + 2; 还会出错，但是概率很小，需要多跑几次 + // int limit = (n << 1) + 3; 就没错了，或者说，概率小到很难重现 + // 为啥？我屌你为啥！ + // 讲了这节课之后，leetcode又增加了测试用例 + // 导致 int limit = (n << 1) + 3; 也会出错 + // 改成如下的+12，可以通过，但是，如果去人为构造错误用例，一定还会出现错误 + int limit = (n << 1) + 12; + int[][][] dp = new int[n][n][limit]; + for (int i = 0; i < n; i++) { + for (int j = 0; j < n; j++) { + Arrays.fill(dp[i][j], -1); + } + } + return process(graph, limit, 2, 1, 1, dp); + } + + // dp[][][] 傻缓存！ + // dp[cat][mouse][turn] == -1 这个状态之前没算过！ + // dp[cat][mouse][turn] == 0 这个状态之前算过！平局！ + // dp[cat][mouse][turn] == 1 这个状态之前算过！老鼠赢！ + // dp[cat][mouse][turn] == 2 这个状态之前算过！猫赢！ + // 固定参数！轮数不要超过limit！如果超过，就算平局！ + public static int process(int[][] graph, int limit, int cat, int mouse, int turn, int[][][] dp) { + if (turn == limit) { + return 0; + } + if (dp[cat][mouse][turn] != -1) { + return dp[cat][mouse][turn]; + } + int ans = 0; + if (cat == mouse) { + ans = 2; + } else if (mouse == 0) { + ans = 1; + } else { + if ((turn & 1) == 1) { // 老鼠回合 + ans = 2; + for (int next : graph[mouse]) { + int p = process(graph, limit, cat, next, turn + 1, dp); + ans = p == 1 ? 1 : (p == 0 ? 0 : ans); + if (ans == 1) { + break; + } + } + } else { // 猫回合 + ans = 1; + for (int next : graph[cat]) { + if (next != 0) { + int p = process(graph, limit, next, mouse, turn + 1, dp); + ans = p == 2 ? 2 : (p == 0 ? 0 : ans); + if (ans == 2) { + break; + } + } + } + } + } + dp[cat][mouse][turn] = ans; + return ans; + } + + // 为了测试 + // 暴力尝试 + public static int right(int[][] graph) { + int n = graph.length; + boolean[][][] path = new boolean[n][n][2]; + return win(graph, 2, 1, 1, path); + } + + // 暴力尝试 + // 返回值：int + // 0 : 平局 + // 1 : 老鼠赢 + // 2 : 猫赢 + // int[][] graph + // 0 : {3, 7, 9} + // 1 : + // 2 : + // 3 : {0, } + // ... + // 7 : {0, } + // ... + // 9 : {0, } + // 猫此时的位置 -> cat + // mouse + // turn == 1 老鼠的回合 turn == 0 猫的回合 + // 当第一次出现，在老鼠的回合，猫在5位置 ，老鼠在7位置 + // path[cat][mouse][1] = false + // 不是第一次出现 path[cat][mouse][1] = true + public static int win(int[][] graph, int cat, int mouse, int turn, boolean[][][] path) { + if (path[cat][mouse][turn]) { // 之前来到过这个状态！平局！ + return 0; + } + // 没来过！ + path[cat][mouse][turn] = true; + int ans = 0; + if (cat == mouse) { + ans = 2; + } else if (mouse == 0) { + ans = 1; + } else { // 游戏继续 + if (turn == 1) { // 老鼠回合 + // 最差情况，是猫赢！ + ans = 2; + for (int next : graph[mouse]) { + int p = win(graph, cat, next, 0, path); + // p的返回值如果是1，代表老鼠赢，那么最差结果ans = 1，直接返回！ + ans = p == 1 ? 1 : (p == 0 ? 0 : ans); + if (ans == 1) { + break; + } + } + } else { // 猫回合 + ans = 1; + for (int next : graph[cat]) { + if (next != 0) { + int p = win(graph, next, mouse, turn ^ 1, path); + ans = p == 2 ? 2 : (p == 0 ? 0 : ans); + if (ans == 2) { + break; + } + } + } + } + } + path[cat][mouse][turn] = false; + return ans; + } + + // 为了测试 + public static int[][] randomGraph(int n) { + ArrayList> g = new ArrayList<>(); + for (int i = 0; i < n; i++) { + g.add(new ArrayList<>()); + } + for (int i = 0; i < n; i++) { + for (int j = i + 1; j < n; j++) { + if (Math.random() > 0.5) { + g.get(i).add(j); + g.get(j).add(i); + } + } + } + int[][] graph = new int[n][]; + for (int i = 0; i < n; i++) { + int m = g.get(i).size(); + graph[i] = new int[m]; + for (int j = 0; j < m; j++) { + graph[i][j] = g.get(i).get(j); + } + } + return graph; + } + + // 为了测试 + public static void main(String[] args) { + System.out.println("证什么证！？对数器万岁!"); + int N = 7; + int testTime = 3000000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * N) + 3; + int[][] graph = randomGraph(n); + int ans1 = catMouseGame1(graph); + int ans2 = catMouseGame2(graph); + if (ans1 != ans2) { + for (int row = 0; row < graph.length; row++) { + System.out.print(row + " : "); + for (int next : graph[row]) { + System.out.print(next + " "); + } + System.out.println(); + } + System.out.println("ans1 : " + ans1); + System.out.println("ans2 : " + ans2); + System.out.println("出错了!"); + break; + } + } + System.out.println("测试结束"); + + // 给你记录一个错误的例子 + int[][] graph = { + // 0 : + { 2, 6, 7 }, + // 1 : + { 3, 4, 5, 7 }, + // 2 : + { 0, 3, 4, 7 }, + // 3 : + { 1, 2, 5, 6 }, + // 4 : + { 1, 2, 5, 7 }, + // 5 : + { 1, 3, 4, 6 }, + // 6 : + { 0, 3, 5, 7 }, + // 7 : + { 0, 1, 2, 4, 6 } }; + System.out.println(catMouseGame1(graph)); + System.out.println(catMouseGame2(graph)); + + } + +} diff --git a/算法周更班/class_2022_01_1_week/Code03_MaximumScoreFromPerformingMultiplicationOperations.java b/算法周更班/class_2022_01_1_week/Code03_MaximumScoreFromPerformingMultiplicationOperations.java new file mode 100644 index 0000000..9eb588b --- /dev/null +++ b/算法周更班/class_2022_01_1_week/Code03_MaximumScoreFromPerformingMultiplicationOperations.java @@ -0,0 +1,58 @@ +package class_2022_01_1_week; + +// 测试链接 : https://leetcode.com/problems/maximum-score-from-performing-multiplication-operations/ +public class Code03_MaximumScoreFromPerformingMultiplicationOperations { + + // B数组消耗完之前，A数组不会耗尽，题目输入保证的！ + // A[left...right] + // B[0..i-1]已经消耗完了！B[i...m-1] + // 直到把B数组消耗完，能获得的最大分数返回 + public static int zuo(int[] A, int[] B, int left, int right) { + int leftAlready = left; + int rightAlready = A.length - right - 1; + int i = leftAlready + rightAlready; + if (i == B.length) { + return 0; + } + // 没消耗完 + int p1 = A[left] * B[i] + zuo(A, B, left + 1, right); + int p2 = A[right] * B[i] + zuo(A, B, left, right - 1); + return Math.max(p1, p2); + } + + public static int maximumScore1(int[] A, int[] B) { + if (A == null || A.length == 0 || B == null || B.length == 0 || A.length < B.length) { + return 0; + } + return process1(A, B, 0, A.length - 1); + } + + public static int process1(int[] A, int[] B, int L, int R) { + int indexB = L + A.length - R - 1; + if (indexB == B.length) { + return 0; + } else { + int p1 = A[L] * B[indexB] + process1(A, B, L + 1, R); + int p2 = A[R] * B[indexB] + process1(A, B, L, R - 1); + return Math.max(p1, p2); + } + } + + public static int maximumScore2(int[] A, int[] B) { + if (A == null || A.length == 0 || B == null || B.length == 0 || A.length < B.length) { + return 0; + } + int N = A.length; + int M = B.length; + int[][] dp = new int[M + 1][M + 1]; + for (int L = M - 1; L >= 0; L--) { + for (int j = L + 1; j <= M; j++) { + int R = N - M + j - 1; + int indexB = L + N - R - 1; + dp[L][j] = Math.max(A[L] * B[indexB] + dp[L + 1][j], A[R] * B[indexB] + dp[L][j - 1]); + } + } + return dp[0][M]; + } + +} diff --git a/算法周更班/class_2022_01_1_week/Code04_MinDistanceFromLeftUpToRightDownWalk4Directions.java b/算法周更班/class_2022_01_1_week/Code04_MinDistanceFromLeftUpToRightDownWalk4Directions.java new file mode 100644 index 0000000..9f2f6ff --- /dev/null +++ b/算法周更班/class_2022_01_1_week/Code04_MinDistanceFromLeftUpToRightDownWalk4Directions.java @@ -0,0 +1,119 @@ +package class_2022_01_1_week; + +import java.util.PriorityQueue; + +// 来自学员问题 +// 给定一个二维数组，其中全是非负数 +// 每一步都可以往上、下、左、右四个方向运动 +// 走过的路径，会沿途累加数字 +// 返回从左下角走到右下角的累加和最小的多少 +public class Code04_MinDistanceFromLeftUpToRightDownWalk4Directions { + + public static int bestWalk1(int[][] map) { + int n = map.length; + int m = map[0].length; + int step = n * m - 1; + return process(map, n, m, step, 0, 0, 0, 0); + } + + public static int process(int[][] map, int n, int m, int limit, int step, int row, int col, int cost) { + if (row < 0 || row == n || col < 0 || col == m || step > limit) { + return Integer.MAX_VALUE; + } + if (row == n - 1 && col == m - 1) { + return cost + map[row][col]; + } + cost += map[row][col]; + int p1 = process(map, n, m, limit, step + 1, row - 1, col, cost); + int p2 = process(map, n, m, limit, step + 1, row + 1, col, cost); + int p3 = process(map, n, m, limit, step + 1, row, col - 1, cost); + int p4 = process(map, n, m, limit, step + 1, row, col + 1, cost); + return Math.min(Math.min(p1, p2), Math.min(p3, p4)); + } + + public static int bestWalk2(int[][] map) { + + int n = map.length; + int m = map[0].length; + // 堆 + // 每一个对象，都是一个小数组 + // {dis, row, col} + // 0 1 2 + PriorityQueue heap = new PriorityQueue<>((a, b) -> a[0] - b[0]); + // X,0,1 已经弹出了！ 以后在遇到(0,1)的事情，不管！ + // poped记录哪些位置弹出，哪些没有！ + boolean[][] poped = new boolean[n][m]; + heap.add(new int[] { map[0][0], 0, 0 }); + int ans = 0; + while (!heap.isEmpty()) { + int[] cur = heap.poll(); + int dis = cur[0]; + int row = cur[1]; + int col = cur[2]; + if (poped[row][col]) { + continue; + } + // 接下来就是要处理这个位置了！ + poped[row][col] = true; + if (row == n - 1 && col == m - 1) { + ans = dis; + break; + } + add(dis, row - 1, col, n, m, map, poped, heap); + add(dis, row + 1, col, n, m, map, poped, heap); + add(dis, row, col - 1, n, m, map, poped, heap); + add(dis, row, col + 1, n, m, map, poped, heap); + } + return ans; + } + + public static void add(int pre, int row, int col, int n, int m, int[][] map, boolean[][] used, + PriorityQueue heap) { + if (row >= 0 && row < n && col >= 0 && col < m && !used[row][col]) { + heap.add(new int[] { pre + map[row][col], row, col }); + } + } + + // 为了测试 + public static int[][] randomMatrix(int n, int m, int v) { + int[][] ans = new int[n][m]; + for (int i = 0; i < n; i++) { + for (int j = 0; j < m; j++) { + ans[i][j] = (int) (Math.random() * v); + } + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int n = 4; + int m = 4; + int v = 10; + int testTime = 10; + System.out.println("功能测试开始"); + for (int i = 0; i < testTime; i++) { + int[][] map = randomMatrix(n, m, v); + int ans1 = bestWalk1(map); + int ans2 = bestWalk2(map); + if (ans1 != ans2) { + System.out.println("出错了!"); + } + } + System.out.println("功能测试结束"); + + n = 1000; + m = 1000; + v = 100; + int[][] map = randomMatrix(n, m, v); + System.out.println("性能测试开始"); + System.out.println("数据规模 : " + n + " * " + m); + System.out.println("值的范围 : 0 ~ " + v); + long start = System.currentTimeMillis(); + bestWalk2(map); + long end = System.currentTimeMillis(); + System.out.println("运行时间(毫秒) : " + (end - start)); + System.out.println("性能测试结束"); + } + +} diff --git a/算法周更班/class_2022_01_1_week/Problem_0913_CatAndMouse.java b/算法周更班/class_2022_01_1_week/Problem_0913_CatAndMouse.java new file mode 100644 index 0000000..a02a782 --- /dev/null +++ b/算法周更班/class_2022_01_1_week/Problem_0913_CatAndMouse.java @@ -0,0 +1,93 @@ +package class_2022_01_1_week; + +import java.util.ArrayList; + +// 本题是Code02_CatAndMoused的补充code +// leetcode官网增加了测试用例 +// 进一步说明，贪心的方式其实是没有道理的 +// 不过原来贪心的方法，把源代码第35行 +// int limit = (n << 1) + 3; +// 改成： +// int limit = (n << 1) + 12; +// 依然继续能通过 +// 但是贪心依然是没道理的，人为构造的话总能找出反例来 +// 所以补充了一个拓扑排序的解法 +// 这个方法，没在课上讲，有空我们补充一下 +public class Problem_0913_CatAndMouse { + + public static final int MOUSE_TURN = 0, CAT_TURN = 1; + public static final int DRAW = 0, MOUSE_WIN = 1, CAT_WIN = 2; + + public int catMouseGame(int[][] graph) { + int n = graph.length; + int[][][] indegree = new int[n][n][2]; + for (int i = 0; i < n; i++) { + for (int j = 1; j < n; j++) { + indegree[i][j][MOUSE_TURN] = graph[i].length; + indegree[i][j][CAT_TURN] = graph[j].length; + } + } + for (int i = 0; i < n; i++) { + for (int node : graph[0]) { + indegree[i][node][CAT_TURN]--; + } + } + int[][][] ans = new int[n][n][2]; + int[][] queue = new int[n * n * 2][]; + int left = 0; + int right = 0; + for (int j = 1; j < n; j++) { + ans[0][j][MOUSE_TURN] = MOUSE_WIN; + ans[0][j][CAT_TURN] = MOUSE_WIN; + queue[right++] = new int[] { 0, j, MOUSE_TURN }; + queue[right++] = new int[] { 0, j, CAT_TURN }; + } + for (int i = 1; i < n; i++) { + ans[i][i][MOUSE_TURN] = CAT_WIN; + ans[i][i][CAT_TURN] = CAT_WIN; + queue[right++] = new int[] { i, i, MOUSE_TURN }; + queue[right++] = new int[] { i, i, CAT_TURN }; + } + while (left != right) { + int[] cur = queue[left++]; + int mouse = cur[0], cat = cur[1], turn = cur[2]; + int curAns = ans[mouse][cat][turn]; + for (int[] prevState : preStatus(graph, mouse, cat, turn)) { + int prevMouse = prevState[0], prevCat = prevState[1], prevTurn = prevState[2]; + if (ans[prevMouse][prevCat][prevTurn] == DRAW) { + boolean canWin = (curAns == MOUSE_WIN && prevTurn == MOUSE_TURN) + || (curAns == CAT_WIN && prevTurn == CAT_TURN); + if (canWin) { + ans[prevMouse][prevCat][prevTurn] = curAns; + queue[right++] = new int[] { prevMouse, prevCat, prevTurn }; + } else { + if (--indegree[prevMouse][prevCat][prevTurn] == 0) { + int lose = prevTurn == MOUSE_TURN ? CAT_WIN : MOUSE_WIN; + ans[prevMouse][prevCat][prevTurn] = lose; + queue[right++] = new int[] { prevMouse, prevCat, prevTurn }; + } + } + } + } + } + return ans[1][2][MOUSE_TURN]; + } + + public ArrayList preStatus(int[][] graph, int mouse, int cat, int turn) { + ArrayList prevStates = new ArrayList(); + int prevTurn = turn == MOUSE_TURN ? CAT_TURN : MOUSE_TURN; + if (prevTurn == MOUSE_TURN) { + for (int prev : graph[mouse]) { + prevStates.add(new int[] { prev, cat, prevTurn }); + } + } else { + for (int prev : graph[cat]) { + if (prev != 0) { + prevStates.add(new int[] { mouse, prev, prevTurn }); + } + } + } + return prevStates; + } + +} diff --git a/算法周更班/class_2022_01_2_week/Code01_StringCounts.java b/算法周更班/class_2022_01_2_week/Code01_StringCounts.java new file mode 100644 index 0000000..dd029a7 --- /dev/null +++ b/算法周更班/class_2022_01_2_week/Code01_StringCounts.java @@ -0,0 +1,13 @@ +package class_2022_01_2_week; + +// 给定一个非常大的List list +// 每一个字符串类似 : "hello,world,have,hello,world" +// 这一个字符串中，有2个hello，2个world，1个have +// 请设计一种多线程处理方案，统计list中每一个字符串，切分出来的单词数量，并且汇总 +// 最终返回一个HashMap表示每个字符串在list中一共出现几次 +public class Code01_StringCounts { + + // 多线程设计 + 算法 + // 本题没有代码实现，会在课上讲述思路 + +} diff --git a/算法周更班/class_2022_01_2_week/Code02_BrickAll.java b/算法周更班/class_2022_01_2_week/Code02_BrickAll.java new file mode 100644 index 0000000..c6c1cc9 --- /dev/null +++ b/算法周更班/class_2022_01_2_week/Code02_BrickAll.java @@ -0,0 +1,49 @@ +package class_2022_01_2_week; + +import java.util.Arrays; + +// 给定一个正数数组arr，其中每个值代表砖块长度 +// 所有砖块等高等宽，只有长度有区别 +// 每一层可以用1块或者2块砖来摆 +// 要求每一层的长度一样 +// 要求必须使用所有的砖块 +// 请问最多摆几层 +// 如果无法做到，返回-1 +public class Code02_BrickAll { + + public static int maxLevels(int[] arr) { + if (arr == null) { + return 0; + } + int n = arr.length; + if (n < 2) { + return n; + } + Arrays.sort(arr); + int p1 = levels(arr, arr[n - 1]); + int p2 = levels(arr, arr[n - 1] + arr[0]); + return Math.max(p1, p2); + } + + // 要求所有砖必须都使用，并且每一层最多两块砖，并且每一层长度都是len + // 返回能摆几层，如果无法做到返回-1 + public static int levels(int[] arr, int len) { + int ans = 0; + int L = 0; + int R = arr.length - 1; + while (L <= R) { + if (arr[R] == len) { + R--; + ans++; + } else if (L < R && arr[L] + arr[R] == len) { + L++; + R--; + ans++; + } else { + return -1; + } + } + return ans; + } + +} diff --git a/算法周更班/class_2022_01_2_week/Code03_StringNumberConvertBinaryAndHexadecimal.java b/算法周更班/class_2022_01_2_week/Code03_StringNumberConvertBinaryAndHexadecimal.java new file mode 100644 index 0000000..fbccc34 --- /dev/null +++ b/算法周更班/class_2022_01_2_week/Code03_StringNumberConvertBinaryAndHexadecimal.java @@ -0,0 +1,119 @@ +package class_2022_01_2_week; + +// 来自兴业数金 +// 给定一个字符串形式的数，比如"3421"或者"-8731" +// 如果这个数不在-32768~32767范围上，那么返回"NODATA" +// 如果这个数在-32768~32767范围上，那么这个数就没有超过16个二进制位所能表达的范围 +// 返回这个数的2进制形式的字符串和16进制形式的字符串，用逗号分割 +public class Code03_StringNumberConvertBinaryAndHexadecimal { + + // 请保证输入的num字符串一定是数字的形式 + public static String convert(String num) { + // 因为-32768~32767所有的数，最多6个字符，所以超过就返回"NODATA" + if (num == null || num.length() == 0 || num.length() > 6) { + return "NODATA"; + } + // 既然长度不超过6，那么转成整数一定不会有问题 + // 当然你也可以自己写这个转化过程，这个是比较简单的 + int n = Integer.valueOf(num); + // 如果转换完成后超过了范围，那么返回"NODATA" + if (n < -32768 || n > 32767) { + return "NODATA"; + } + // 接下来n就是一个在范围上的数字 + // 我们要取出16位信息(info)，这包括： + // 提取出n的14位~0位的信息 : 也就是(n & 65535) + // 提取出第15位符号位信息 : 如果n<0，第15位就是1，如果n>=0第15位就是0 + // 然后把(15位)和(14位~0位)，或在一起 + // 比如5323，32位二进制形式如下： + // 00000000000000000001010011001011 + // 14位~0位是： + // _________________001010011001011 + // 15位符号位应该是0： + // ________________0_______________ + // 所以info应该是： + // ________________0001010011001011 + // + // 再比如-6725，32位二进制形式如下： + // 11111111111111111110010110111011 + // 14位~0位是： + // _________________110010110111011 + // 15位符号位应该是1： + // ________________1_______________ + // 所以info应该是： + // ________________1110010110111011 + int info = (n < 0 ? (1 << 15) : 0) | (n & 65535); + // 此时info的15位~0位，就是我们要的；info的更高位完全没用 + StringBuilder builder = new StringBuilder(); + // 依次提取二进制信息很简单 + for (int i = 15; i >= 0; i--) { + builder.append((info & (1 << i)) != 0 ? '1' : '0'); + } + builder.append(",0x"); + // 依次提取16进制的时候，每4位提取 + // 依次提取4位信息的时候 + // 0000 -> 0 -> '0' + // 0001 -> 1 -> '1' + // ... + // 1001 -> 9 -> '9' + // 1010 -> 10 -> 'a' + // 1011 -> 11 -> 'b' + // 1100 -> 12 -> 'c' + // 1101 -> 13 -> 'd' + // 1110 -> 14 -> 'e' + // 1111 -> 15 -> 'f' + for (int i = 12; i >= 0; i -= 4) { + // 1111 << 12 + // ( info & (1111 << 12) ) >> 12 + // ( info & (1111 << 8)) >> 8 + // ( info & (1111 << 4)) >> 4 + // ( info & (1111 << 0)) >> 0 + int cur = (info & (15 << i)) >> i; + if (cur < 10) { + builder.append(cur); + } else { + switch (cur) { + case 10: + builder.append('a'); + break; + case 11: + builder.append('b'); + break; + case 12: + builder.append('c'); + break; + case 13: + builder.append('d'); + break; + case 14: + builder.append('e'); + break; + default: + builder.append('f'); + } + } + } + return builder.toString(); + } + + public static void main(String[] args) { + String num1 = "0"; + System.out.println(convert(num1)); + + String zuo = "457"; + System.out.println(convert(zuo)); + + String num2 = "-32768"; + System.out.println(convert(num2)); + + String num3 = "32768"; + System.out.println(convert(num3)); + + String num4 = "32767"; + System.out.println(convert(num4)); + + String num5 = "-1"; + System.out.println(convert(num5)); + } + +} diff --git a/算法周更班/class_2022_01_2_week/Code04_MinimumOperationsToMakeTheArrayKIncreasing.java b/算法周更班/class_2022_01_2_week/Code04_MinimumOperationsToMakeTheArrayKIncreasing.java new file mode 100644 index 0000000..06aebcc --- /dev/null +++ b/算法周更班/class_2022_01_2_week/Code04_MinimumOperationsToMakeTheArrayKIncreasing.java @@ -0,0 +1,49 @@ +package class_2022_01_2_week; + +// 来自Leetcode周赛第5题 +// 测试链接 : https://leetcode.com/problems/minimum-operations-to-make-the-array-k-increasing/ +public class Code04_MinimumOperationsToMakeTheArrayKIncreasing { + + public static int kIncreasing(int[] arr, int k) { + int n = arr.length; + // a / b = 向下取整 + // a / b 向上取整 -> (a + b - 1) / b + int[] help = new int[(n + k - 1) / k]; + int ans = 0; + for (int i = 0; i < k; i++) { + ans += need(arr, help, n, i, k); + } + return ans; + } + + // arr[start , start + k, start + 2k, start + 3k,....] + // 辅助数组help，为了求最长递增子序列，需要开辟的空间，具体看体系学习班 + // 上面的序列，要改几个数，能都有序！ + public static int need(int[] arr, int[] help, int n, int start, int k) { + int j = 0; + int size = 0; + for (; start < n; start += k, j++) { + size = insert(help, size, arr[start]); + } + return j - size; + } + + public static int insert(int[] help, int size, int num) { + int l = 0; + int r = size - 1; + int m = 0; + int ans = size; + while (l <= r) { + m = (l + r) / 2; + if (help[m] > num) { + ans = m; + r = m - 1; + } else { + l = m + 1; + } + } + help[ans] = num; + return ans == size ? size + 1 : size; + } + +} diff --git a/算法周更班/class_2022_01_2_week/Code05_MagicTowSubarrayMakeMaxSum.java b/算法周更班/class_2022_01_2_week/Code05_MagicTowSubarrayMakeMaxSum.java new file mode 100644 index 0000000..d169640 --- /dev/null +++ b/算法周更班/class_2022_01_2_week/Code05_MagicTowSubarrayMakeMaxSum.java @@ -0,0 +1,129 @@ +package class_2022_01_2_week; + +// 来自美团 +// 小美有一个长度为n的数组，为了使得这个数组的和尽量大，她向会魔法的小团进行求助 +// 小团可以选择数组中至多两个不相交的子数组，并将区间里的数全都变为原来的10倍 +// 小团想知道他的魔法最多可以帮助小美将数组的和变大到多少? +public class Code05_MagicTowSubarrayMakeMaxSum { + + // 比较暴力的方法 + // 用于对数器 + public static int maxSum1(int[] arr) { + int n = arr.length; + int[] presum = presum(arr); + int ans = maxSumAtMostOneRangeMagic(presum, 0, n - 1); + for (int split = 0; split < n - 1; split++) { + ans = Math.max(ans, + maxSumAtMostOneRangeMagic(presum, 0, split) + maxSumAtMostOneRangeMagic(presum, split + 1, n - 1)); + } + return ans; + } + + public static int[] presum(int[] arr) { + int n = arr.length; + int[] presum = new int[n]; + presum[0] = arr[0]; + for (int i = 1; i < n; i++) { + presum[i] = presum[i - 1] + arr[i]; + } + return presum; + } + + public static int sum(int[] presum, int l, int r) { + return l > r ? 0 : (l == 0 ? presum[r] : (presum[r] - presum[l - 1])); + } + + public static int maxSumAtMostOneRangeMagic(int[] presum, int l, int r) { + if (l > r) { + return 0; + } + int ans = sum(presum, l, r); + for (int s = l; s <= r; s++) { + for (int e = s; e <= r; e++) { + ans = Math.max(ans, sum(presum, l, s - 1) + sum(presum, s, e) * 10 + sum(presum, e + 1, r)); + } + } + return ans; + } + + // 正式方法，时间复杂度O(N) + public static int maxSum2(int[] arr) { + int n = arr.length; + if (n == 0) { + return 0; + } + if (n == 1) { + return Math.max(arr[0], arr[0] * 10); + } + // dp[i] + // 1) arr[0...i]原始累加和 + // 2) dp[i-1] + arr[i] + // 3) magic[i] + + // : arr[0..i]范围上，可以没有10倍区域、或者有10倍区域但是最多有一个的情况下， + // 最大累加和是多少？ + // 可能性1：就是没有10倍区域，那就是arr[0..i]的累加和, 这个好弄！ + // + // 可能性2：有一个10倍区域 + // a : arr[i]不在10倍区域里，但是之前可能有，那么就是dp[i-1] + arr[i] + // + // b : arr[i]在10倍区域里 + // 甲：arr[0..i-1]没有10倍区域，arr[i]自己10倍，arr[0..i-1] + 10 * arr[i] + // 乙：arr[0..i-1]中i-1位置在10倍区域里，arr[i]也在10倍区域里 + // magic[i] : magic[i] ..i i + // 对于乙，要求知道magic[j]的信息 + // magic[j]：arr[0..j]范围上，j一定要在10倍区域里，并且只有一个10倍区域的情况下，最大累加和 + // 可能性1：只有arr[j]是10倍，arr[0..j-1]没有10倍 + // 可能性2：magic[j-1] + 10 * arr[j] + int sum = arr[n - 1]; + int magic = sum * 10; + int[] right = new int[n]; + right[n - 1] = Math.max(sum, sum * 10); + for (int i = n - 2; i >= 0; i--) { + magic = 10 * arr[i] + Math.max(sum, magic); + sum += arr[i]; + right[i] = Math.max(Math.max(sum, right[i + 1] + arr[i]), magic); + } + int ans = right[0]; + sum = arr[0]; + magic = sum * 10; + int dp = Math.max(sum, sum * 10); + ans = Math.max(ans, dp + right[1]); + for (int i = 1; i < n - 1; i++) { + magic = 10 * arr[i] + Math.max(sum, magic); + sum += arr[i]; + dp = Math.max(Math.max(sum, dp + arr[i]), magic); + ans = Math.max(ans, dp + right[i + 1]); + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int len, int value) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * value) - (int) (Math.random() * value); + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int len = 30; + int value = 100; + int testTime = 500000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * len) + 1; + int[] arr = randomArray(n, value); + int ans1 = maxSum1(arr); + int ans2 = maxSum2(arr); + if (ans1 != ans2) { + System.out.println("出错了!"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_01_2_week/Code06_QuietSum.java b/算法周更班/class_2022_01_2_week/Code06_QuietSum.java new file mode 100644 index 0000000..faa1584 --- /dev/null +++ b/算法周更班/class_2022_01_2_week/Code06_QuietSum.java @@ -0,0 +1,290 @@ +package class_2022_01_2_week; + +// 给定一个非负数组arr，学生依次坐在0~N-1位置，每个值表示学生的安静值 +// 如果在i位置安置插班生，那么i位置的安静值变成0，同时任何同学都会被影响到而减少安静值 +// 同学安静值减少的量: N - 这个同学到插班生的距离 +// 但是减到0以下的话，当做0处理 +// 返回一个和arr等长的ans数组，ans[i]表示如果把插班生安排在i位置，所有学生的安静值的和 +// 比如 : arr = {3,4,2,1,5}，应该返回{4,3,2,3,4} +// 比如 : arr = {10,1,10,10,10}，应该返回{24,27,20,20,22} +// arr长度 <= 10^5 +// arr中值 <= 2 * 10^5 +public class Code06_QuietSum { + + // 为了测试 + // 彻底暴力的方法 + public static long[] quiet1(int[] arr) { + if (arr == null || arr.length == 0) { + return new long[0]; + } + int n = arr.length; + long[] ans = new long[n]; + for (int i = 0; i < n; i++) { + long sum = 0; + for (int j = 0; j < i; j++) { + sum += Math.max(0, arr[j] - (n - Math.abs(i - j))); + } + for (int j = i + 1; j < n; j++) { + sum += Math.max(0, arr[j] - (n - Math.abs(i - j))); + } + ans[i] = sum; + } + return ans; + } + + // 时间复杂度O(N * logN)的方法 + public static long[] quiet2(int[] arr) { + if (arr == null || arr.length == 0) { + return new long[0]; + } + int n = arr.length; + SBTree sbt = new SBTree(); + long[] ans = new long[n]; + // arr[n-1] + n - 1 - n + // + sbt.add(arr[n - 1] - 1); + for (int i = n - 2; i >= 0; i--) { + long moreCount = sbt.moreCount(i); + long moreSum = sbt.moreSum(i); + ans[i] = moreSum - (moreCount * i); + sbt.add(arr[i] + i - n); + } + sbt = new SBTree(); + sbt.add(arr[0] - 1); + for (int i = 1, j = n - 2; i < n; i++, j--) { + long moreCount = sbt.moreCount(j); + long moreSum = sbt.moreSum(j); + ans[i] += moreSum - (moreCount * j); + sbt.add(arr[i] + j - n); + } + return ans; + } + + public static class SBTNode { + public int value; + // 和业务无关 + // 不同key的size，纯粹为了树的平衡调整 + public int size; + // 和业务有关 + // 加过几个数字 + public int all; + // 和业务有关 + // 子树的累加和 + public long sum; + public SBTNode l; + public SBTNode r; + + public SBTNode(int v) { + value = v; + size = 1; + all = 1; + sum = v; + } + } + + // add count(?) sum(?) logN复杂度！ + // 有序表提供的性能，size - balanced - tree + public static class SBTree { + + private SBTNode root; + + private SBTNode rightRotate(SBTNode cur) { + int curCount = cur.all - (cur.l != null ? cur.l.all : 0) - (cur.r != null ? cur.r.all : 0); + long curSum = cur.sum - (cur.l != null ? cur.l.sum : 0) - (cur.r != null ? cur.r.sum : 0); + SBTNode leftNode = cur.l; + cur.l = leftNode.r; + leftNode.r = cur; + leftNode.size = cur.size; + leftNode.all = cur.all; + leftNode.sum = cur.sum; + cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1; + cur.all = (cur.l != null ? cur.l.all : 0) + (cur.r != null ? cur.r.all : 0) + curCount; + cur.sum = (cur.l != null ? cur.l.sum : 0) + (cur.r != null ? cur.r.sum : 0) + curSum; + return leftNode; + } + + private SBTNode leftRotate(SBTNode cur) { + int curCount = cur.all - (cur.l != null ? cur.l.all : 0) - (cur.r != null ? cur.r.all : 0); + long curSum = cur.sum - (cur.l != null ? cur.l.sum : 0) - (cur.r != null ? cur.r.sum : 0); + SBTNode rightNode = cur.r; + cur.r = rightNode.l; + rightNode.l = cur; + rightNode.size = cur.size; + rightNode.all = cur.all; + rightNode.sum = cur.sum; + cur.size = (cur.l != null ? cur.l.size : 0) + (cur.r != null ? cur.r.size : 0) + 1; + cur.all = (cur.l != null ? cur.l.all : 0) + (cur.r != null ? cur.r.all : 0) + curCount; + cur.sum = (cur.l != null ? cur.l.sum : 0) + (cur.r != null ? cur.r.sum : 0) + curSum; + return rightNode; + } + + private SBTNode maintain(SBTNode cur) { + if (cur == null) { + return null; + } + long leftSize = cur.l != null ? cur.l.size : 0; + long leftLeftSize = cur.l != null && cur.l.l != null ? cur.l.l.size : 0; + long leftRightSize = cur.l != null && cur.l.r != null ? cur.l.r.size : 0; + long rightSize = cur.r != null ? cur.r.size : 0; + long rightLeftSize = cur.r != null && cur.r.l != null ? cur.r.l.size : 0; + long rightRightSize = cur.r != null && cur.r.r != null ? cur.r.r.size : 0; + if (leftLeftSize > rightSize) { + cur = rightRotate(cur); + cur.r = maintain(cur.r); + cur = maintain(cur); + } else if (leftRightSize > rightSize) { + cur.l = leftRotate(cur.l); + cur = rightRotate(cur); + cur.l = maintain(cur.l); + cur.r = maintain(cur.r); + cur = maintain(cur); + } else if (rightRightSize > leftSize) { + cur = leftRotate(cur); + cur.l = maintain(cur.l); + cur = maintain(cur); + } else if (rightLeftSize > leftSize) { + cur.r = rightRotate(cur.r); + cur = leftRotate(cur); + cur.l = maintain(cur.l); + cur.r = maintain(cur.r); + cur = maintain(cur); + } + return cur; + } + + public void add(int v) { + root = add(root, v, contains(root, v)); + } + + private boolean contains(SBTNode cur, int v) { + if (cur == null) { + return false; + } else if (cur.value == v) { + return true; + } else if (cur.value > v) { + return contains(cur.l, v); + } else { + return contains(cur.r, v); + } + } + + private SBTNode add(SBTNode cur, int v, boolean contains) { + if (cur == null) { + return new SBTNode(v); + } else { + if (!contains) { + cur.size++; + } + cur.all++; + cur.sum += v; + if (cur.value == v) { + return cur; + } else if (cur.value > v) { + cur.l = add(cur.l, v, contains); + } else { + cur.r = add(cur.r, v, contains); + } + return maintain(cur); + } + } + + public long moreCount(int num) { + return moreCount(root, num); + } + + private long moreCount(SBTNode cur, int num) { + if (cur == null) { + return 0; + } + if (cur.value <= num) { + return moreCount(cur.r, num); + } else { // cur.value > num + return cur.all - (cur.l != null ? cur.l.all : 0) + moreCount(cur.l, num); + } + } + + public long moreSum(int num) { + return moreSum(root, num); + } + + private long moreSum(SBTNode cur, int num) { + if (cur == null) { + return 0; + } + if (cur.value <= num) { + return moreSum(cur.r, num); + } else { // cur.value > num + return cur.sum - (cur.l != null ? cur.l.sum : 0) + moreSum(cur.l, num); + } + } + + } + + // 为了测试 + public static int[] randomArray(int len, int v) { + int[] ans = new int[len]; + for (int i = 0; i < len; i++) { + ans[i] = (int) (Math.random() * (v + 1)); + } + return ans; + } + + public static boolean sameArray(long[] arr1, long[] arr2) { + if (arr1.length != arr2.length) { + return false; + } + int n = arr1.length; + for (int i = 0; i < n; i++) { + if (arr1[i] != arr2[i]) { + return false; + } + } + return true; + } + + // 为了测试 + public static void main(String[] args) { + int N = 1000; + int V = 5000; + int testTime = 10000; + System.out.println("功能测试开始"); + for (int i = 0; i < testTime; i++) { + int size = (int) (Math.random() * N) + 1; + int[] arr = randomArray(size, V); + long[] ans1 = quiet1(arr); + long[] ans2 = quiet2(arr); + if (!sameArray(ans1, ans2)) { + System.out.println("出错了！"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + for (long num : ans1) { + System.out.print(num + " "); + } + System.out.println(); + for (long num : ans2) { + System.out.print(num + " "); + } + System.out.println(); + break; + } + } + System.out.println("功能测试结束"); + + System.out.println("性能测试开始"); + N = 200000; + V = 500000; + long start; + long end; + int[] arr = randomArray(N, V); + start = System.currentTimeMillis(); + quiet2(arr); + end = System.currentTimeMillis(); + System.out.println("测试数组长度 : " + N); + System.out.println("测试数组数值最大值 : " + V); + System.out.println("运行时间（毫秒） : " + (end - start)); + System.out.println("性能测试结束"); + } + +} diff --git a/算法周更班/class_2022_01_3_week/Code01_AStarAlgorithm.java b/算法周更班/class_2022_01_3_week/Code01_AStarAlgorithm.java new file mode 100644 index 0000000..0116371 --- /dev/null +++ b/算法周更班/class_2022_01_3_week/Code01_AStarAlgorithm.java @@ -0,0 +1,195 @@ +package class_2022_01_3_week; + +import java.util.PriorityQueue; + +// A*算法 +// 过程和Dijskra高度相处 +// 有到终点的预估函数 +// 只要预估值<=客观上最优距离，就是对的 +// 预估函数是一种吸引力： +// 1）合适的吸引力可以提升算法的速度 +// 2）吸引力“过强”会出现错误 +// 讲述A*算法 +// 预估终点距离选择曼哈顿距离 +// 要求先在体系学习班图的章节听过"Dijkstra算法" +public class Code01_AStarAlgorithm { + + // Dijkstra算法 + // map[i][j] == 0 代表障碍 + // map[i][j] > 0 代表通行代价 + public static int minDistance1(int[][] map, int startX, int startY, int targetX, int targetY) { + if (map[startX][startY] == 0 || map[targetX][targetY] == 0) { + return Integer.MAX_VALUE; + } + int n = map.length; + int m = map[0].length; + // heap小根堆 + // [20,1,7] + PriorityQueue heap = new PriorityQueue<>((a, b) -> a[0] - b[0]); + // 1,7已经处理过了，closed[1][7] = true + boolean[][] closed = new boolean[n][m]; + heap.add(new int[] { map[startX][startY], startX, startY }); + int ans = Integer.MAX_VALUE; + while (!heap.isEmpty()) { + int[] cur = heap.poll(); + int dis = cur[0]; + int row = cur[1]; + int col = cur[2]; + if (closed[row][col]) { + continue; + } + closed[row][col] = true; + if (row == targetX && col == targetY) { + ans = dis; + break; + } + add1(dis, row - 1, col, n, m, map, closed, heap); + add1(dis, row + 1, col, n, m, map, closed, heap); + add1(dis, row, col - 1, n, m, map, closed, heap); + add1(dis, row, col + 1, n, m, map, closed, heap); + } + return ans; + } + + public static void add1(int pre, int row, int col, int n, int m, int[][] map, boolean[][] closed, + PriorityQueue heap) { + if (row >= 0 && row < n && col >= 0 && col < m && map[row][col] != 0 && !closed[row][col]) { + heap.add(new int[] { pre + map[row][col], row, col }); + } + } + + // A*算法 + // map[i][j] == 0 代表障碍 + // map[i][j] > 0 代表通行代价 + public static int minDistance2(int[][] map, int startX, int startY, int targetX, int targetY) { + if (map[startX][startY] == 0 || map[targetX][targetY] == 0) { + return Integer.MAX_VALUE; + } + int n = map.length; + int m = map[0].length; + // [20,1,7] + // [20,?,1,7] + PriorityQueue heap = new PriorityQueue<>((a, b) -> (a[0] + a[1]) - (b[0] + b[1])); + boolean[][] closed = new boolean[n][m]; + heap.add(new int[] { map[startX][startY], distance(startX, startY, targetX, targetY), startX, startY }); + int ans = Integer.MAX_VALUE; + while (!heap.isEmpty()) { + int[] cur = heap.poll(); + int fromDistance = cur[0]; + int row = cur[2]; + int col = cur[3]; + if (closed[row][col]) { + continue; + } + closed[row][col] = true; + if (row == targetX && col == targetY) { + ans = fromDistance; + break; + } + add2(fromDistance, row - 1, col, targetX, targetY, n, m, map, closed, heap); + add2(fromDistance, row + 1, col, targetX, targetY, n, m, map, closed, heap); + add2(fromDistance, row, col - 1, targetX, targetY, n, m, map, closed, heap); + add2(fromDistance, row, col + 1, targetX, targetY, n, m, map, closed, heap); + } + return ans; + } + + public static void add2(int pre, int row, int col, int targetX, int targetY, int n, int m, int[][] map, + boolean[][] closed, PriorityQueue heap) { + if (row >= 0 && row < n && col >= 0 && col < m && map[row][col] != 0 && !closed[row][col]) { + heap.add(new int[] { pre + map[row][col], distance(row, col, targetX, targetY), row, col }); + } + + } + + // 曼哈顿距离 + public static int distance(int curX, int curY, int targetX, int targetY) { + return Math.abs(targetX - curX) + Math.abs(targetY - curY); + } + + // 为了测试 + // map[i][j] == 0 代表障碍 + // map[i][j] > 0 代表通行代价 + public static int[][] randomMap(int len, int value) { + int[][] ans = new int[len][len]; + for (int i = 0; i < len; i++) { + for (int j = 0; j < len; j++) { + if (Math.random() < 0.2) { + ans[i][j] = 0; + } else { + ans[i][j] = (int) (Math.random() * value); + } + } + } + return ans; + } + + // 为了测试 + public static void printMap(int[][] map) { + for (int i = 0; i < map.length; i++) { + for (int j = 0; j < map[0].length; j++) { + System.out.print((map[i][j] == 0 ? "X" : map[i][j]) + " "); + } + System.out.println(); + } + } + + public static void main(String[] args) { + int len = 100; + int value = 50; + int testTime = 10000; + System.out.println("功能测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * len) + 2; + int[][] map = randomMap(n, value); + int startX = (int) (Math.random() * n); + int startY = (int) (Math.random() * n); + int targetX = (int) (Math.random() * n); + int targetY = (int) (Math.random() * n); + int ans1 = minDistance1(map, startX, startY, targetX, targetY); + int ans2 = minDistance2(map, startX, startY, targetX, targetY); + if (ans1 != ans2) { + System.out.println("出错了!"); + printMap(map); + System.out.println(startX + "," + startY); + System.out.println(targetX + "," + targetY); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("功能测试结束"); + + int[][] map = new int[4000][4000]; + for (int i = 0; i < map.length; i++) { + for (int j = 0; j < map[0].length; j++) { + map[i][j] = (int) (Math.random() * 10); + } + } + int startX = 0; + int startY = 0; + int targetX = 3456; + int targetY = 3728; + + long start = System.currentTimeMillis(); + minDistance2(map, startX, startY, targetX, targetY); + long end = System.currentTimeMillis(); + System.out.println("运行时间(毫秒) : " + (end - start)); + +// System.out.println("性能测试开始"); +// int n = 10000; +// int v = 500; +// int[][] map = randomMap(n, v); +// int startX = 0; +// int startY = 0; +// int targetX = n - 1; +// int targetY = n - 1; +// System.out.println("数据规模 : " + n + " * " + n); +// long start = System.currentTimeMillis(); +// minDistance2(map, startX, startY, targetX, targetY); +// long end = System.currentTimeMillis(); +// System.out.println("运行时间(毫秒) : " + (end - start)); +// System.out.println("性能测试结束"); + } + +} diff --git a/算法周更班/class_2022_01_3_week/Code02_EscapeALargeMaze.java b/算法周更班/class_2022_01_3_week/Code02_EscapeALargeMaze.java new file mode 100644 index 0000000..1482d8d --- /dev/null +++ b/算法周更班/class_2022_01_3_week/Code02_EscapeALargeMaze.java @@ -0,0 +1,74 @@ +package class_2022_01_3_week; + +import java.util.HashSet; +import java.util.LinkedList; +import java.util.Queue; + +// 在一个10^6 * 10^6的网格中， +// source = [sx, sy]是出发位置，target = [tx, ty]是目标位置 +// 数组blocked是封锁的方格列表， +// blocked[i] = [xi, yi] 表示(xi, yi)的方格是禁止通行的 +// 每次移动都可以走上、下、左、右四个方向 +// 但是来到的位置不能在封锁列表blocked上 +// 同时不允许走出网格 +// 如果从source能到达target返回 true。否则返回false。 +// 测试链接 : https://leetcode.com/problems/escape-a-large-maze/ +// 本题最强的优化是一个剪枝 +public class Code02_EscapeALargeMaze { + + public static long offset = 1000000; + + public boolean isEscapePossible(int[][] blocked, int[] source, int[] target) { + int n = blocked.length; + int maxPoints = n * (n - 1) / 2; + HashSet blockSet = new HashSet<>(); + for (int i = 0; i < n; i++) { + blockSet.add((long) blocked[i][0] * offset + blocked[i][1]); + } + return bfs(source[0], source[1], target[0], target[1], maxPoints, blockSet) + && bfs(target[0], target[1], source[0], source[1], maxPoints, blockSet); + } + + public static boolean bfs(int fromX, int fromY, int toX, int toY, int maxPoints, HashSet blockSet) { + HashSet visited = new HashSet<>(); + Queue queue = new LinkedList<>(); + visited.add((long) fromX * offset + fromY); + queue.add((long) fromX * offset + fromY); + while (!queue.isEmpty() && (visited.size() <= maxPoints)) { + long cur = queue.poll(); + long curX = cur / offset; + long curY = cur - curX * offset; + if (findAndAdd(curX - 1, curY, toX, toY, blockSet, visited, queue) + || findAndAdd(curX + 1, curY, toX, toY, blockSet, visited, queue) + || findAndAdd(curX, curY - 1, toX, toY, blockSet, visited, queue) + || findAndAdd(curX, curY + 1, toX, toY, blockSet, visited, queue)) { + return true; + } + } + return visited.size() > maxPoints; + } + + // 来到的点，(row, col) + // 要寻找的目标点，toX， toY + // HashSet blockSet存着不能走的格子！障碍点！ + // HashSet visited, Queue queue 为了宽度优先遍历服务的！ + // visited，已经处理过的点，请不要重复的放入queue + // 如果已经到达了(toX， toY) + public static boolean findAndAdd(long row, long col, + int toX, int toY, HashSet blockSet, + HashSet visited, Queue queue) { + if (row < 0 || row == offset || col < 0 || col == offset) { + return false; + } + if (row == toX && col == toY) { + return true; + } + long value = row * offset + col; + if (!blockSet.contains(value) && !visited.contains(value)) { + visited.add(value); + queue.add(value); + } + return false; + } + +} diff --git a/算法周更班/class_2022_01_3_week/Code03_ShortestSubarrayWithSumAtLeastK.java b/算法周更班/class_2022_01_3_week/Code03_ShortestSubarrayWithSumAtLeastK.java new file mode 100644 index 0000000..59d12d3 --- /dev/null +++ b/算法周更班/class_2022_01_3_week/Code03_ShortestSubarrayWithSumAtLeastK.java @@ -0,0 +1,74 @@ +package class_2022_01_3_week; + +// 来自字节跳动 +// 给定一个数组arr，其中的值有可能正、负、0 +// 给定一个正数k +// 返回累加和>=k的所有子数组中，最短的子数组长度 +// 本题测试链接 : https://leetcode.com/problems/shortest-subarray-with-sum-at-least-k/ +public class Code03_ShortestSubarrayWithSumAtLeastK { + + public static int shortestSubarray1(int[] arr, int k) { + if (arr == null || arr.length < 1) { + return -1; + } + int n = arr.length + 1; + long[] sum = new long[n]; + for (int i = 1; i < n; i++) { + sum[i] = sum[i - 1] + arr[i - 1]; + } + int[] stack = new int[n]; + int size = 1; + int ans = Integer.MAX_VALUE; + for (int i = 1; i < n; i++) { + int mostRight = mostRight(sum, stack, size, sum[i] - k); + ans = Math.min(ans, mostRight == -1 ? Integer.MAX_VALUE : (i - mostRight)); + while (size > 0 && sum[stack[size - 1]] >= sum[i]) { + size--; + } + stack[size++] = i; + } + return ans == Integer.MAX_VALUE ? -1 : ans; + } + + public static int mostRight(long[] sum, int[] stack, int size, long aim) { + int l = 0; + int r = size - 1; + int m = 0; + int ans = -1; + while (l <= r) { + m = (l + r) / 2; + if (sum[stack[m]] <= aim) { + ans = stack[m]; + l = m + 1; + } else { + r = m - 1; + } + } + return ans; + } + + public static int shortestSubarray2(int[] arr, int K) { + int N = arr.length; + long[] sum = new long[N + 1]; + for (int i = 0; i < N; i++) { + sum[i + 1] = sum[i] + arr[i]; + } + int ans = Integer.MAX_VALUE; + int[] dq = new int[N + 1]; + int l = 0; + int r = 0; + for (int i = 0; i < N + 1; i++) { + // 头部开始，符合条件的，从头部弹出！ + while (l != r && sum[i] - sum[dq[l]] >= K) { + ans = Math.min(ans, i - dq[l++]); + } + // 尾部开始，前缀和比当前的前缀和大于等于的，从尾部弹出！ + while (l != r && sum[dq[r - 1]] >= sum[i]) { + r--; + } + dq[r++] = i; + } + return ans != Integer.MAX_VALUE ? ans : -1; + } + +} diff --git a/算法周更班/class_2022_01_4_week/Code01_BuyThingsAboutCollocation.java b/算法周更班/class_2022_01_4_week/Code01_BuyThingsAboutCollocation.java new file mode 100644 index 0000000..d52ffb3 --- /dev/null +++ b/算法周更班/class_2022_01_4_week/Code01_BuyThingsAboutCollocation.java @@ -0,0 +1,172 @@ +package class_2022_01_4_week; + +// things是一个N*3的二维数组，商品有N件，商品编号从1~N +// 比如things[3] = [300, 2, 6] +// 代表第3号商品：价格300，重要度2，它是6号商品的附属商品 +// 再比如things[6] = [500, 3, 0] +// 代表第6号商品：价格500，重要度3，它不是任何附属，它是主商品 +// 每件商品的收益是价格*重要度，花费就是价格 +// 如果一个商品是附属品，那么只有它附属的主商品购买了，它才能被购买 +// 任何一个附属商品，只会有1个主商品 +// 任何一个主商品的附属商品数量，不会超过2件 +// 主商品和附属商品的层级最多有2层 +// 给定二维数组things、钱数money，返回整体花费不超过money的情况下，最大的收益总和 +// 测试链接 : https://www.nowcoder.com/practice/f9c6f980eeec43ef85be20755ddbeaf4 +// 请把如下的代码的主类名改为"Main", 可以直接通过 +import java.util.ArrayList; +import java.util.Scanner; + +public class Code01_BuyThingsAboutCollocation { + +// // index.....货自由选择！剩余的钱数是rest， +// // 返回：在不花超的情况下，返回最大的价值 +// public static int f(int[] prices, int[] values, int index, int rest) { +// if(rest < 0) { +// return -1; +// } +// // rest >= 0 +// if(index == prices.length) { // 没货了！ +// return 0; +// } +// // rest >=0 有货！ +// // 可能1 ：index.... 自由选择，不要index位置的货！ +// int p1 = f(prices, values, index + 1, rest); +// // 可能性2：index.... 自由选择，要index位置的货！ +// int p2 = -1; +// int next = f(prices, values, index + 1, rest - prices[index]); +// if(next != -1) { +// p2 = values[index] + next; +// } +// return Math.max(p1, p2); +// } +// +// // 商品组 主商品，重要度3，价格9 附件！重要度6，价格7 +// // 0 : [ [3,9], [6,7], [4, 3] ] +// // 1 : [ [4,7] ] 没有附件，只有主商品 +// // 2 : [ [5,100] , [2,1] ] +// +// public static int p(int[][][] matrix, int index, int rest) { +// if(rest < 0) { +// return -1; +// } +// if(index == matrix.length) { +// return 0; +// } +// // 有商品组！也有钱>=0; +// int[][] team = matrix[index]; +// if(team.length == 1) { // 要 、不要 +// int p1 = p(matrix, index + 1, rest); +// int p2 = -1; +// int next = p(matrix, index + 1, rest - team[0][1]); +// if(next != -1) { +// p2 = team[0][0] * team[0][1] + next; +// } +// return Math.max(p1, p2); +// }else if(team.length == 2) { // a b 不要 a ab +// int[] a = team[0]; +// int[] b = team[1]; +// int p1 = p(matrix, index + 1, rest); +// +// int p2 = -1;// 只要a +// int next2 = p(matrix, index + 1, rest - a[1]); +// if(next2 != -1) { +// p2 = a[0] * a[1] + next2; +// } +// +// int p3 = -1;// 要 ab +// int next3= p(matrix, index + 1, rest - a[1] - b[1]); +// if(next3 != -1) { +// p3 = a[0] * a[1] + b[0] * b[1] + next3; +// } +// return Math.max(p1, Math.max(p2, p3)); +// }else { // a : b c 不要 a ab ac abc +// +// } +// +// +// +// } + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + while (sc.hasNext()) { + int money = sc.nextInt(); + int size = sc.nextInt(); + ArrayList> things = new ArrayList<>(); + things.add(new ArrayList<>()); + for (int i = 0; i < size; i++) { + ArrayList cur = new ArrayList<>(); + cur.add(new int[] { sc.nextInt(), sc.nextInt(), sc.nextInt() }); + things.add(cur); + } + int n = clean(things, size); + int ans = maxScore(things, n, money); + System.out.println(ans); + } + sc.close(); + } + + public static int clean(ArrayList> things, int size) { + for (int i = 1; i <= size; i++) { + int[] cur = things.get(i).get(0); + if (cur[2] != 0) { + things.get(i).clear(); + things.get(cur[2]).add(cur); + } + } + int n = 0; + for (int i = 0; i <= size; i++) { + if (!things.get(i).isEmpty()) { + things.set(n++, things.get(i)); + } + } + return n; + } + + public static int maxScore(ArrayList> things, int n, int money) { + int[][] dp = new int[n][money + 1]; + for (int i = 0; i < n; i++) { + for (int j = 0; j <= money; j++) { + dp[i][j] = -2; + } + } + return process(things, n, 0, money, dp); + } + + public static int process(ArrayList> things, int n, int index, int rest, int[][] dp) { + if (rest < 0) { + return -1; + } + if (index == n) { + return 0; + } + if (dp[index][rest] != -2) { + return dp[index][rest]; + } + ArrayList project = things.get(index); + int[] a = project.get(0); + int[] b = project.size() > 1 ? project.get(1) : null; + int[] c = project.size() > 2 ? project.get(2) : null; + int p1 = process(things, n, index + 1, rest, dp); + int p2 = process(things, n, index + 1, rest - a[0], dp); + if (p2 != -1) { + p2 += a[0] * a[1]; + } + int p3 = b != null ? process(things, n, index + 1, rest - a[0] - b[0], dp) : -1; + if (p3 != -1) { + p3 += a[0] * a[1] + b[0] * b[1]; + } + int p4 = c != null ? process(things, n, index + 1, rest - a[0] - c[0], dp) : -1; + if (p4 != -1) { + p4 += a[0] * a[1] + c[0] * c[1]; + } + int p5 = c != null ? process(things, n, index + 1, rest - a[0] - b[0] - c[0], dp) : -1; + if (p5 != -1) { + p5 += a[0] * a[1] + b[0] * b[1] + c[0] * c[1]; + } + int ans = Math.max(Math.max(Math.max(p1, p2), Math.max(p3, p4)), p5); + dp[index][rest] = ans; + return ans; + } + +} diff --git a/算法周更班/class_2022_01_4_week/Code02_SplitToMArraysMinScore.java b/算法周更班/class_2022_01_4_week/Code02_SplitToMArraysMinScore.java new file mode 100644 index 0000000..fd6ac74 --- /dev/null +++ b/算法周更班/class_2022_01_4_week/Code02_SplitToMArraysMinScore.java @@ -0,0 +1,178 @@ +package class_2022_01_4_week; + +import java.util.ArrayList; +import java.util.LinkedList; + +// 来自美团 +// 小团去参加军训，军训快要结束了，长官想要把大家一排n个人分成m组，然后让每组分别去参加阅兵仪式 +// 只能选择相邻的人一组，不能随意改变队伍中人的位置 +// 阅兵仪式上会进行打分，其中有一个奇怪的扣分点是每组的最大差值，即每组最大值减去最小值 +// 长官想要让这分成的m组总扣分量最小，即这m组分别的极差之和最小 +// 长官正在思索如何安排中，就让小团来帮帮他吧 +public class Code02_SplitToMArraysMinScore { + + +// public static int splitArrayMinScore(int[] arr, int k) { +// // arr[L...R] +// int[][] record = creatRecord(arr); +// return f(arr, arr.length - 1, k, record, .); +// } +// +// +// +// public static int[][] creatRecord(int[] arr) { +// +// // arr[L...R] 最大值 - 最小值 是多少？ +// int n = arr.length; +// int[][] ans = new int[n][n]; +// for() { +// for() { +// +// } +// } +// +// // ans[3][6] = arr[3...6] 最大值 - 最小值 是多少？ +// +// } +// +// // arr[0......7] 8 +// // arr[0...3] 4 +// // arr[0..index] index + 1 +// // arr[0....index]的数，一定要分成part组 +// // 返回最小的扣分量 +// public static int f(int[] arr, int index, int part, int[][] record) { +// if (index + 1 <= part) { +// return 0; +// } +// // 数字的个数 > 划分部分的 +// // 0...index part组 +// // 0..7 3份！ +// // 最后一份，最右的一份: 7..7 +// // 最后一份，最右的一份: 6..7 之前 0...5 2份 +// // 最后一份，最右的一份: 5..7 +// // 最后一份，最右的一份: 4..7 +// // ... +// // 最后一份，最右的一份: 0..7 +// int ans = Integer.MAX_VALUE; +// for (int rightTeamFirst = index; rightTeamFirst >= 0; rightTeamFirst--) { +// int rightTeamCost = record[rightTeamFirst][index]; +// int preCost = f(arr, rightTeamFirst - 1, part - 1, record); +// int curAllCost = preCost + rightTeamCost; +// ans = Math.min(ans, curAllCost); +// } +// return ans; +// } + + // 暴力方法 + // 为了验证 + public static int minScore1(int[] arr, int m) { + if (m == 0) { + return 0; + } + return process(arr, 1, m, arr[0], arr[0]); + } + + public static int process(int[] arr, int index, int rest, int preMin, int preMax) { + if (index == arr.length) { + return rest == 1 ? (preMax - preMin) : -1; + } + int p1 = process(arr, index + 1, rest, Math.min(preMin, arr[index]), Math.max(preMax, arr[index])); + int p2next = process(arr, index + 1, rest - 1, arr[index], arr[index]); + int p2 = p2next == -1 ? -1 : (preMax - preMin + p2next); + if (p1 == -1) { + return p2; + } + if (p2 == -1) { + return p1; + } + return Math.min(p1, p2); + } + + public static int score(ArrayList> sets) { + int ans = 0; + for (LinkedList set : sets) { + if (set.isEmpty()) { + return Integer.MAX_VALUE; + } + int max = Integer.MIN_VALUE; + int min = Integer.MAX_VALUE; + for (int num : set) { + max = Math.max(max, num); + min = Math.min(min, num); + } + ans += max - min; + } + return ans; + } + + // 正式方法 + // 时间复杂度O(M * N * N) + // 特别难的一个结论 : 这道题不能利用四边形不等式，我试了 + // 这个特别难的结论不要求必须掌握，因为四边形不等式的技巧非常冷门 + public static int minScore2(int[] arr, int m) { + if (m == 0) { + return 0; + } + int n = arr.length; + int[][] score = new int[n][n]; + for (int i = 0; i < n; i++) { + int max = arr[i]; + int min = arr[i]; + score[i][i] = max - min; + for (int j = i + 1; j < n; j++) { + max = Math.max(max, arr[j]); + min = Math.min(min, arr[j]); + score[i][j] = max - min; + } + } + int[][] dp = new int[m + 1][n]; + for (int i = 0; i < n; i++) { + dp[1][i] = score[0][i]; + } + for (int split = 2; split <= m; split++) { + for (int i = split; i < n; i++) { + dp[split][i] = dp[split - 1][i]; + for (int j = 1; j <= i; j++) { + dp[split][i] = Math.min(dp[split][i], dp[split - 1][j - 1] + score[j][i]); + } + } + } + return dp[m][n - 1]; + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * v); + } + return arr; + } + + public static void main(String[] args) { + int len = 15; + int value = 50; + int testTime = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * len) + 1; + int[] arr = randomArray(n, value); + int m = (int) (Math.random() * n) + 1; + int ans1 = minScore1(arr, m); + int ans2 = minScore2(arr, m); + if (ans1 != ans2) { + System.out.println("出错了!"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println("m : " + m); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_01_4_week/Code03_RandomPickWithBlacklist.java b/算法周更班/class_2022_01_4_week/Code03_RandomPickWithBlacklist.java new file mode 100644 index 0000000..55f6365 --- /dev/null +++ b/算法周更班/class_2022_01_4_week/Code03_RandomPickWithBlacklist.java @@ -0,0 +1,52 @@ +package class_2022_01_4_week; + +import java.util.Arrays; +import java.util.HashMap; + +// 给定一个包含 [0，n) 中不重复整数的黑名单 blacklist +// 写一个函数从 [0, n) 中返回一个不在 blacklist 中的随机整数 +// 对它进行优化使其尽量少调用系统方法 Math.random() +// 1 <= n <= 1000000000 +// 0 <= blacklist.length < min(100000, N) +// 测试链接: https://leetcode.com/problems/random-pick-with-blacklist/ +public class Code03_RandomPickWithBlacklist { + + class Solution { + + // 0~99 所有的数字都可以随机，可能有若干黑名单！ + // 填到洞里去，size -> 0~size-1 是最后调整的下标范围，紧实的！ + private int size; + + // 13 -> 99 + // 27 -> 98 + private HashMap convert = new HashMap<>(); + + public Solution(int n, int[] blackArray) { + Arrays.sort(blackArray); + int m = blackArray.length; + // [34, 56, 78, 103, ..., 980] + // 0 1 2 3 100 + for (int i = 0; i < m && blackArray[i] < n; i++) { + for (n--; n > blackArray[i]; n--) { + if (n == blackArray[m - 1]) { + m--; + } else { + convert.put(blackArray[i], n); + break; + } + } + } + size = n; + } + + public int pick() { + int ans = (int) (Math.random() * size); + // 13 -> 99 + // 17 -> 98 + // 35 -> 35 + return convert.containsKey(ans) ? convert.get(ans) : ans; + } + + } + +} \ No newline at end of file diff --git a/算法周更班/class_2022_01_4_week/Code04_BattleshipsInABoard.java b/算法周更班/class_2022_01_4_week/Code04_BattleshipsInABoard.java new file mode 100644 index 0000000..3708220 --- /dev/null +++ b/算法周更班/class_2022_01_4_week/Code04_BattleshipsInABoard.java @@ -0,0 +1,21 @@ +package class_2022_01_4_week; + +// 来自米哈游 +// 测试链接 : https://leetcode.com/problems/battleships-in-a-board/ +public class Code04_BattleshipsInABoard { + + public static int countBattleships(char[][] m) { + int ans = 0; + for (int i = 0; i < m.length; i++) { + for (int j = 0; j < m[0].length; j++) { + if ((m[i][j] == 'X') + && (i == 0 || m[i - 1][j] != 'X') + && (j == 0 || m[i][j - 1] != 'X')) { + ans++; + } + } + } + return ans; + } + +} diff --git a/算法周更班/class_2022_02_2_week/Code01_24Game.java b/算法周更班/class_2022_02_2_week/Code01_24Game.java new file mode 100644 index 0000000..af75442 --- /dev/null +++ b/算法周更班/class_2022_02_2_week/Code01_24Game.java @@ -0,0 +1,97 @@ +package class_2022_02_2_week; + +// 测试链接 : https://leetcode.com/problems/24-game/ +public class Code01_24Game { + + public static boolean judgePoint24(int[] cards) { + if (cards == null || cards.length == 0) { + return false; + } + int n = cards.length; + Number[] arr = new Number[n]; + for (int i = 0; i < n; i++) { + arr[i] = new Number(cards[i], 1); + } + return judge(arr, cards.length); + } + + // arr中，有效的范围arr[0...size-1] ... 再往右，都无效了，不用看了！ + public static boolean judge(Number[] arr, int size) { + if (size == 1) { + return arr[0].numerator == 24 && arr[0].denominator == 1; + } + for (int i = 0; i < size; i++) { + for (int j = i + 1; j < size; j++) { + Number inum = arr[i]; + Number jnum = arr[j]; + arr[j] = arr[size - 1]; + arr[i] = add(inum, jnum); + if (judge(arr, size - 1)) { + return true; + } + arr[i] = minus(inum, jnum); + if (judge(arr, size - 1)) { + return true; + } + arr[i] = minus(jnum, inum); + if (judge(arr, size - 1)) { + return true; + } + arr[i] = multiply(inum, jnum); + if (judge(arr, size - 1)) { + return true; + } + arr[i] = divide(inum, jnum); + if (arr[i] != null && judge(arr, size - 1)) { + return true; + } + arr[i] = divide(jnum, inum); + if (arr[i] != null && judge(arr, size - 1)) { + return true; + } + arr[i] = inum; + arr[j] = jnum; + } + } + return false; + } + + public static class Number { + public int numerator; + public int denominator; + + public Number(int n, int d) { + numerator = n; + denominator = d; + } + } + + public static Number add(Number a, Number b) { + return simple(a.numerator * b.denominator + b.numerator * a.denominator, a.denominator * b.denominator); + } + + public static Number minus(Number a, Number b) { + return simple(a.numerator * b.denominator - b.numerator * a.denominator, a.denominator * b.denominator); + } + + public static Number multiply(Number a, Number b) { + return simple(a.numerator * b.numerator, a.denominator * b.denominator); + } + + public static Number divide(Number a, Number b) { + return b.numerator == 0 ? null : simple(a.numerator * b.denominator, a.denominator * b.numerator); + } + + public static Number simple(int up, int down) { + if (up == 0) { + return new Number(0, 1); + } + int gcd = Math.abs(gcd(up, down)); + return new Number(up / gcd, down / gcd); + } + + public static int gcd(int a, int b) { + return b == 0 ? a : gcd(b, a % b); + } + +} diff --git a/算法周更班/class_2022_02_2_week/Code02_DesignBitset.java b/算法周更班/class_2022_02_2_week/Code02_DesignBitset.java new file mode 100644 index 0000000..f5842f9 --- /dev/null +++ b/算法周更班/class_2022_02_2_week/Code02_DesignBitset.java @@ -0,0 +1,95 @@ +package class_2022_02_2_week; + +// 测试链接 : https://leetcode-cn.com/problems/design-bitset/ +public class Code02_DesignBitset { + + class Bitset { + + // size int 32 + // size / 32 向上取整 + private int[] bits; + private final int size; + private int zeros; + private int ones; + private boolean reverse; + + public Bitset(int n) { + bits = new int[(n + 31) / 32]; + size = n; + zeros = n; + ones = 0; + reverse = false; + } + + // 把idx位置的状态，如果是0，变成1；如果是1，没有变化！ + public void fix(int idx) { + int index = idx / 32; + int bit = idx % 32; + if (!reverse) { + if ((bits[index] & (1 << bit)) == 0) { + zeros--; + ones++; + bits[index] |= (1 << bit); + } + } else { + if ((bits[index] & (1 << bit)) != 0) { + zeros--; + ones++; + bits[index] ^= (1 << bit); + } + } + } + + // 1 > 0 + public void unfix(int idx) { + int index = idx / 32; + int bit = idx % 32; + if (!reverse) { + if ((bits[index] & (1 << bit)) != 0) { + ones--; + zeros++; + bits[index] ^= (1 << bit); + } + } else { + if ((bits[index] & (1 << bit)) == 0) { + ones--; + zeros++; + bits[index] |= (1 << bit); + } + } + } + + // 0变1，1变0 + public void flip() { + reverse = !reverse; + int tmp = zeros; + zeros = ones; + ones = tmp; + } + + // 是不是所有的位都是1 + public boolean all() { + return ones == size; + } + + // 是不是至少有1位是1 + public boolean one() { + return ones > 0; + } + + // 返回1的数量！ + public int count() { + return ones; + } + + public String toString() { + StringBuilder builder = new StringBuilder(); + for (int i = 0; i < size; i++) { + int status = bits[i / 32] & (1 << (i % 32)); + builder.append(reverse ? (status == 0 ? '1' : '0') : (status == 0 ? '0' : '1')); + } + return builder.toString(); + } + } + +} diff --git a/算法周更班/class_2022_02_2_week/Code03_FindKthSmallestPairDistance.java b/算法周更班/class_2022_02_2_week/Code03_FindKthSmallestPairDistance.java new file mode 100644 index 0000000..225af64 --- /dev/null +++ b/算法周更班/class_2022_02_2_week/Code03_FindKthSmallestPairDistance.java @@ -0,0 +1,64 @@ +package class_2022_02_2_week; + +import java.util.Arrays; + +// 测试链接 : https://leetcode.com/problems/find-k-th-smallest-pair-distance/ +public class Code03_FindKthSmallestPairDistance { + +// public static int smallestDistancePair(int[] nums, int k) { +// int n = nums.length; +// Arrays.sort(nums); +// int l = 0; +// int r = nums[n - 1] - nums[0]; +// int ans = 0; +// while (l <= r) { +// int cnt = 0; +// int m = l + ((r - l) >> 1); +// for (int i = 0, j = 0; i < n; i++) { +// while (j < n && nums[j] <= nums[i] + m) { +// j++; +// } +// cnt += j - i - 1; +// } +// if (cnt >= k) { +// ans = m; +// r = m - 1; +// } else { +// l = m + 1; +// } +// } +// return ans; +// } + + public static int smallestDistancePair(int[] nums, int k) { + int n = nums.length; + Arrays.sort(nums); + int l = 0; + int r = nums[n - 1] - nums[0]; + int ans = 0; + while (l <= r) { + int dis = l + ((r - l) >> 1); + int cnt = f(nums, dis); + if (cnt >= k) { + ans = dis; + r = dis - 1; + } else { + l = dis + 1; + } + } + return ans; + } + + // <= dis的数字对，有几个，返回 + public static int f(int[] arr, int dis) { + int cnt = 0; + for (int l = 0, r = 0; l < arr.length; l++) { + while (r < arr.length && arr[r] <= arr[l] + dis) { + r++; + } + cnt += r - l - 1; + } + return cnt; + } + +} diff --git a/算法周更班/class_2022_02_2_week/Code04_ReachingPoints.java b/算法周更班/class_2022_02_2_week/Code04_ReachingPoints.java new file mode 100644 index 0000000..b4f2e39 --- /dev/null +++ b/算法周更班/class_2022_02_2_week/Code04_ReachingPoints.java @@ -0,0 +1,38 @@ +package class_2022_02_2_week; + +// 测试链接 : https://leetcode.com/problems/reaching-points/ +public class Code04_ReachingPoints { + + // 会超时，但是揭示了大思路 + public static boolean reachingPoints1(int sx, int sy, int tx, int ty) { + while (tx != ty) { + if (tx < ty) { + ty -= tx; + } else { + tx -= ty; + } + if (sx == tx && sy == ty) { + return true; + } + } + return false; + } + + // 对大体思路的优化 + // s ( 5, 10) + // t (100, 65) + public static boolean reachingPoints2(int sx, int sy, int tx, int ty) { + while (sx < tx && sy < ty) { + if (tx < ty) { + ty %= tx; + } else { + tx %= ty; + } + } + // 1) startx >= tx + // 2) starty >= ty + return (sx == tx && sy <= ty && (ty - sy) % sx == 0) + || (sy == ty && sx <= tx && (tx - sx) % sy == 0); + } + +} diff --git a/算法周更班/class_2022_02_2_week/Code05_RecoverTheOriginalArray.java b/算法周更班/class_2022_02_2_week/Code05_RecoverTheOriginalArray.java new file mode 100644 index 0000000..43c665b --- /dev/null +++ b/算法周更班/class_2022_02_2_week/Code05_RecoverTheOriginalArray.java @@ -0,0 +1,48 @@ +package class_2022_02_2_week; + +import java.util.Arrays; + +// 测试链接 : https://leetcode.com/problems/recover-the-original-array/ +public class Code05_RecoverTheOriginalArray { + + public static int[] recoverArray(int[] nums) { + Arrays.sort(nums); + int n = nums.length; + // nums[0] -> 小数组的第0个 + int m = n >> 1; + // 谁是大数组的第0个？不知道，试！first位置的数！ + for (int first = 1; first <= m; first++) { + // d = 2 * k; k正数！ + int d = nums[first] - nums[0]; + if (d > 0 && (d & 1) == 0) { + // 试图生成原始数组！ans！ + int[] ans = new int[m]; + int i = 0; + boolean[] set = new boolean[n]; + int k = d >> 1; + int l = 0; + int r = first; + while (r < n) { + while (set[l]) { + l++; + } + if (l == r) { + r++; + } else if (nums[r] - nums[l] > d) { + break; + } else if (nums[r] - nums[l] < d) { + r++; + } else { + set[r++] = true; + ans[i++] = nums[l++] + k; + } + } + if (i == m) { + return ans; + } + } + } + return null; + } + +} diff --git a/算法周更班/class_2022_02_3_week/Code01_CheapestFlightsWithinKStops.java b/算法周更班/class_2022_02_3_week/Code01_CheapestFlightsWithinKStops.java new file mode 100644 index 0000000..5833260 --- /dev/null +++ b/算法周更班/class_2022_02_3_week/Code01_CheapestFlightsWithinKStops.java @@ -0,0 +1,75 @@ +package class_2022_02_3_week; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.HashMap; +import java.util.Map.Entry; + +// 测试链接 : https://leetcode.com/problems/cheapest-flights-within-k-stops/ +public class Code01_CheapestFlightsWithinKStops { + + // 类似宽度优先遍历 + // n，城市数量，0...n-1 + // flights = { {0,7,100}, {15,3,300} .... } + // src -> 源点 + // dst -> 目标点 + // k -> 最多能中转几次 + public static int findCheapestPrice1(int n, int[][] flights, int src, int dst, int k) { + + // 图 + // 0 : {1, 500} {7, 20} {13, 70} + ArrayList> graph = new ArrayList<>(); + for (int i = 0; i < n; i++) { + graph.add(new ArrayList()); + } + for (int[] line : flights) { + // [0, 7 , 200] + // 0 : { {7, 200} } + graph.get(line[0]).add(new int[] { line[1], line[2] }); + } + // distance[i] : 从源点src出发的情况下，到达i号点，最少费用是多少 + int[] cost = new int[n]; + Arrays.fill(cost, Integer.MAX_VALUE); + cost[src] = 0; + HashMap curMap = new HashMap<>(); + curMap.put(src, 0); + for (int i = 0; i <= k; i++) { + HashMap nextMap = new HashMap<>(); + for (Entry entry : curMap.entrySet()) { + // A -> from 100 + // from -> ???? + int from = entry.getKey(); + int preCost = entry.getValue(); + for (int[] line : graph.get(from)) { + int to = line[0]; + int curCost = line[1]; + cost[to] = Math.min(cost[to], preCost + curCost); + nextMap.put(to, + Math.min(nextMap.getOrDefault(to, Integer.MAX_VALUE), + preCost + curCost)); + } + } + curMap = nextMap; + } + return cost[dst] == Integer.MAX_VALUE ? -1 : cost[dst]; + } + + // Bellman Ford + // 快！ + public static int findCheapestPrice2(int n, int[][] flights, int src, int dst, int k) { + int[] cost = new int[n]; + Arrays.fill(cost, Integer.MAX_VALUE); + cost[src] = 0; + for (int i = 0; i <= k; i++) { + int[] next = Arrays.copyOf(cost, n); + for (int[] f : flights) { + if (cost[f[0]] != Integer.MAX_VALUE) { + next[f[1]] = Math.min(next[f[1]], cost[f[0]] + f[2]); + } + } + cost = next; + } + return cost[dst] == Integer.MAX_VALUE ? -1 : cost[dst]; + } + +} diff --git a/算法周更班/class_2022_02_3_week/Code02_MinimumNumberOfDaysToEatNOranges.java b/算法周更班/class_2022_02_3_week/Code02_MinimumNumberOfDaysToEatNOranges.java new file mode 100644 index 0000000..ff78efd --- /dev/null +++ b/算法周更班/class_2022_02_3_week/Code02_MinimumNumberOfDaysToEatNOranges.java @@ -0,0 +1,35 @@ +package class_2022_02_3_week; + +import java.util.HashMap; + +// 来自学员看到的腾讯面试 +// 测试链接 : https://leetcode.com/problems/minimum-number-of-days-to-eat-n-oranges/ +public class Code02_MinimumNumberOfDaysToEatNOranges { + + // 所有的答案都填在这个表里 + // 这个表对所有的过程共用 + public static HashMap dp = new HashMap<>(); + + public static int minDays(int n) { + if (n <= 1) { + return n; + } + if (dp.containsKey(n)) { + return dp.get(n); + } + // 1) 吃掉一个橘子 + // 2) 如果n能被2整除，吃掉一半的橘子，剩下一半 + // 3) 如果n能被3正数，吃掉三分之二的橘子，剩下三分之一 + // 因为方法2）和3），是按比例吃橘子，所以必然会非常快 + // 所以，决策如下： + // 可能性1：为了使用2）方法，先把橘子吃成2的整数倍，然后直接干掉一半，剩下的n/2调用递归 + // 即，n % 2 + 1 + minDays(n/2) + // 可能性2：为了使用3）方法，先把橘子吃成3的整数倍，然后直接干掉三分之二，剩下的n/3调用递归 + // 即，n % 3 + 1 + minDays(n/3) + // 至于方法1)，完全是为了这两种可能性服务的，因为能按比例吃，肯定比一个一个吃快(显而易见的贪心) + int ans = Math.min(n % 2 + 1 + minDays(n / 2), n % 3 + 1 + minDays(n / 3)); + dp.put(n, ans); + return ans; + } + +} diff --git a/算法周更班/class_2022_02_3_week/Code03_RobotBoundedInCircle.java b/算法周更班/class_2022_02_3_week/Code03_RobotBoundedInCircle.java new file mode 100644 index 0000000..b13dd5e --- /dev/null +++ b/算法周更班/class_2022_02_3_week/Code03_RobotBoundedInCircle.java @@ -0,0 +1,40 @@ +package class_2022_02_3_week; + +// 测试链接 : https://leetcode.com/problems/robot-bounded-in-circle/ +public class Code03_RobotBoundedInCircle { + + public static boolean isRobotBounded(String ins) { + int row = 0; + int col = 0; + int direction = 0; // 0 1 2 3 + char[] str = ins.toCharArray(); + for (char cur : str) { + if (cur == 'R') { + direction = right(direction); + } else if (cur == 'L') { + direction = left(direction); + } else { + row = row(direction, row); + col = col(direction, col); + } + } + return row == 0 && col == 0 || direction != 0; + } + + public static int left(int direction) { + return direction == 0 ? 3 : (direction - 1); + } + + public static int right(int direction) { + return direction == 3 ? 0 : (direction + 1); + } + + public static int row(int direction, int r) { + return (direction == 1 || direction == 3) ? r : (r + (direction == 0 ? 1 : -1)); + } + + public static int col(int direction, int c) { + return (direction == 0 || direction == 2) ? c : (c + (direction == 1 ? 1 : -1)); + } + +} diff --git a/算法周更班/class_2022_02_3_week/Code04_MaxTeamNumber.java b/算法周更班/class_2022_02_3_week/Code04_MaxTeamNumber.java new file mode 100644 index 0000000..75a5ecc --- /dev/null +++ b/算法周更班/class_2022_02_3_week/Code04_MaxTeamNumber.java @@ -0,0 +1,94 @@ +package class_2022_02_3_week; + +import java.util.Arrays; + +// 来自微软 +// 给定一个数组arr，一个正数num，一个正数k +// 可以把arr中的某些数字拿出来组成一组，要求该组中的最大值减去最小值<=num +// 且该组数字的个数一定要正好等于k +// 每个数字只能选择进某一组，不能进多个组 +// 返回arr中最多有多少组 +public class Code04_MaxTeamNumber { + + // 对数器方法 + public static int maxTeams1(int[] arr, int num, int k) { + Arrays.sort(arr); + return process1(arr, 0, new int[arr.length], 0, num, k); + } + + public static int process1(int[] arr, int index, int[] path, int size, int num, int k) { + if (index == arr.length) { + if (size % k != 0) { + return 0; + } else { + for (int start = 0; start < size; start += k) { + if (path[start + k - 1] - path[start] > num) { + return 0; + } + } + return size / k; + } + } else { + int p1 = process1(arr, index + 1, path, size, num, k); + path[size] = arr[index]; + int p2 = process1(arr, index + 1, path, size + 1, num, k); + return Math.max(p1, p2); + } + } + + // 正式方法 + // 时间复杂度O(N * logN) + public static int maxTeams2(int[] arr, int num, int k) { + int n = arr.length; + if (k > n) { + return 0; + } + Arrays.sort(arr); + int[] dp = new int[n]; + dp[k - 1] = arr[k - 1] - arr[0] <= num ? 1 : 0; + for (int i = k; i < n; i++) { + int p1 = dp[i - 1]; + int p2 = (arr[i] - arr[i - k + 1] <= num ? 1 : 0) + dp[i - k]; + dp[i] = Math.max(p1, p2); + } + return dp[n - 1]; + } + + // 为了测试 + public static int[] randomArray(int len, int value) { + int[] ans = new int[len]; + for (int i = 0; i < len; i++) { + ans[i] = (int) (Math.random() * value); + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int n = 18; + int v = 50; + int testTimes = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + int len = (int) (Math.random() * n) + 1; + int[] arr = randomArray(len, v); + int num = (int) (Math.random() * v) + 1; + int k = (int) (Math.random() * len) + 1; + int ans1 = maxTeams1(arr, num, k); + int ans2 = maxTeams2(arr, num, k); + if (ans1 != ans2) { + for (int number : arr) { + System.out.print(number + " "); + } + System.out.println(); + System.out.println("num : " + num); + System.out.println("k : " + k); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_02_3_week/Code05_StoneGameIX.java b/算法周更班/class_2022_02_3_week/Code05_StoneGameIX.java new file mode 100644 index 0000000..86015f7 --- /dev/null +++ b/算法周更班/class_2022_02_3_week/Code05_StoneGameIX.java @@ -0,0 +1,16 @@ +package class_2022_02_3_week; + +// 测试链接 : https://leetcode.com/problems/stone-game-ix/ +public class Code05_StoneGameIX { + + public static boolean stoneGameIX(int[] stones) { + int[] counts = new int[3]; + for (int num : stones) { + counts[num % 3]++; + } + return counts[0] % 2 == 0 + ? counts[1] != 0 && counts[2] != 0 + : Math.abs(counts[1] - counts[2]) > 2; + } + +} diff --git a/算法周更班/class_2022_02_4_week/Code01_SplitSameNumberWays.java b/算法周更班/class_2022_02_4_week/Code01_SplitSameNumberWays.java new file mode 100644 index 0000000..122d3ad --- /dev/null +++ b/算法周更班/class_2022_02_4_week/Code01_SplitSameNumberWays.java @@ -0,0 +1,36 @@ +package class_2022_02_4_week; + +// 来自微软 +// 比如，str = "ayxbx" +// 有以下4种切法 : a | yxbx、ay | xbx、ayx | bx、ayxb | x +// 其中第1、3、4种切法符合：x和y的个数，至少在左右两块中的一块里有相同的数量 +// 所以返回3 +// 给定一个字符串str，长度为N +// 你有N-1种划分方法，把str切成左右两半，返回有几种切法满足： +// x和y的个数，至少在左右两块中的一块里有相同的数量 +public class Code01_SplitSameNumberWays { + + public static int splitSameNumberWays(char[] str) { + if (str == null || str.length == 0) { + return 0; + } + int xAll = 0; + int yAll = 0; + for (char c : str) { + xAll += c == 'x' ? 1 : 0; + yAll += c == 'y' ? 1 : 0; + } + int leftX = str[0] == 'x' ? 1 : 0; + int leftY = str[0] == 'y' ? 1 : 0; + int ans = 0; + for (int i = 1; i < str.length; i++) { + if (leftX == leftY || (xAll - leftX) == (yAll - leftY)) { + ans++; + } + leftX += str[i] == 'x' ? 1 : 0; + leftY += str[i] == 'y' ? 1 : 0; + } + return ans; + } + +} diff --git a/算法周更班/class_2022_02_4_week/Code02_NearBiggerNoSameNeighbour.java b/算法周更班/class_2022_02_4_week/Code02_NearBiggerNoSameNeighbour.java new file mode 100644 index 0000000..5ad0e47 --- /dev/null +++ b/算法周更班/class_2022_02_4_week/Code02_NearBiggerNoSameNeighbour.java @@ -0,0 +1,69 @@ +package class_2022_02_4_week; + +// 来自微软 +// 给定一个正数num，要返回一个大于num的数，并且每一位和相邻位的数字不能相等 +// 返回达标的数字中，最小的那个 +// 10^9 +public class Code02_NearBiggerNoSameNeighbour { + + public static int near(int num) { + // num = 174 + // "0175" + // num = 899 + // "0900" + // num = 999 + // "01000" + char[] raw = ("0" + String.valueOf(num + 1)).toCharArray(); + process(raw); + return Integer.valueOf(String.valueOf(raw)); + } + + public static void process(char[] raw) { + for (int i = 1; i < raw.length; i++) { + if (raw[i - 1] == raw[i]) { + addOne(raw, i); + for (int j = i + 1; j < raw.length; j++) { + raw[j] = '0'; + } + process(raw); + return; + } + } + } + + // 99..... + // +1 + //100 + public static void addOne(char[] r, int i) { + boolean up = true; + while (up && r[i] == '9') { + r[i--] = '0'; + } + r[i]++; + } + + public static void main(String[] args) { + char[] test = new char[] { '0', '1', '2', '3' }; + + System.out.println(Integer.valueOf(String.valueOf(test))); + + int num1 = 55; + System.out.println(near(num1)); + + int num2 = 1765; + System.out.println(near(num2)); + + int num3 = 98; + System.out.println(near(num3)); + + int num4 = 44432; + System.out.println(near(num4)); + + int num5 = 3298; + System.out.println(near(num5)); + + int num6 = 9999998; + System.out.println(near(num6)); + } + +} diff --git a/算法周更班/class_2022_02_4_week/Code03_PartitionArrayForMaximumSum.java b/算法周更班/class_2022_02_4_week/Code03_PartitionArrayForMaximumSum.java new file mode 100644 index 0000000..43a70ac --- /dev/null +++ b/算法周更班/class_2022_02_4_week/Code03_PartitionArrayForMaximumSum.java @@ -0,0 +1,53 @@ +package class_2022_02_4_week; + +import java.util.Arrays; + +// 测试链接 : https://leetcode.com/problems/partition-array-for-maximum-sum/ +public class Code03_PartitionArrayForMaximumSum { + + public static int maxSumAfterPartitioning1(int[] arr, int k) { + int[] dp = new int[arr.length]; + Arrays.fill(dp, -1); + return process1(arr, k, arr.length - 1, dp); + } + + // 永远不变的固定参数 : arr, k + // 可变参数 : index + // 缓存 : dp + // process含义 : arr[0...index]最优划分搞出的最大和是多少，返回 + public static int process1(int[] arr, int k, int index, int[] dp) { + if (index == -1) { + return 0; + } + if (dp[index] != -1) { + return dp[index]; + } + int max = Integer.MIN_VALUE; + int ans = Integer.MIN_VALUE; + for (int i = index, j = 1; i >= 0 && j <= k; i--, j++) { + max = Math.max(max, arr[i]); + ans = Math.max(ans, process1(arr, k, i - 1, dp) + (index - i + 1) * max); + } + dp[index] = ans; + return ans; + } + + public static int maxSumAfterPartitioning2(int[] arr, int k) { + if (arr == null || arr.length == 0) { + return 0; + } + int n = arr.length; + int[] dp = new int[n]; + dp[0] = arr[0]; + for (int i = 1; i < n; i++) { + dp[i] = arr[i] + dp[i - 1]; + int max = arr[i]; + for (int j = i - 1; j >= 0 && (i - j + 1) <= k; j--) { + max = Math.max(max, arr[j]); + dp[i] = Math.max(dp[i], max * (i - j + 1) + (j - 1 >= 0 ? dp[j - 1] : 0)); + } + } + return dp[n - 1]; + } + +} diff --git a/算法周更班/class_2022_02_4_week/Code04_NumberOfDescendingTriples.java b/算法周更班/class_2022_02_4_week/Code04_NumberOfDescendingTriples.java new file mode 100644 index 0000000..18f929c --- /dev/null +++ b/算法周更班/class_2022_02_4_week/Code04_NumberOfDescendingTriples.java @@ -0,0 +1,134 @@ +package class_2022_02_4_week; + +import java.util.Arrays; + +// 返回一个数组中，所有降序三元组的数量 +// 比如 : {5, 3, 4, 2, 1} +// 所有降序三元组为 : +// {5, 3, 2}、{5, 3, 1}、{5, 4, 2}、{5, 4, 1} +// {5, 2, 1}、{3, 2, 1}、{4, 2, 1} +// 所以返回数量7 +public class Code04_NumberOfDescendingTriples { + + // 暴力方法 + // 对数器 + public static int num1(int[] arr) { + if (arr == null || arr.length < 3) { + return 0; + } + return process(arr, 0, new int[3], 0); + } + + public static int process(int[] arr, int index, int[] path, int size) { + if (size == 3) { + return path[0] > path[1] && path[1] > path[2] ? 1 : 0; + } + int ans = 0; + if (index < arr.length) { + ans = process(arr, index + 1, path, size); + path[size] = arr[index]; + ans += process(arr, index + 1, path, size + 1); + } + return ans; + } + + // 正式方法 + // 时间复杂度O(N * logN) + // 利用index tree + public static int num2(int[] arr) { + if (arr == null || arr.length < 3) { + return 0; + } + int n = arr.length; + int[] sorted = Arrays.copyOf(arr, n); + Arrays.sort(sorted); + int max = Integer.MIN_VALUE; + for (int i = 0; i < n; i++) { + arr[i] = rank(sorted, arr[i]); + max = Math.max(max, arr[i]); + } + IndexTree it2 = new IndexTree(max); + IndexTree it3 = new IndexTree(max); + int ans = 0; + for (int i = n - 1; i >= 0; i--) { + ans += arr[i] == 1 ? 0 : it3.sum(arr[i] - 1); + it2.add(arr[i], 1); + it3.add(arr[i], arr[i] == 1 ? 0 : it2.sum(arr[i] - 1)); + } + return ans; + } + + // 返回>=num, 最左位置 + public static int rank(int[] sorted, int num) { + int l = 0; + int r = sorted.length - 1; + int m = 0; + int ans = 0; + while (l <= r) { + m = (l + r) / 2; + if (sorted[m] >= num) { + ans = m; + r = m - 1; + } else { + l = m + 1; + } + } + return ans + 1; + } + + // 下标从1开始 + public static class IndexTree { + + private int[] tree; + private int N; + + // 0位置弃而不用 + public IndexTree(int size) { + N = size; + tree = new int[N + 1]; + } + + // 1~index 累加和是多少？ + public int sum(int index) { + int ret = 0; + while (index > 0) { + ret += tree[index]; + index -= index & -index; + } + return ret; + } + + public void add(int index, int d) { + while (index <= N) { + tree[index] += d; + index += index & -index; + } + } + } + + // 为了测试 + public static int[] randomArray(int len, int value) { + int[] arr = new int[len]; + for (int i = 0; i < arr.length; i++) { + arr[i] = (int) (Math.random() * value) - (int) (Math.random() * value); + } + return arr; + } + + public static void main(String[] args) { + int len = 20; + int value = 100; + int testTimes = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + int[] arr = randomArray(len, value); + int ans1 = num1(arr); + int ans2 = num2(arr); + if (ans1 != ans2) { + System.out.println("出错了!"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_02_4_week/Code05_GroupsOfStrings.java b/算法周更班/class_2022_02_4_week/Code05_GroupsOfStrings.java new file mode 100644 index 0000000..9a77b32 --- /dev/null +++ b/算法周更班/class_2022_02_4_week/Code05_GroupsOfStrings.java @@ -0,0 +1,195 @@ +package class_2022_02_4_week; + +import java.util.HashMap; + +// 测试链接 : https://leetcode.com/problems/groups-of-strings/ +public class Code05_GroupsOfStrings { + + // 可能会超时，或者打败比例很低 + // 因为常数优化不到位 + public static int[] groupStrings1(String[] words) { + int n = words.length; + // 0 1 2 ... n-1 + UnionFind uf = new UnionFind(n); + int[] strs = new int[n]; + // abd -> 0..01011 7 + // 0..01011 key value 7 + HashMap stands = new HashMap<>(); + for (int i = 0; i < n; i++) { + int status = 0; + for (char c : words[i].toCharArray()) { + status |= 1 << (c - 'a'); + } + strs[i] = status; + if (stands.containsKey(status)) { + uf.union(stands.get(status), i); + } else { + stands.put(status, i); + } + } + for (int i = 0; i < n; i++) { + // 一个字符串，状态 + int status = strs[i]; + for (int j = 0; j < 26; j++) { + // 001101 + // a + // 001101 + // b + // 001111 + // c + // 001101 + // z + //.. + uf.union(i, stands.get(status | (1 << j))); + } + // 有的字符，减少一遍 + for (int j = 0; j < 26; j++) { + if ((status & (1 << j)) != 0) { + uf.union(i, stands.get(status ^ (1 << j))); + } + } + for (int has = 0; has < 26; has++) { + if ((status & (1 << has)) != 0) { + status ^= 1 << has; + for (int replace = 0; replace < 26; replace++) { + uf.union(i, stands.get(status | (1 << replace))); + } + status |= 1 << has; + } + } + } + return new int[] { uf.sets(), uf.maxSize() }; + } + + // 肯定通过 + // 打败比例达标 + // 优化了常数时间 + public static int[] groupStrings2(String[] words) { + int n = words.length; + UnionFind uf = new UnionFind(n); + int[] strs = new int[n]; + HashMap stands = new HashMap<>(); + for (int i = 0; i < n; i++) { + int status = 0; + for (char c : words[i].toCharArray()) { + status |= 1 << (c - 'a'); + } + strs[i] = status; + if (stands.containsKey(status)) { + uf.union(stands.get(status), i); + } else { + stands.put(status, i); + } + } + for (int i = 0; i < n; i++) { + int yes = strs[i]; + int no = (~yes) & ((1 << 26) - 1); + int tmpYes = yes; + int tmpNo = no; + int rightOneYes = 0; + int rightOneNo = 0; + + + // 0....0 0110011 + // + // 0....0 0110011 + // 0....0 0000001 -> 用 + + // 0....0 0110010 + // 0....0 0000010 -> 用 + + // 0....0 0110000 + + while (tmpYes != 0) { + rightOneYes = tmpYes & (-tmpYes); + uf.union(i, stands.get(yes ^ rightOneYes)); + tmpYes ^= rightOneYes; + } + + + + + + // tmpNo = 该去试试什么添加！ + while(tmpNo != 0) { + rightOneNo = tmpNo & (-tmpNo); + uf.union(i, stands.get(yes | rightOneNo)); + tmpNo ^= rightOneNo; + } + tmpYes = yes; + while (tmpYes != 0) { + rightOneYes = tmpYes & (-tmpYes); + tmpNo = no; + while (tmpNo != 0) { + rightOneNo = tmpNo & (-tmpNo); + uf.union(i, stands.get((yes ^ rightOneYes) | rightOneNo)); + tmpNo ^= rightOneNo; + } + tmpYes ^= rightOneYes; + } + } + return new int[] { uf.sets(), uf.maxSize() }; + } + + public static class UnionFind { + private int[] parent; + private int[] size; + private int[] help; + + public UnionFind(int N) { + parent = new int[N]; + size = new int[N]; + help = new int[N]; + for (int i = 0; i < N; i++) { + parent[i] = i; + size[i] = 1; + } + } + + private int find(int i) { + int hi = 0; + while (i != parent[i]) { + help[hi++] = i; + i = parent[i]; + } + for (hi--; hi >= 0; hi--) { + parent[help[hi]] = i; + } + return i; + } + + public void union(Integer i, Integer j) { + if (i == null || j == null) { + return; + } + int f1 = find(i); + int f2 = find(j); + if (f1 != f2) { + if (size[f1] >= size[f2]) { + size[f1] += size[f2]; + parent[f2] = f1; + } else { + size[f2] += size[f1]; + parent[f1] = f2; + } + } + } + + public int sets() { + int ans = 0; + for (int i = 0; i < parent.length; i++) { + ans += parent[i] == i ? 1 : 0; + } + return ans; + } + + public int maxSize() { + int ans = 0; + for (int s : size) { + ans = Math.max(ans, s); + } + return ans; + } + } + +} diff --git a/算法周更班/class_2022_03_1_week/Code01_StronglyConnectedComponents.java b/算法周更班/class_2022_03_1_week/Code01_StronglyConnectedComponents.java new file mode 100644 index 0000000..a5886d4 --- /dev/null +++ b/算法周更班/class_2022_03_1_week/Code01_StronglyConnectedComponents.java @@ -0,0 +1,105 @@ +package class_2022_03_1_week; + +import java.util.ArrayList; + +// tarjan算法求有向图的强连通分量 +public class Code01_StronglyConnectedComponents { + + public static class StronglyConnectedComponents { + public ArrayList> nexts; + public int n; + public int[] stack; + public int stackSize; + public int[] dfn; + public int[] low; + public int cnt; + public int[] scc; + public int sccn; + + // 请保证点的编号从1开始，不从0开始 + // 注意： + // 如果edges里有0、1、2...n这些点，那么容器edges的大小为n+1 + // 但是0点是弃而不用的，所以1..n才是有效的点，所以有效大小是n + public StronglyConnectedComponents(ArrayList> edges) { + nexts = edges; + init(); + scc(); + } + + private void init() { + n = nexts.size(); + stack = new int[n]; + stackSize = 0; + dfn = new int[n]; + low = new int[n]; + cnt = 0; + scc = new int[n]; + sccn = 0; + n--; + } + + private void scc() { + for (int i = 1; i <= n; i++) { + if (dfn[i] == 0) { + tarjan(i); + } + } + } + + // low[] + // dfn[] + // stack[] + // int stackSize + // boolean isStack[] + // int cnt; + // int sccn; + // scc[] + private void tarjan(int p) { + low[p] = dfn[p] = ++cnt; + stack[stackSize++] = p; + for (int q : nexts.get(p)) { + // q 当前p的每一个孩子 + if (dfn[q] == 0) { + tarjan(q); + } + // q 肯定遍历过 1) 遍历过，结算了！2）遍历过，没结算 + if (scc[q] == 0) { // scc[q]!=0 q已经属于某个集团了！不能用来更新 + low[p] = Math.min(low[p], low[q]); + } + } + if (low[p] == dfn[p]) { + sccn++; + int top = 0; + do { + top = stack[--stackSize]; + scc[top] = sccn; + } while (top != p); + } + } + + public int[] getScc() { + return scc; + } + + public int getSccn() { + return sccn; + } + + public ArrayList> getShortGraph() { + ArrayList> shortGraph = new ArrayList>(); + for (int i = 0; i <= sccn; i++) { + shortGraph.add(new ArrayList()); + } + for (int u = 1; u <= n; u++) { + for (int v : nexts.get(u)) { + if (scc[u] != scc[v]) { + shortGraph.get(scc[u]).add(scc[v]); + } + } + } + return shortGraph; + } + + } + +} diff --git a/算法周更班/class_2022_03_1_week/Code02_NetworkOfSchools.java b/算法周更班/class_2022_03_1_week/Code02_NetworkOfSchools.java new file mode 100644 index 0000000..4c280cc --- /dev/null +++ b/算法周更班/class_2022_03_1_week/Code02_NetworkOfSchools.java @@ -0,0 +1,148 @@ +package class_2022_03_1_week; + +// 强连通分量练习题目 +// N个学校之间有单向的网络，每个学校得到一套软件后，可以通过单向网络向周边的学校传输 +// 问题1：初始至少需要向多少个学校发放软件，使得网络内所有的学校最终都能得到软件 +// 问题2：至少需要添加几条传输线路(边)，使任意向一个学校发放软件后 +// 经过若干次传送，网络内所有的学校最终都能得到软件 +// 2 <= N <= 1000 +// 从题意中抽象出的算法模型, 给定一个有向图，求： +// 1) 至少要选几个顶点，才能做到从这些顶点出发，可以到达全部顶点 +// 2) 至少要加多少条边，才能使得从任何一个顶点出发，都能到达全部顶点 +// 测试链接 : http://poj.org/problem?id=1236 +// 注册一下 -> 页面上点击"submit" -> 语言选择java +// 然后把如下代码粘贴进去, 把主类名改成"Main", 可以直接通过 +import java.util.ArrayList; +import java.util.Scanner; + +public class Code02_NetworkOfSchools { + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + while (sc.hasNext()) { + int n = sc.nextInt(); + ArrayList> edges = new ArrayList>(); + for (int i = 0; i <= n; i++) { + edges.add(new ArrayList()); + } + for (int from = 1; from <= n; from++) { + int to = 0; + while ((to = sc.nextInt()) != 0) { + edges.get(from).add(to); + } + } + StronglyConnectedComponents scc = new StronglyConnectedComponents(edges); + int sccn = scc.getSccn(); + int[] in = new int[sccn + 1]; + int[] out = new int[sccn + 1]; + ArrayList> dag = scc.getShortGraph(); + for (int i = 1; i <= sccn; i++) { + for (int j : dag.get(i)) { + out[i]++; + in[j]++; + } + } + int zeroIn = 0; + int zeroOut = 0; + for (int i = 1; i <= sccn; i++) { + if (in[i] == 0) { + zeroIn++; + } + if (out[i] == 0) { + zeroOut++; + } + } + System.out.println(zeroIn); + System.out.println(sccn == 1 ? 0 : Math.max(zeroIn, zeroOut)); + } + sc.close(); + } + + public static class StronglyConnectedComponents { + public ArrayList> nexts; + public int n; + public int[] stack; + public int stackSize; + public int[] dfn; + public int[] low; + public int cnt; + public int[] scc; + public int sccn; + + // 请保证点的编号从1开始，不从0开始 + // 注意： + // 如果edges里有0、1、2...n这些点，那么容器edges的大小为n+1 + // 但是0点是弃而不用的，所以1..n才是有效的点，所以有效大小是n + public StronglyConnectedComponents(ArrayList> edges) { + nexts = edges; + init(); + scc(); + } + + private void init() { + n = nexts.size(); + stack = new int[n]; + stackSize = 0; + dfn = new int[n]; + low = new int[n]; + cnt = 0; + scc = new int[n]; + sccn = 0; + n--; + } + + private void scc() { + for (int i = 1; i <= n; i++) { + if (dfn[i] == 0) { + tarjan(i); + } + } + } + + private void tarjan(int p) { + low[p] = dfn[p] = ++cnt; + stack[stackSize++] = p; + for (int q : nexts.get(p)) { + if (dfn[q] == 0) { + tarjan(q); + } + if (scc[q] == 0) { + low[p] = Math.min(low[p], low[q]); + } + } + if (low[p] == dfn[p]) { + sccn++; + int top = 0; + do { + top = stack[--stackSize]; + scc[top] = sccn; + } while (top != p); + } + } + + public int[] getScc() { + return scc; + } + + public int getSccn() { + return sccn; + } + + public ArrayList> getShortGraph() { + ArrayList> shortGraph = new ArrayList>(); + for (int i = 0; i <= sccn; i++) { + shortGraph.add(new ArrayList()); + } + for (int u = 1; u <= n; u++) { + for (int v : nexts.get(u)) { + if (scc[u] != scc[v]) { + shortGraph.get(scc[u]).add(scc[v]); + } + } + } + return shortGraph; + } + + } + +} diff --git a/算法周更班/class_2022_03_1_week/Code03_PopularCows.java b/算法周更班/class_2022_03_1_week/Code03_PopularCows.java new file mode 100644 index 0000000..1851176 --- /dev/null +++ b/算法周更班/class_2022_03_1_week/Code03_PopularCows.java @@ -0,0 +1,150 @@ +package class_2022_03_1_week; + +// 强连通分量练习题目 +// A -> B，表示A认为B是红人 +// A -> B -> C，表示A认为B是红人，B认为C是红人，规定“认为”关系有传递性，所以A也认为C是红人 +// 给定一张有向图，方式是给定M个有序对(A, B) +// (A, B)表示A认为B是红人，该关系具有传递性 +// 给定的有序对中可能包含(A, B)和(B, C)，但不包含(A,C) +// 求被其他所有人认为是红人的总数。 +// 测试链接 : http://poj.org/problem?id=2186 +// 注册一下 -> 页面上点击"submit" -> 语言选择java +// 然后把如下代码粘贴进去, 把主类名改成"Main", 可以直接通过 +import java.util.ArrayList; +import java.util.Scanner; + +public class Code03_PopularCows { + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + while (sc.hasNext()) { + int n = sc.nextInt(); + int m = sc.nextInt(); + ArrayList> edges = new ArrayList>(); + for (int i = 0; i <= n; i++) { + edges.add(new ArrayList()); + } + for (int i = 0; i < m; i++) { + int from = sc.nextInt(); + int to = sc.nextInt(); + edges.get(from).add(to); + } + StronglyConnectedComponents connectedComponents = new StronglyConnectedComponents(edges); + int sccn = connectedComponents.getSccn(); + if (sccn == 1) { + System.out.println(n); + } else { + ArrayList> dag = connectedComponents.getShortGraph(); + int zeroOut = 0; + int outScc = 0; + for (int i = 1; i <= sccn; i++) { + if (dag.get(i).size() == 0) { + zeroOut++; + outScc = i; + } + } + if (zeroOut > 1) { + System.out.println(0); + } else { + int[] scc = connectedComponents.getScc(); + int ans = 0; + for (int i = 1; i <= n; i++) { + if (scc[i] == outScc) { + ans++; + } + } + System.out.println(ans); + } + } + } + sc.close(); + } + + public static class StronglyConnectedComponents { + public ArrayList> nexts; + public int n; + public int[] stack; + public int stackSize; + public int[] dfn; + public int[] low; + public int cnt; + public int[] scc; + public int sccn; + + // 请保证点的编号从1开始，不从0开始 + // 注意： + // 如果edges里有0、1、2...n这些点，那么容器edges的大小为n+1 + // 但是0点是弃而不用的，所以1..n才是有效的点，所以有效大小是n + public StronglyConnectedComponents(ArrayList> edges) { + nexts = edges; + init(); + scc(); + } + + private void init() { + n = nexts.size(); + stack = new int[n]; + stackSize = 0; + dfn = new int[n]; + low = new int[n]; + cnt = 0; + scc = new int[n]; + sccn = 0; + n--; + } + + private void scc() { + for (int i = 1; i <= n; i++) { + if (dfn[i] == 0) { + tarjan(i); + } + } + } + + private void tarjan(int p) { + low[p] = dfn[p] = ++cnt; + stack[stackSize++] = p; + for (int q : nexts.get(p)) { + if (dfn[q] == 0) { + tarjan(q); + } + if (scc[q] == 0) { + low[p] = Math.min(low[p], low[q]); + } + } + if (low[p] == dfn[p]) { + sccn++; + int top = 0; + do { + top = stack[--stackSize]; + scc[top] = sccn; + } while (top != p); + } + } + + public int[] getScc() { + return scc; + } + + public int getSccn() { + return sccn; + } + + public ArrayList> getShortGraph() { + ArrayList> shortGraph = new ArrayList>(); + for (int i = 0; i <= sccn; i++) { + shortGraph.add(new ArrayList()); + } + for (int u = 1; u <= n; u++) { + for (int v : nexts.get(u)) { + if (scc[u] != scc[v]) { + shortGraph.get(scc[u]).add(scc[v]); + } + } + } + return shortGraph; + } + + } + +} diff --git a/算法周更班/class_2022_03_1_week/Code04_IgniteMinBombs.java b/算法周更班/class_2022_03_1_week/Code04_IgniteMinBombs.java new file mode 100644 index 0000000..fab1a24 --- /dev/null +++ b/算法周更班/class_2022_03_1_week/Code04_IgniteMinBombs.java @@ -0,0 +1,328 @@ +package class_2022_03_1_week; + +import java.util.ArrayList; +import java.util.TreeMap; + +// 来自亚马逊 +// 帖子链接 : https://www.nowcoder.com/discuss/826182 +// 这道题是帖子里第2题 +// 在一个地图上有若干个炸弹，每个炸弹会呈现十字型引爆 +// 每个炸弹都有其当量值，这个值决定了这个炸弹的爆炸半径 +// 如果一个炸弹被引爆时，有其它炸弹在其爆炸半径内，那么其它炸弹也会爆炸 +// 请问使地图上所有炸弹爆炸所需的最少人为引爆次数。 +// 例如： +// 0,0,0,0,0 +// 0,0,0,1,0 +// 0,0,0,0,0 +// 上图中val为1的单元是一个炸弹，人为引爆后地图变成下面的样子： +// 0, 0, 0,-1, 0 +// 0, 0,-1,-1,-1 +// 0, 0, 0,-1, 0 +// 题目并没有给数据量，面经题目的通病 +public class Code04_IgniteMinBombs { + + // 暴力方法 + // 为了测试 + public static int minBombs1(int[][] map) { + int n = map.length; + int m = 0; + for (int i = 0; i < n; i++) { + for (int j = 0; j < n; j++) { + m += map[i][j] == 0 ? 0 : 1; + } + } + int[][] bombs = new int[m][2]; + m = 0; + for (int i = 0; i < n; i++) { + for (int j = 0; j < n; j++) { + if (map[i][j] != 0) { + bombs[m][0] = i; + bombs[m++][1] = j; + } + } + } + int[] arr = new int[m]; + for (int i = 0; i < m; i++) { + arr[i] = i; + } + return process1(arr, 0, bombs, map); + } + + public static int process1(int[] arr, int index, int[][] bombs, int[][] map) { + int ans = Integer.MAX_VALUE; + if (index == arr.length) { + ans = orderIgnite(arr, bombs, map); + } else { + for (int i = index; i < arr.length; i++) { + swap(arr, index, i); + ans = Math.min(ans, process1(arr, index + 1, bombs, map)); + swap(arr, index, i); + } + } + return ans; + } + + public static void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } + + public static int orderIgnite(int[] arr, int[][] bombs, int[][] map) { + int[][] copy = copyMap(map); + int ans = 0; + for (int i : arr) { + int row = bombs[i][0]; + int col = bombs[i][1]; + if (copy[row][col] != -1) { + ans++; + burn(copy, row, col, copy[row][col]); + } + } + return ans; + } + + public static int[][] copyMap(int[][] map) { + int n = map.length; + int[][] ans = new int[n][n]; + for (int i = 0; i < n; i++) { + for (int j = 0; j < n; j++) { + ans[i][j] = map[i][j]; + } + } + return ans; + } + + public static void burn(int[][] map, int i, int j, int v) { + map[i][j] = -1; + ArrayList queue = new ArrayList<>(); + for (int row = i - 1, cnt = 1; row >= 0 && cnt <= v; row--, cnt++) { + if (map[row][j] > 0) { + queue.add(new int[] { row, j, map[row][j] }); + } + map[row][j] = -1; + } + for (int row = i + 1, cnt = 1; row < map.length && cnt <= v; row++, cnt++) { + if (map[row][j] > 0) { + queue.add(new int[] { row, j, map[row][j] }); + } + map[row][j] = -1; + } + for (int col = j - 1, cnt = 1; col >= 0 && cnt <= v; col--, cnt++) { + if (map[i][col] > 0) { + queue.add(new int[] { i, col, map[i][col] }); + } + map[i][col] = -1; + } + for (int col = j + 1, cnt = 1; col < map.length && cnt <= v; col++, cnt++) { + if (map[i][col] > 0) { + queue.add(new int[] { i, col, map[i][col] }); + } + map[i][col] = -1; + } + for (int[] next : queue) { + burn(map, next[0], next[1], next[2]); + } + } + + // 正式方法 + // 用到有序表 + 强连通分量 + public static int minBombs2(int[][] map) { + int n = map.length; + ArrayList> rowTreeMaps = new ArrayList<>(); + ArrayList> colTreeMaps = new ArrayList<>(); + for (int i = 0; i < n; i++) { + rowTreeMaps.add(new TreeMap<>()); + colTreeMaps.add(new TreeMap<>()); + } + int m = 0; + for (int i = 0; i < n; i++) { + for (int j = 0; j < n; j++) { + if (map[i][j] != 0) { + m++; + rowTreeMaps.get(i).put(j, m); + colTreeMaps.get(j).put(i, m); + } + } + } + ArrayList> edges = new ArrayList<>(); + for (int i = 0; i <= m; i++) { + edges.add(new ArrayList<>()); + } + for (int i = 0; i < n; i++) { + for (int j = 0; j < n; j++) { + if (map[i][j] != 0) { + TreeMap rowTreeMap = rowTreeMaps.get(i); + TreeMap colTreeMap = colTreeMaps.get(j); + int from = rowTreeMap.get(j); + int col = j - 1; + while (rowTreeMap.floorKey(col) != null && j - rowTreeMap.floorKey(col) <= map[i][j]) { + col = rowTreeMap.floorKey(col); + edges.get(from).add(rowTreeMap.get(col)); + col--; + } + col = j + 1; + while (rowTreeMap.ceilingKey(col) != null && rowTreeMap.ceilingKey(col) - j <= map[i][j]) { + col = rowTreeMap.ceilingKey(col); + edges.get(from).add(rowTreeMap.get(col)); + col++; + } + int row = i - 1; + while (colTreeMap.floorKey(row) != null && i - colTreeMap.floorKey(row) <= map[i][j]) { + row = colTreeMap.floorKey(row); + edges.get(from).add(colTreeMap.get(row)); + row--; + } + row = i + 1; + while (colTreeMap.ceilingKey(row) != null && colTreeMap.ceilingKey(row) - i <= map[i][j]) { + row = colTreeMap.ceilingKey(row); + edges.get(from).add(colTreeMap.get(row)); + row++; + } + } + } + } + StronglyConnectedComponents scc = new StronglyConnectedComponents(edges); + int sccn = scc.getSccn(); + int[] in = new int[sccn + 1]; + int[] out = new int[sccn + 1]; + ArrayList> dag = scc.getShortGraph(); + for (int i = 1; i <= sccn; i++) { + for (int j : dag.get(i)) { + out[i]++; + in[j]++; + } + } + int zeroIn = 0; + for (int i = 1; i <= sccn; i++) { + if (in[i] == 0) { + zeroIn++; + } + } + return zeroIn; + } + + public static class StronglyConnectedComponents { + public ArrayList> nexts; + public int n; + public int[] stack; + public int stackSize; + public int[] dfn; + public int[] low; + public int cnt; + public int[] scc; + public int sccn; + + // 请保证点的编号从1开始，不从0开始 + // 注意： + // 如果edges里有0、1、2...n这些点，那么容器edges的大小为n+1 + // 但是0点是弃而不用的，所以1..n才是有效的点，所以有效大小是n + public StronglyConnectedComponents(ArrayList> edges) { + nexts = edges; + init(); + scc(); + } + + private void init() { + n = nexts.size(); + stack = new int[n]; + stackSize = 0; + dfn = new int[n]; + low = new int[n]; + cnt = 0; + scc = new int[n]; + sccn = 0; + n--; + } + + private void scc() { + for (int i = 1; i <= n; i++) { + if (dfn[i] == 0) { + tarjan(i); + } + } + } + + private void tarjan(int p) { + low[p] = dfn[p] = ++cnt; + stack[stackSize++] = p; + for (int q : nexts.get(p)) { + if (dfn[q] == 0) { + tarjan(q); + } + if (scc[q] == 0) { + low[p] = Math.min(low[p], low[q]); + } + } + if (low[p] == dfn[p]) { + sccn++; + int top = 0; + do { + top = stack[--stackSize]; + scc[top] = sccn; + } while (top != p); + } + } + + public int[] getScc() { + return scc; + } + + public int getSccn() { + return sccn; + } + + public ArrayList> getShortGraph() { + ArrayList> shortGraph = new ArrayList>(); + for (int i = 0; i <= sccn; i++) { + shortGraph.add(new ArrayList()); + } + for (int u = 1; u <= n; u++) { + for (int v : nexts.get(u)) { + if (scc[u] != scc[v]) { + shortGraph.get(scc[u]).add(scc[v]); + } + } + } + return shortGraph; + } + + } + + // 为了测试 + public static int[][] randomMatrix(int n, int m, int v) { + int[][] map = new int[n][n]; + for (int i = 0; i < m; i++) { + map[(int) (Math.random() * n)][(int) (Math.random() * n)] = (int) (Math.random() * v) + 1; + } + return map; + } + + // 为了测试 + public static void main(String[] args) { + int n = 8; + int m = 8; + int v = 5; + int testTime = 1000; + System.out.println("测试开始"); + for (int k = 0; k < testTime; k++) { + int[][] map = randomMatrix(n, m, v); + int ans1 = minBombs1(map); + int ans2 = minBombs2(map); + if (ans1 != ans2) { + for (int i = 0; i < n; i++) { + for (int j = 0; j < n; j++) { + System.out.print(map[i][j] + " "); + } + System.out.println(); + } + System.out.println(ans1); + System.out.println(ans2); + System.out.println("出错了!"); + } + + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_03_2_week/Code01_MeetingCheck.java b/算法周更班/class_2022_03_2_week/Code01_MeetingCheck.java new file mode 100644 index 0000000..0757332 --- /dev/null +++ b/算法周更班/class_2022_03_2_week/Code01_MeetingCheck.java @@ -0,0 +1,202 @@ +package class_2022_03_2_week; + +import java.util.Arrays; + +// 来自字节飞书团队 +// 在字节跳动，大家都使用飞书的日历功能进行会议室的预订，遇到会议高峰时期， +// 会议室就可能不够用，现在请你实现一个算法，判断预订会议时是否有空的会议室可用。 +// 为简化问题，这里忽略会议室的大小，认为所有的会议室都是等价的， +// 只要空闲就可以容纳任意的会议，并且： +// 1. 所有的会议预订都是当日预订当日的时段 +// 2. 会议时段是一个左闭右开的时间区间，精确到分钟 +// 3. 每个会议室刚开始都是空闲状态，同一时间一个会议室只能进行一场会议 +// 4. 会议一旦预订成功就会按时进行 +// 比如上午11点到中午12点的会议即[660, 720) +// 给定一个会议室总数m +// 一个预定事件由[a,b,c]代表 : +// a代表预定动作的发生时间，早来早得; b代表会议的召开时间; c代表会议的结束时间 +// 给定一个n*3的二维数组，即可表示所有预定事件 +// 返回一个长度为n的boolean类型的数组，表示每一个预定时间是否成功 +public class Code01_MeetingCheck { + + public static boolean[] reserveMeetings(int m, int[][] meetings) { + // 会议的总场次 + int n = meetings.length; + // 开头时间，结尾时间 + int[] ranks = new int[n << 1]; + for (int i = 0; i < n; i++) { + ranks[i] = meetings[i][1]; + ranks[i + n] = meetings[i][2] - 1; + } + Arrays.sort(ranks); + // 0 : [6, 100, 200] + // 1 : [4, 30, 300] + // 30,1 100,2 200,3 300,4 + // [0,6,2,3] + // [1,4,1,4] + // + // 0 T/F , 1, T/ 2, + + // [1,4,1,4] [0,6,2,3] .... + int[][] reMeetings = new int[n][4]; + int max = 0; + for (int i = 0; i < n; i++) { + reMeetings[i][0] = i; + reMeetings[i][1] = meetings[i][0]; + reMeetings[i][2] = rank(ranks, meetings[i][1]); + reMeetings[i][3] = rank(ranks, meetings[i][2] - 1); + max = Math.max(max, reMeetings[i][3]); + } + SegmentTree st = new SegmentTree(max); + Arrays.sort(reMeetings, (a, b) -> a[1] - b[1]); + boolean[] ans = new boolean[n]; + for (int[] meeting : reMeetings) { + if (st.queryMax(meeting[2], meeting[3]) < m) { + ans[meeting[0]] = true; + st.add(meeting[2], meeting[3], 1); + } + } + return ans; + } + + // 返回>=num, 最左位置 + public static int rank(int[] sorted, int num) { + int l = 0; + int r = sorted.length - 1; + int m = 0; + int ans = 0; + while (l <= r) { + m = (l + r) / 2; + if (sorted[m] >= num) { + ans = m; + r = m - 1; + } else { + l = m + 1; + } + } + return ans + 1; + } + + public static class SegmentTree { + private int n; + private int[] max; + private int[] lazy; + + public SegmentTree(int maxSize) { + n = maxSize; + max = new int[n << 2]; + lazy = new int[n << 2]; + } + + private void pushUp(int rt) { + max[rt] = Math.max(max[rt << 1], max[rt << 1 | 1]); + } + + private void pushDown(int rt, int ln, int rn) { + if (lazy[rt] != 0) { + lazy[rt << 1] += lazy[rt]; + max[rt << 1] += lazy[rt]; + lazy[rt << 1 | 1] += lazy[rt]; + max[rt << 1 | 1] += lazy[rt]; + lazy[rt] = 0; + } + } + + public void add(int L, int R, int C) { + add(L, R, C, 1, n, 1); + } + + private void add(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + max[rt] += C; + lazy[rt] += C; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + add(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + add(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + public int queryMax(int L, int R) { + return queryMax(L, R, 1, n, 1); + } + + private int queryMax(int L, int R, int l, int r, int rt) { + if (L <= l && r <= R) { + return max[rt]; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + int ans = 0; + if (L <= mid) { + ans = Math.max(ans, queryMax(L, R, l, mid, rt << 1)); + } + if (R > mid) { + ans = Math.max(ans, queryMax(L, R, mid + 1, r, rt << 1 | 1)); + } + return ans; + } + + } + + // 为了测试线段树 + public static class Right { + public int[] arr; + + public Right(int maxSize) { + arr = new int[maxSize + 1]; + } + + public void add(int L, int R, int C) { + for (int i = L; i <= R; i++) { + arr[i] += C; + } + } + + public int queryMax(int L, int R) { + int ans = 0; + for (int i = L; i <= R; i++) { + ans = Math.max(ans, arr[i]); + } + return ans; + } + + } + + // 测试线段树的对数器 + public static void main(String[] args) { + int N = 50; + int V = 10; + int testTimes1 = 1000; + int testTimes2 = 1000; + System.out.println("测试线段树开始"); + for (int i = 0; i < testTimes1; i++) { + int n = (int) (Math.random() * N) + 1; + SegmentTree st = new SegmentTree(n); + Right right = new Right(n); + for (int j = 0; j < testTimes2; j++) { + int a = (int) (Math.random() * n) + 1; + int b = (int) (Math.random() * n) + 1; + int L = Math.min(a, b); + int R = Math.max(a, b); + if (Math.random() < 0.5) { + int C = (int) (Math.random() * V); + st.add(L, R, C); + right.add(L, R, C); + } else { + if (st.queryMax(L, R) != right.queryMax(L, R)) { + System.out.println("出错了!"); + } + } + } + } + System.out.println("测试线段树结束"); + } + +} diff --git a/算法周更班/class_2022_03_2_week/Code02_StringCheck.java b/算法周更班/class_2022_03_2_week/Code02_StringCheck.java new file mode 100644 index 0000000..5524a47 --- /dev/null +++ b/算法周更班/class_2022_03_2_week/Code02_StringCheck.java @@ -0,0 +1,65 @@ +package class_2022_03_2_week; + +import java.util.Arrays; + +// 来自字节飞书团队 +// 小歪每次会给你两个字符串： +// 笔记s1和关键词s2，请你写一个函数 +// 判断s2的排列之一是否是s1的子串 +// 如果是，返回true +// 否则，返回false +public class Code02_StringCheck { + + public static boolean check1(String s1, String s2) { + if (s1.length() < s2.length()) { + return false; + } + char[] str2 = s2.toCharArray(); + Arrays.sort(str2); + s2 = String.valueOf(str2); + for (int L = 0; L < s1.length(); L++) { + for (int R = L; R < s1.length(); R++) { + char[] cur = s1.substring(L, R + 1).toCharArray(); + Arrays.sort(cur); + String curSort = String.valueOf(cur); + if (curSort.equals(s2)) { + return true; + } + } + } + return false; + } + + public static boolean check2(String s1, String s2) { + if (s1.length() < s2.length()) { + return false; + } + char[] str2 = s2.toCharArray(); + int[] count = new int[256]; + for (int i = 0; i < str2.length; i++) { + count[str2[i]]++; + } + int M = str2.length; + char[] st1 = s1.toCharArray(); + int inValidTimes = 0; + int R = 0; + for (; R < M; R++) { + if (count[st1[R]]-- <= 0) { + inValidTimes++; + } + } + for (; R < st1.length; R++) { + if (inValidTimes == 0) { + return true; + } + if (count[st1[R]]-- <= 0) { + inValidTimes++; + } + if (count[st1[R - M]]++ < 0) { + inValidTimes--; + } + } + return inValidTimes == 0; + } + +} diff --git a/算法周更班/class_2022_03_2_week/Code03_AiFill.java b/算法周更班/class_2022_03_2_week/Code03_AiFill.java new file mode 100644 index 0000000..a71cb85 --- /dev/null +++ b/算法周更班/class_2022_03_2_week/Code03_AiFill.java @@ -0,0 +1,112 @@ +package class_2022_03_2_week; + +import java.util.ArrayList; +import java.util.HashMap; +import java.util.List; +import java.util.TreeSet; + +// 来自字节飞书团队 +// 语法补全功能，比如"as soon as possible" +// 当我们识别到"as soon as"时, 基本即可判定用户需要键入"possible" +// 设计一个统计词频的模型，用于这个功能 +// 类似(prefix, next word)这样的二元组 +// 比如一个上面的句子"as soon as possible" +// 有产生如下的二元组(as, soon, 1)、(as soon, as, 1)、(as soon as, possible, 1) +// 意思是这一个句子产生了如下的统计： +// 当前缀为"as"，接下来的单词是"soon"，有了1个期望点 +// 当前缀为"as soon"，接下来的单词是"as"，有了1个期望点 +// 当前缀为"as soon as"，接下来的单词是"possible"，有了1个期望点 +// 那么如果给你很多的句子，当然就可以产生很多的期望点，同一个前缀下，同一个next word的期望点可以累加 +// 现在给你n个句子，让你来建立统计 +// 然后给你m个句子，作为查询 +// 最后给你k，表示每个句子作为前缀的情况下，词频排在前k名的联想 +// 返回m个结果，每个结果最多k个单词 +public class Code03_AiFill { + + public static class TrieNode { + public String word; + public int times; + public HashMap nextNodes; + public TreeSet nextRanks; + + public TrieNode(String w) { + word = w; + times = 1; + nextNodes = new HashMap<>(); + nextRanks = new TreeSet<>((a, b) -> a.times != b.times ? (b.times - a.times) : a.word.compareTo(b.word)); + } + + } + + public static class AI { + public TrieNode root; + public int topk; + + public AI(List sentences, int k) { + root = new TrieNode(""); + topk = k; + for (String sentence : sentences) { + fill(sentence); + } + } + + public void fill(String sentence) { + TrieNode cur = root; + TrieNode next = null; + for (String word : sentence.split(" ")) { + if (!cur.nextNodes.containsKey(word)) { + next = new TrieNode(word); + cur.nextNodes.put(word, next); + cur.nextRanks.add(next); + } else { + next = cur.nextNodes.get(word); + cur.nextRanks.remove(next); + next.times++; + cur.nextRanks.add(next); + } + cur = next; + } + } + + public List suggest(String sentence) { + List ans = new ArrayList<>(); + TrieNode cur = root; + for (String word : sentence.split(" ")) { + if (!cur.nextNodes.containsKey(word)) { + return ans; + } else { + cur = cur.nextNodes.get(word); + } + } + for (TrieNode n : cur.nextRanks) { + ans.add(n.word); + if (ans.size() == topk) { + break; + } + } + return ans; + } + + } + + public static void main(String[] args) { + ArrayList sentences = new ArrayList<>(); + sentences.add("i think you are good"); + sentences.add("i think you are fine"); + sentences.add("i think you are good man"); + int k = 2; + AI ai = new AI(sentences, k); + for (String ans : ai.suggest("i think you are")) { + System.out.println(ans); + } + System.out.println("====="); + ai.fill("i think you are fucking good"); + ai.fill("i think you are fucking great"); + ai.fill("i think you are fucking genius"); + for (String ans : ai.suggest("i think you are")) { + System.out.println(ans); + } + System.out.println("====="); + } + +} diff --git a/算法周更班/class_2022_03_2_week/Code04_SameTeams.java b/算法周更班/class_2022_03_2_week/Code04_SameTeams.java new file mode 100644 index 0000000..65ec1b0 --- /dev/null +++ b/算法周更班/class_2022_03_2_week/Code04_SameTeams.java @@ -0,0 +1,62 @@ +package class_2022_03_2_week; + +import java.util.ArrayList; +import java.util.HashMap; + +// 来自字节飞书团队 +// 假设数组a和数组b为两组信号 +// 1) length(b) <= length(a) +// 2) 对于任意0<=i indices; + public HashMap nexts; + + public TrieNode() { + indices = new ArrayList<>(); + nexts = new HashMap<>(); + } + + } + + public static int[] sameTeamsArray(int[][] bs, int[][] as) { + int m = bs.length; + TrieNode root = new TrieNode(); + TrieNode cur = null; + for (int i = 0; i < m; i++) { + int k = bs[i].length; + cur = root; + for (int j = 1; j < k; j++) { + int diff = bs[i][j] - bs[i][j - 1]; + if (!cur.nexts.containsKey(diff)) { + cur.nexts.put(diff, new TrieNode()); + } + cur = cur.nexts.get(diff); + } + cur.indices.add(i); + } + int[] ans = new int[m]; + int n = as.length; + for (int i = 0; i < n; i++) { + int k = as[i].length; + cur = root; + for (int j = 1; j < k; j++) { + int diff = as[i][j] - as[i][j - 1]; + if (!cur.nexts.containsKey(diff)) { + break; + } + cur = cur.nexts.get(diff); + for (int index : cur.indices) { + ans[index]++; + } + } + } + return ans; + } + +} diff --git a/算法周更班/class_2022_03_2_week/Code05_NumberOfDivisibleByM.java b/算法周更班/class_2022_03_2_week/Code05_NumberOfDivisibleByM.java new file mode 100644 index 0000000..03e500a --- /dev/null +++ b/算法周更班/class_2022_03_2_week/Code05_NumberOfDivisibleByM.java @@ -0,0 +1,69 @@ +package class_2022_03_2_week; + +// 来自微软 +// 给定一个数组arr，给定一个正数M +// 如果arr[i] + arr[j]可以被M整除，并且i < j，那么(i,j)叫做一个M整除对 +// 返回arr中M整除对的总数量 +public class Code05_NumberOfDivisibleByM { + + public static int num1(int[] arr, int m) { + int n = arr.length; + int ans = 0; + for (int i = 0; i < n; i++) { + for (int j = i + 1; j < n; j++) { + if (Math.abs(arr[i] + arr[j]) % m == 0) { + ans++; + } + } + } + return ans; + } + + public static int num2(int[] arr, int m) { + int n = arr.length; + int[] cnts = new int[m]; + int ans = 0; + for (int i = n - 1; i >= 0; i--) { + int cur = (arr[i] % m + m) % m; + ans += cnts[(m - cur) % m]; + cnts[cur]++; + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * v) - (int) (Math.random() * v); + } + return arr; + } + + public static void main(String[] args) { + int len = 50; + int value = 50; + int testTime = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * len) + 1; + int[] arr = randomArray(n, value); + int m = (int) (Math.random() * value) + 1; + int ans1 = num1(arr, m); + int ans2 = num2(arr, m); + if (ans1 != ans2) { + System.out.println("出错了!"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println("m = " + m); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_03_2_week/Code06_JobMinDays.java b/算法周更班/class_2022_03_2_week/Code06_JobMinDays.java new file mode 100644 index 0000000..fc36ecc --- /dev/null +++ b/算法周更班/class_2022_03_2_week/Code06_JobMinDays.java @@ -0,0 +1,100 @@ +package class_2022_03_2_week; + +// 来自微软 +// 给定一个正数数组arr，长度为N，依次代表N个任务的难度，给定一个正数k +// 你只能从0任务开始，依次处理到N-1号任务结束，就是一定要从左往右处理任务 +// 只不过，难度差距绝对值不超过k的任务，可以在一天之内都完成 +// 返回完成所有任务的最少天数 +public class Code06_JobMinDays { + + public static int minDays1(int[] arr, int k) { + int n = arr.length; + int[] dp = new int[n]; + dp[0] = 1; + for (int i = 1; i < n; i++) { + dp[i] = dp[i - 1] + 1; + int min = arr[i]; + int max = arr[i]; + for (int j = i - 1; j >= 0; j--) { + min = Math.min(min, arr[j]); + max = Math.max(max, arr[j]); + if (max - min <= k) { + dp[i] = Math.min(dp[i], 1 + (j - 1 >= 0 ? dp[j - 1] : 0)); + } else { + break; + } + } + } + return dp[n - 1]; + } + + public static int minDays2(int[] arr, int k) { + int n = arr.length; + int[] dp = new int[n]; + int[] windowMax = new int[n]; + int[] windowMin = new int[n]; + int maxL = 0; + int maxR = 0; + int minL = 0; + int minR = 0; + int L = 0; + for (int i = 0; i < n; i++) { + while (maxL < maxR && arr[windowMax[maxR - 1]] <= arr[i]) { + maxR--; + } + windowMax[maxR++] = i; + while (minL < minR && arr[windowMin[minR - 1]] >= arr[i]) { + minR--; + } + windowMin[minR++] = i; + while (arr[windowMax[maxL]] - arr[windowMin[minL]] > k) { + if (windowMax[maxL] == L) { + maxL++; + } + if (windowMin[minL] == L) { + minL++; + } + L++; + } + dp[i] = 1 + (L - 1 >= 0 ? dp[L - 1] : 0); + } + return dp[n - 1]; + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * v); + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int len = 50; + int value = 20; + int testTime = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * len) + 1; + int[] arr = randomArray(n, value); + int k = (int) (Math.random() * value); + int ans1 = minDays1(arr, k); + int ans2 = minDays2(arr, k); + if (ans1 != ans2) { + System.out.println("出错了!"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println("k = " + k); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_03_2_week/Code07_MinWaitingTime.java b/算法周更班/class_2022_03_2_week/Code07_MinWaitingTime.java new file mode 100644 index 0000000..90cbeed --- /dev/null +++ b/算法周更班/class_2022_03_2_week/Code07_MinWaitingTime.java @@ -0,0 +1,95 @@ +package class_2022_03_2_week; + +import java.util.PriorityQueue; + +// 来自谷歌 +// 给定一个数组arr，长度为n +// 表示n个服务员，每个人服务一个人的时间 +// 给定一个正数m，表示有m个人等位 +// 如果你是刚来的人，请问你需要等多久？ +// 假设：m远远大于n，比如n<=1000, m <= 10的9次方，该怎么做？ +public class Code07_MinWaitingTime { + + public static int minWaitingTime1(int[] arr, int m) { + if (arr == null || arr.length == 0) { + return -1; + } + PriorityQueue heap = new PriorityQueue<>((a, b) -> (a[0] - b[0])); + int n = arr.length; + for (int i = 0; i < n; i++) { + heap.add(new int[] { 0, arr[i] }); + } + for (int i = 0; i < m; i++) { + int[] cur = heap.poll(); + cur[0] += cur[1]; + heap.add(cur); + } + return heap.peek()[0]; + } + + public static int minWaitingTime2(int[] arr, int m) { + if (arr == null || arr.length == 0) { + return -1; + } + int best = Integer.MAX_VALUE; + for (int num : arr) { + best = Math.min(best, num); + } + int left = 0; + int right = best * m; + int mid = 0; + int near = 0; + while (left <= right) { + mid = (left + right) / 2; + int cover = 0; + for (int num : arr) { + cover += (mid / num) + 1; + } + if (cover >= m + 1) { + near = mid; + right = mid - 1; + } else { + left = mid + 1; + } + } + return near; + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * v) + 1; + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int len = 50; + int value = 30; + int mMax = 3000; + int testTime = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * len) + 1; + int[] arr = randomArray(n, value); + int m = (int) (Math.random() * mMax); + int ans1 = minWaitingTime1(arr, m); + int ans2 = minWaitingTime2(arr, m); + if (ans1 != ans2) { + System.out.println("出错了!"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println("m : " + m); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_03_2_week/Code08_TimeNSpace1LowestCommonAncestor.java b/算法周更班/class_2022_03_2_week/Code08_TimeNSpace1LowestCommonAncestor.java new file mode 100644 index 0000000..f4e91aa --- /dev/null +++ b/算法周更班/class_2022_03_2_week/Code08_TimeNSpace1LowestCommonAncestor.java @@ -0,0 +1,111 @@ +package class_2022_03_2_week; + +// 如何时间复杂度O(N)，额外空间复杂度O(1)，解决最低公共祖先问题? +// 测试链接 : https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/ +public class Code08_TimeNSpace1LowestCommonAncestor { + + // 这个类不要提交 + public static class TreeNode { + public int val; + public TreeNode left; + public TreeNode right; + } + + // 提交以下的方法 + // 该方法亮点在于：时间复杂度O(N)，额外空间复杂度O(1) + public static TreeNode lowestCommonAncestor(TreeNode head, TreeNode o1, TreeNode o2) { + if (findFirst(o1.left, o1, o2) != null || findFirst(o1.right, o1, o2) != null) { + return o1; + } + if (findFirst(o2.left, o1, o2) != null || findFirst(o2.right, o1, o2) != null) { + return o2; + } + TreeNode leftAim = findFirst(head, o1, o2); + TreeNode cur = head; + TreeNode mostRight = null; + TreeNode ans = null; + while (cur != null) { + mostRight = cur.left; + if (mostRight != null) { + while (mostRight.right != null && mostRight.right != cur) { + mostRight = mostRight.right; + } + if (mostRight.right == null) { + mostRight.right = cur; + cur = cur.left; + continue; + } else { + mostRight.right = null; + if (findLeftAim(cur.left, leftAim)) { + if (ans == null && findFirst(leftAim.right, o1, o2) != null) { + ans = leftAim; + } + leftAim = cur; + } + } + } + cur = cur.right; + } + return ans != null ? ans : (findFirst(leftAim.right, o1, o2) != null ? leftAim : head); + } + + public static boolean findLeftAim(TreeNode head, TreeNode leftAim) { + TreeNode tail = reverseEdge(head); + TreeNode cur = tail; + boolean ans = false; + while (cur != null) { + if (cur == leftAim) { + ans = true; + } + cur = cur.right; + } + reverseEdge(tail); + return ans; + } + + public static TreeNode reverseEdge(TreeNode from) { + TreeNode pre = null; + TreeNode next = null; + while (from != null) { + next = from.right; + from.right = pre; + pre = from; + from = next; + } + return pre; + } + + public static TreeNode findFirst(TreeNode head, TreeNode o1, TreeNode o2) { + if (head == null) { + return null; + } + TreeNode cur = head; + TreeNode mostRight = null; + TreeNode first = null; + while (cur != null) { + mostRight = cur.left; + if (mostRight != null) { + while (mostRight.right != null && mostRight.right != cur) { + mostRight = mostRight.right; + } + if (mostRight.right == null) { + if (first == null && (cur == o1 || cur == o2)) { + first = cur; + } + mostRight.right = cur; + cur = cur.left; + continue; + } else { + mostRight.right = null; + } + } else { + if (first == null && (cur == o1 || cur == o2)) { + first = cur; + } + } + cur = cur.right; + } + return first; + } + +} diff --git a/算法周更班/class_2022_03_3_week/Code01_LongestUncontinuousSet.java b/算法周更班/class_2022_03_3_week/Code01_LongestUncontinuousSet.java new file mode 100644 index 0000000..c32f6e2 --- /dev/null +++ b/算法周更班/class_2022_03_3_week/Code01_LongestUncontinuousSet.java @@ -0,0 +1,100 @@ +package class_2022_03_3_week; + +import java.util.Arrays; + +// 来自美团 +// 给定一个数组arr，你可以随意挑选其中的数字 +// 但是你挑选的数中，任何两个数a和b，必须Math.abs(a - b) > 1 +// 返回你最多能挑选几个数 +public class Code01_LongestUncontinuousSet { + + // 暴力方法 + // 为了验证 + public static int longestUncontinuous1(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + Arrays.sort(arr); + return process1(arr, 0, new int[arr.length], 0); + } + + public static int process1(int[] arr, int i, int[] path, int j) { + if (i == arr.length) { + for (int k = 1; k < j; k++) { + if (path[k - 1] + 1 >= path[k]) { + return 0; + } + } + return j; + } else { + int p1 = process1(arr, i + 1, path, j); + path[j] = arr[i]; + int p2 = process1(arr, i + 1, path, j + 1); + return Math.max(p1, p2); + } + } + + // 最优解 + public static int longestUncontinuous2(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + Arrays.sort(arr); + int n = arr.length; + int size = 1; + for (int i = 1; i < n; i++) { + if (arr[i] != arr[size - 1]) { + arr[size++] = arr[i]; + } + } + int[] dp = new int[size]; + dp[0] = 1; + int ans = 1; + for (int i = 1; i < size; i++) { + dp[i] = 1; + if (arr[i] - arr[i - 1] > 1) { + dp[i] = 1 + dp[i - 1]; + } + if (i - 2 >= 0 && arr[i] - arr[i - 2] > 1) { + dp[i] = Math.max(dp[i], 1 + dp[i - 2]); + } + ans = Math.max(ans, dp[i]); + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * v) - (int) (Math.random() * v); + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int len = 10; + int value = 20; + int testTime = 2000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * len) + 1; + int[] arr = randomArray(n, value); + int ans1 = longestUncontinuous1(arr); + int ans2 = longestUncontinuous2(arr); + if (ans1 != ans2) { + System.out.println("出错了!"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_03_3_week/Code02_CutDouFu.java b/算法周更班/class_2022_03_3_week/Code02_CutDouFu.java new file mode 100644 index 0000000..d0a695b --- /dev/null +++ b/算法周更班/class_2022_03_3_week/Code02_CutDouFu.java @@ -0,0 +1,54 @@ +package class_2022_03_3_week; + +import java.util.Arrays; + +// 来自美团 +// 有一块10000 * 10000 * 10000的立方体豆腐 +// 豆腐的前左下角放在(0,0,0)点，豆腐的后右上角放在(10000,10000,10000)点 +// 下面给出切法的数据结构 +// [a,b] +// a = 1，表示x = b处，一把无穷大的刀平行于yz面贯穿豆腐切过去 +// a = 2，表示y = b处，一把无穷大的刀平行于xz面贯穿豆腐切过去 +// a = 3，表示z = b处，一把无穷大的刀平行于xy面贯穿豆腐切过去 +// a = 1 or 2 or 3，0<=b<=10000 +// 给定一个n*2的二维数组，表示切了n刀 +// 返回豆腐中最大的一块体积是多少 +public class Code02_CutDouFu { + + public static long maxCut(int[][] cuts) { + if (cuts == null || cuts.length == 0) { + return 10000L * 10000L * 10000L; + } + // 0 类型 + // 1 切哪了 + Arrays.sort(cuts, (a, b) -> a[0] != b[0] ? (a[0] - b[0]) : (a[1] - b[1])); + int n = cuts.length; + int i = 0; + int xMaxDiff = 0; + int pre = 0; + while (i < n && cuts[i][0] == 1) { + xMaxDiff = Math.max(xMaxDiff, cuts[i][1] - pre); + pre = cuts[i][1]; + i++; + } + xMaxDiff = Math.max(xMaxDiff, 10000 - pre); + int yMaxDiff = 0; + pre = 0; + while (i < n && cuts[i][0] == 2) { + yMaxDiff = Math.max(yMaxDiff, cuts[i][1] - pre); + pre = cuts[i][1]; + i++; + } + yMaxDiff = Math.max(yMaxDiff, 10000 - pre); + int zMaxDiff = 0; + pre = 0; + while (i < n && cuts[i][0] == 3) { + zMaxDiff = Math.max(zMaxDiff, cuts[i][1] - pre); + pre = cuts[i][1]; + i++; + } + zMaxDiff = Math.max(zMaxDiff, 10000 - pre); + return (long) xMaxDiff * (long) yMaxDiff * (long) zMaxDiff; + } + +} diff --git a/算法周更班/class_2022_03_3_week/Code03_MaxSumOnReverseArray.java b/算法周更班/class_2022_03_3_week/Code03_MaxSumOnReverseArray.java new file mode 100644 index 0000000..4bb6815 --- /dev/null +++ b/算法周更班/class_2022_03_3_week/Code03_MaxSumOnReverseArray.java @@ -0,0 +1,105 @@ +package class_2022_03_3_week; + +// 来自美团 +// 最大子段和是 +// 一个经典问题，即对于一个数组找出其和最大的子数组。 +// 现在允许你在求解该问题之前翻转这个数組的连续一段 +// 如翻转(1,2,3,4,5,6)的第三个到第五个元素組成的子数组得到的是(1,2,5,4,3,6)， +// 则翻转后该数组的最大子段和最大能达到多少？ +// 来自字节 +// 几乎一样的题，来自字节笔试第4题 +// 给定两个数組values和numbers， +// values[i]表示i号宝石的单品价值 +// numbers[i]表示i号宝石的数量 +// i号宝石的总价值 = values[i] * numbers[i] +// 如果有一种魔法，可以翻转任何区间L...R的宝石，也就是改变L..R的宝石排列，变成逆序的 +// 求在允许用一次魔法的情况下，任取一段连续区间，能达到的最大价值 +// 这两个问法解法都几乎一样，区别无非是: +// 美团的: 可进行一次翻转情况下，子数组最大累加和 +// 字节的: 可进行一次翻转情况下，子数组最大价值和 +public class Code03_MaxSumOnReverseArray { + + public static int maxSumReverse1(int[] arr) { + int ans = Integer.MIN_VALUE; + for (int L = 0; L < arr.length; L++) { + for (int R = L; R < arr.length; R++) { + reverse(arr, L, R); + ans = Math.max(ans, maxSum(arr)); + reverse(arr, L, R); + } + } + return ans; + } + + public static void reverse(int[] arr, int L, int R) { + while (L < R) { + int tmp = arr[L]; + arr[L++] = arr[R]; + arr[R--] = tmp; + } + } + + public static int maxSum(int[] arr) { + int pre = arr[0]; + int max = arr[0]; + for (int i = 1; i < arr.length; i++) { + pre = Math.max(arr[i], arr[i] + pre); + max = Math.max(max, pre); + } + return max; + } + + public static int maxSumReverse2(int[] arr) { + int n = arr.length; + int[] prefix = new int[n]; + prefix[n - 1] = arr[n - 1]; + for (int i = n - 2; i >= 0; i--) { + prefix[i] = arr[i] + Math.max(0, prefix[i + 1]); + } + int ans = prefix[0]; + int suffix = arr[0]; + int maxSuffix = suffix; + for (int i = 1; i < n; i++) { + ans = Math.max(ans, maxSuffix + prefix[i]); + suffix = arr[i] + Math.max(0, suffix); + maxSuffix = Math.max(maxSuffix, suffix); + } + ans = Math.max(ans, maxSuffix); + return ans; + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * v) - (int) (Math.random() * v); + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int len = 50; + int value = 20; + int testTime = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * len) + 1; + int[] arr = randomArray(n, value); + int ans1 = maxSumReverse1(arr); + int ans2 = maxSumReverse2(arr); + if (ans1 != ans2) { + System.out.println("出错了!"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_03_3_week/Code04_ArrangeAddGetMax.java b/算法周更班/class_2022_03_3_week/Code04_ArrangeAddGetMax.java new file mode 100644 index 0000000..f7b0de6 --- /dev/null +++ b/算法周更班/class_2022_03_3_week/Code04_ArrangeAddGetMax.java @@ -0,0 +1,150 @@ +package class_2022_03_3_week; + +import java.util.Arrays; + +// 来自美团 +// void add(int L, int R, int C)代表在arr[L...R]上每个数加C +// int get(int L, int R)代表查询arr[L...R]上的累加和 +// 假设你可以在所有操作开始之前，重新排列arr +// 请返回每一次get查询的结果都加在一起最大能是多少 +// 输入参数: +// int[] arr : 原始数组 +// int[][] ops，二维数组每一行解释如下: +// [a,b,c]，如果数组有3个数，表示调用add(a,b,c) +// [a,b]，如果数组有2个数，表示调用get(a,b) +// a和b表示arr范围，范围假设从1开始，不从0开始 +// 输出： +// 假设你可以在开始时重新排列arr，返回所有get操作返回值累计和最大是多少？ +public class Code04_ArrangeAddGetMax { + + public static int maxGets(int[] arr, int[][] ops) { + int n = arr.length; + SegmentTree getTree = new SegmentTree(n); + for (int[] op : ops) { + if (op.length == 2) { + getTree.add(op[0], op[1], 1); + } + } + int[][] getCnts = new int[n][2]; + for (int i = 1, j = 0; i <= n; i++, j++) { + getCnts[j][0] = j; + getCnts[j][1] = getTree.get(i, i); + } + Arrays.sort(getCnts, (a, b) -> a[1] - b[1]); + Arrays.sort(arr); + int[] reArrange = new int[n]; + for (int i = 0; i < n; i++) { + reArrange[getCnts[i][0]] = arr[i]; + } + SegmentTree st = new SegmentTree(reArrange); + int ans = 0; + for (int[] op : ops) { + if (op.length == 3) { + st.add(op[0], op[1], op[2]); + } else { + ans += st.get(op[0], op[1]); + } + } + return ans; + } + + public static class SegmentTree { + private int n; + private int[] arr; + private int[] sum; + private int[] lazy; + + public SegmentTree(int size) { + n = size + 1; + sum = new int[n << 2]; + lazy = new int[n << 2]; + n--; + } + + public SegmentTree(int[] origin) { + n = origin.length + 1; + arr = new int[n]; + for (int i = 1; i < n; i++) { + arr[i] = origin[i - 1]; + } + sum = new int[n << 2]; + lazy = new int[n << 2]; + build(1, --n, 1); + } + + private void build(int l, int r, int rt) { + if (l == r) { + sum[rt] = arr[l]; + return; + } + int mid = (l + r) >> 1; + build(l, mid, rt << 1); + build(mid + 1, r, rt << 1 | 1); + pushUp(rt); + } + + private void pushUp(int rt) { + sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; + } + + private void pushDown(int rt, int ln, int rn) { + if (lazy[rt] != 0) { + lazy[rt << 1] += lazy[rt]; + sum[rt << 1] += lazy[rt] * ln; + lazy[rt << 1 | 1] += lazy[rt]; + sum[rt << 1 | 1] += lazy[rt] * rn; + lazy[rt] = 0; + } + } + + public void add(int L, int R, int C) { + add(L, R, C, 1, n, 1); + } + + private void add(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + sum[rt] += C * (r - l + 1); + lazy[rt] += C; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + add(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + add(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + public int get(int L, int R) { + return query(L, R, 1, n, 1); + } + + private int query(int L, int R, int l, int r, int rt) { + if (L <= l && r <= R) { + return sum[rt]; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + int ans = 0; + if (L <= mid) { + ans += query(L, R, l, mid, rt << 1); + } + if (R > mid) { + ans += query(L, R, mid + 1, r, rt << 1 | 1); + } + return ans; + } + + } + + public static void main(String[] args) { + int[] arr = { 1, 2, 3, 4, 5 }; + int[][] ops = { { 1, 3 }, { 2, 4 }, { 1, 5 } }; + System.out.println(maxGets(arr, ops)); + + } + +} diff --git a/算法周更班/class_2022_03_3_week/Code05_EatFish.java b/算法周更班/class_2022_03_3_week/Code05_EatFish.java new file mode 100644 index 0000000..7ab2b89 --- /dev/null +++ b/算法周更班/class_2022_03_3_week/Code05_EatFish.java @@ -0,0 +1,106 @@ +package class_2022_03_3_week; + +// 来自bilibili +// 现在有N条鱼，每条鱼的体积为Ai，从左到右排列，数组arr给出 +// 每一轮，左边的大鱼一定会吃掉右边比自己小的第一条鱼， +// 并且每条鱼吃比自己小的鱼的事件是同时发生的。 +// 返回多少轮之后，鱼的数量会稳定 +// 注意：6 6 3 3 +// 第一轮过后 : +// 对于两个6来说，右边比自己小的第一条鱼都是第1个3，所以只有这个3被吃掉， +// 数组变成 : 6 6 3（第2个3） +// 第二轮过后 : 6 6 +// 返回2 +public class Code05_EatFish { + + public static int minTurns1(int[] arr) { + int ans = 0; + for (;; ans++) { + int[] rest = eatRest(arr); + if (arr.length == rest.length) { + break; + } + arr = rest; + } + return ans; + } + + public static int[] eatRest(int[] arr) { + if (arr.length == 0) { + return new int[0]; + } + int n = arr.length; + boolean[] delete = new boolean[n]; + int len = n; + for (int i = 0; i < n; i++) { + for (int j = i + 1; j < n; j++) { + if (arr[i] > arr[j]) { + if (!delete[j]) { + delete[j] = true; + len--; + } + break; + } + } + } + int[] rest = new int[len]; + for (int i = 0, j = 0; i < n; i++) { + if (!delete[i]) { + rest[j++] = arr[i]; + } + } + return rest; + } + + public static int minTurns2(int[] arr) { + int n = arr.length; + int[][] stack = new int[n][2]; + int stackSize = 0; + int ans = 0; + for (int i = n - 1; i >= 0; i--) { + int curAns = 0; + while (stackSize > 0 && stack[stackSize - 1][0] < arr[i]) { + curAns = Math.max(curAns + 1, stack[--stackSize][1]); + } + stack[stackSize][0] = arr[i]; + stack[stackSize++][1] = curAns; + ans = Math.max(ans, curAns); + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * v); + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int len = 50; + int value = 20; + int testTime = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * len) + 1; + int[] arr = randomArray(n, value); + int ans1 = minTurns1(arr); + int ans2 = minTurns2(arr); + if (ans1 != ans2) { + System.out.println("出错了!"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_03_3_week/Code06_FinancialProduct.java b/算法周更班/class_2022_03_3_week/Code06_FinancialProduct.java new file mode 100644 index 0000000..05a6219 --- /dev/null +++ b/算法周更班/class_2022_03_3_week/Code06_FinancialProduct.java @@ -0,0 +1,51 @@ +package class_2022_03_3_week; + +import java.util.Arrays; + +// 来自银联编程比赛 +// 某公司计划推出一批投资项目。 product[i] = price 表示第 i 个理财项目的投资金额 price 。 +// 客户在按需投资时，需要遵循以下规则： +// 客户在首次对项目 product[i] 投资时，需要投入金额 price +// 对已完成首次投资的项目 product[i] 可继续追加投入， +// 但追加投入的金额需小于上一次对该项目的投入(追加投入为大于 0 的整数) +// 为控制市场稳定，每人交易次数不得大于 limit。(首次投资和追加投入均记作 1 次交易) +// 若对所有理财项目中最多进行 limit 次交易，使得投入金额总和最大，请返回这个最大值的总和。 +// 测试链接 : https://leetcode-cn.com/contest/cnunionpay-2022spring/problems/I4mOGz/ +public class Code06_FinancialProduct { + + public static long mod = 1000000007L; + + public int maxInvestment(int[] arr, int limit) { + Arrays.sort(arr); + int n = arr.length; + long ans = 0; + int r = n - 1; + int l = r; + while (limit > 0 && r != -1) { + while (l >= 0 && arr[l] == arr[r]) { + l--; + } + int big = arr[r]; + int small = l == -1 ? 0 : arr[l]; + int teams = n - l - 1; + int all = (big - small) * teams; + if (limit >= all) { + ans += get(big, small + 1, teams); + ans %= mod; + limit -= all; + } else { + int a = limit / teams; + ans += get(big, big - a + 1, teams) + (long) (big - a) * (long) (limit % teams); + ans %= mod; + limit = 0; + } + r = l; + } + return (int) (ans % mod); + } + + public static long get(long up, long down, long num) { + return num * ((up + down) * (up - down + 1) / 2); + } + +} diff --git a/算法周更班/class_2022_03_3_week/Code07_CoopDevelop.java b/算法周更班/class_2022_03_3_week/Code07_CoopDevelop.java new file mode 100644 index 0000000..d931fa6 --- /dev/null +++ b/算法周更班/class_2022_03_3_week/Code07_CoopDevelop.java @@ -0,0 +1,51 @@ +package class_2022_03_3_week; + +import java.util.HashMap; + +// 来自银联编程比赛 +// 为了不断提高用户使用的体验，开发团队正在对产品进行全方位的开发和优化。 +// 已知开发团队共有若干名成员，skills[i] 表示第 i 名开发人员掌握技能列表。 +// 如果两名成员各自拥有至少一门对方未拥有的技能，则这两名成员可以「合作开发」。 +// 请返回当前有多少对开发成员满足「合作开发」的条件。 +// 由于答案可能很大，请你返回答案对 10^9 + 7 取余的结果。 +// 测试链接 : https://leetcode-cn.com/contest/cnunionpay-2022spring/problems/lCh58I/ +public class Code07_CoopDevelop { + + public static long mod = 1000000007L; + + public static int coopDevelop(int[][] skills) { + int n = skills.length; + // key : 子集 + // value : 个数 + HashMap noFullSetsNums = new HashMap<>(); + for (int[] people : skills) { + fillNoFullMap(people, 0, 0, true, noFullSetsNums); + } + HashMap cntsNums = new HashMap<>(); + long minus = 0L; + for (int[] people : skills) { + long status = 0L; + for (int skill : people) { + status = (status << 10) | skill; + } + minus += noFullSetsNums.getOrDefault(status, 0L); + minus += cntsNums.getOrDefault(status, 0L); + cntsNums.put(status, cntsNums.getOrDefault(status, 0L) + 1); + } + long ans = (long) n * (long) (n - 1) / 2L; + return (int) ((ans - minus) % mod); + } + + public static void fillNoFullMap(int[] people, int i, long status, boolean full, + HashMap noFullSetsNums) { + if (i == people.length) { + if (!full) { + noFullSetsNums.put(status, noFullSetsNums.getOrDefault(status, 0L) + 1); + } + } else { + fillNoFullMap(people, i + 1, status, false, noFullSetsNums); + fillNoFullMap(people, i + 1, (status << 10) | people[i], full, noFullSetsNums); + } + } + +} diff --git a/算法周更班/class_2022_03_4_week/Code01_ArrangeJob.java b/算法周更班/class_2022_03_4_week/Code01_ArrangeJob.java new file mode 100644 index 0000000..8cd7e9b --- /dev/null +++ b/算法周更班/class_2022_03_4_week/Code01_ArrangeJob.java @@ -0,0 +1,66 @@ +package class_2022_03_4_week; + +import java.util.Arrays; +import java.util.PriorityQueue; + +// 来自学员的考试 +// 来自华为 +// 给定一个n*2的二维数组，表示有n个任务 +// 一个信息是任务能够开始做的时间，另一个信息是任务的结束期限，后者一定大于前者，且数值上都是正数 +// 你作为单线程的人，不能并行处理任务，但是每个任务都只需要一个单位时间完成 +// 你需要将所有任务的执行时间，位于开始做的时间和最后期限之间 +// 返回你能否做到这一点 +public class Code01_ArrangeJob { + + // 1 开 7 + // 5 闭 end没有用！ + public static class TimePoint { + // 时间 + public int time; + public int end; + // add = true time 任务的添加时间 + // add = false time 任务的结束时间 + public boolean add; + + public TimePoint(int t, int e, boolean a) { + time = t; + end = e; + add = a; + } + + } + + public static boolean canDo(int[][] jobs) { + if (jobs == null || jobs.length < 2) { + return true; + } + int n = jobs.length; + TimePoint[] arr = new TimePoint[n << 1]; + for (int i = 0; i < n; i++) { + arr[i] = new TimePoint(jobs[i][0], jobs[i][1], true); + arr[i + n] = new TimePoint(jobs[i][1], jobs[i][1], false); + } + Arrays.sort(arr, (a, b) -> a.time - b.time); + PriorityQueue heap = new PriorityQueue<>(); + // 经过一个一个的时间点，遭遇事件：添加时间、检查时间 + for (int i = 0, lastTime = arr[0].time; i < arr.length; i++) { + if (arr[i].add) { + heap.add(arr[i].end); + } else { // 检查时间 + int curTime = arr[i].time; + for (int j = lastTime; j < curTime; j++) { + if (heap.isEmpty()) { + break; + } + heap.poll(); + } + if (heap.peek() <= curTime) { + return false; + } + lastTime = curTime; + } + } + return true; + } + +} diff --git a/算法周更班/class_2022_03_4_week/Code02_BuyGoodsHaveDiscount.java b/算法周更班/class_2022_03_4_week/Code02_BuyGoodsHaveDiscount.java new file mode 100644 index 0000000..5d768a8 --- /dev/null +++ b/算法周更班/class_2022_03_4_week/Code02_BuyGoodsHaveDiscount.java @@ -0,0 +1,89 @@ +package class_2022_03_4_week; + +// 来自字节内部训练营 +// 某公司游戏平台的夏季特惠开始了，你决定入手一些游戏。现在你一共有X元的预算。 +// 该平台上所有的 n 个游戏均有折扣，标号为 i 的游戏的原价a_i元，现价只要b_i元 +// 也就是说该游戏可以优惠 a_i - b_i，并且你购买该游戏能获得快乐值为 w_i +// 由于优惠的存在，你可能做出一些冲动消费导致最终买游戏的总费用超过预算， +// 只要满足 : 获得的总优惠金额不低于超过预算的总金额 +// 那在心理上就不会觉得吃亏。 +// 现在你希望在心理上不觉得吃亏的前提下，获得尽可能多的快乐值。 +// 测试链接 : https://leetcode-cn.com/problems/tJau2o/ +// 提交以下的code，将主类名字改成"Main" +// 可以直接通过 +import java.util.Scanner; + +public class Code02_BuyGoodsHaveDiscount { + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + while (sc.hasNext()) { + int n = sc.nextInt(); + int money = sc.nextInt(); + int[] costs = new int[n]; + long[] values = new long[n]; + int size = 0; + long ans = 0; + for (int i = 0; i < n; i++) { + // 打折前 + int pre = sc.nextInt(); + // 打折后 + int pos = sc.nextInt(); + // 满足度 + int happy = sc.nextInt(); + // 节省的钱(save) = 打折前(pre) - 打折后(pos) + int save = pre - pos; + // 带来的好处(well) = 节省的钱 - 打折后(pos) + int well = save - pos; + // 比如，一件"一定要买的商品": + // 预算 = 100，商品原价 = 10，打折后 = 3 + // 那么好处 = (10 - 3) - 3 = 4 + // 所以，这件商品把预算增加到了104，一定要买 + // 接下来，比如一件"需要考虑的商品"，预算 = 104，商品原价 = 10，打折后 = 8 + // 那么好处 = (10 - 8) - 8 = -6 + // 这件商品，就花掉6元！ + // 也就是说，以后花的不是打折后的值，是"坏处" + int cost = -well; + if (well >= 0) { + money += well; + ans += happy; + } else { + costs[size] = cost; + values[size++] = happy; + } + + } + long[][] dp = new long[size + 1][money + 1]; + for (int a = 0; a <= size; a++) { + for (int b = 0; b <= money; b++) { + dp[a][b] = -2; + } + } + ans += process(costs, values, size, 0, money, dp); + System.out.println(ans); + } + sc.close(); + } + + public static long process(int[] costs, long[] values, int size, int i, int money, long[][] dp) { + if (money < 0) { + return -1; + } + if (i == size) { + return 0; + } + if (dp[i][money] != -2) { + return dp[i][money]; + } + long p1 = process(costs, values, size, i + 1, money, dp); + long p2 = -1; + long next = process(costs, values, size, i + 1, money - costs[i], dp); + if (next != -1) { + p2 = values[i] + next; + } + long ans = Math.max(p1, p2); + dp[i][money] = ans; + return ans; + } + +} \ No newline at end of file diff --git a/算法周更班/class_2022_03_4_week/Code03_MinTowNumberSumABS.java b/算法周更班/class_2022_03_4_week/Code03_MinTowNumberSumABS.java new file mode 100644 index 0000000..1a46cf2 --- /dev/null +++ b/算法周更班/class_2022_03_4_week/Code03_MinTowNumberSumABS.java @@ -0,0 +1,159 @@ +package class_2022_03_4_week; + +import java.util.Arrays; + +// 来自学员问题 +// 给定一个数组arr，可能有正、有负、有0，无序 +// 只能挑选两个数字，想尽量让两个数字加起来的绝对值尽量小 +// 返回可能的最小的值 +public class Code03_MinTowNumberSumABS { + + public static int minSumABS1(int[] arr) { + if (arr == null || arr.length < 2) { + return -1; + } + int ans = Integer.MAX_VALUE; + for (int i = 0; i < arr.length; i++) { + for (int j = i + 1; j < arr.length; j++) { + ans = Math.min(ans, Math.abs(arr[i] + arr[j])); + } + } + return ans; + } + + public static int minSumABS2(int[] arr) { + if (arr == null || arr.length < 2) { + return -1; + } + Arrays.sort(arr); + int n = arr.length; + int split = -1; + for (int i = 0; i < n; i++) { + if (arr[i] >= 0) { + split = i; + break; + } + } + if (split == 0) { + return arr[0] + arr[1]; + } + if (split == -1) { + return -arr[n - 2] - arr[n - 1]; + } + int ans = Integer.MAX_VALUE; + if (split + 1 < n) { + ans = arr[split] + arr[split + 1]; + } + if (split - 2 >= 0) { + ans = Math.min(ans, -arr[split - 1] - arr[split - 2]); + } + for (int i = 0; i < split; i++) { + ans = Math.min(ans, Math.abs(arr[i] + near(arr, split, -arr[i]))); + } + return ans; + } + + // arr[start...]是有序的 + // 返回离num最近的数字 + public static int near(int[] arr, int start, int num) { + int l = start; + int r = arr.length - 1; + int m = 0; + int ans = -1; + while (l <= r) { + m = (l + r) / 2; + if (arr[m] <= num) { + ans = m; + l = m + 1; + } else { + r = m - 1; + } + } + if (ans == -1) { + return arr[start]; + } else { + if (ans == arr.length - 1) { + return arr[arr.length - 1]; + } else { + if (Math.abs(arr[ans] - num) <= Math.abs(arr[ans + 1] - num)) { + return arr[ans]; + } else { + return arr[ans + 1]; + } + } + } + } + + public static int minSumABS3(int[] arr) { + if (arr == null || arr.length < 2) { + return -1; + } + Arrays.sort(arr); + int n = arr.length; + int split = -1; + for (int i = 0; i < n; i++) { + if (arr[i] >= 0) { + split = i; + break; + } + } + if (split == 0) { + return arr[0] + arr[1]; + } + if (split == -1) { + return -arr[n - 2] - arr[n - 1]; + } + int ans = Integer.MAX_VALUE; + if (split + 1 < n) { + ans = arr[split] + arr[split + 1]; + } + if (split - 2 >= 0) { + ans = Math.min(ans, -arr[split - 1] - arr[split - 2]); + } + int r = n - 1; + for (int l = 0; l < split; l++) { + ans = Math.min(ans, Math.abs(arr[l] + arr[r])); + while (r - 1 >= split && Math.abs(arr[l] + arr[r]) >= Math.abs(arr[l] + arr[r - 1])) { + ans = Math.min(ans, Math.abs(arr[l] + arr[--r])); + } + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * v) - (int) (Math.random() * v); + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int len = 50; + int value = 500; + int testTime = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * len) + 1; + int[] arr = randomArray(n, value); + int ans1 = minSumABS1(arr); + int ans2 = minSumABS2(arr); + int ans3 = minSumABS3(arr); + if (ans1 != ans2 || ans1 != ans3) { + System.out.println("出错了!"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println(ans1); + System.out.println(ans2); + System.out.println(ans3); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_03_4_week/Code04_JumpToTargets.java b/算法周更班/class_2022_03_4_week/Code04_JumpToTargets.java new file mode 100644 index 0000000..88a286e --- /dev/null +++ b/算法周更班/class_2022_03_4_week/Code04_JumpToTargets.java @@ -0,0 +1,45 @@ +package class_2022_03_4_week; + +// 来自字节 +// 一开始在0位置，每一次都可以向左或者向右跳 +// 第i次能向左或者向右跳严格的i步 +// 请问从0到x位置，至少跳几次可以到达 +// 字节考的问题其实就是这个问题 +// leetcode测试链接 : https://leetcode.com/problems/reach-a-number/ +public class Code04_JumpToTargets { + + public static int reachNumber(long target) { + if (target == 0) { + return 0; + } + target = Math.abs(target); + long l = 0; + long r = target; + long m = 0; + long near = 0; + while (l <= r) { + m = (l + r) / 2; + if (sum(m) >= target) { + near = m; + r = m - 1; + } else { + l = m + 1; + } + } + if (sum(near) == target) { + return (int)near; + } + if (((sum(near) - target) & 1) == 1) { + near++; + } + if (((sum(near) - target) & 1) == 1) { + near++; + } + return (int)near; + } + + public static long sum(long n) { + return (n * (n + 1)) / 2; + } + +} diff --git a/算法周更班/class_2022_03_4_week/Code05_HowManyWaysFromBottomToTop.java b/算法周更班/class_2022_03_4_week/Code05_HowManyWaysFromBottomToTop.java new file mode 100644 index 0000000..2816a0e --- /dev/null +++ b/算法周更班/class_2022_03_4_week/Code05_HowManyWaysFromBottomToTop.java @@ -0,0 +1,81 @@ +package class_2022_03_4_week; + +// 来自理想汽车 +// a -> b，代表a在食物链中被b捕食 +// 给定一个有向无环图，返回 +// 这个图中从最初级动物到最顶级捕食者的食物链有几条 +// 线上测试链接 : https://www.luogu.com.cn/problem/P4017 +// 以下代码都提交，提交时把主类名改成"Main"即可 +// 注意：洛谷测试平台对java提交非常不友好，空间限制可能会卡住，C++的提交就完全不会 +// 所以提交时如果显示失败，就多提交几次，一定是能成功的 +// 这道题本身就是用到拓扑排序，没什么特别的 +// 但是为了提交能通过，逼迫我在空间上做的优化值得好好说一下，可以推广到其他题目 +import java.util.Arrays; +import java.util.Scanner; + +public class Code05_HowManyWaysFromBottomToTop { + + public static int[] in = new int[5001]; + public static boolean[] out = new boolean[5001]; + public static int[] lines = new int[5001]; + public static int[] headEdge = new int[5001]; + public static int[] queue = new int[5001]; + public static int mod = 80112002; + public static int n = 0; + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + while (sc.hasNext()) { + Arrays.fill(in, 0); + Arrays.fill(out, false); + Arrays.fill(lines, 0); + Arrays.fill(headEdge, 0); + n = sc.nextInt(); + int m = sc.nextInt(); + int[] preEdge = new int[m + 1]; + int[] edgesTo = new int[m + 1]; + for (int i = 1; i <= m; i++) { + int from = sc.nextInt(); + int to = sc.nextInt(); + edgesTo[i] = to; + preEdge[i] = headEdge[from]; + headEdge[from] = i; + out[from] = true; + in[to]++; + } + System.out.println(howManyWays(preEdge, edgesTo)); + } + sc.close(); + } + + public static int howManyWays(int[] preEdge, int[] edgesTo) { + int ql = 0; + int qr = 0; + for (int i = 1; i <= n; i++) { + if (in[i] == 0) { + queue[qr++] = i; + lines[i] = 1; + } + } + while (ql < qr) { + int cur = queue[ql++]; + int edge = headEdge[cur]; + while (edge != 0) { + int next = edgesTo[edge]; + lines[next] = (lines[next] + lines[cur]) % mod; + if (--in[next] == 0) { + queue[qr++] = next; + } + edge = preEdge[edge]; + } + } + int ans = 0; + for (int i = 1; i <= n; i++) { + if (!out[i]) { + ans = (ans + lines[i]) % mod; + } + } + return ans; + } + +} diff --git a/算法周更班/class_2022_03_4_week/Code06_LongestContinuousTrees.java b/算法周更班/class_2022_03_4_week/Code06_LongestContinuousTrees.java new file mode 100644 index 0000000..0e77cc9 --- /dev/null +++ b/算法周更班/class_2022_03_4_week/Code06_LongestContinuousTrees.java @@ -0,0 +1,21 @@ +package class_2022_03_4_week; + +// 来自学员问题 +// 给定一个数字n，表示一开始有编号1~n的树木，列成一条直线 +// 给定一个有序数组arr，表示现在哪些树已经没了，arr[i]一定在[1,n]范围 +// 给定一个数字m，表示你可以补种多少棵树 +// 返回补种之后，最长的连续树木，有多少棵 +public class Code06_LongestContinuousTrees { + + public static int longestTrees(int n, int m, int[] arr) { + int ans = 0; + int start = 1; + for (int i = 0, j = m; j < arr.length; i++, j++) { + ans = Math.max(ans, arr[j] - start); + start = arr[i] + 1; + } + ans = Math.max(ans, n - start + 1); + return ans; + } + +} diff --git a/算法周更班/class_2022_03_4_week/Code07_IrregularSudoku.java b/算法周更班/class_2022_03_4_week/Code07_IrregularSudoku.java new file mode 100644 index 0000000..0c2e7f4 --- /dev/null +++ b/算法周更班/class_2022_03_4_week/Code07_IrregularSudoku.java @@ -0,0 +1,127 @@ +package class_2022_03_4_week; + +// 来自网易 +// 不规则数独问题 +// 3*3填数独 +// 每一行要填1~3 +// 每一列要填1~3 +// 3*3的区域会拆分成不规则的三个集团区域 +// 每个集团区域3个格子 +// 每个集团的区域都一定是一个连在一起的整体，可能不规则 +// 每个集团内要填1~3 +// 如果只有一个解返回"Unique"，如果有多个解返回"Multiple"，如果没有解返回"No" +// 解析请看，大厂刷题班，28节，leetcode原题，数独那两个题 +// 本题就是改变一下桶的归属而已 +public class Code07_IrregularSudoku { + + // sudoku[i][j] == 0，代表这个位置没有数字，需要填 + // sudoku[i][j] != 0，代表这个位置有数字，不需要填 + // map[0] = {0,0}、{0,1}、{1,0} 代表0集团拥有的三个点 + // map[i] = {a,b}、{c,d}、{e,f} 代表i集团拥有的三个点 + public static String solution(int[][] sudoku, int[][][] map) { + boolean[][] row = new boolean[3][4]; + boolean[][] col = new boolean[3][4]; + boolean[][] bucket = new boolean[3][4]; + int[][] own = new int[3][3]; + for (int i = 0; i < 3; i++) { + for (int[] arr : map[i]) { + own[arr[0]][arr[1]] = i; + } + } + for (int i = 0; i < 3; i++) { + for (int j = 0; j < 3; j++) { + if (sudoku[i][j] != 0) { + row[i][sudoku[i][j]] = true; + col[j][sudoku[i][j]] = true; + bucket[own[i][j]][sudoku[i][j]] = true; + } + } + } + int ans = process(sudoku, 0, 0, row, col, bucket, own); + return ans == 0 ? "No" : (ans == 1 ? "Unique" : "Multiple"); + } + + public static int process(int[][] sudoku, int i, int j, boolean[][] row, boolean[][] col, boolean[][] bucket, + int[][] own) { + if (i == 3) { + return 1; + } + int nexti = j != 2 ? i : i + 1; + int nextj = j != 2 ? j + 1 : 0; + if (sudoku[i][j] != 0) { + return process(sudoku, nexti, nextj, row, col, bucket, own); + } else { + int ans = 0; + int bid = own[i][j]; + for (int num = 1; num <= 3; num++) { + if ((!row[i][num]) && (!col[j][num]) && (!bucket[bid][num])) { + row[i][num] = true; + col[j][num] = true; + bucket[bid][num] = true; + sudoku[i][j] = num; + ans += process(sudoku, nexti, nextj, row, col, bucket, own); + row[i][num] = false; + col[j][num] = false; + bucket[bid][num] = false; + sudoku[i][j] = 0; + if (ans > 1) { + return ans; + } + } + } + return ans; + } + } + + public static void main(String[] args) { + int[][] sudoku1 = { + { 0, 2, 0 }, + { 1, 0, 2 }, + { 0, 0, 0 } + }; + int[][][] map1 = { + { { 0, 0 }, { 0, 1 }, { 1, 0 } }, + { { 0, 2 }, { 1, 1 }, { 1, 2 } }, + { { 2, 0 }, { 2, 1 }, { 2, 2 } } + }; + System.out.println(solution(sudoku1, map1)); + + int[][] sudoku2 = { + { 0, 0, 3 }, + { 0, 0, 0 }, + { 0, 0, 0 } + }; + int[][][] map2 = { + { { 0, 0 }, { 1, 0 }, { 1, 1 } }, + { { 0, 1 }, { 0, 2 }, { 1, 2 } }, + { { 2, 0 }, { 2, 1 }, { 2, 2 } } + }; + System.out.println(solution(sudoku2, map2)); + + int[][] sudoku3 = { + { 0, 0, 3 }, + { 1, 0, 0 }, + { 0, 0, 2 } + }; + int[][][] map3 = { + { { 0, 0 }, { 1, 0 }, { 1, 1 } }, + { { 0, 1 }, { 0, 2 }, { 1, 2 } }, + { { 2, 0 }, { 2, 1 }, { 2, 2 } } + }; + System.out.println(solution(sudoku3, map3)); + + int[][] sudoku4 = { + { 3, 0, 3 }, + { 1, 0, 0 }, + { 0, 0, 2 } + }; + int[][][] map4 = { + { { 0, 0 }, { 1, 0 }, { 1, 1 } }, + { { 0, 1 }, { 0, 2 }, { 1, 2 } }, + { { 2, 0 }, { 2, 1 }, { 2, 2 } } + }; + System.out.println(solution(sudoku4, map4)); + + } + +} diff --git a/算法周更班/class_2022_03_4_week/Code08_EggXtoY.java b/算法周更班/class_2022_03_4_week/Code08_EggXtoY.java new file mode 100644 index 0000000..424af25 --- /dev/null +++ b/算法周更班/class_2022_03_4_week/Code08_EggXtoY.java @@ -0,0 +1,82 @@ +package class_2022_03_4_week; + +// 来自学员问题 +// 大妈一开始手上有x个鸡蛋，她想让手上的鸡蛋数量变成y +// 操作1 : 从仓库里拿出1个鸡蛋到手上，x变成x+1个 +// 操作2 : 如果手上的鸡蛋数量是3的整数倍，大妈可以直接把三分之二的鸡蛋放回仓库，手里留下三分之一 +// 返回从x到y的最小操作次数 +// 1 <= x,y <= 10^18 +// 练一下，用平凡解做限制 +public class Code08_EggXtoY { + + // 彻底贪心！ + public static int minTimes1(int x, int y) { + if (x <= y) { + return y - x; + } + // 0 0 + // 1 2 + // 2 1 + int mod = x % 3; + // 鸡蛋拿到3的整数倍，需要耗费的行动点数 + int need = mod == 0 ? 0 : (mod == 1 ? 2 : 1); + return need + 1 + minTimes1((x + 2) / 3, y); + } + + public static int minTimes2(int x, int y) { + if (x <= y) { + return y - x; + } + int limit = minTimes1(x, y); + int[][] dp = new int[x + limit + 1][limit + 1]; + return process(x, y, 0, limit, dp); + } + + // 当前鸡蛋数量cur，目标aim + // 之前已经用了多少行动点，pre + // limit : 一定行动点，超过limit，不需要尝试了！ + public static int process(int cur, int aim, int pre, int limit, int[][] dp) { + if (pre > limit) { + return Integer.MAX_VALUE; + } + if (dp[cur][pre] != 0) { + return dp[cur][pre]; + } + int ans = 0; + if (cur == aim) { + ans = pre; + } else { + int p1 = process(cur + 1, aim, pre + 1, limit, dp); + int p2 = Integer.MAX_VALUE; + if (cur % 3 == 0) { + p2 = process(cur / 3, aim, pre + 1, limit, dp); + } + ans = Math.min(p1, p2); + } + dp[cur][pre] = ans; + return ans; + } + + public static void main(String[] args) { + int max = 3000; + int testTime = 500; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int x = (int) (Math.random() * max) + 1; + int y = (int) (Math.random() * max) + 1; + int ans1 = minTimes1(x, y); + int ans2 = minTimes2(x, y); + if (ans1 != ans2) { + System.out.println("出错了!"); + System.out.println("x = " + x); + System.out.println("y = " + y); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + + } + +} diff --git a/算法周更班/class_2022_03_5_week/Code01_KMAlgorithm.java b/算法周更班/class_2022_03_5_week/Code01_KMAlgorithm.java new file mode 100644 index 0000000..adc3ced --- /dev/null +++ b/算法周更班/class_2022_03_5_week/Code01_KMAlgorithm.java @@ -0,0 +1,166 @@ +// 注意本文件中，graph不是邻接矩阵的含义，而是一个二部图 +// 在长度为N的邻接矩阵matrix中，所有的点有N个，matrix[i][j]表示点i到点j的距离或者权重 +// 而在二部图graph中，所有的点有2*N个，行所对应的点有N个，列所对应的点有N个 +// 而且认为，行所对应的点之间是没有路径的，列所对应的点之间也是没有路径的！ + +package class_2022_03_5_week; + +import java.util.Arrays; + +// km算法 +// O(N^3)，最大匹配问题，最优的解！ +public class Code01_KMAlgorithm { + + // 暴力解 + public static int right(int[][] graph) { + int N = graph.length; + int[] to = new int[N]; + for (int i = 0; i < N; i++) { + to[i] = 1; + } + return process(0, to, graph); + } + + public static int process(int from, int[] to, int[][] graph) { + if (from == graph.length) { + return 0; + } + int ans = 0; + for (int i = 0; i < to.length; i++) { + if (to[i] == 1) { + to[i] = 0; + ans = Math.max(ans, graph[from][i] + process(from + 1, to, graph)); + to[i] = 1; + } + } + return ans; + } + + public static int km(int[][] graph) { + int N = graph.length; + int[] match = new int[N]; + int[] lx = new int[N]; + int[] ly = new int[N]; + // dfs过程中，碰过的点！ + boolean[] x = new boolean[N]; + boolean[] y = new boolean[N]; + // 降低的预期！ + // 公主上，打一个，降低预期的值，只维持最小！ + int[] slack = new int[N]; + int invalid = Integer.MAX_VALUE; + for (int i = 0; i < N; i++) { + match[i] = -1; + lx[i] = -invalid; + for (int j = 0; j < N; j++) { + lx[i] = Math.max(lx[i], graph[i][j]); + } + ly[i] = 0; + } + for (int from = 0; from < N; from++) { + for (int i = 0; i < N; i++) { + slack[i] = invalid; + } + Arrays.fill(x, false); + Arrays.fill(y, false); + // dfs() : from王子，能不能不降预期，匹配成功！ + // 能：dfs返回true！ + // 不能：dfs返回false！ + while (!dfs(from, x, y, lx, ly, match, slack, graph)) { + // 刚才的dfs，失败了！ + // 需要拿到，公主的slack里面，预期下降幅度的最小值！ + int d = invalid; + for (int i = 0; i < N; i++) { + if (!y[i] && slack[i] < d) { + d = slack[i]; + } + } + // 按照最小预期来调整预期 + for (int i = 0; i < N; i++) { + if (x[i]) { + lx[i] = lx[i] - d; + } + if (y[i]) { + ly[i] = ly[i] + d; + } + } + Arrays.fill(x, false); + Arrays.fill(y, false); + // 然后回到while里，再次尝试 + } + } + int ans = 0; + for (int i = 0; i < N; i++) { + ans += (lx[i] + ly[i]); + } + return ans; + } + + // from, 当前的王子 + // x，王子碰没碰过 + // y, 公主碰没碰过 + // lx，所有王子的预期 + // ly, 所有公主的预期 + // match，所有公主，之前的分配，之前的爷们！ + // slack，连过，但没允许的公主，最小下降的幅度 + // map，报价，所有王子对公主的报价 + // 返回，from号王子，不降预期能不能配成！ + public static boolean dfs(int from, boolean[] x, boolean[] y, int[] lx, int[] ly, int[] match, int[] slack, + int[][] map) { + int N = map.length; + x[from] = true; + for (int to = 0; to < N; to++) { + if (!y[to]) { // 只有没dfs过的公主，才会去尝试 + int d = lx[from] + ly[to] - map[from][to]; + if (d != 0) {// 如果当前的路不符合预期，更新公主的slack值 + slack[to] = Math.min(slack[to], d); + } else { // 如果当前的路符合预期，尝试直接拿下，或者抢夺让之前的安排倒腾去 + y[to] = true; + if (match[to] == -1 || dfs(match[to], x, y, lx, ly, match, slack, map)) { + match[to] = from; + return true; + } + } + } + } + return false; + } + + // 为了测试 + public static int[][] randomGraph(int N, int V) { + int[][] graph = new int[N][N]; + for (int i = 0; i < N; i++) { + for (int j = i + 1; j < N; j++) { + int num = (int) (Math.random() * V); + graph[i][j] = num; + graph[j][i] = num; + } + } + return graph; + } + + // 为了测试 + public static void main(String[] args) { + int N = 10; + int V = 20; + int testTime = 100; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int[][] graph = randomGraph(N, V); + int ans1 = right(graph); + int ans2 = km(graph); + if (ans1 != ans2) { + System.out.println("Oops!"); + for (int r = 0; r < graph.length; r++) { + for (int c = 0; c < graph.length; c++) { + System.out.print(graph[r][c] + " "); + } + System.out.println(); + } + System.out.println(ans1); + System.out.println(ans2); + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_03_5_week/Code02_ToAllSpace.java b/算法周更班/class_2022_03_5_week/Code02_ToAllSpace.java new file mode 100644 index 0000000..00e499a --- /dev/null +++ b/算法周更班/class_2022_03_5_week/Code02_ToAllSpace.java @@ -0,0 +1,213 @@ +package class_2022_03_5_week; + +import java.util.Arrays; + +// 来自微软 +// 在N*N的正方形棋盘中，有N*N个棋子，那么每个格子正好可以拥有一个棋子 +// 但是现在有些棋子聚集到一个格子上了，比如： +// 2 0 3 +// 0 1 0 +// 3 0 0 +// 如上的二维数组代表，一共3*3个格子 +// 但是有些格子有2个棋子、有些有3个、有些有1个、有些没有 +// 请你用棋子移动的方式，让每个格子都有一个棋子 +// 每个棋子可以上、下、左、右移动，每移动一步算1的代价 +// 返回最小的代价 +public class Code02_ToAllSpace { + + // 暴力解 + // 作为对数器 + public static int minDistance1(int[][] map) { + int n = 0; + int m = 0; + for (int i = 0; i < map.length; i++) { + for (int j = 0; j < map[0].length; j++) { + n += Math.max(0, map[i][j] - 1); + m += map[i][j] == 0 ? 1 : 0; + } + } + if (n != m || n == 0) { + return 0; + } + int[][] nodes = new int[n][2]; + int[][] space = new int[m][2]; + n = 0; + m = 0; + for (int i = 0; i < map.length; i++) { + for (int j = 0; j < map[0].length; j++) { + for (int k = 2; k <= map[i][j]; k++) { + nodes[n][0] = i; + nodes[n++][1] = j; + } + if (map[i][j] == 0) { + space[m][0] = i; + space[m++][1] = j; + } + } + } + return process1(nodes, 0, space); + } + + public static int process1(int[][] nodes, int index, int[][] space) { + int ans = 0; + if (index == nodes.length) { + for (int i = 0; i < nodes.length; i++) { + ans += distance(nodes[i], space[i]); + } + } else { + ans = Integer.MAX_VALUE; + for (int i = index; i < nodes.length; i++) { + swap(nodes, index, i); + ans = Math.min(ans, process1(nodes, index + 1, space)); + swap(nodes, index, i); + } + } + return ans; + } + + public static void swap(int[][] nodes, int i, int j) { + int[] tmp = nodes[i]; + nodes[i] = nodes[j]; + nodes[j] = tmp; + } + + public static int distance(int[] a, int[] b) { + return Math.abs(a[0] - b[0]) + Math.abs(a[1] - b[1]); + } + + // 正式方法 + // KM算法 + public static int minDistance2(int[][] map) { + int n = 0; + int m = 0; + for (int i = 0; i < map.length; i++) { + for (int j = 0; j < map[0].length; j++) { + n += Math.max(0, map[i][j] - 1); + m += map[i][j] == 0 ? 1 : 0; + } + } + if (n != m || n == 0) { + return 0; + } + int[][] nodes = new int[n][2]; + int[][] space = new int[m][2]; + n = 0; + m = 0; + for (int i = 0; i < map.length; i++) { + for (int j = 0; j < map[0].length; j++) { + for (int k = 2; k <= map[i][j]; k++) { + nodes[n][0] = i; + nodes[n++][1] = j; + } + if (map[i][j] == 0) { + space[m][0] = i; + space[m++][1] = j; + } + } + } + int[][] graph = new int[n][n]; + for (int i = 0; i < n; i++) { + for (int j = 0; j < n; j++) { + graph[i][j] = -distance(nodes[i], space[j]); + } + } + return -km(graph); + } + + public static int km(int[][] graph) { + int N = graph.length; + int[] match = new int[N]; + int[] lx = new int[N]; + int[] ly = new int[N]; + boolean[] x = new boolean[N]; + boolean[] y = new boolean[N]; + int[] slack = new int[N]; + int invalid = Integer.MAX_VALUE; + for (int i = 0; i < N; i++) { + match[i] = -1; + lx[i] = -invalid; + for (int j = 0; j < N; j++) { + lx[i] = Math.max(lx[i], graph[i][j]); + } + ly[i] = 0; + } + for (int from = 0; from < N; from++) { + for (int i = 0; i < N; i++) { + slack[i] = invalid; + } + Arrays.fill(x, false); + Arrays.fill(y, false); + while (!dfs(from, x, y, lx, ly, match, slack, graph)) { + int d = invalid; + for (int i = 0; i < N; i++) { + if (!y[i] && slack[i] < d) { + d = slack[i]; + } + } + for (int i = 0; i < N; i++) { + if (x[i]) { + lx[i] = lx[i] - d; + } + if (y[i]) { + ly[i] = ly[i] + d; + } + } + Arrays.fill(x, false); + Arrays.fill(y, false); + } + } + int ans = 0; + for (int i = 0; i < N; i++) { + ans += (lx[i] + ly[i]); + } + return ans; + } + + public static boolean dfs(int from, boolean[] x, boolean[] y, int[] lx, int[] ly, int[] match, int[] slack, + int[][] map) { + int N = map.length; + x[from] = true; + for (int to = 0; to < N; to++) { + if (!y[to]) { + int d = lx[from] + ly[to] - map[from][to]; + if (d != 0) { + slack[to] = Math.min(slack[to], d); + } else { + y[to] = true; + if (match[to] == -1 || dfs(match[to], x, y, lx, ly, match, slack, map)) { + match[to] = from; + return true; + } + } + } + } + return false; + } + + // 为了测试 + public static int[][] randomValidMatrix(int len) { + int[][] ans = new int[len][len]; + int all = len * len; + for (int i = 1; i <= all; i++) { + ans[(int) (Math.random() * len)][(int) (Math.random() * len)]++; + } + return ans; + } + + public static void main(String[] args) { + int len = 4; + int testTimes = 1000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + int[][] map = randomValidMatrix(len); + int ans1 = minDistance1(map); + int ans2 = minDistance2(map); + if (ans1 != ans2) { + System.out.println("出错了!"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_03_5_week/Code03_MaximumAndSumOfArray.java b/算法周更班/class_2022_03_5_week/Code03_MaximumAndSumOfArray.java new file mode 100644 index 0000000..433d74d --- /dev/null +++ b/算法周更班/class_2022_03_5_week/Code03_MaximumAndSumOfArray.java @@ -0,0 +1,90 @@ +package class_2022_03_5_week; + +import java.util.Arrays; + +// 测试链接 : https://leetcode.com/problems/maximum-and-sum-of-array/ +public class Code03_MaximumAndSumOfArray { + + public static int maximumANDSum(int[] arr, int m) { + m <<= 1; + int[][] graph = new int[m][m]; + for (int i = 0; i < arr.length; i++) { + for (int j = 0, num = 1; j < m; num++, j += 2) { + graph[i][j] = arr[i] & num; + graph[i][j + 1] = arr[i] & num; + } + } + return km(graph); + } + + public static int km(int[][] graph) { + int N = graph.length; + int[] match = new int[N]; + int[] lx = new int[N]; + int[] ly = new int[N]; + boolean[] x = new boolean[N]; + boolean[] y = new boolean[N]; + int[] slack = new int[N]; + int invalid = Integer.MAX_VALUE; + for (int i = 0; i < N; i++) { + match[i] = -1; + lx[i] = -invalid; + for (int j = 0; j < N; j++) { + lx[i] = Math.max(lx[i], graph[i][j]); + } + ly[i] = 0; + } + for (int from = 0; from < N; from++) { + for (int i = 0; i < N; i++) { + slack[i] = invalid; + } + Arrays.fill(x, false); + Arrays.fill(y, false); + while (!dfs(from, x, y, lx, ly, match, slack, graph)) { + int d = invalid; + for (int i = 0; i < N; i++) { + if (!y[i] && slack[i] < d) { + d = slack[i]; + } + } + for (int i = 0; i < N; i++) { + if (x[i]) { + lx[i] = lx[i] - d; + } + if (y[i]) { + ly[i] = ly[i] + d; + } + } + Arrays.fill(x, false); + Arrays.fill(y, false); + } + } + int ans = 0; + for (int i = 0; i < N; i++) { + ans += (lx[i] + ly[i]); + } + return ans; + } + + public static boolean dfs(int from, boolean[] x, boolean[] y, int[] lx, int[] ly, int[] match, int[] slack, + int[][] map) { + int N = map.length; + x[from] = true; + for (int to = 0; to < N; to++) { + if (!y[to]) { + int d = lx[from] + ly[to] - map[from][to]; + if (d != 0) { + slack[to] = Math.min(slack[to], d); + } else { + y[to] = true; + if (match[to] == -1 || dfs(match[to], x, y, lx, ly, match, slack, map)) { + match[to] = from; + return true; + } + } + } + } + return false; + } + +} diff --git a/算法周更班/class_2022_03_5_week/Code04_KillAllSameTime.java b/算法周更班/class_2022_03_5_week/Code04_KillAllSameTime.java new file mode 100644 index 0000000..051368c --- /dev/null +++ b/算法周更班/class_2022_03_5_week/Code04_KillAllSameTime.java @@ -0,0 +1,18 @@ +package class_2022_03_5_week; + +// 来自网易 +// 我军一起干掉敌人的最少移动数 +// km算法的又一个题 +// 给定一个矩阵int[][] matrix +// matrix[i][j] == -2，代表此处(i,j)有山脉，无法通行 +// matrix[i][j] == -1，代表此处(i,j)是一个敌军 +// matrix[i][j] == 0，代表此处(i,j)是空地，可以自由行动 +// matrix[i][j] > 0，代表此处(i,j)是一个我军，行动能力就是matrix[i][j] +// 我军只能上、下、左、右移动，只可以穿过同样是我军的地点和空地的地点，但是最多移动matrix[i][j]步 +// 任何一个我军都不能穿过山脉，任何一个我军可以来到敌军的位置，表示消灭了敌军，但是如果这么做了，这个我军就不能再移动了 +// 你可以任意决定所有我军的行动策略，每一步你都可以随意选择一个友军移动随意方向的，但是必须合法。 +// 如果你可以让所有我军在消耗完自身的行动能力之前，消灭所有的敌军，请返回总距离的最小值 +// 如果你就是无法消灭所有敌军，返回-1 +public class Code04_KillAllSameTime { + +} diff --git a/算法周更班/class_2022_04_1_week/Code01_FourNumbersMinusOne.java b/算法周更班/class_2022_04_1_week/Code01_FourNumbersMinusOne.java new file mode 100644 index 0000000..006bd9d --- /dev/null +++ b/算法周更班/class_2022_04_1_week/Code01_FourNumbersMinusOne.java @@ -0,0 +1,58 @@ +package class_2022_04_1_week; + +import java.util.Arrays; + +// 来自阿里笔试 +// 牛牛今年上幼儿园了，老师叫他学习减法 +// 老师给了他5个数字，他每次操作可以选择其中的4个数字减1 +// 减一之后的数字不能小于0，因为幼儿园的牛牛还没有接触过负数 +// 现在牛牛想知道，自己最多可以进行多少次这样的操作 +// 扩展问题来自leetcode 2141，掌握了这个题原始问题就非常简单了 +// leetcode测试链接 : https://leetcode.com/problems/maximum-running-time-of-n-computers/ +public class Code01_FourNumbersMinusOne { + + public static long maxRunTime(int n, int[] arr) { + Arrays.sort(arr); + int size = arr.length; + long[] sums = new long[size]; + sums[0] = arr[0]; + for (int i = 1; i < size; i++) { + sums[i] = sums[i - 1] + arr[i]; + } + long l = 0; + long m = 0; + long r = sums[size - 1] / n; + long ans = -1; + while (l <= r) { + m = (l + r) / 2; + if (ok(arr, sums, m, n)) { + ans = m; + l = m + 1; + } else { + r = m - 1; + } + } + return ans; + } + + public static boolean ok(int[] arr, long[] sum, long time, int num) { + int l = 0; + int m = 0; + int r = arr.length - 1; + int left = arr.length; + // >= time 最左 + while (l <= r) { + m = (l + r) / 2; + if (arr[m] >= time) { + left = m; + r = m - 1; + } else { + l = m + 1; + } + } + num -= arr.length - left; + long rest = left == 0 ? 0 : sum[left - 1]; + return time * (long) num <= rest; + } + +} diff --git a/算法周更班/class_2022_04_1_week/Code02_MaxOrSmallestSubarray.java b/算法周更班/class_2022_04_1_week/Code02_MaxOrSmallestSubarray.java new file mode 100644 index 0000000..702618b --- /dev/null +++ b/算法周更班/class_2022_04_1_week/Code02_MaxOrSmallestSubarray.java @@ -0,0 +1,116 @@ +package class_2022_04_1_week; + +// 来自学员问题 +// 找到非负数组中拥有"最大或的结果"的最短子数组 +public class Code02_MaxOrSmallestSubarray { + + public static int longest1(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int max = 0; + for (int num : arr) { + max |= num; + } + if (max == 0) { + return 1; + } + int n = arr.length; + int ans = n; + for (int i = 0; i < n; i++) { + int cur = 0; + for (int j = i; j < n; j++) { + cur |= arr[j]; + if (cur == max) { + ans = Math.min(ans, j - i + 1); + break; + } + } + } + return ans; + } + + public static int longest2(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int max = 0; + for (int num : arr) { + max |= num; + } + if (max == 0) { + return 1; + } + int n = arr.length; + int ans = n; + int[] counts = new int[32]; + int l = 0; + int cur = 0; + for (int r = 0; r < n; r++) { + cur = add(counts, arr[r]); + while (cur == max) { + cur = delete(counts, arr[l++]); + } + if (l > 0) { + cur = add(counts, arr[--l]); + } + if (cur == max) { + ans = Math.min(ans, r - l + 1); + } + } + return ans; + } + + public static int add(int[] counts, int num) { + int ans = 0; + for (int i = 0; i < 32; i++) { + counts[i] += (num & (1 << i)) != 0 ? 1 : 0; + ans |= (counts[i] > 0 ? 1 : 0) << i; + } + return ans; + } + + public static int delete(int[] counts, int num) { + int ans = 0; + for (int i = 0; i < 32; i++) { + counts[i] -= (num & (1 << i)) != 0 ? 1 : 0; + ans |= (counts[i] > 0 ? 1 : 0) << i; + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * v); + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int len = 50; + int value = 50000; + int testTime = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * len) + 1; + int[] arr = randomArray(n, value); + int ans1 = longest1(arr); + int ans2 = longest2(arr); + if (ans1 != ans2) { + System.out.println("出错了!"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_04_1_week/Code03_ArrangeMeetingPosCancelPre.java b/算法周更班/class_2022_04_1_week/Code03_ArrangeMeetingPosCancelPre.java new file mode 100644 index 0000000..ea560b7 --- /dev/null +++ b/算法周更班/class_2022_04_1_week/Code03_ArrangeMeetingPosCancelPre.java @@ -0,0 +1,219 @@ +package class_2022_04_1_week; + +import java.util.ArrayList; +import java.util.Arrays; + +// 来自通维数码 +// 每个会议给定开始和结束时间 +// 后面的会议如果跟前面的会议有任何冲突，完全取消冲突的、之前的会议，安排当前的 +// 给定一个会议数组，返回安排的会议列表 +public class Code03_ArrangeMeetingPosCancelPre { + + // 比较暴力的解 + // 为了对数器来验证 + public static ArrayList arrange1(int[][] meetings) { + int max = 0; + for (int[] meeting : meetings) { + max = Math.max(max, meeting[1]); + } + boolean[] occupy = new boolean[max + 1]; + ArrayList ans = new ArrayList<>(); + for (int i = meetings.length - 1; i >= 0; i--) { + int[] cur = meetings[i]; + boolean add = true; + for (int j = cur[0]; j < cur[1]; j++) { + if (occupy[j]) { + add = false; + break; + } + } + if (add) { + ans.add(cur); + } + for (int j = cur[0]; j < cur[1]; j++) { + occupy[j] = true; + } + } + return ans; + } + + // 最优解 + // 会议有N个，时间复杂度O(N*logN) + public static ArrayList arrange2(int[][] meetings) { + int n = meetings.length; + // n << 1 -> n*2 + int[] rank = new int[n << 1]; + for (int i = 0; i < meetings.length; i++) { + rank[i] = meetings[i][0]; // 会议开头点 + rank[i + n] = meetings[i][1] - 1; // 会议的结束点 + } + Arrays.sort(rank); + // n*2 + SegmentTree st = new SegmentTree(n << 1); + // 哪些会议安排了，放入到ans里去！ + ArrayList ans = new ArrayList<>(); + // 从右往左遍历，意味着，后出现的会议，先看看能不能安排 + for (int i = meetings.length - 1; i >= 0; i--) { + // cur 当前会议 + int[] cur = meetings[i]; + // cur[0] = 17万 -> 6 + int from = rank(rank, cur[0]); + // cur[1] = 90 -> 89 -> 2 + int to = rank(rank, cur[1] - 1); + if (st.sum(from, to) == 0) { + ans.add(cur); + } + st.add(from, to, 1); + } + return ans; + } + + public static int rank(int[] rank, int num) { + int l = 0; + int r = rank.length - 1; + int m = 0; + int ans = 0; + while (l <= r) { + m = (l + r) / 2; + if (rank[m] >= num) { + ans = m; + r = m - 1; + } else { + l = m + 1; + } + } + return ans + 1; + } + + public static class SegmentTree { + private int n; + private int[] sum; + private int[] lazy; + + public SegmentTree(int size) { + n = size + 1; + sum = new int[n << 2]; + lazy = new int[n << 2]; + n--; + } + + private void pushUp(int rt) { + sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; + } + + private void pushDown(int rt, int ln, int rn) { + if (lazy[rt] != 0) { + lazy[rt << 1] += lazy[rt]; + sum[rt << 1] += lazy[rt] * ln; + lazy[rt << 1 | 1] += lazy[rt]; + sum[rt << 1 | 1] += lazy[rt] * rn; + lazy[rt] = 0; + } + } + + public void add(int L, int R, int C) { + add(L, R, C, 1, n, 1); + } + + private void add(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + sum[rt] += C * (r - l + 1); + lazy[rt] += C; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + add(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + add(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + public int sum(int L, int R) { + return query(L, R, 1, n, 1); + } + + private int query(int L, int R, int l, int r, int rt) { + if (L <= l && r <= R) { + return sum[rt]; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + int ans = 0; + if (L <= mid) { + ans += query(L, R, l, mid, rt << 1); + } + if (R > mid) { + ans += query(L, R, mid + 1, r, rt << 1 | 1); + } + return ans; + } + + } + + // 为了测试 + public static int[][] randomMeeting(int len, int time) { + int[][] meetings = new int[len][2]; + for (int i = 0; i < len; i++) { + int a = (int) (Math.random() * (time + 1)); + int b = (int) (Math.random() * (time + 1)); + if (a == b) { + b++; + } + meetings[i][0] = Math.min(a, b); + meetings[i][1] = Math.max(a, b); + } + return meetings; + } + + // 为了测试 + public static int[][] copyMeetings(int[][] meetings) { + int len = meetings.length; + int[][] ans = new int[len][2]; + for (int i = 0; i < len; i++) { + ans[i][0] = meetings[i][0]; + ans[i][1] = meetings[i][1]; + } + return ans; + } + + // 为了测试 + public static boolean equal(ArrayList arr1, ArrayList arr2) { + if (arr1.size() != arr2.size()) { + return false; + } + for (int i = 0; i < arr1.size(); i++) { + int[] a = arr1.get(i); + int[] b = arr2.get(i); + if (a[0] != b[0] || a[1] != b[1]) { + return false; + } + } + return true; + } + + public static void main(String[] args) { + int n = 100; + int t = 5000; + int testTime = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * n) + 1; + int[][] meetings1 = randomMeeting(len, t); + int[][] meetings2 = copyMeetings(meetings1); + ArrayList ans1 = arrange1(meetings1); + ArrayList ans2 = arrange2(meetings2); + if (!equal(ans1, ans2)) { + System.out.println("出错了!"); + System.out.println(ans1.size()); + System.out.println(ans2.size()); + System.out.println("===="); + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_04_1_week/Code04_MaxScoreMoveInBoard.java b/算法周更班/class_2022_04_1_week/Code04_MaxScoreMoveInBoard.java new file mode 100644 index 0000000..a7a4935 --- /dev/null +++ b/算法周更班/class_2022_04_1_week/Code04_MaxScoreMoveInBoard.java @@ -0,0 +1,88 @@ +package class_2022_04_1_week; + +// 来自小红书 +// 小红书第一题： +// 薯队长从北向南穿过一片红薯地（南北长M，东西宽N），红薯地被划分为1x1的方格， +// 他可以从北边的任何一个格子出发，到达南边的任何一个格子， +// 但每一步只能走到东南、正南、西南方向的三个格子之一， +// 而且不能跨出红薯地，他可以获得经过的格子上的所有红薯，请问他可以获得最多的红薯个数。 +public class Code04_MaxScoreMoveInBoard { + +// // 目前来到(row, col)的位置 +// // 只能往不越界的三个方向移动：西南、正南、东南 +// // 一旦遇到最南格子，停 +// // 返回这个过程中，最大的收成 +// public static int f(int[][] board, int row, int col) { +// if (col < 0 || col == board[0].length) { +// return 0; +// } +// if (row == board.length - 1) { +// return board[row][col]; +// } +// int p1 = f(board, row + 1, col - 1); +// int p2 = f(board, row + 1, col); +// int p3 = f(board, row + 1, col + 1); +// return board[row][col] + Math.max(p1, Math.max(p2, p3)); +// } + + public static int maxScore(int[][] map) { + int ans = 0; + for (int col = 0; col < map[0].length; col++) { + ans = Math.max(ans, process(map, 0, col)); + } + return ans; + } + + public static int process(int[][] map, int row, int col) { + if (col < 0 || col == map[0].length) { + return -1; + } + if (row == map.length - 1) { + return map[row][col]; + } + int cur = map[row][col]; + int next1 = process(map, row + 1, col - 1); + int next2 = process(map, row + 1, col); + int next3 = process(map, row + 1, col + 1); + return cur + Math.max(Math.max(next1, next2), next3); + } + + public static int maxScore2(int[][] map) { + int ans = 0; + int n = map.length; + int m = map[0].length; + int[][] dp = new int[n][m]; + for (int i = 0; i < n; i++) { + for (int j = 0; j < m; j++) { + dp[i][j] = -2; // -2表示，这个格子没算过！ + } + } + for (int col = 0; col < map[0].length; col++) { + ans = Math.max(ans, process2(map, 0, col, dp)); + } + return ans; + } + + public static int process2(int[][] map, int row, int col, int[][] dp) { + if (col < 0 || col == map[0].length) { + return -1; + } + if (dp[row][col] != -2) { + return dp[row][col]; + } + // 继续算！ + int ans = 0; + if (row == map.length - 1) { + ans = map[row][col]; + } else { + int cur = map[row][col]; + int next1 = process2(map, row + 1, col - 1, dp); + int next2 = process2(map, row + 1, col, dp); + int next3 = process2(map, row + 1, col + 1, dp); + ans = cur + Math.max(Math.max(next1, next2), next3); + } + dp[row][col] = ans; + return ans; + } + +} diff --git a/算法周更班/class_2022_04_1_week/Code05_PickKnumbersNearTowNumberMaxDiff.java b/算法周更班/class_2022_04_1_week/Code05_PickKnumbersNearTowNumberMaxDiff.java new file mode 100644 index 0000000..519ea65 --- /dev/null +++ b/算法周更班/class_2022_04_1_week/Code05_PickKnumbersNearTowNumberMaxDiff.java @@ -0,0 +1,57 @@ +package class_2022_04_1_week; + +import java.util.Arrays; + +// 小红书第二题： +// 薯队长最近在参加了一个活动，主办方提供了N个礼物以供挑选， +// 每个礼物有一个价值，范围在0 ~ 10^9之间， +// 薯队长可以从中挑选k个礼物 +// 返回：其中价值最接近的两件礼物之间相差值尽可能大的结果 +// 我们课上讲 +// +// 小红书第三题： +// 薯队长最近在玩一个游戏，这个游戏桌上会有一排不同颜色的方块， +// 每次薯队长可以选择一个方块，便可以消除这个方块以及和他左右相临的 +// 若干的颜色相同的方块，而每次消除的方块越多，得分越高。 +// 具体来说，桌上有以个方块排成一排 (1 <= N <= 200）， +// 每个方块有一个颜色，用1~N之间的一个整数表示，相同的数宇代表相同的颜色， +// 每次消除的时候，会把连续的K个相同颜色的方块消除，并得到K*K的分数， +// 直到所有方块都消除。显然，不同的消除顺序得分不同，薯队长希望您能告诉他，这个游戏最多能得到多少分 +// 体系学习班，代码46节，视频在47节，消箱子原题，RemoveBoxes +public class Code05_PickKnumbersNearTowNumberMaxDiff { + + public static int maxNear(int[] arr, int k) { + if (arr.length < k) { + return -1; + } + Arrays.sort(arr); + int n = arr.length; + int l = 0; + int r = arr[n - 1] - arr[0]; + int m = 0; + int ans = 0; + while (l <= r) { + m = (l + r) / 2; + if (yeah(arr, k, m)) { + ans = m; + l = m + 1; + } else { + r = m - 1; + } + } + return ans; + } + + public static boolean yeah(int[] arr, int k, int limit) { + int last = arr[0]; + int pick = 1; + for (int i = 1; i < arr.length; i++) { + if (arr[i] - last >= limit) { + pick++; + last = arr[i]; + } + } + return pick >= k; + } + +} diff --git a/算法周更班/class_2022_04_1_week/Code06_TopMinSubsquenceSum.java b/算法周更班/class_2022_04_1_week/Code06_TopMinSubsquenceSum.java new file mode 100644 index 0000000..67ad8c5 --- /dev/null +++ b/算法周更班/class_2022_04_1_week/Code06_TopMinSubsquenceSum.java @@ -0,0 +1,114 @@ +package class_2022_04_1_week; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.PriorityQueue; + +// 给定一个数组arr，含有n个数字，都是非负数 +// 给定一个正数k +// 返回所有子序列中，累加和最小的前k个子序列累加和 +// 假设K不大，怎么算最快？ +public class Code06_TopMinSubsquenceSum { + + public static int[] topMinSum1(int[] arr, int k) { + ArrayList allAns = new ArrayList<>(); + process(arr, 0, 0, allAns); + allAns.sort((a, b) -> a.compareTo(b)); + int[] ans = new int[k]; + for (int i = 0; i < k; i++) { + ans[i] = allAns.get(i); + } + return ans; + } + + public static void process(int[] arr, int index, int sum, ArrayList ans) { + if (index == arr.length) { + ans.add(sum); + } else { + process(arr, index + 1, sum, ans); + process(arr, index + 1, sum + arr[index], ans); + } + } + + public static int[] topMinSum2(int[] arr, int k) { + Arrays.sort(arr); + // (最右的下标，集合的累加和) + PriorityQueue heap = new PriorityQueue<>((a, b) -> a[1] - b[1]); + heap.add(new int[] { 0, arr[0] }); + int[] ans = new int[k]; + // ans[0] = 0 + // 0 1 2 k-1 + // k个！ + for (int i = 1; i < k; i++) { + int[] cur = heap.poll(); + // (7, 100) + // 左 ：8, 100 - arr[7] + arr[8] + // 右 ：8， 100 + arr[8] + int last = cur[0]; + int sum = cur[1]; + ans[i] = sum; + if (last + 1 < arr.length) { + heap.add(new int[] { last + 1, sum - arr[last] + arr[last + 1] }); + heap.add(new int[] { last + 1, sum + arr[last + 1] }); + } + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int len, int value) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * value); + } + return arr; + } + + // 为了测试 + public static boolean equals(int[] ans1, int[] ans2) { + if (ans1.length != ans2.length) { + return false; + } + for (int i = 0; i < ans1.length; i++) { + if (ans1[i] != ans2[i]) { + return false; + } + } + return true; + } + + // 为了测试 + public static void main(String[] args) { + int n = 10; + int v = 40; + int testTime = 5000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * n) + 1; + int[] arr = randomArray(len, v); + int k = (int) (Math.random() * ((1 << len) - 1)) + 1; + int[] ans1 = topMinSum1(arr, k); + int[] ans2 = topMinSum2(arr, k); + if (!equals(ans1, ans2)) { + System.out.println("出错了！"); + System.out.print("arr : "); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println("k : " + k); + for (int num : ans1) { + System.out.print(num + " "); + } + System.out.println(); + for (int num : ans2) { + System.out.print(num + " "); + } + System.out.println(); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_04_1_week/Code07_TopMaxSubsquenceSum.java b/算法周更班/class_2022_04_1_week/Code07_TopMaxSubsquenceSum.java new file mode 100644 index 0000000..0312656 --- /dev/null +++ b/算法周更班/class_2022_04_1_week/Code07_TopMaxSubsquenceSum.java @@ -0,0 +1,124 @@ +package class_2022_04_1_week; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.PriorityQueue; + +// 来自Amazon +// 给定一个数组arr，含有n个数字，可能有正、有负、有0 +// 给定一个正数k +// 返回所有子序列中，累加和最大的前k个子序列累加和 +// 假设K不大，怎么算最快？ +public class Code07_TopMaxSubsquenceSum { + + public static int[] topMaxSum1(int[] arr, int k) { + ArrayList allAns = new ArrayList<>(); + process(arr, 0, 0, allAns); + allAns.sort((a, b) -> a.compareTo(b)); + int[] ans = new int[k]; + for (int i = allAns.size() - 1, j = 0; j < k; i--, j++) { + ans[j] = allAns.get(i); + } + return ans; + } + + public static void process(int[] arr, int index, int sum, ArrayList ans) { + if (index == arr.length) { + ans.add(sum); + } else { + process(arr, index + 1, sum, ans); + process(arr, index + 1, sum + arr[index], ans); + } + } + + public static int[] topMaxSum2(int[] arr, int k) { + int sum = 0; + for (int i = 0; i < arr.length; i++) { + if (arr[i] >= 0) { + sum += arr[i]; + } else { + arr[i] = -arr[i]; + } + } + int[] ans = topMinSum(arr, k); + for (int i = 0; i < ans.length; i++) { + ans[i] = sum - ans[i]; + } + return ans; + } + + public static int[] topMinSum(int[] arr, int k) { + Arrays.sort(arr); + PriorityQueue heap = new PriorityQueue<>((a, b) -> a[1] - b[1]); + heap.add(new int[] { 0, arr[0] }); + int[] ans = new int[k]; + for (int i = 1; i < k; i++) { + int[] cur = heap.poll(); + int last = cur[0]; + int sum = cur[1]; + ans[i] = sum; + if (last + 1 < arr.length) { + heap.add(new int[] { last + 1, sum - arr[last] + arr[last + 1] }); + heap.add(new int[] { last + 1, sum + arr[last + 1] }); + } + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int len, int value) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * value) + 1; + } + return arr; + } + + // 为了测试 + public static boolean equals(int[] ans1, int[] ans2) { + if (ans1.length != ans2.length) { + return false; + } + for (int i = 0; i < ans1.length; i++) { + if (ans1[i] != ans2[i]) { + return false; + } + } + return true; + } + + // 为了测试 + public static void main(String[] args) { + int n = 10; + int v = 40; + int testTime = 5000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * n) + 1; + int[] arr = randomArray(len, v); + int k = (int) (Math.random() * ((1 << len) - 1)) + 1; + int[] ans1 = topMaxSum1(arr, k); + int[] ans2 = topMaxSum2(arr, k); + if (!equals(ans1, ans2)) { + System.out.println("出错了！"); + System.out.print("arr : "); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println("k : " + k); + for (int num : ans1) { + System.out.print(num + " "); + } + System.out.println(); + for (int num : ans2) { + System.out.print(num + " "); + } + System.out.println(); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_04_2_week/Code01_SumOfValuesAboutPrimes.java b/算法周更班/class_2022_04_2_week/Code01_SumOfValuesAboutPrimes.java new file mode 100644 index 0000000..93a9468 --- /dev/null +++ b/算法周更班/class_2022_04_2_week/Code01_SumOfValuesAboutPrimes.java @@ -0,0 +1,114 @@ +package class_2022_04_2_week; + +import java.math.BigInteger; +import java.util.ArrayList; +import java.util.HashMap; + +// 来自携程 +// 给出n个数字，你可以任选其中一些数字相乘，相乘之后得到的新数字x +// x的价值是x的不同质因子的数量 +// 返回所有选择数字的方案中，得到的x的价值之和 +public class Code01_SumOfValuesAboutPrimes { + + // 工具！ + // 返回num质数因子列表(去重) + // 时间复杂度，根号(num) + public static ArrayList primes(long num) { + ArrayList ans = new ArrayList<>(); + for (long i = 2; i * i <= num && num > 1; i++) { + if (num % i == 0) { + ans.add(i); + while (num % i == 0) { + num /= i; + } + } + } + if (num != 1) { + ans.add(num); + } + return ans; + } + + public static long sumOfValues1(int[] arr) { + return process1(arr, 0, 1); + } + + public static long process1(int[] arr, int index, long pre) { + if (index == arr.length) { + return (long) primes(pre).size(); + } + long p1 = process1(arr, index + 1, pre); + long p2 = process1(arr, index + 1, pre * (long) arr[index]); + return p1 + p2; + } + + public static long sumOfValues2(int[] arr) { + // key : 某个质数因子 + // value : 有多少个数含有这个因子 + HashMap cntMap = new HashMap<>(); + for (int num : arr) { + for (long factor : primes(num)) { + cntMap.put(factor, cntMap.getOrDefault(factor, 0L) + 1L); + } + } + int n = arr.length; + long ans = 0; + // count ：含有这个因子的数，有多少个 + // others : 不含有这个因子的数，有多少个 + for (long count : cntMap.values()) { + long others = n - count; + ans += (power(2, count) - 1) * power(2, others); + } + return ans; + } + + public static long power(long num, long n) { + if (n == 0L) { + return 1L; + } + long ans = 1L; + while (n > 0) { + if ((n & 1) != 0) { + ans *= num; + } + num *= num; + n >>= 1; + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int len, int value) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * value) + 1; + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int n = 10; + int v = 20; + int testTime = 5000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int[] arr = randomArray(n, v); + long ans1 = sumOfValues1(arr); + long ans2 = sumOfValues2(arr); + if (ans1 != ans2) { + System.out.println("出错了！"); + BigInteger all = new BigInteger("1"); + for (int num : arr) { + all = all.multiply(new BigInteger(String.valueOf(num))); + } + System.out.println("所有数都乘起来 : " + all.toString()); + System.out.println("长整型最大范围 : " + Long.MAX_VALUE); + System.out.println("可以看看上面的打印，看看是不是因为溢出了"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_04_2_week/Code02_MinDistanceFromLeftUpToRightDown.java b/算法周更班/class_2022_04_2_week/Code02_MinDistanceFromLeftUpToRightDown.java new file mode 100644 index 0000000..12c6b77 --- /dev/null +++ b/算法周更班/class_2022_04_2_week/Code02_MinDistanceFromLeftUpToRightDown.java @@ -0,0 +1,125 @@ +package class_2022_04_2_week; + +import java.util.PriorityQueue; + +// 来自网易 +// 3.27笔试 +// 一个二维矩阵，上面只有 0 和 1，只能上下左右移动 +// 如果移动前后的元素值相同，则耗费 1 ，否则耗费 2。 +// 问从左上到右下的最小耗费 +public class Code02_MinDistanceFromLeftUpToRightDown { + + // 一个错误的贪心 + // 网上帖子最流行的解答，看似对，其实不行 + public static int bestWalk1(int[][] map) { + int n = map.length; + int m = map[0].length; + int[][] dp = new int[n][m]; + for (int i = 1; i < m; i++) { + dp[0][i] = dp[0][i - 1] + (map[0][i - 1] == map[0][i] ? 1 : 2); + } + for (int i = 1; i < n; i++) { + dp[i][0] = dp[i - 1][0] + (map[i - 1][0] == map[i][0] ? 1 : 2); + } + for (int i = 1; i < n; i++) { + for (int j = 1; j < m; j++) { + dp[i][j] = dp[i - 1][j] + (map[i - 1][j] == map[i][j] ? 1 : 2); + dp[i][j] = Math.min(dp[i][j], dp[i][j - 1] + (map[i][j - 1] == map[i][j] ? 1 : 2)); + } + } + return dp[n - 1][m - 1]; + } + + // 正确的解法 + // Dijskra + public static int bestWalk2(int[][] map) { + int n = map.length; + int m = map[0].length; + // 小根堆：[代价，行，列] + // 根据代价，谁代价小，谁放在堆的上面 + PriorityQueue heap = new PriorityQueue<>((a, b) -> a[0] - b[0]); + // poped[i][j] == true 已经弹出过了！不要再处理，直接忽略！ + // poped[i][j] == false 之间(i,j)没弹出过！要处理 + boolean[][] poped = new boolean[n][m]; + heap.add(new int[] { 0, 0, 0 }); + int ans = 0; + while (!heap.isEmpty()) { + // 当前弹出了，[代价，行，列]，当前位置 + int[] cur = heap.poll(); + int dis = cur[0]; + int row = cur[1]; + int col = cur[2]; + if (poped[row][col]) { + continue; + } + // 第一次弹出！ + poped[row][col] = true; + if (row == n - 1 && col == m - 1) { + ans = dis; + break; + } + add(dis, row - 1, col, map[row][col], n, m, map, poped, heap); + add(dis, row + 1, col, map[row][col], n, m, map, poped, heap); + add(dis, row, col - 1, map[row][col], n, m, map, poped, heap); + add(dis, row, col + 1, map[row][col], n, m, map, poped, heap); + } + return ans; + } + + // preDistance ： 之前的距离 + // int row, int col ： 当前要加入的是什么位置 + // preValue : 前一个格子是什么值， + // int n, int m ：边界，固定参数 + // map: 每一个格子的值，都在map里 + // boolean[][] poped : 当前位置如果是弹出过的位置，要忽略！ + // PriorityQueue heap : 小根堆 + public static void add(int preDistance, + int row, int col, int preValue, int n, int m, + int[][] map, boolean[][] poped, + PriorityQueue heap) { + if (row >= 0 && row < n && col >= 0 && col < m + && !poped[row][col]) { + heap.add(new int[] { + preDistance + (map[row][col] == preValue ? 1 : 2), + row, col }); + } + } + + // 为了测试 + public static int[][] randomMatrix(int n, int m) { + int[][] ans = new int[n][m]; + for (int i = 0; i < n; i++) { + for (int j = 0; j < m; j++) { + ans[i][j] = (int) (Math.random() * 2); + } + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int n = 100; + int m = 100; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int[][] map = randomMatrix(n, m); + int ans1 = bestWalk1(map); + int ans2 = bestWalk2(map); + if (ans1 != ans2) { + System.out.println("出错了!"); + for (int[] arr : map) { + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + } + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_04_2_week/Code03_MaxSumDividedBy7.java b/算法周更班/class_2022_04_2_week/Code03_MaxSumDividedBy7.java new file mode 100644 index 0000000..0412aab --- /dev/null +++ b/算法周更班/class_2022_04_2_week/Code03_MaxSumDividedBy7.java @@ -0,0 +1,82 @@ +package class_2022_04_2_week; + +// 来自美团 +// 3.26笔试 +// 给定一个非负数组，任意选择数字，使累加和最大且为7的倍数，返回最大累加和 +// n比较大，10的5次方 +public class Code03_MaxSumDividedBy7 { + + public static int maxSum1(int[] arr) { + return process1(arr, 0, 0); + } + + public static int process1(int[] arr, int index, int pre) { + if (index == arr.length) { + return pre % 7 == 0 ? pre : 0; + } + int p1 = process1(arr, index + 1, pre); + int p2 = process1(arr, index + 1, pre + arr[index]); + return Math.max(p1, p2); + } + + public static int maxSum2(int[] arr) { + if (arr == null || arr.length == 0) { + return 0; + } + int n = arr.length; + int[][] dp = new int[n][7]; + for (int i = 0; i < n; i++) { + for (int j = 1; j < 7; j++) { + dp[i][j] = -1; + } + } + dp[0][arr[0] % 7] = arr[0]; + for (int i = 1; i < n; i++) { + // 当前arr[i] % 7 的余数 + int curMod = arr[i] % 7; + for (int j = 0; j < 7; j++) { + dp[i][j] = dp[i - 1][j]; + int findMod = (7 - curMod + j) % 7; + if (dp[i - 1][findMod] != -1) { + dp[i][j] = Math.max(dp[i][j], dp[i - 1][findMod] + arr[i]); + } + } + } + return dp[n - 1][0] == -1 ? 0 : dp[n - 1][0]; + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * v) + 1; + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int len = 12; + int value = 30; + int testTime = 2000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * len) + 1; + int[] arr = randomArray(n, value); + int ans1 = maxSum1(arr); + int ans2 = maxSum2(arr); + if (ans1 != ans2) { + System.out.println("出错了!"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_04_2_week/Code04_AllJobFinishTime.java b/算法周更班/class_2022_04_2_week/Code04_AllJobFinishTime.java new file mode 100644 index 0000000..3e8ca27 --- /dev/null +++ b/算法周更班/class_2022_04_2_week/Code04_AllJobFinishTime.java @@ -0,0 +1,48 @@ +package class_2022_04_2_week; + +import java.util.ArrayList; +import java.util.LinkedList; +import java.util.Queue; + +// 来自美团 +// 3.26笔试 +// 给定一个正数n, 表示有0~n-1号任务 +// 给定一个长度为n的数组time，time[i]表示i号任务做完的时间 +// 给定一个二维数组matrix +// matrix[j] = {a, b} 代表：a任务想要开始，依赖b任务的完成 +// 只要能并行的任务都可以并行，但是任何任务只有依赖的任务完成，才能开始 +// 返回一个长度为n的数组ans，表示每个任务完成的时间 +// 输入可以保证没有循环依赖 +public class Code04_AllJobFinishTime { + + public static int[] finishTime(int n, int[] time, int[][] matrix) { + ArrayList> nexts = new ArrayList<>(); + for (int i = 0; i < n; i++) { + nexts.add(new ArrayList<>()); + } + int[] in = new int[n]; + for (int[] line : matrix) { + nexts.get(line[1]).add(line[0]); + in[line[0]]++; + } + Queue zeroInQueue = new LinkedList<>(); + int[] ans = new int[n]; + for (int i = 0; i < n; i++) { + if (in[i] == 0) { + zeroInQueue.add(i); + } + } + while (!zeroInQueue.isEmpty()) { + int cur = zeroInQueue.poll(); + ans[cur] += time[cur]; + for (int next : nexts.get(cur)) { + ans[next] = Math.max(ans[next], ans[cur]); + if (--in[next] == 0) { + zeroInQueue.add(next); + } + } + } + return ans; + } + +} diff --git a/算法周更班/class_2022_04_2_week/Code05_TowLongestSubarraySame01Number.java b/算法周更班/class_2022_04_2_week/Code05_TowLongestSubarraySame01Number.java new file mode 100644 index 0000000..1941890 --- /dev/null +++ b/算法周更班/class_2022_04_2_week/Code05_TowLongestSubarraySame01Number.java @@ -0,0 +1,95 @@ +package class_2022_04_2_week; + +import java.util.HashMap; + +// 来自百度 +// 给出一个长度为n的01串，现在请你找到两个区间， +// 使得这两个区间中，1的个数相等，0的个数也相等 +// 这两个区间可以相交，但是不可以完全重叠，即两个区间的左右端点不可以完全一样 +// 现在请你找到两个最长的区间，满足以上要求。 +public class Code05_TowLongestSubarraySame01Number { + + public static int longest1(int[] arr) { + HashMap> map = new HashMap<>(); + for (int i = 0; i < arr.length; i++) { + int zero = 0; + int one = 0; + for (int j = i; j < arr.length; j++) { + zero += arr[j] == 0 ? 1 : 0; + one += arr[j] == 1 ? 1 : 0; + map.putIfAbsent(zero, new HashMap<>()); + map.get(zero).put(one, map.get(zero).getOrDefault(one, 0) + 1); + } + } + int ans = 0; + for (int zeros : map.keySet()) { + for (int ones : map.get(zeros).keySet()) { + int num = map.get(zeros).get(ones); + if (num > 1) { + ans = Math.max(ans, zeros + ones); + } + } + } + return ans; + } + + public static int longest2(int[] arr) { + int leftZero = -1; + int rightZero = -1; + int leftOne = -1; + int rightOne = -1; + for (int i = 0; i < arr.length; i++) { + if (arr[i] == 0) { + leftZero = i; + break; + } + } + for (int i = 0; i < arr.length; i++) { + if (arr[i] == 1) { + leftOne = i; + break; + } + } + for (int i = arr.length - 1; i >= 0; i--) { + if (arr[i] == 0) { + rightZero = i; + break; + } + } + for (int i = arr.length - 1; i >= 0; i--) { + if (arr[i] == 1) { + rightOne = i; + break; + } + } + int p1 = rightZero - leftZero; + int p2 = rightOne - leftOne; + return Math.max(p1, p2); + } + + public static int[] randomArray(int len) { + int[] ans = new int[len]; + for (int i = 0; i < len; i++) { + ans[i] = (int) (Math.random() * 2); + } + return ans; + } + + public static void main(String[] args) { + int n = 500; + int testTime = 200; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int size = (int) (Math.random() * n) + 2; + int[] arr = randomArray(size); + int ans1 = longest1(arr); + int ans2 = longest2(arr); + if (ans1 != ans2) { + System.out.println("出错了!"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_04_2_week/Code06_PerfectPairNumber.java b/算法周更班/class_2022_04_2_week/Code06_PerfectPairNumber.java new file mode 100644 index 0000000..32d3212 --- /dev/null +++ b/算法周更班/class_2022_04_2_week/Code06_PerfectPairNumber.java @@ -0,0 +1,53 @@ +package class_2022_04_2_week; + +// 来自阿里 +// x = { a, b, c, d } +// y = { e, f, g, h } +// x、y两个小数组长度都是4 +// 如果有: a + e = b + f = c + g = d + h +// 那么说x和y是一个完美对 +// 题目给定N个小数组，每个小数组长度都是K +// 返回这N个小数组中，有多少完美对 +// 本题测试链接 : https://www.nowcoder.com/practice/f5a3b5ab02ed4202a8b54dfb76ad035e +// 提交如下代码，把主类名改成Main +// 可以直接通过 +import java.util.HashMap; +import java.util.Scanner; + +public class Code06_PerfectPairNumber { + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + while (sc.hasNext()) { + int n = sc.nextInt(); + int m = sc.nextInt(); + int[][] matrix = new int[n][m]; + for (int i = 0; i < n; i++) { + for (int j = 0; j < m; j++) { + matrix[i][j] = sc.nextInt(); + } + } + long ans = perfectPairs(matrix); + System.out.println(ans); + } + sc.close(); + } + + public static long perfectPairs(int[][] matrix) { + long ans = 0; + // key : 字符串 特征，差值特征 : "_5_-2_6_9" + HashMap counts = new HashMap<>(); + for (int[] arr : matrix) { + StringBuilder self = new StringBuilder(); + StringBuilder minus = new StringBuilder(); + for (int i = 1; i < arr.length; i++) { + self.append("_" + (arr[i] - arr[i - 1])); + minus.append("_" + (arr[i - 1] - arr[i])); + } + ans += counts.getOrDefault(minus.toString(), 0); + counts.put(self.toString(), counts.getOrDefault(self.toString(), 0) + 1); + } + return ans; + } + +} \ No newline at end of file diff --git a/算法周更班/class_2022_04_2_week/Code07_MaxMoneyMostMin.java b/算法周更班/class_2022_04_2_week/Code07_MaxMoneyMostMin.java new file mode 100644 index 0000000..61b6ed6 --- /dev/null +++ b/算法周更班/class_2022_04_2_week/Code07_MaxMoneyMostMin.java @@ -0,0 +1,236 @@ +package class_2022_04_2_week; + +import java.util.Arrays; +import java.util.HashMap; +import java.util.LinkedList; + +// 来自快手 +// 某公司年会上，大家要玩一食发奖金游戏，一共有n个员工， +// 每个员工都有建设积分和捣乱积分 +// 他们需要排成一队，在队伍最前面的一定是老板，老板也有建设积分和捣乱积分 +// 排好队后，所有员工都会获得各自的奖金， +// 该员工奖金 = 排在他前面所有人的建设积分乘积 / 该员工自己的捣乱积分，向下取整 +// 为了公平(放屁)，老板希望 : 让获得奖金最高的员工，所获得的奖金尽可能少 +// 所以想请你帮他重新排一下队伍，返回奖金最高的员工获得的、尽可能少的奖金数额 +// 快手考试的时候，给定的数据量，全排列的代码也能过的！ +// 1 <= n <= 1000, 1<= 积分 <= 10000; +public class Code07_MaxMoneyMostMin { + + // 暴力方法 + // 为了验证 + // a : 老板的贡献积分 + // b : 老板的捣乱积分 + // value[i] : i号员工的贡献积分 + // trouble[i] : i号员工的捣乱积分 + // 返回 : 奖金最高的员工获得的、尽可能少的奖金数额 + public static long mostMin1(int a, int b, int[] value, int[] trouble) { + return process1(a, value, trouble, 0); + } + + public static long process1(int boss, int[] value, int[] trouble, int index) { + if (index == value.length) { + long valueAll = boss; + long ans = 0; + for (int i = 0; i < value.length; i++) { + ans = Math.max(ans, valueAll / trouble[i]); + valueAll *= value[i]; + } + return ans; + } else { + long ans = Long.MAX_VALUE; + for (int i = index; i < value.length; i++) { + swap(value, trouble, i, index); + ans = Math.min(ans, process1(boss, value, trouble, index + 1)); + swap(value, trouble, i, index); + } + return ans; + } + } + + public static void swap(int[] value, int[] trouble, int i, int j) { + int tmp = value[i]; + value[i] = value[j]; + value[j] = tmp; + tmp = trouble[i]; + trouble[i] = trouble[j]; + trouble[j] = tmp; + } + + // 正式方法 + // 所有员工数量为N + // 假设所有员工建设积分乘起来为M + // 时间复杂度O(N * logN * logM) + public static long mostMin2(int a, int b, int[] value, int[] trouble) { + int n = value.length; + long l = 0; + long r = 0; + long valueAll = a; + long[][] staff = new long[n][2]; + for (int i = 0; i < n; i++) { + r = Math.max(r, valueAll / trouble[i]); + valueAll *= value[i]; + staff[i][0] = (long) value[i] * (long) trouble[i]; + staff[i][1] = value[i]; + } + Arrays.sort(staff, (x, y) -> (y[0] >= x[0] ? 1 : -1)); + long m = 0; + long ans = 0; + while (l <= r) { + m = l + ((r - l) >> 1); + if (yeah(valueAll, staff, m)) { + ans = m; + r = m - 1; + } else { + l = m + 1; + } + } + return ans; + } + + // staff长度为N，时间复杂度O(N * logN) + public static boolean yeah(long all, long[][] staff, long limit) { + int n = staff.length; + SegmentTree st = new SegmentTree(n); + HashMap> map = new HashMap<>(); + for (int i = 0, index = 1; i < n; i++, index++) { + int value = (int) staff[i][1]; + st.update(index, value); + if (!map.containsKey(value)) { + map.put(value, new LinkedList<>()); + } + map.get(value).addLast(index); + } + for (int k = 0; k < n; k++) { + int right = boundary(staff, all, limit); + if (right == 0) { + return false; + } + int max = st.max(right); + if (max == 0) { + return false; + } + int index = map.get(max).pollFirst(); + st.update(index, 0); + all /= max; + } + return true; + } + + public static int boundary(long[][] staff, long all, long limit) { + int l = 0; + int r = staff.length - 1; + int m = 0; + int ans = -1; + while (l <= r) { + m = l + ((r - l) >> 1); + if (all / staff[m][0] <= limit) { + ans = m; + l = m + 1; + } else { + r = m - 1; + } + } + return ans + 1; + } + + public static class SegmentTree { + private int n; + private int[] max; + private int[] update; + + public SegmentTree(int maxSize) { + n = maxSize + 1; + max = new int[n << 2]; + update = new int[n << 2]; + Arrays.fill(update, -1); + } + + public void update(int index, int c) { + update(index, index, c, 1, n, 1); + } + + public int max(int right) { + return max(1, right, 1, n, 1); + } + + private void pushUp(int rt) { + max[rt] = Math.max(max[rt << 1], max[rt << 1 | 1]); + } + + private void pushDown(int rt, int ln, int rn) { + if (update[rt] != -1) { + update[rt << 1] = update[rt]; + max[rt << 1] = update[rt]; + update[rt << 1 | 1] = update[rt]; + max[rt << 1 | 1] = update[rt]; + update[rt] = -1; + } + } + + private void update(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + max[rt] = C; + update[rt] = C; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + update(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + update(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + private int max(int L, int R, int l, int r, int rt) { + if (L <= l && r <= R) { + return max[rt]; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + int ans = 0; + if (L <= mid) { + ans = Math.max(ans, max(L, R, l, mid, rt << 1)); + } + if (R > mid) { + ans = Math.max(ans, max(L, R, mid + 1, r, rt << 1 | 1)); + } + return ans; + } + + } + + // 为了测试 + public static int[] randomArray(int len, int value) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * value) + 1; + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int n = 9; + int v = 50; + int testTime = 5000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int a = (int) (Math.random() * v) + 1; + int b = (int) (Math.random() * v) + 1; + int len = (int) (Math.random() * n); + int[] value = randomArray(len, v); + int[] trouble = randomArray(len, v); + long ans1 = mostMin1(a, b, value, trouble); + long ans2 = mostMin2(a, b, value, trouble); + if (ans1 != ans2) { + System.out.println("出错了！"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_04_3_week/Code01_MaxOneNumbers.java b/算法周更班/class_2022_04_3_week/Code01_MaxOneNumbers.java new file mode 100644 index 0000000..6348b83 --- /dev/null +++ b/算法周更班/class_2022_04_3_week/Code01_MaxOneNumbers.java @@ -0,0 +1,79 @@ +package class_2022_04_3_week; + +// 小红书 +// 3.13 笔试 +// 数组里有0和1，一定要翻转一个区间，翻转：0变1，1变0 +// 请问翻转后可以使得1的个数最多是多少？ +public class Code01_MaxOneNumbers { + + public static int maxOneNumbers1(int[] arr) { + int ans = 0; + for (int l = 0; l < arr.length; l++) { + for (int r = l; r < arr.length; r++) { + reverse(arr, l, r); + ans = Math.max(ans, oneNumbers(arr)); + reverse(arr, l, r); + } + } + return ans; + } + + public static void reverse(int[] arr, int l, int r) { + for (int i = l; i <= r; i++) { + arr[i] ^= 1; + } + } + + public static int oneNumbers(int[] arr) { + int ans = 0; + for (int num : arr) { + ans += num; + } + return ans; + } + + public static int maxOneNumbers2(int[] arr) { + int ans = 0; + for (int num : arr) { + ans += num; + } + for (int i = 0; i < arr.length; i++) { + arr[i] = arr[i] == 0 ? 1 : -1; + } + int max = Integer.MIN_VALUE; + int cur = 0; + for (int i = 0; i < arr.length; i++) { + cur += arr[i]; + max = Math.max(max, cur); + cur = cur < 0 ? 0 : cur; + } + return ans + max; + } + + // 为了测试 + public static int[] randomArray(int n) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * 2); + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int N = 100; + int testTime = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * N) + 1; + int[] arr = randomArray(n); + int ans1 = maxOneNumbers1(arr); + int ans2 = maxOneNumbers2(arr); + if (ans1 != ans2) { + System.out.println("出错了！"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_04_3_week/Code02_RMQ.java b/算法周更班/class_2022_04_3_week/Code02_RMQ.java new file mode 100644 index 0000000..e6d5c5c --- /dev/null +++ b/算法周更班/class_2022_04_3_week/Code02_RMQ.java @@ -0,0 +1,118 @@ +package class_2022_04_3_week; + +// 小红书 +// 3.13 笔试 +// 给定一个数组，想随时查询任何范围上的最大值 +// 如果只是根据初始数组建立、并且以后没有修改， +// 那么RMQ方法比线段树方法好实现，时间复杂度O(N*logN)，额外空间复杂度O(N*logN) +public class Code02_RMQ { + + public static class RMQ { + public int[][] max; + + // 下标一定要从1开始，没有道理！就是约定俗成！ + public RMQ(int[] arr) { + // 长度！ + int n = arr.length; + // 2的几次方，可以拿下n + int k = power2(n); + // n*logn + max = new int[n + 1][k + 1]; + for (int i = 1; i <= n; i++) { + // i 0：从下标i开始，往下连续的2的0次方个数，中，最大值 + // 1...1个 + // 2...1个 + // 3...1个 + max[i][0] = arr[i - 1]; + } + for (int j = 1; (1 << j) <= n; j++) { + // i...连续的、2的1次方个数，这个范围，最大值 + // i...连续的、2的2次方个数，这个范围，最大值 + // i...连续的、2的3次方个数，这个范围，最大值 + for (int i = 1; i + (1 << j) - 1 <= n; i++) { + // max[10][3] + // 下标10开始，连续的8个数，最大值是多少 + // 1) max[10][2] + // 2) max[14][2] + max[i][j] = Math.max( + max[i][j - 1], + max[i + (1 << (j - 1))][j - 1]); + } + } + } + + public int max(int l, int r) { + // l...r -> r - l + 1 -> 2的哪个次方最接近它！ + int k = power2(r - l + 1); + return Math.max(max[l][k], max[r - (1 << k) + 1][k]); + } + + private int power2(int m) { + int ans = 0; + while ((1 << ans) <= (m >> 1)) { + ans++; + } + return ans; + } + + } + + // 为了测试 + public static class Right { + public int[][] max; + + public Right(int[] arr) { + int n = arr.length; + max = new int[n + 1][n + 1]; + for (int l = 1; l <= n; l++) { + max[l][l] = arr[l - 1]; + for (int r = l + 1; r <= n; r++) { + max[l][r] = Math.max(max[l][r - 1], arr[r - 1]); + } + } + } + + public int max(int l, int r) { + return max[l][r]; + } + + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * v); + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int N = 150; + int V = 200; + int testTimeOut = 20000; + int testTimeIn = 200; + System.out.println("测试开始"); + for (int i = 0; i < testTimeOut; i++) { + int n = (int) (Math.random() * N) + 1; + int[] arr = randomArray(n, V); + int m = arr.length; + RMQ rmq = new RMQ(arr); + Right right = new Right(arr); + for (int j = 0; j < testTimeIn; j++) { + int a = (int) (Math.random() * m) + 1; + int b = (int) (Math.random() * m) + 1; + int l = Math.min(a, b); + int r = Math.max(a, b); + int ans1 = rmq.max(l, r); + int ans2 = right.max(l, r); + if (ans1 != ans2) { + System.out.println("出错了!"); + } + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_04_3_week/Code03_ValidSortedArrayWays.java b/算法周更班/class_2022_04_3_week/Code03_ValidSortedArrayWays.java new file mode 100644 index 0000000..309219f --- /dev/null +++ b/算法周更班/class_2022_04_3_week/Code03_ValidSortedArrayWays.java @@ -0,0 +1,125 @@ +package class_2022_04_3_week; + +import java.util.Arrays; + +// 来自腾讯音乐 +// 原本数组中都是大于0、小于等于k的数字，是一个单调不减的数组 +// 其中可能有相等的数字，总体趋势是递增的 +// 但是其中有些位置的数被替换成了0，我们需要求出所有的把0替换的方案数量： +// 1）填充的每一个数可以大于等于前一个数，小于等于后一个数 +// 2）填充的每一个数不能大于k +public class Code03_ValidSortedArrayWays { + + // 动态规划 + public static long ways1(int[] nums, int k) { + int n = nums.length; + // dp[i][j] : 一共i个格子，随意填，但是不能降序，j种数可以选 + long[][] dp = new long[n + 1][k + 1]; + for (int i = 1; i <= n; i++) { + dp[i][1] = 1; + } + for (int i = 1; i <= k; i++) { + dp[1][i] = i; + } + for (int i = 2; i <= n; i++) { + for (int j = 2; j <= k; j++) { + dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; + } + } + long res = 1; + for (int i = 0, j = 0; i < nums.length; i++) { + if (nums[i] == 0) { + j = i + 1; + while (j < nums.length && nums[j] == 0) { + j++; + } + int leftValue = i - 1 >= 0 ? nums[i - 1] : 1; + int rightValue = j < nums.length ? nums[j] : k; + res *= dp[j - i][rightValue - leftValue + 1]; + i = j; + } + } + return res; + } + + // 数学方法 + // a ~ b范围的数字随便选，可以选重复的数，一共选m个 + // 选出有序序列的方案数：C ( m, b - a + m ) + public static long ways2(int[] nums, int k) { + long res = 1; + for (int i = 0, j = 0; i < nums.length; i++) { + if (nums[i] == 0) { + j = i + 1; + while (j < nums.length && nums[j] == 0) { + j++; + } + int leftValue = i - 1 >= 0 ? nums[i - 1] : 1; + int rightValue = j < nums.length ? nums[j] : k; + int numbers = j - i; + res *= c(rightValue - leftValue + numbers, numbers); + i = j; + } + } + return res; + } + + // 从一共a个数里，选b个数，方法数是多少 + public static long c(int a, int b) { + if (a == b) { + return 1; + } + long x = 1; + long y = 1; + for (int i = b + 1, j = 1; i <= a; i++, j++) { + x *= i; + y *= j; + long gcd = gcd(x, y); + x /= gcd; + y /= gcd; + } + return x / y; + } + + public static long gcd(long m, long n) { + return n == 0 ? m : gcd(n, m % n); + } + + // 为了测试 + public static int[] randomArray(int n, int k) { + int[] ans = new int[n]; + for (int i = 0; i < n; i++) { + ans[i] = (int) (Math.random() * k) + 1; + } + Arrays.sort(ans); + for (int i = 0; i < n; i++) { + ans[i] = Math.random() < 0.5 ? 0 : ans[i]; + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int N = 20; + int K = 30; + int testTimes = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + int n = (int) (Math.random() * N) + 1; + int k = (int) (Math.random() * K) + 1; + int[] arr = randomArray(n, k); + long ans1 = ways1(arr, k); + long ans2 = ways2(arr, k); + if (ans1 != ans2) { + System.out.println("出错了"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println(ans1); + System.out.println(ans2); + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_04_3_week/Code04_SumEvenSubNumber.java b/算法周更班/class_2022_04_3_week/Code04_SumEvenSubNumber.java new file mode 100644 index 0000000..b30658b --- /dev/null +++ b/算法周更班/class_2022_04_3_week/Code04_SumEvenSubNumber.java @@ -0,0 +1,84 @@ +package class_2022_04_3_week; + +// 来自学员问题 +// 总长度为n的数组中，所有长度为k的子序列里，有多少子序列的和为偶数 +public class Code04_SumEvenSubNumber { + + public static int number1(int[] arr, int k) { + if (arr == null || arr.length == 0 || k < 1 || k > arr.length) { + return 0; + } + return process1(arr, 0, k, 0); + } + + public static int process1(int[] arr, int index, int rest, int sum) { + if (index == arr.length) { + return rest == 0 && (sum & 1) == 0 ? 1 : 0; + } else { + return process1(arr, index + 1, rest, sum) + process1(arr, index + 1, rest - 1, sum + arr[index]); + } + } + + public static int number2(int[] arr, int k) { + if (arr == null || arr.length == 0 || k < 1 || k > arr.length) { + return 0; + } + int n = arr.length; + // even[i][j] : 在前i个数的范围上(0...i-1)，一定选j个数，加起来是偶数的子序列个数 + // odd[i][j] : 在前i个数的范围上(0...i-1)，一定选j个数，加起来是奇数的子序列个数 + int[][] even = new int[n + 1][k + 1]; + int[][] odd = new int[n + 1][k + 1]; + for (int i = 0; i <= n; i++) { + // even[0][0] = 1; + // even[1][0] = 1; + // even[2][0] = 1; + // even[n][0] = 1; + even[i][0] = 1; + } + for (int i = 1; i <= n; i++) { + for (int j = 1; j <= Math.min(i, k); j++) { + even[i][j] = even[i - 1][j]; + odd[i][j] = odd[i - 1][j]; + even[i][j] += (arr[i - 1] & 1) == 0 ? even[i - 1][j - 1] : odd[i - 1][j - 1]; + odd[i][j] += (arr[i - 1] & 1) == 0 ? odd[i - 1][j - 1] : even[i - 1][j - 1]; + } + } + return even[n][k]; + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] ans = new int[n]; + for (int i = 0; i < n; i++) { + ans[i] = (int) (Math.random() * v); + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int N = 20; + int V = 30; + int testTimes = 3000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + int n = (int) (Math.random() * N) + 1; + int[] arr = randomArray(n, V); + int k = (int) (Math.random() * n) + 1; + int ans1 = number1(arr, k); + int ans2 = number2(arr, k); + if (ans1 != ans2) { + System.out.println("出错了"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println(k); + System.out.println(ans1); + System.out.println(ans2); + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_04_3_week/Code05_ModKSubstringNumbers.java b/算法周更班/class_2022_04_3_week/Code05_ModKSubstringNumbers.java new file mode 100644 index 0000000..c69cec2 --- /dev/null +++ b/算法周更班/class_2022_04_3_week/Code05_ModKSubstringNumbers.java @@ -0,0 +1,83 @@ +package class_2022_04_3_week; + +// 来自微众 +// 4.11笔试 +// 给定n位长的数字字符串和正数k，求该子符串能被k整除的子串个数 +// (n<=1000，k<=100) +public class Code05_ModKSubstringNumbers { + + // 暴力方法 + // 为了验证 + public static int modWays1(String s, int k) { + int n = s.length(); + int ans = 0; + for (int i = 0; i < n; i++) { + for (int j = i; j < n; j++) { + if (Long.valueOf(s.substring(i, j + 1)) % k == 0) { + ans++; + } + } + } + return ans; + } + + // 正式方法 + // 时间复杂度O(N * k) + public static int modWays2(String s, int k) { + int[] cur = new int[k]; + // 帮忙迁移 + int[] next = new int[k]; + // 0...i 整体余几？ + int mod = 0; + // 答案：统计有多少子串的值%k == 0 + int ans = 0; + for (char cha : s.toCharArray()) { + for (int i = 0; i < k; i++) { + // i -> 10个 + // (i * 10) % k + next[(i * 10) % k] += cur[i]; + cur[i] = 0; + } + int[] tmp = cur; + cur = next; + next = tmp; + mod = (mod * 10 + (cha - '0')) % k; + ans += (mod == 0 ? 1 : 0) + cur[mod]; + cur[mod]++; + } + return ans; + } + + // 为了测试 + public static String randomNumber(int n) { + char[] ans = new char[n]; + for (int i = 0; i < n; i++) { + ans[i] = (char) ((int) (Math.random() * 10) + '0'); + } + return String.valueOf(ans); + } + + // 为了测试 + public static void main(String[] args) { + int N = 18; + int K = 20; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + String str = randomNumber((int) (Math.random() * N) + 1); + int k = (int) (Math.random() * K) + 1; + int ans1 = modWays1(str, k); + int ans2 = modWays2(str, k); + if (ans1 != ans2) { + System.out.println("出错了!"); + System.out.println(str); + System.out.println(k); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_05_1_week/Code01_JumMinSameValue.java b/算法周更班/class_2022_05_1_week/Code01_JumMinSameValue.java new file mode 100644 index 0000000..8274de1 --- /dev/null +++ b/算法周更班/class_2022_05_1_week/Code01_JumMinSameValue.java @@ -0,0 +1,79 @@ +package class_2022_05_1_week; + +import java.util.ArrayList; +import java.util.HashMap; + +// 来自蔚来汽车 +// 给你一个整数数组 arr ，你一开始在数组的第一个元素处（下标为 0）。 +// 每一步，你可以从下标 i 跳到下标 i + 1 、i - 1 或者 j ： +// i + 1 需满足：i + 1 < arr.length +// i - 1 需满足：i - 1 >= 0 +// j 需满足：arr[i] == arr[j] 且 i != j +// 请你返回到达数组最后一个元素的下标处所需的 最少操作次数 。 +// 注意：任何时候你都不能跳到数组外面。 +// leetcode测试链接 : https://leetcode-cn.com/problems/jump-game-iv/ +public class Code01_JumMinSameValue { + + public static int minJumps(int[] arr) { + int n = arr.length; + // 为了找某个值，有哪些位置，能快一些 + // key : 某个值9， + // value : 列表：0，7，19 + HashMap> valueIndex = new HashMap<>(); + for (int i = 0; i < n; i++) { + if (!valueIndex.containsKey(arr[i])) { + valueIndex.put(arr[i], new ArrayList<>()); + } + valueIndex.get(arr[i]).add(i); + } + // i会有哪些展开：左，右，i通过自己的值，能蹦到哪些位置上去 + // 宽度优先遍历，遍历过的位置，不希望重复处理 + // visited[i] == false：i位置，之前没来过，可以处理 + // visited[i] == true : i位置，之前来过，可以跳过 + boolean[] visited = new boolean[n]; + int[] queue = new int[n]; + int l = 0; + int r = 0; + // 0位置加到队列里去 + queue[r++] = 0; + visited[0] = true; + int jump = 0; + // 宽度优先遍历 + // 一次，遍历一整层！ + // 该技巧，多次出现！ + while (l != r) { // 队列里还有东西的意思！ + // 此时的r记录！ + // 0 1 2 | 3 4 5 6 7 8 + // 当前层的终止位置 + int tmp = r; + for (; l < tmp; l++) { // 遍历当前层！ + int cur = queue[l]; + if (cur == n - 1) { + return jump; + } + if (cur + 1 < n && !visited[cur + 1]) { + visited[cur + 1] = true; + queue[r++] = cur + 1; + } + // cur > 0 cur - 1 >=0 + if (cur > 0 && !visited[cur - 1]) { + visited[cur - 1] = true; + queue[r++] = cur - 1; + } + // i -> 9 + // 值同样为9的那些位置，也能去 + for (int next : valueIndex.get(arr[cur])) { + if (!visited[next]) { + visited[next] = true; + queue[r++] = next; + } + } + // 重要优化！ + valueIndex.get(arr[cur]).clear(); + } + jump++; + } + return -1; + } + +} diff --git a/算法周更班/class_2022_05_1_week/Code02_WhoWin21Balls.java b/算法周更班/class_2022_05_1_week/Code02_WhoWin21Balls.java new file mode 100644 index 0000000..6be690a --- /dev/null +++ b/算法周更班/class_2022_05_1_week/Code02_WhoWin21Balls.java @@ -0,0 +1,160 @@ +package class_2022_05_1_week; + +// 来自微众 +// 人工智能岗 +// 一开始有21个球，甲和乙轮流拿球，甲先、乙后 +// 每个人在自己的回合，一定要拿不超过3个球，不能不拿 +// 最终谁的总球数为偶数，谁赢 +// 请问谁有必胜策略 +public class Code02_WhoWin21Balls { + + // balls = 21 + // ball是奇数 + public static String win(int balls) { + return process(0, balls, 0, 0); + } + + // 憋递归！ + // turn 谁的回合！ + // turn == 0 甲回合 + // turn == 1 乙回合 + // rest剩余球的数量 + // 之前，jiaBalls、yiBalls告诉你！ + // 当前，根据turn，知道是谁的回合！ + // 当前，还剩多少球，rest + // 返回：谁会赢！ + public static String process(int turn, int rest, int jia, int yi) { + if (rest == 0) { + return (jia & 1) == 0 ? "甲" : "乙"; + } + // rest > 0, 还剩下球！ + if (turn == 0) { // 甲的回合！ + // 甲，自己赢！甲赢！ + for (int pick = 1; pick <= Math.min(rest, 3); pick++) { + // pick 甲当前做的选择 + if (process(1, rest - pick, jia + pick, yi).equals("甲")) { + return "甲"; + } + } + return "乙"; + } else { + for (int pick = 1; pick <= Math.min(rest, 3); pick++) { + // pick 甲当前做的选择 + if (process(0, rest - pick, jia, yi + pick).equals("乙")) { + return "乙"; + } + } + return "甲"; + } + } + + // 我们补充一下设定，假设一开始的球数量不是21，是任意的正数 + // 如果最终两个人拿的都是偶数，认为无人获胜，平局 + // 如果最终两个人拿的都是奇数，认为无人获胜，平局 + // rest代表目前剩下多少球 + // cur == 0 代表目前是甲行动 + // cur == 1 代表目前是乙行动 + // first == 0 代表目前甲所选的球数，加起来是偶数 + // first == 1 代表目前甲所选的球数，加起来是奇数 + // second == 0 代表目前乙所选的球数，加起来是偶数 + // second == 1 代表目前乙所选的球数，加起来是奇数 + // 返回选完了rest个球，谁会赢，只会返回"甲"、"乙"、"平" + // win1方法，就是彻底暴力的做所有尝试，并且返回最终的胜利者 + // 在甲的回合，甲会尝试所有的可能，以保证自己会赢，如果自己怎么都不会赢，那也要尽量平局，如果这个也不行，只能对方赢 + // 在乙的回合，乙会尝试所有的可能，以保证自己会赢，如果自己怎么都不会赢，那也要尽量平局，如果这个也不行，只能对方赢 + // 算法和数据结构体系学习班，视频39章节，牛羊吃草问题，就是类似这种递归 + public static String win1(int rest, int cur, int first, int second) { + if (rest == 0) { + if (first == 0 && second == 1) { + return "甲"; + } + if (first == 1 && second == 0) { + return "乙"; + } + return "平"; + } + if (cur == 0) { // 甲行动 + String bestAns = "乙"; + for (int pick = 1; pick <= Math.min(3, rest); pick++) { + String curAns = win1(rest - pick, 1, first ^ (pick & 1), second); + if (curAns.equals("甲")) { + bestAns = "甲"; + break; + } + if (curAns.equals("平")) { + bestAns = "平"; + } + } + return bestAns; + } else { // 乙行动 + String bestAns = "甲"; + for (int pick = 1; pick <= Math.min(3, rest); pick++) { + String curAns = win1(rest - pick, 0, first, second ^ (pick & 1)); + if (curAns.equals("乙")) { + bestAns = "乙"; + break; + } + if (curAns.equals("平")) { + bestAns = "平"; + } + } + return bestAns; + } + } + + // 下面的win2方法，仅仅是把win1方法，做了记忆化搜索 + // 变成了动态规划 + public static String[][][][] dp = new String[5000][2][2][2]; + + public static String win2(int rest, int cur, int first, int second) { + if (rest == 0) { + if (first == 0 && second == 1) { + return "甲"; + } + if (first == 1 && second == 0) { + return "乙"; + } + return "平"; + } + if (dp[rest][cur][first][second] != null) { + return dp[rest][cur][first][second]; + } + if (cur == 0) { // 甲行动 + String bestAns = "乙"; + for (int pick = 1; pick <= Math.min(3, rest); pick++) { + String curAns = win2(rest - pick, 1, first ^ (pick & 1), second); + if (curAns.equals("甲")) { + bestAns = "甲"; + break; + } + if (curAns.equals("平")) { + bestAns = "平"; + } + } + dp[rest][cur][first][second] = bestAns; + return bestAns; + } else { // 乙行动 + String bestAns = "甲"; + for (int pick = 1; pick <= Math.min(3, rest); pick++) { + String curAns = win2(rest - pick, 0, first, second ^ (pick & 1)); + if (curAns.equals("乙")) { + bestAns = "乙"; + break; + } + if (curAns.equals("平")) { + bestAns = "平"; + } + } + dp[rest][cur][first][second] = bestAns; + return bestAns; + } + } + + // 为了测试 + public static void main(String[] args) { + for (int balls = 1; balls <= 500; balls += 2) { + System.out.println("球数为 " + balls + " 时 , 赢的是 " + win(balls)); + } + } + +} \ No newline at end of file diff --git a/算法周更班/class_2022_05_1_week/Code03_FindDuplicateOnlyOne.java b/算法周更班/class_2022_05_1_week/Code03_FindDuplicateOnlyOne.java new file mode 100644 index 0000000..ecbc931 --- /dev/null +++ b/算法周更班/class_2022_05_1_week/Code03_FindDuplicateOnlyOne.java @@ -0,0 +1,114 @@ +package class_2022_05_1_week; + +import java.util.Arrays; +import java.util.HashSet; + +// 来自学员问题，真实大厂面试题 +// 1、2、3...n-1、n、n、n+1、n+2... +// 在这个序列中，只有一个数字有重复(n) +// 这个序列是无序的，找到重复数字n +// 这个序列是有序的，找到重复数字n +public class Code03_FindDuplicateOnlyOne { + + // 为了测试 + // 绝对正确，但是直接遍历+哈希表，没有得分的方法 + public static int right(int[] arr) { + HashSet set = new HashSet<>(); + for (int num : arr) { + if (set.contains(num)) { + return num; + } + set.add(num); + } + return -1; + } + + // 符合题目要求的、无序数组，找重复数 + // 时间复杂度O(N)，额外空间复杂度O(1) + public static int findDuplicate(int[] arr) { + if (arr == null || arr.length < 2) { + return -1; + } + int slow = arr[0]; + int fast = arr[arr[0]]; + while (slow != fast) { + slow = arr[slow]; + fast = arr[arr[fast]]; + } + // slow == fast + fast = 0; + while (slow != fast) { + fast = arr[fast]; + slow = arr[slow]; + } + // 再相遇！一个结论 + return slow; + } + + // 符合题目要求的、有序数组，找重复数 + // 时间复杂度O(logN)，额外空间复杂度O(1) + public static int findDuplicateSorted(int[] arr) { + if (arr == null || arr.length < 2) { + return -1; + } + int l = 0; + int r = arr.length - 1; + int m = 0; + int ans = -1; + while (l <= r) { + m = (l + r) / 2; + if ((m - 1 >= 0 && arr[m - 1] == arr[m]) || (m + 1 < arr.length && arr[m + 1] == arr[m])) { + ans = arr[m]; + break; + } + if (m - l == arr[m] - arr[l]) { + l = m + 1; + } else { + r = m - 1; + } + } + return ans; + } + + // 为了测试 + public static int[] randomArray(int n) { + int[] ans = new int[n + 1]; + for (int i = 0; i < n; i++) { + ans[i] = i + 1; + } + ans[n] = (int) (Math.random() * n) + 1; + for (int i = n; i > 0; i--) { + int j = (int) (Math.random() * (i + 1)); + int tmp = ans[i]; + ans[i] = ans[j]; + ans[j] = tmp; + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int N = 10; + int testTime = 5000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int[] arr = randomArray((int) (Math.random() * N) + 1); + if (right(arr) != findDuplicate(arr)) { + System.out.println("未排序情况出错!"); + } + Arrays.sort(arr); + if (right(arr) != findDuplicateSorted(arr)) { + System.out.println("排序情况出错!"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println(right(arr)); + System.out.println(findDuplicateSorted(arr)); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_05_1_week/Code04_SumOfQuadraticSum.java b/算法周更班/class_2022_05_1_week/Code04_SumOfQuadraticSum.java new file mode 100644 index 0000000..0af7de4 --- /dev/null +++ b/算法周更班/class_2022_05_1_week/Code04_SumOfQuadraticSum.java @@ -0,0 +1,113 @@ +package class_2022_05_1_week; + +// 来自学员问题，蓝桥杯练习题 +// f(i) : i的所有因子，每个因子都平方之后，累加起来 +// 比如f(10) = 1平方 + 2平方 + 5平方 + 10平方 = 1 + 4 + 25 + 100 = 130 +// 给定一个数n，求f(1) + f(2) + .. + f(n) +// n <= 10的9次方 +// O(n)的方法都会超时！低于它的！ +// O(根号N)的方法，就过了，一个思路 +// O(log N)的方法， +public class Code04_SumOfQuadraticSum { + + // 暴力方法 + public static long sum1(long n) { + int[] cnt = new int[(int) n + 1]; + for (int num = 1; num <= n; num++) { + for (int j = 1; j <= num; j++) { + if (num % j == 0) { + cnt[j]++; + } + } + } + long ans = 0; + for (long i = 1; i <= n; i++) { + ans += i * i * (long) (cnt[(int) i]); + } + return ans; + } + + // 正式方法 + // 时间复杂度O(开平方根N + 开平方根N * logN) + public static long sum2(long n) { + // 100 -> 10 + // 200 -> 14 + long sqrt = (long) Math.pow((double) n, 0.5); + long ans = 0; + for (long i = 1; i <= sqrt; i++) { + ans += i * i * (n / i); + } + // 后半段 + // 给你一个个数，二分出几个因子，处在这个个数上！ + // 由最大个数(根号N), 开始二分 + for (long k = n / (sqrt + 1); k >= 1; k--) { + ans += sumOfLimitNumber(n, k); + } + return ans; + } + + // 平方和公式n(n+1)(2n+1)/6 + public static long sumOfLimitNumber(long v, long n) { + long r = cover(v, n); + long l = cover(v, n + 1); + return ((r * (r + 1) * ((r << 1) + 1) + - l * (l + 1) * ((l << 1) + 1)) * n) + / 6; + } + + public static long cover(long v, long n) { + long l = 1; + long r = v; + long m = 0; + long ans = 0; + while (l <= r) { + m = (l + r) / 2; + if (m * n <= v) { + ans = m; + l = m + 1; + } else { + r = m - 1; + } + } + return ans; + } + + // 实验 + // 解法来自观察 + // 打表（暴力） + // f(1) + ... + f(n) + public static void test(int n) { + int[] cnt = new int[n + 1]; + for (int num = 1; num <= n; num++) { + for (int j = 1; j <= num; j++) { + if (num % j == 0) { + cnt[j]++; + } + } + } + for (int i = 1; i <= n; i++) { + System.out.println("因子 : " + i + ", 个数 : " + cnt[i]); + } + } + + public static void main(String[] args) { + +// test(100); + + System.out.println("测试开始"); + for (long i = 1; i < 1000; i++) { + if (sum1(i) != sum2(i)) { + System.out.println("出错了!"); + } + } + System.out.println("测试结束"); + + long n = 50000000000L; // 5 * 10的10次方 + long start = System.currentTimeMillis(); + sum2(n); + long end = System.currentTimeMillis(); + System.out.println("大样本测试，n = " + n); + System.out.println("运行时间 : " + (end - start) + " ms"); + } + +} diff --git a/算法周更班/class_2022_05_1_week/Code05_PalindromeStringNoLessKLenNoOverlapingMaxParts.java b/算法周更班/class_2022_05_1_week/Code05_PalindromeStringNoLessKLenNoOverlapingMaxParts.java new file mode 100644 index 0000000..31f67a2 --- /dev/null +++ b/算法周更班/class_2022_05_1_week/Code05_PalindromeStringNoLessKLenNoOverlapingMaxParts.java @@ -0,0 +1,131 @@ +package class_2022_05_1_week; + +// 来自optiver +// 给定一个字符串str，和一个正数k +// 你可以随意的划分str成多个子串， +// 目的是找到在某一种划分方案中，有尽可能多的回文子串，长度>=k，并且没有重合 +// 返回有几个回文子串 +public class Code05_PalindromeStringNoLessKLenNoOverlapingMaxParts { + + // 暴力尝试 + // 为了测试 + // 可以改成动态规划，但不是最优解 + public static int max1(String s, int k) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = s.toCharArray(); + return process1(str, 0, k); + } + + public static int process1(char[] str, int index, int k) { + if (str.length - index < k) { + return 0; + } + int ans = process1(str, index + 1, k); + for (int i = index + k - 1; i < str.length; i++) { + if (isPalindrome(str, index, i)) { + ans = Math.max(ans, 1 + process1(str, i + 1, k)); + } + } + return ans; + } + + public static boolean isPalindrome(char[] str, int L, int R) { + while (L < R) { + if (str[L++] != str[R--]) { + return false; + } + } + return true; + } + + // 最优解 + // 时间复杂度O(N) + public static int max2(String s, int k) { + if (s == null || s.length() == 0) { + return 0; + } + char[] str = manacherString(s); + int[] p = new int[str.length]; + int ans = 0; + int next = 0; + // k == 5 回文串长度要 >= 5 + // next == 0 + // 0.... 8 第一块！ + // next -> 9 + // 9.....17 第二块！ + // next -> 18 + // 18....23 第三块 + // next一直到最后! + while ((next = manacherFind(str, p, next, k)) != -1) { + next = str[next] == '#' ? next : (next + 1); + ans++; + } + return ans; + } + + public static char[] manacherString(String s) { + char[] str = s.toCharArray(); + char[] ans = new char[s.length() * 2 + 1]; + int index = 0; + for (int i = 0; i != ans.length; i++) { + ans[i] = (i & 1) == 0 ? '#' : str[index++]; + } + return ans; + } + + // s[l...]字符串只在这个范围上，且s[l]一定是'#' + // 从下标l开始，之前都不算，一旦有某个中心回文半径>k，马上返回右边界 + public static int manacherFind(char[] s, int[] p, int l, int k) { + int c = l - 1; + int r = l - 1; + int n = s.length; + for (int i = l; i < s.length; i++) { + p[i] = r > i ? Math.min(p[2 * c - i], r - i) : 1; + while (i + p[i] < n && i - p[i] > l - 1 && s[i + p[i]] == s[i - p[i]]) { + if (++p[i] > k) { + return i + k; + } + } + if (i + p[i] > r) { + r = i + p[i]; + c = i; + } + } + return -1; + } + + // 为了测试 + public static String randomString(int n, int r) { + char[] str = new char[(int) (Math.random() * n)]; + for (int i = 0; i < str.length; i++) { + str[i] = (char) ((int) (Math.random() * r) + 'a'); + } + return String.valueOf(str); + } + + // 为了测试 + public static void main(String[] args) { + int n = 20; + int r = 3; + int testTime = 50000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + String str = randomString(n, r); + int k = (int) (Math.random() * str.length()) + 1; + int ans1 = max1(str, k); + int ans2 = max2(str, k); + if (ans1 != ans2) { + System.out.println(str); + System.out.println(k); + System.out.println(ans1); + System.out.println(ans2); + System.out.println("出错了!"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_05_2_week/Code01_TwoObjectMaxValue.java b/算法周更班/class_2022_05_2_week/Code01_TwoObjectMaxValue.java new file mode 100644 index 0000000..be8c0f8 --- /dev/null +++ b/算法周更班/class_2022_05_2_week/Code01_TwoObjectMaxValue.java @@ -0,0 +1,155 @@ +package class_2022_05_2_week; + +import java.util.Arrays; + +// 来自字节 +// 5.6笔试 +// 给定N件物品，每个物品有重量(w[i])、有价值(v[i]) +// 只能最多选两件商品，重量不超过bag，返回价值最大能是多少？ +// N <= 10^5, w[i] <= 10^5, v[i] <= 10^5, bag <= 10^5 +// 本题的关键点：什么数据范围都很大，唯独只需要最多选两件商品，这个可以利用一下 +public class Code01_TwoObjectMaxValue { + + // 暴力方法 + // 为了验证而写的方法 + public static int max1(int[] w, int[] v, int bag) { + return process1(w, v, 0, 2, bag); + } + + public static int process1(int[] w, int[] v, int index, int restNumber, int restWeight) { + if (restNumber < 0 || restWeight < 0) { + return -1; + } + if (index == w.length) { + return 0; + } + int p1 = process1(w, v, index + 1, restNumber, restWeight); + int p2 = -1; + int next = process1(w, v, index + 1, restNumber - 1, restWeight - w[index]); + if (next != -1) { + p2 = v[index] + next; + } + return Math.max(p1, p2); + } + + // 正式方法 + // 时间复杂度O(N * logN) + public static int max2(int[] w, int[] v, int bag) { + int n = w.length; + int[][] arr = new int[n][2]; + for (int i = 0; i < n; i++) { + arr[i][0] = w[i]; + arr[i][1] = v[i]; + } + // O(N * logN) + Arrays.sort(arr, (a, b) -> (a[0] - b[0])); + // 重量从轻到重，依次标号1、2、3、4.... + // 价值依次被构建成了RMQ结构 + // O(N * logN) + RMQ rmq = new RMQ(arr); + int ans = 0; + // N * logN + for (int i = 0, j = 1; i < n && arr[i][0] <= bag; i++, j++) { + // 当前来到0号货物，RMQ结构1号 + // 当前来到i号货物，RMQ结构i+1号 + // 查询重量的边界，重量 边界 <= bag - 当前货物的重量 + // 货物数组中，找到 <= 边界，最右的位置i + // RMQ，位置 i + 1 + int right = right(arr, bag - arr[i][0]) + 1; + int rest = 0; + // j == i + 1，当前的货物，在RMQ里的下标 + if (right == j) { + rest = rmq.max(1, right - 1); + } else if (right < j) { + rest = rmq.max(1, right); + } else { // right > j + rest = Math.max(rmq.max(1, j - 1), rmq.max(j + 1, right)); + } + ans = Math.max(ans, arr[i][1] + rest); + } + return ans; + } + + public static int right(int[][] arr, int limit) { + int l = 0; + int r = arr.length - 1; + int m = 0; + int ans = -1; + while (l <= r) { + m = (l + r) / 2; + if (arr[m][0] <= limit) { + ans = m; + l = m + 1; + } else { + r = m - 1; + } + } + return ans; + } + + public static class RMQ { + public int[][] max; + + public RMQ(int[][] arr) { + int n = arr.length; + int k = power2(n); + max = new int[n + 1][k + 1]; + for (int i = 1; i <= n; i++) { + max[i][0] = arr[i - 1][1]; + } + for (int j = 1; (1 << j) <= n; j++) { + for (int i = 1; i + (1 << j) - 1 <= n; i++) { + max[i][j] = Math.max(max[i][j - 1], max[i + (1 << (j - 1))][j - 1]); + } + } + } + + public int max(int l, int r) { + if (r < l) { + return 0; + } + int k = power2(r - l + 1); + return Math.max(max[l][k], max[r - (1 << k) + 1][k]); + } + + private int power2(int m) { + int ans = 0; + while ((1 << ans) <= (m >> 1)) { + ans++; + } + return ans; + } + + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = (int) (Math.random() * v); + } + return arr; + } + + // 为了测试 + public static void main(String[] args) { + int N = 12; + int V = 20; + int testTimes = 5000; + System.out.println("测试开始"); + for (int i = 0; i < testTimes; i++) { + int n = (int) (Math.random() * N) + 1; + int[] w = randomArray(n, V); + int[] v = randomArray(n, V); + int bag = (int) (Math.random() * V * 3); + int ans1 = max1(w, v, bag); + int ans2 = max2(w, v, bag); + if (ans1 != ans2) { + System.out.println("出错了!"); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_05_2_week/Code02_ModifyOneNumberModXWays.java b/算法周更班/class_2022_05_2_week/Code02_ModifyOneNumberModXWays.java new file mode 100644 index 0000000..ec32159 --- /dev/null +++ b/算法周更班/class_2022_05_2_week/Code02_ModifyOneNumberModXWays.java @@ -0,0 +1,84 @@ +package class_2022_05_2_week; + +// 来自网易 +// 小红拿到了一个长度为N的数组arr，她准备只进行一次修改 +// 可以将数组中任意一个数arr[i]，修改为不大于P的正数（修改后的数必须和原数不同) +// 并使得所有数之和为X的倍数 +// 小红想知道，一共有多少种不同的修改方案 +// 1 <= N, X <= 10^5 +// 1 <= arr[i], P <= 10^9 +public class Code02_ModifyOneNumberModXWays { + + public static int ways1(int[] arr, int p, int x) { + long sum = 0; + for (int num : arr) { + sum += num; + } + int ans = 0; + for (int num : arr) { + sum -= num; + for (int v = 1; v <= p; v++) { + if (v != num) { + if ((sum + v) % x == 0) { + ans++; + } + } + } + sum += num; + } + return ans; + } + + public static int ways2(int[] arr, int p, int x) { + long sum = 0; + for (int num : arr) { + sum += num; + } + int ans = 0; + for (int num : arr) { + ans += cnt(p, x, num, (x - (int) ((sum - num) % x)) % x); + } + return ans; + } + + // 当前数字num + // 1~p以内，不能是num的情况下，% x == mod的数字有几个 + // O(1) + public static int cnt(int p, int x, int num, int mod) { + // p/x 至少有几个 + // (p % x) >= mod ? 1 : 0 + // 在不考虑变出来的数，是不是num的情况下，算一下有几个数，符合要求 + int ans = (p / x) + ((p % x) >= mod ? 1 : 0) - (mod == 0 ? 1 : 0); + // 不能等于num！ + return ans - ((num <= p && num % x == mod) ? 1 : 0); + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] ans = new int[n]; + for (int i = 0; i < n; i++) { + ans[i] = (int) (Math.random() * v) + 1; + } + return ans; + } + + public static void main(String[] args) { + int len = 100; + int value = 100; + int testTime = 100000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * len) + 1; + int[] arr = randomArray(n, value); + int p = (int) (Math.random() * value) + 1; + int x = (int) (Math.random() * value) + 1; + int ans1 = ways1(arr, p, x); + int ans2 = ways2(arr, p, x); + if (ans1 != ans2) { + System.out.println("出错了！"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_05_2_week/Code03_SortedSubsequenceMaxSum.java b/算法周更班/class_2022_05_2_week/Code03_SortedSubsequenceMaxSum.java new file mode 100644 index 0000000..967b8ef --- /dev/null +++ b/算法周更班/class_2022_05_2_week/Code03_SortedSubsequenceMaxSum.java @@ -0,0 +1,162 @@ +package class_2022_05_2_week; + +import java.util.Arrays; + +// 来自字节 +// 一共有n个人，从左到右排列，依次编号0~n-1 +// h[i]是第i个人的身高 +// v[i]是第i个人的分数 +// 要求从左到右选出一个子序列，在这个子序列中的人，从左到右身高是不下降的 +// 返回所有符合要求的子序列中，分数最大累加和是多大 +// n <= 10的5次方, 1 <= h[i] <= 10的9次方, 1 <= v[i] <= 10的9次方 +public class Code03_SortedSubsequenceMaxSum { + + // 为了测试 + // 绝对正确的暴力方法 + public static int right(int[] h, int[] v) { + return process(h, v, 0, 0); + } + + public static int process(int[] h, int[] v, int index, int preValue) { + if (index == h.length) { + return 0; + } + int p1 = process(h, v, index + 1, preValue); + int p2 = h[index] >= preValue ? (v[index] + process(h, v, index + 1, h[index])) : 0; + return Math.max(p1, p2); + } + + // 正式方法 + // 时间复杂度O(N * logN) + public static int maxSum(int[] h, int[] v) { + int n = h.length; + int[] rank = new int[n]; + for (int i = 0; i < n; i++) { + rank[i] = h[i]; + } + Arrays.sort(rank); + SegmentTree st = new SegmentTree(n); + for (int i = 0; i < n; i++) { + int height = rank(rank, h[i]); + // 1~height max + st.update(height, st.max(height) + v[i]); + } + return st.max(n); + } + + // [150, 152, 160, 175] 160 + // 1 2 3 4 + // 3 + public static int rank(int[] rank, int num) { + int l = 0; + int r = rank.length - 1; + int m = 0; + int ans = 0; + while (l <= r) { + m = (l + r) / 2; + if (rank[m] >= num) { + ans = m; + r = m - 1; + } else { + l = m + 1; + } + } + return ans + 1; + } + + public static class SegmentTree { + private int n; + private int[] max; + private int[] update; + + public SegmentTree(int maxSize) { + n = maxSize + 1; + max = new int[n << 2]; + update = new int[n << 2]; + Arrays.fill(update, -1); + } + + public void update(int index, int c) { + update(index, index, c, 1, n, 1); + } + + public int max(int right) { + return max(1, right, 1, n, 1); + } + + private void pushUp(int rt) { + max[rt] = Math.max(max[rt << 1], max[rt << 1 | 1]); + } + + private void pushDown(int rt, int ln, int rn) { + if (update[rt] != -1) { + update[rt << 1] = update[rt]; + max[rt << 1] = update[rt]; + update[rt << 1 | 1] = update[rt]; + max[rt << 1 | 1] = update[rt]; + update[rt] = -1; + } + } + + private void update(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + max[rt] = C; + update[rt] = C; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + update(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + update(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + private int max(int L, int R, int l, int r, int rt) { + if (L <= l && r <= R) { + return max[rt]; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + int ans = 0; + if (L <= mid) { + ans = Math.max(ans, max(L, R, l, mid, rt << 1)); + } + if (R > mid) { + ans = Math.max(ans, max(L, R, mid + 1, r, rt << 1 | 1)); + } + return ans; + } + + } + + // 为了测试 + public static int[] randomArray(int n, int v) { + int[] ans = new int[n]; + for (int i = 0; i < n; i++) { + ans[i] = (int) (Math.random() * v) + 1; + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int N = 30; + int V = 100; + int testTime = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * N) + 1; + int[] h = randomArray(n, V); + int[] v = randomArray(n, V); + if (right(h, v) != maxSum(h, v)) { + System.out.println("出错了!"); + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_05_2_week/Code04_OneEdgeMagicMinPathSum.java b/算法周更班/class_2022_05_2_week/Code04_OneEdgeMagicMinPathSum.java new file mode 100644 index 0000000..c2e88d5 --- /dev/null +++ b/算法周更班/class_2022_05_2_week/Code04_OneEdgeMagicMinPathSum.java @@ -0,0 +1,173 @@ +package class_2022_05_2_week; + +import java.util.ArrayList; +import java.util.PriorityQueue; + +// 来自网易 +// 给出一个有n个点，m条有向边的图 +// 你可以施展魔法，把有向边，变成无向边 +// 比如A到B的有向边，权重为7。施展魔法之后，A和B通过该边到达彼此的代价都是7。 +// 求，允许施展一次魔法的情况下，1到n的最短路，如果不能到达，输出-1。 +// n为点数, 每条边用(a,b,v)表示，含义是a到b的这条边，权值为v +// 点的数量 <= 10^5，边的数量 <= 2 * 10^5，1 <= 边的权值 <= 10^6 +public class Code04_OneEdgeMagicMinPathSum { + + // 为了测试 + // 相对暴力的解 + // 尝试每条有向边，都变一次无向边，然后跑一次dijkstra算法 + // 那么其中一定有最好的答案 + public static int min1(int n, int[][] roads) { + int ans = Integer.MAX_VALUE; + for (int i = 0; i < roads.length; i++) { + ArrayList> graph = new ArrayList<>(); + for (int j = 0; j <= n; j++) { + graph.add(new ArrayList<>()); + } + graph.get(roads[i][1]).add(new int[] { roads[i][0], roads[i][2] }); + for (int[] r : roads) { + graph.get(r[0]).add(new int[] { r[1], r[2] }); + } + ans = Math.min(ans, dijkstra1(n, graph)); + } + return ans == Integer.MAX_VALUE ? -1 : ans; + } + + public static int dijkstra1(int n, ArrayList> graph) { + PriorityQueue heap = new PriorityQueue<>((a, b) -> a[1] - b[1]); + boolean[] visited = new boolean[n + 1]; + heap.add(new int[] { 1, 0 }); + int ans = Integer.MAX_VALUE; + while (!heap.isEmpty()) { + int[] cur = heap.poll(); + if (cur[0] == n) { + ans = cur[1]; + break; + } + if (visited[cur[0]]) { + continue; + } + visited[cur[0]] = true; + for (int[] edge : graph.get(cur[0])) { + int to = edge[0]; + int weight = edge[1]; + if (!visited[to]) { + heap.add(new int[] { to, cur[1] + weight }); + } + } + } + return ans; + } + + // 最优解 + // 时间复杂度O(N * logN) + // N <= 2 * 10^5 + public static int min2(int n, int[][] roads) { + ArrayList> graph = new ArrayList<>(); + for (int i = 0; i <= n; i++) { + graph.add(new ArrayList<>()); + } + for (int[] r : roads) { + graph.get(r[0]).add(new int[] { 0, r[1], r[2] }); + graph.get(r[1]).add(new int[] { 1, r[0], r[2] }); + } + PriorityQueue heap = new PriorityQueue<>((a, b) -> a[2] - b[2]); + boolean[][] visited = new boolean[2][n + 1]; + // a -> 0,a 1,a + // boolean[] visted = new boolean[n+1] + // visted[i] == true 去过了！从队列里弹出来过了！以后别碰了！ + // visted[i] == false 没去过！第一次从队列里弹出来！当前要处理！ + // 0,1,0 -> 之前没有走过魔法路，当前来到1号出发点，代价是0 + heap.add(new int[] { 0, 1, 0 }); + int ans = Integer.MAX_VALUE; + while (!heap.isEmpty()) { + int[] cur = heap.poll(); + if (visited[cur[0]][cur[1]]) { + continue; + } + visited[cur[0]][cur[1]] = true; + if (cur[1] == n) { + ans = Math.min(ans, cur[2]); + if (visited[0][n] && visited[1][n]) { + break; + } + } + for (int[] edge : graph.get(cur[1])) { + // 当前来到cur + // 之前有没有走过魔法路径：cur[0] == 0 ，没走过！cur[0] = 1, 走过了 + // 当前来到的点是啥，cur[1]，点编号！ + // 之前的总代价是啥？cur[2] + // cur，往下，能走的，所有的路在哪？ + // 当前的路，叫edge + // 当前的路，是不是魔法路！edge[0] = 0 , 不是魔法路 + // edge[0] == 1，是魔法路 + // cur[0] + edge[0] == 0 + // 路 ：0 5 20 + // 当前路，不是魔法路，去往的点是5号点，该路权重是20 + // 路 ：1 7 13 + // 当前路，是魔法路，去往的点是7号点，该路权重是13 + if (cur[0] + edge[0] == 0) { + if (!visited[0][edge[1]]) { + heap.add(new int[] { 0, edge[1], cur[2] + edge[2] }); + } + } + // cur[0] + edge[0] == 1 + // 0 1 + // 1 0 + if (cur[0] + edge[0] == 1) { + if (!visited[1][edge[1]]) { + heap.add(new int[] { 1, edge[1], cur[2] + edge[2] }); + } + } + // 1 1 == 2 + } + } + return ans == Integer.MAX_VALUE ? -1 : ans; + } + + // 为了测试 + public static int[][] randomRoads(int n, int v) { + int m = (int) (Math.random() * (n * (n - 1) / 2)) + 1; + int[][] roads = new int[m][3]; + for (int i = 0; i < m; i++) { + roads[i][0] = (int) (Math.random() * n) + 1; + roads[i][1] = (int) (Math.random() * n) + 1; + roads[i][2] = (int) (Math.random() * v) + 1; + } + return roads; + } + + // 为了测试 + public static void main(String[] args) { + int N = 20; + int V = 30; + int testTime = 20000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * N) + 1; + int[][] roads = randomRoads(n, V); + if (min1(n, roads) != min2(n, roads)) { + System.out.println("出错了!"); + } + } + System.out.println("测试结束"); + + // 点的数量10^5 + int n = 100000; + // 边的数量2 * 10^5 + int m = 200000; + // 时间复杂度很明显和边的权重是没关系的 + // 所以这里设置小一点，防止出现溢出的解 + int v = 100; + int[][] roads = new int[m][3]; + for (int i = 0; i < m; i++) { + roads[i][0] = (int) (Math.random() * n) + 1; + roads[i][1] = (int) (Math.random() * n) + 1; + roads[i][2] = (int) (Math.random() * v) + 1; + } + long start = System.currentTimeMillis(); + System.out.println("运行结果 : " + min2(n, roads)); + long end = System.currentTimeMillis(); + System.out.println("运行时间(毫秒) : " + (end - start)); + } + +} diff --git a/算法周更班/class_2022_05_2_week/Code05_RedAndWhiteSquares.java b/算法周更班/class_2022_05_2_week/Code05_RedAndWhiteSquares.java new file mode 100644 index 0000000..040b33c --- /dev/null +++ b/算法周更班/class_2022_05_2_week/Code05_RedAndWhiteSquares.java @@ -0,0 +1,211 @@ +package class_2022_05_2_week; + +// 来自网易 +// 小红拿到了一个大立方体，该大立方体由1*1*1的小方块拼成，初始每个小方块都是白色。 +// 小红可以每次选择一个小方块染成红色 +// 每次小红可能选择同一个小方块重复染色 +// 每次染色以后，你需要帮小红回答出当前的白色连通块数 +// 如果两个小方块共用同一个面，且颜色相同，则它们是连通的 +// 给定n、m、h，表示大立方体的长、宽、高 +// 给定k次操作，每一次操作用(a, b, c)表示在大立方体的该位置进行染色 +// 返回长度为k的数组，表示每一次操作后，白色方块的连通块数 +// n * m * h <= 10 ^ 5，k <= 10 ^ 5 +public class Code05_RedAndWhiteSquares { + + // 暴力方法 + // 时间复杂度(k * n * m * h); + public static int[] blocks1(int n, int m, int h, int[][] ops) { + int k = ops.length; + int[][][] cube = new int[n][m][h]; + int value = 1; + int[] ans = new int[k]; + for (int i = 0; i < k; i++) { + cube[ops[i][0]][ops[i][1]][ops[i][2]] = -1; + for (int x = 0; x < n; x++) { + for (int y = 0; y < m; y++) { + for (int z = 0; z < h; z++) { + if (cube[x][y][z] != -1 && cube[x][y][z] != value) { + ans[i]++; + infect(cube, x, y, z, value); + } + } + } + } + value++; + } + return ans; + } + + public static void infect(int[][][] cube, int a, int b, int c, int change) { + if (a < 0 || a == cube.length || b < 0 || b == cube[0].length || c < 0 || c == cube[0][0].length + || cube[a][b][c] == -1 || cube[a][b][c] == change) { + return; + } + cube[a][b][c] = change; + infect(cube, a - 1, b, c, change); + infect(cube, a + 1, b, c, change); + infect(cube, a, b - 1, c, change); + infect(cube, a, b + 1, c, change); + infect(cube, a, b, c - 1, change); + infect(cube, a, b, c + 1, change); + } + + // 最优解 + // O(k + n * m * h) + public static int[] blocks2(int n, int m, int h, int[][] ops) { + int k = ops.length; + int[][][] red = new int[n][m][h]; + for (int[] op : ops) { + red[op[0]][op[1]][op[2]]++; + } + UnionFind uf = new UnionFind(n, m, h, red); + int[] ans = new int[k]; + for (int i = k - 1; i >= 0; i--) { + ans[i] = uf.sets; + int x = ops[i][0]; + int y = ops[i][1]; + int z = ops[i][2]; + if (--red[x][y][z] == 0) { + // x, y ,z 这个格子，变白，建立自己的小集合 + // 然后6个方向，集合该合并合并 + uf.finger(x, y, z); + } + } + return ans; + } + + public static class UnionFind { + public int n; + public int m; + public int h; + public int[] father; + public int[] size; + public int[] help; + public int sets; + + public UnionFind(int a, int b, int c, int[][][] red) { + n = a; + m = b; + h = c; + int len = n * m * h; + father = new int[len]; + size = new int[len]; + help = new int[len]; + for (int x = 0; x < n; x++) { + for (int y = 0; y < m; y++) { + for (int z = 0; z < h; z++) { + if (red[x][y][z] == 0) { + finger(x, y, z); + } + } + } + } + } + + public void finger(int x, int y, int z) { + // x，y，z + // 一维数值 + int i = index(x, y, z); + father[i] = i; + size[i] = 1; + sets++; + union(i, x - 1, y, z); + union(i, x + 1, y, z); + union(i, x, y - 1, z); + union(i, x, y + 1, z); + union(i, x, y, z - 1); + union(i, x, y, z + 1); + } + + private int index(int x, int y, int z) { + return z * n * m + y * n + x; + } + + private void union(int i, int x, int y, int z) { + if (x < 0 || x == n || y < 0 || y == m || z < 0 || z == h) { + return; + } + int j = index(x, y, z); + if (size[j] == 0) { + return; + } + i = find(i); + j = find(j); + if (i != j) { + if (size[i] >= size[j]) { + father[j] = i; + size[i] += size[j]; + } else { + father[i] = j; + size[j] += size[i]; + } + sets--; + } + } + + private int find(int i) { + int s = 0; + while (i != father[i]) { + help[s++] = i; + i = father[i]; + } + while (s > 0) { + father[help[--s]] = i; + } + return i; + } + + } + + // 为了测试 + public static int[][] randomOps(int n, int m, int h) { + int size = (int) (Math.random() * (n * m * h)) + 1; + int[][] ans = new int[size][3]; + for (int i = 0; i < size; i++) { + ans[i][0] = (int) (Math.random() * n); + ans[i][1] = (int) (Math.random() * m); + ans[i][2] = (int) (Math.random() * h); + } + return ans; + } + + // 为了测试 + public static void main(String[] args) { + int size = 10; + int testTime = 5000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int n = (int) (Math.random() * size) + 1; + int m = (int) (Math.random() * size) + 1; + int h = (int) (Math.random() * size) + 1; + int[][] ops = randomOps(n, m, h); + int[] ans1 = blocks1(n, m, h, ops); + int[] ans2 = blocks2(n, m, h, ops); + for (int j = 0; j < ops.length; j++) { + if (ans1[j] != ans2[j]) { + System.out.println("出错了!"); + } + } + } + System.out.println("测试结束"); + + // 立方体达到10^6规模 + int n = 100; + int m = 100; + int h = 100; + int len = n * m * h; + // 操作条数达到10^6规模 + int[][] ops = new int[len][3]; + for (int i = 0; i < len; i++) { + ops[i][0] = (int) (Math.random() * n); + ops[i][1] = (int) (Math.random() * m); + ops[i][2] = (int) (Math.random() * h); + } + long start = System.currentTimeMillis(); + blocks2(n, m, h, ops); + long end = System.currentTimeMillis(); + System.out.println("运行时间(毫秒) : " + (end - start)); + + } + +} diff --git a/算法周更班/class_2022_05_3_week/Code01_MaxNumberUnderLimit.java b/算法周更班/class_2022_05_3_week/Code01_MaxNumberUnderLimit.java new file mode 100644 index 0000000..924ac09 --- /dev/null +++ b/算法周更班/class_2022_05_3_week/Code01_MaxNumberUnderLimit.java @@ -0,0 +1,188 @@ +package class_2022_05_3_week; + +import java.util.Arrays; + +// 来自字节 +// 输入: +// 去重数组arr，里面的数只包含0~9 +// limit，一个数字 +// 返回: +// 要求比limit小的情况下，能够用arr拼出来的最大数字 +public class Code01_MaxNumberUnderLimit { + + public static int tmp = 0; + + // 暴力尝试的方法 + public static int maxNumber1(int[] arr, int limit) { + tmp = 0; + Arrays.sort(arr); + limit--; + int offset = 1; + while (offset <= limit / 10) { + offset *= 10; + } + process1(arr, 0, offset, limit); + if (tmp == 0) { + int rest = 0; + offset /= 10; + while (offset > 0) { + rest += arr[arr.length - 1] * offset; + offset /= 10; + } + return rest; + } + return tmp; + } + + public static void process1(int[] arr, int num, int offset, int limit) { + if (offset == 0) { + if (num <= limit) { + tmp = Math.max(tmp, num); + } + } else { + for (int cur : arr) { + process1(arr, num * 10 + cur, offset / 10, limit); + } + } + } + + // 正式方法 + public static int maxNumber2(int[] arr, int limit) { + // arr里面是不重复的数字，且只包含0~9 + Arrays.sort(arr); + limit--; + // <= limit 且最大的数字 + // 68886 + // 10000 + // 为了取数而设计的！ + // 457 + // 100 + int offset = 1; + while (offset <= limit / 10) { + offset *= 10; + } + int ans = process2(arr, limit, offset); + if (ans != -1) { + return ans; + } else { + offset /= 10; + int rest = 0; + while (offset > 0) { + rest += arr[arr.length - 1] * offset; + offset /= 10; + } + return rest; + } + } + + // 可以选哪些数字，都在arr里，arr是有序的，[3,6,8,9] + // limit : <= limit 且尽量的大！ 68886 + // offset : 10000 + // 1000 + // 100 + // offset 下标用！ + public static int process2(int[] arr, int limit, int offset) { + // 之前的数字和limit完全一样，且limit所有数字都一样 + if (offset == 0) { + return limit; + } + // 当前的数字 + // 68886 + // 10000 + // 6 + int cur = (limit / offset) % 10; + // 6 arr中 <=6 最近的！ + int near = near(arr, cur); + if (near == -1) { + return -1; + } else if (arr[near] == cur) { // 找出来的数字，真的和当前数字cur一样! + int ans = process2(arr, limit, offset / 10); + if (ans != -1) { + return ans; + } else if (near > 0) { // 虽然后续没成功，但是我自己还能下降！还能选更小的数字 + near--; + return (limit / (offset * 10)) * offset * 10 + (arr[near] * offset) + rest(arr, offset / 10); + } else { // 后续没成功，我自己也不能再下降了！宣告失败，往上返回！ + return -1; + } + } else { // arr[near] < cur + return (limit / (offset * 10)) * offset * 10 + (arr[near] * offset) + rest(arr, offset / 10); + } + } + + // 比如offset = 100 + // 一共3位数 + // 那么就把arr中最大的数字x，拼成xxx，返回 + // 比如offset = 10000 + // 一共5位数 + // 那么就把arr中最大的数字x，拼成xxxxx，返回 + public static int rest(int[] arr, int offset) { + int rest = 0; + while (offset > 0) { + rest += arr[arr.length - 1] * offset; + offset /= 10; + } + return rest; + } + + // 在有序数组arr中，找到<=num，且最大的数字，在arr中的位置返回 + // 如果所有数字都大于num，返回-1 + // [3,6,9] num = 4 3 + // [5,7,9] num = 4 -1 + public static int near(int[] arr, int num) { + int l = 0; + int r = arr.length - 1; + int m = 0; + int near = -1; + while (l <= r) { + m = (l + r) / 2; + if (arr[m] <= num) { + near = m; + l = m + 1; + } else { + r = m - 1; + } + } + return near; + } + + // 为了测试 + public static int[] randomArray() { + int[] arr = new int[(int) (Math.random() * 10) + 1]; + boolean[] cnt = new boolean[10]; + for (int i = 0; i < arr.length; i++) { + do { + arr[i] = (int) (Math.random() * 10); + } while (cnt[arr[i]]); + cnt[arr[i]] = true; + } + return arr; + } + + public static void main(String[] args) { + int max = 3000; + int testTime = 100; + System.out.println("测试开始"); + for (int i = 0; i < max; i++) { + int[] arr = randomArray(); + for (int j = 0; j < testTime; j++) { + int ans1 = maxNumber1(arr, i); + int ans2 = maxNumber2(arr, i); + if (ans1 != ans2) { + System.out.println("出错了!"); + System.out.println("数组为 ："); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println("数字为 ：" + i); + System.out.println(ans1); + System.out.println(ans2); + } + } + } + System.out.println("测试结束"); + + } + +} diff --git a/算法周更班/class_2022_05_3_week/Code02_RemoveNumbersNotIncreasingAll.java b/算法周更班/class_2022_05_3_week/Code02_RemoveNumbersNotIncreasingAll.java new file mode 100644 index 0000000..9c09f2f --- /dev/null +++ b/算法周更班/class_2022_05_3_week/Code02_RemoveNumbersNotIncreasingAll.java @@ -0,0 +1,162 @@ +package class_2022_05_3_week; + +// 来自京东 +// 4.2笔试 +// 给定一个数组arr，长度为N，arr中所有的值都在1~K范围上 +// 你可以删除数字，目的是让arr的最长递增子序列长度小于K +// 返回至少删除几个数字能达到目的 +// N <= 10^4，K <= 10^2 +public class Code02_RemoveNumbersNotIncreasingAll { + + // 暴力方法 + // 为了验证 + public static int minRemove1(int[] arr, int k) { + return process1(arr, 0, new int[arr.length], 0, k); + } + + public static int process1(int[] arr, int index, int[] path, int size, int k) { + if (index == arr.length) { + return lengthOfLIS(path, size) < k ? (arr.length - size) : Integer.MAX_VALUE; + } else { + int p1 = process1(arr, index + 1, path, size, k); + path[size] = arr[index]; + int p2 = process1(arr, index + 1, path, size + 1, k); + return Math.min(p1, p2); + } + } + + public static int lengthOfLIS(int[] arr, int size) { + if (size == 0) { + return 0; + } + int[] ends = new int[size]; + ends[0] = arr[0]; + int right = 0; + int l = 0; + int r = 0; + int m = 0; + int max = 1; + for (int i = 1; i < size; i++) { + l = 0; + r = right; + while (l <= r) { + m = (l + r) / 2; + if (arr[i] > ends[m]) { + l = m + 1; + } else { + r = m - 1; + } + } + right = Math.max(right, l); + ends[l] = arr[i]; + max = Math.max(max, l + 1); + } + return max; + } + + // arr[0...index-1]上，选择了一些数字，之前的决定！ + // len长度了！len = 3 ： 1 2 3 + // arr[index....]是能够决定的，之前的，已经不能再决定了 + // 返回：让最终保留的数字，凑不足k长度的情况下，至少要删几个！ + public static int zuo(int[] arr, int index, int len, int k) { + if (len == k) { + return Integer.MAX_VALUE; + } + // 凑的(1...len)还不到(1...k) + if (index == arr.length) { + return 0; + } + // 没凑到 < k, 有数字! + int cur = arr[index]; + // 可能性1：保留 + // 可能性2：删除 + // 1...3 3 + if (len >= cur || len + 1 < cur) { + return zuo(arr, index + 1, len, k); + } + // 1..3 4 + // len + 1 == cur + // 可能性1：保留 + int p1 = zuo(arr, index + 1, len + 1, k); + // 可能性2：删除 + int p2 = Integer.MAX_VALUE; + int next2 = zuo(arr, index + 1, len, k); + if(next2 != Integer.MAX_VALUE) { + p2 = 1 + next2; + } + return Math.min(p1, p2); + } + + // 正式方法 + // 时间复杂度O(N*K) + public static int minRemove2(int[] arr, int k) { + int n = arr.length; + int[][] dp = new int[n][k]; + for (int i = 0; i < n; i++) { + for (int j = 0; j < k; j++) { + dp[i][j] = -1; + } + } + return process2(arr, k, 0, 0, dp); + } + + public static int process2(int[] arr, int k, int index, int range, int[][] dp) { + if (range == k) { + return Integer.MAX_VALUE; + } + if (index == arr.length) { + return 0; + } + if (dp[index][range] != -1) { + return dp[index][range]; + } + int ans = 0; + if (arr[index] == range + 1) { + int p1 = process2(arr, k, index + 1, range, dp); + p1 += p1 != Integer.MAX_VALUE ? 1 : 0; + int p2 = process2(arr, k, index + 1, range + 1, dp); + ans = Math.min(p1, p2); + } else { + ans = process2(arr, k, index + 1, range, dp); + } + dp[index][range] = ans; + return ans; + } + + // 为了验证 + public static int[] randomArray(int len, int k) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * k) + 1; + } + return arr; + } + + // 为了验证 + public static void main(String[] args) { + int N = 15; + int K = 6; + int testTime = 10000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int len = (int) (Math.random() * N) + 1; + int k = (int) (Math.random() * K) + 1; + int[] arr = randomArray(len, k); + int ans1 = minRemove1(arr, k); + int ans2 = minRemove2(arr, k); + if (ans1 != ans2) { + System.out.println("出错了！"); + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println("k : " + k); + System.out.println("ans1 : " + ans1); + System.out.println("ans2 : " + ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_05_3_week/Code03_NumberOfCannon.java b/算法周更班/class_2022_05_3_week/Code03_NumberOfCannon.java new file mode 100644 index 0000000..6e482c1 --- /dev/null +++ b/算法周更班/class_2022_05_3_week/Code03_NumberOfCannon.java @@ -0,0 +1,97 @@ +package class_2022_05_3_week; + +import java.util.TreeMap; +import java.util.TreeSet; + +// 来自学员问题 +// 给定一个数组arr，表示从早到晚，依次会出现的导弹的高度 +// 大炮打导弹的时候，如果一旦大炮定了某个高度去打，那么这个大炮每次打的高度都必须下降一点 +// 1) 如果只有一个大炮，返回最多能拦截多少导弹 +// 2) 如果所有的导弹都必须拦截，返回最少的大炮数量 +public class Code03_NumberOfCannon { + + public static int numOfCannon(int[] arr) { + // key : 某个大炮打的结尾数值 + // value : 有多少个大炮有同样的结尾数值 + // 比如： + // 一共有A、B、C三个大炮 + // 如果A大炮此时打的高度是17，B大炮此时打的高度是7，C大炮此时打的高度是13 + // 那么在表中： + // 7, 1 + // 13, 1 + // 17, 1 + // 如果A大炮此时打的高度是13，B大炮此时打的高度是7，C大炮此时打的高度是13 + // 那么在表中： + // 7, 1 + // 13, 2 + TreeMap ends = new TreeMap<>(); + for (int num : arr) { + if (ends.ceilingKey(num + 1) == null) { + ends.put(Integer.MAX_VALUE, 1); + } + int ceilKey = ends.ceilingKey(num + 1); + if (ends.get(ceilKey) > 1) { + ends.put(ceilKey, ends.get(ceilKey) - 1); + } else { + ends.remove(ceilKey); + } + ends.put(num, ends.getOrDefault(num, 0) + 1); + } + int ans = 0; + for (int value : ends.values()) { + ans += value; + } + return ans; + } + + public static void main(String[] args) { + + // 有序表来说 + // add + // remove + // ceiling + // <= floor + // O(logN)! + TreeSet set = new TreeSet<>(); + + set.add(17); + set.add(20); + set.add(25); + + // >= 23 + System.out.println(set.ceiling(26)); + + // 有序表是去重的，key去重 + // A ：99 + // B : 99 + // C : 99 + + TreeMap map = new TreeMap<>(); + map.put(99, 3); + // 76 + + if (map.ceilingKey(76) == null) { + // 没有大炮可以打76 + // 新开一门大炮，打76 + // 这个新跑，只能打75~ + map.put(75, map.getOrDefault(75, 0) + 1); + } else { // 之前有大炮可以打76，不需要新开一门炮！ + int key = map.ceilingKey(76); + // 99 -1 75 +1 + + if(map.get(key) > 1) { + map.put(key, map.get(key) - 1); + }else { + map.remove(key); + } + + map.put(75, map.getOrDefault(75, 0) + 1); + } + + +// +// int[] arr = { 15, 7, 14, 6, 5, 13, 5, 10, 9 }; +// System.out.println(numOfCannon(arr)); + } + +} diff --git a/算法周更班/class_2022_05_3_week/Code04_MinJumpUsePre.java b/算法周更班/class_2022_05_3_week/Code04_MinJumpUsePre.java new file mode 100644 index 0000000..692109d --- /dev/null +++ b/算法周更班/class_2022_05_3_week/Code04_MinJumpUsePre.java @@ -0,0 +1,154 @@ +package class_2022_05_3_week; + +import java.util.Arrays; + +// 来自学员问题 +// 为了给刷题的同学一些奖励，力扣团队引入了一个弹簧游戏机 +// 游戏机由 N 个特殊弹簧排成一排，编号为 0 到 N-1 +// 初始有一个小球在编号 0 的弹簧处。若小球在编号为 i 的弹簧处 +// 通过按动弹簧，可以选择把小球向右弹射 jump[i] 的距离，或者向左弹射到任意左侧弹簧的位置 +// 也就是说，在编号为 i 弹簧处按动弹簧， +// 小球可以弹向 0 到 i-1 中任意弹簧或者 i+jump[i] 的弹簧（若 i+jump[i]>=N ，则表示小球弹出了机器） +// 小球位于编号 0 处的弹簧时不能再向左弹。 +// 为了获得奖励，你需要将小球弹出机器。 +// 请求出最少需要按动多少次弹簧，可以将小球从编号 0 弹簧弹出整个机器，即向右越过编号 N-1 的弹簧。 +// 测试链接 : https://leetcode-cn.com/problems/zui-xiao-tiao-yue-ci-shu/ +public class Code04_MinJumpUsePre { + + // 宽度优先遍历 + // N*logN + public int minJump(int[] jump) { + int n = jump.length; + int[] queue = new int[n]; + int l = 0; + int r = 0; + queue[r++] = 0; + IndexTree it = new IndexTree(n); + // 1...n初始化的时候 每个位置填上1 + for (int i = 1; i < n; i++) { + it.add(i, 1); + } + int step = 0; + while (l != r) { // 队列里面还有东西 + // tmp记录了当前层的终止位置！ + int tmp = r; + // 当前层的所有节点，都去遍历! + for (; l < tmp; l++) { + int cur = queue[l]; + int forward = cur + jump[cur]; + if (forward >= n) { + return step + 1; + } + if (it.value(forward) != 0) { + queue[r++] = forward; + it.add(forward, -1); + } + // cur + // 1....cur-1 cur + while (it.sum(cur - 1) != 0) { + int find = find(it, cur - 1); + it.add(find, -1); + queue[r++] = find; + } + } + step++; + } + return -1; + } + + public static int find(IndexTree it, int right) { + int left = 0; + int mid = 0; + int find = 0; + while (left <= right) { + mid = (left + right) / 2; + if (it.sum(mid) > 0) { + find = mid; + right = mid - 1; + } else { + left = mid + 1; + } + } + return find; + } + + public static class IndexTree { + + private int[] tree; + private int N; + + public IndexTree(int size) { + N = size; + tree = new int[N + 1]; + } + + public int value(int index) { + if (index == 0) { + return sum(0); + } else { + return sum(index) - sum(index - 1); + } + } + + public int sum(int i) { + int index = i + 1; + int ret = 0; + while (index > 0) { + ret += tree[index]; + index -= index & -index; + } + return ret; + } + + public void add(int i, int d) { + int index = i + 1; + while (index <= N) { + tree[index] += d; + index += index & -index; + } + } + + } + + // 感谢黄汀同学 + // 弄出了时间复杂度O(N)的过程 + // 和大厂刷题班，第10节，jump game类似 + public int minJump2(int[] jump) { + int N = jump.length; + int ans = N; + int next = jump[0]; + if (next >= N) { + return 1; + } + if (next + jump[next] >= N) { + return 2; + } + // dp[i] : 来到i位置，最少跳几步？ + int[] dp = new int[N + 1]; + Arrays.fill(dp, N); + // dis[i] : <= i步的情况下，最远能跳到哪？ + int[] dis = new int[N]; + // 如果从0开始向前跳，<=1步的情况下，最远当然能到next + dis[1] = next; + // 如果从0开始向前跳，<=2步的情况下，最远可能比next + jump[next]要远， + // 这里先设置，以后可能更新 + dis[2] = next + jump[next]; + dp[next + jump[next]] = 2; + int step = 1; + for (int i = 1; i < N; i++) { + if (i > dis[step]) { + step++; + } + dp[i] = Math.min(dp[i], step + 1); + next = i + jump[i]; + if (next >= N) { + ans = Math.min(ans, dp[i] + 1); + } else if (dp[next] > dp[i] + 1) { + dp[next] = dp[i] + 1; + dis[dp[next]] = Math.max(dis[dp[next]], next); + } + } + return ans; + } + +} diff --git a/算法周更班/class_2022_05_4_week/Code01_SomeDPFromVT.java b/算法周更班/class_2022_05_4_week/Code01_SomeDPFromVT.java new file mode 100644 index 0000000..7f25965 --- /dev/null +++ b/算法周更班/class_2022_05_4_week/Code01_SomeDPFromVT.java @@ -0,0 +1,246 @@ +package class_2022_05_4_week; + +// 来自弗吉尼亚理工大学(VT)，算法考试卷 +// 精选了还可以的几道题 +// 这些都是简单难度的动态规划，是面试中最常见的难度 +// 这几个题都有一些非常小的常见技巧可说 +public class Code01_SomeDPFromVT { + +// // arr[i][0] : 有趣值 +// // arr[i][1] : 进攻值 +// // arr[index...]所有的方案自由选择 +// // 必须让restFunny、restOffense值 <= 0 +// // 返回最小的方案数量(index...) +// public static int process(int[][] arr, int index, int restFunny, int restOffense) { +// if (restFunny <= 0 && restOffense <= 0) { +// return 0; +// } +// // 有的值，还没扣完 +// if (index == arr.length) { +// return Integer.MAX_VALUE; // 无效值 +// } +// // 有的值还没扣完 但是还有方案可选 +// // index号方案 不要 要 +// int p1 = process(arr, index + 1, restFunny, restOffense); +// // 要用index号方案 +// int p2 = Integer.MAX_VALUE; +// int next = process(arr, index + 1, restFunny - arr[index][0], restOffense - arr[index][1]); +// if (next != Integer.MAX_VALUE) { +// p2 = 1 + next; +// } +// return Math.min(p1, p2); +// } + + // 题目1 + // 方案1 : {7, 10} + // xxxx : {a , b} + // 1 2 3 4 + // FunnyGoal = 100 + // OffenseGoal = 130 + // 找到一个最少方案数，让FunnyGoal、OffenseGoal，都大于等于 + // 定义如下尝试过程 + // 贴纸数组stickers + // stickers[i][0] : i号贴纸的Funny值 + // stickers[i][1] : i号贴纸的Offense值 + // index....所有的贴纸，随便选择。index之前的贴纸不能选择， + // 在让restFunny和restOffense都小于等于0的要求下，返回最少的贴纸数量 + public static int minStickers1(int[][] stickers, int funnyGoal, int offenseGoal) { + return process1(stickers, 0, funnyGoal, offenseGoal); + } + + public static int process1(int[][] stickers, int index, int restFunny, int restOffense) { + if (restFunny <= 0 && restOffense <= 0) { + return 0; + } + if (index == stickers.length) { + return Integer.MAX_VALUE; + } + // 不选当前的贴纸 + int p1 = process1(stickers, index + 1, restFunny, restOffense); + // 选当前贴纸 + int p2 = Integer.MAX_VALUE; + int next2 = process1(stickers, index + 1, Math.max(0, restFunny - stickers[index][0]), + Math.max(0, restOffense - stickers[index][1])); + if (next2 != Integer.MAX_VALUE) { + p2 = next2 + 1; + } + return Math.min(p1, p2); + } + + // 改动态规划 + public static int minStickers2(int[][] stickers, int funnyGoal, int offenseGoal) { + int[][][] dp = new int[stickers.length][funnyGoal + 1][offenseGoal + 1]; + for (int i = 0; i < stickers.length; i++) { + for (int j = 0; j <= funnyGoal; j++) { + for (int k = 0; k <= offenseGoal; k++) { + dp[i][j][k] = -1; + } + } + } + return process2(stickers, 0, funnyGoal, offenseGoal, dp); + } + + public static int process2(int[][] stickers, int index, int restFunny, int restOffense, int[][][] dp) { + if (restFunny <= 0 && restOffense <= 0) { + return 0; + } + if (index == stickers.length) { + return Integer.MAX_VALUE; + } + if (dp[index][restFunny][restOffense] != -1) { + return dp[index][restFunny][restOffense]; + } + // 不选当前的贴纸 + int p1 = process2(stickers, index + 1, restFunny, restOffense, dp); + // 选当前贴纸 + int p2 = Integer.MAX_VALUE; + int next2 = process2(stickers, index + 1, Math.max(0, restFunny - stickers[index][0]), + Math.max(0, restOffense - stickers[index][1]), dp); + if (next2 != Integer.MAX_VALUE) { + p2 = next2 + 1; + } + int ans = Math.min(p1, p2); + dp[index][restFunny][restOffense] = ans; + return ans; + } + + // 严格位置依赖的动态规划 + public static int minStickers3(int[][] stickers, int funnyGoal, int offenseGoal) { + int n = stickers.length; + int[][][] dp = new int[n + 1][funnyGoal + 1][offenseGoal + 1]; + for (int f = 0; f <= funnyGoal; f++) { + for (int o = 0; o <= offenseGoal; o++) { + if (f != 0 || o != 0) { + dp[n][f][o] = Integer.MAX_VALUE; + } + } + } + for (int i = n - 1; i >= 0; i--) { + for (int f = 0; f <= funnyGoal; f++) { + for (int o = 0; o <= offenseGoal; o++) { + if (f != 0 || o != 0) { + int p1 = dp[i + 1][f][o]; + int p2 = Integer.MAX_VALUE; + int next2 = dp[i + 1][Math.max(0, f - stickers[i][0])][Math.max(0, o - stickers[i][1])]; + if (next2 != Integer.MAX_VALUE) { + p2 = next2 + 1; + } + dp[i][f][o] = Math.min(p1, p2); + } + } + } + } + return dp[0][funnyGoal][offenseGoal]; + } + + // 题目2 + // 绳子总长度为M + // 100 -> M + // (6, 100) (7,23) (10,34) -> arr + // 每一个长度的绳子对应一个价格，比如(6, 10)表示剪成长度为6的绳子，对应价格10 + // 可以重复切出某个长度的绳子 + // 定义递归如下： + // 所有可以切出来的长度 对应 价值都在数组ropes里 + // ropes[i] = {6, 10} 代表i方案为：切出长度为6的绳子，可以卖10元 + // index....所有的方案，随便选择。index之前的方案，不能选择 + // 返回最大的价值 + // 自己去改动态规划 + // arr[i][0] -> i号方案能切多少长度 + // arr[i][1] -> 切出来这个长度，就能获得的价值 + // arr[index....]自由选择，绳子还剩restLen长度 + // 返回，最大价值 + public static int maxValue(int[][] arr, int index, int restLen) { + if (restLen <= 0 || index == arr.length) { + return 0; + } + // 绳子还有剩余、且还有方案 + // index号方案 + // 不选 + int p1 = maxValue(arr, index + 1, restLen); + // 选 + int p2 = 0; + if (arr[index][0] <= restLen) { // 剩余绳子够长，才能选当前方案 + p2 = arr[index][1] + maxValue(arr, index, restLen - arr[index][0]); + } + return Math.max(p1, p2); + } + +// public static int maxValue(int[][] ropes, int index, int restLen) { +// if (restLen <= 0) { +// return 0; +// } +// // 当前index方案，就是不考虑 +// int p1 = maxValue(ropes, index + 1, restLen); +// // 当前index方案，考虑，然后因为可以重复选，所以注意下面的子过程调用 +// int p2 = -1; +// if (ropes[index][0] <= restLen) { +// // 当前index方案，选了一份 +// // 但是下面依然可以重复的选index方案 +// // 所以子过程里的index不增加，只是剩余长度减少 +// p2 = ropes[index][1] + maxValue(ropes, index, restLen - ropes[index][0]); +// } +// return Math.max(p1, p2); +// } + + // 题目3 + // 每一个序列都是[a,b]的形式，a < b + // 序列连接的方式为，前一个序列的b，要等于后一个序列的a + // 比如 : [3, 7]、[7, 13]、[13, 26]这三个序列就可以依次连接 + // 给定若干个序列，求最大连接的数量 + // 定义尝试过程如下 + // arr[i] = {4, 9}表示，第i个序列4开始，9结束 + // pre : 代表选择的上一个序列，的，index是多少 + // 比如选择的上一个序列如果是(4,9)，是第5个序列，那么pre==5 + // 特别注意：如果从来没有选过序列，那么pre == -1 + // 这个函数含义 : + // index....所有的序列，随便选择。index之前的序列，不能选择 + // 上一个选择的序列，是pre号，如果pre==-1,说明之前没有选择过序列 + // 返回题目要求的那种连接方式下，最大的序列数量 + // [5,13] [1,19] [2, 3] [79, 81] ... + // [1,19] [2, 3] [5, 13] [79, 81] + // arr[i][0] : 开头 + // arr[i][1] : 结尾 + // arr已经根据开头排过序了！ + // preEnd index + // [1, 3] [2, 4] [4, 7] + // 0 1 2 + // maxLen(0, -1) + // 0（选） -> maxLen(1, 0) + // 在arr[index...]选择序列，之前选的，离index最近的序列，位置在preIndex + // 请返回，index...能链接起来的，序列数量的最大值 + public static int maxLen(int[][] arr, int index, int preIndex) { + if (index == arr.length) { + return 0; + } + // 还有序列可以选 + // index号序列 + // 不选 + int p1 = maxLen(arr, index + 1, preIndex); + // 选 + int p2 = 0; + // [3,17] index(9,24) + if (arr[preIndex][1] == arr[index][0]) { // 才能选 + p2 = 1 + maxLen(arr, index + 1, index); + } + return Math.max(p1, p2); + } + + // O(N^2) + public static int maxNumberSubsequence(int[][] arr, int index, int pre) { + if (index == arr.length) { + return 0; + } + // 就是不要当前序列 + int p1 = maxNumberSubsequence(arr, index + 1, pre); + // 要当前序列 + int p2 = -1; + if (pre == -1 || arr[pre][1] == arr[index][0]) { + p2 = 1 + maxNumberSubsequence(arr, index + 1, index); + } + return Math.max(p1, p2); + } + + // O(N) + + +} diff --git a/算法周更班/class_2022_05_4_week/Code02_MinSetForEveryRange.java b/算法周更班/class_2022_05_4_week/Code02_MinSetForEveryRange.java new file mode 100644 index 0000000..28d2617 --- /dev/null +++ b/算法周更班/class_2022_05_4_week/Code02_MinSetForEveryRange.java @@ -0,0 +1,48 @@ +package class_2022_05_4_week; + +import java.util.Arrays; +import java.util.HashSet; + +// 给定区间的范围[xi,yi]，xi<=yi，且都是正整数 +// 找出一个坐标集合set，set中有若干个数字 +// set要和每个给定的区间，有交集 +// 求set的最少需要几个数 +// 比如给定区间 : [5, 8] [1, 7] [2, 4] [1, 9] +// set最小可以是: {2, 6}或者{2, 5}或者{4, 5} +public class Code02_MinSetForEveryRange { + + public static int minSet(int[][] ranges) { + int n = ranges.length; + // events[i] = {a, b, c} + // a == 0, 表示这是一个区间的开始事件，这个区间结束位置是b + // a == 1, 表示这是一个区间的结束事件，b的值没有意义 + // c表示这个事件的时间点，不管是开始事件还是结束事件，都会有c这个值 + int[][] events = new int[n << 1][3]; + for (int i = 0; i < n; i++) { + // [3, 7] + // (0,7,3) + // (1,X,7) + events[i][0] = 0; + events[i][1] = ranges[i][1]; + events[i][2] = ranges[i][0]; + events[i + n][0] = 1; + events[i + n][2] = ranges[i][1]; + } + Arrays.sort(events, (a, b) -> a[2] - b[2]); + // 容器 + HashSet tmp = new HashSet<>(); + int ans = 0; + for (int[] event : events) { + if (event[0] == 0) { + tmp.add(event[1]); + } else { + if (tmp.contains(event[2])) { + ans++; + tmp.clear(); + } + } + } + return ans; + } + +} diff --git a/算法周更班/class_2022_05_4_week/Code03_MaxIncreasingSubarrayCanDeleteContinuousPart.java b/算法周更班/class_2022_05_4_week/Code03_MaxIncreasingSubarrayCanDeleteContinuousPart.java new file mode 100644 index 0000000..6eeb8ec --- /dev/null +++ b/算法周更班/class_2022_05_4_week/Code03_MaxIncreasingSubarrayCanDeleteContinuousPart.java @@ -0,0 +1,221 @@ +package class_2022_05_4_week; + +import java.util.Arrays; + +// 来自字节 +// 5.6笔试 +// 给定一个数组arr，长度为n，最多可以删除一个连续子数组， +// 求剩下的数组，严格连续递增的子数组最大长度 +// n <= 10^6 +public class Code03_MaxIncreasingSubarrayCanDeleteContinuousPart { + + // 暴力方法 + // 为了验证 + public static int maxLen1(int[] arr) { + int ans = max(arr); + int n = arr.length; + for (int L = 0; L < n; L++) { + for (int R = L; R < n; R++) { + int[] cur = delete(arr, L, R); + ans = Math.max(ans, max(cur)); + } + } + return ans; + } + + public static int[] delete(int[] arr, int L, int R) { + int n = arr.length; + int[] ans = new int[n - (R - L + 1)]; + int index = 0; + for (int i = 0; i < L; i++) { + ans[index++] = arr[i]; + } + for (int i = R + 1; i < n; i++) { + ans[index++] = arr[i]; + } + return ans; + } + + public static int max(int[] arr) { + if (arr.length == 0) { + return 0; + } + int ans = 1; + int cur = 1; + for (int i = 1; i < arr.length; i++) { + if (arr[i] > arr[i - 1]) { + cur++; + } else { + cur = 1; + } + ans = Math.max(ans, cur); + } + return ans; + } + + // 正式方法 + // 时间复杂度O(N*logN) + public static int maxLen2(int[] arr) { + if (arr.length == 0) { + return 0; + } + int n = arr.length; + int[] sorted = new int[n]; + for (int i = 0; i < n; i++) { + sorted[i] = arr[i]; + } + Arrays.sort(sorted); + SegmentTree st = new SegmentTree(n); + st.update(rank(sorted, arr[0]), 1); + int[] dp = new int[n]; + dp[0] = 1; + int ans = 1; + // 一个数字也不删！长度！ + int cur = 1; + for (int i = 1; i < n; i++) { + int rank = rank(sorted, arr[i]); + // (dp[i - 1] + 1) + int p1 = arr[i - 1] < arr[i] ? (dp[i - 1] + 1) : 1; +// // rank : 就是当前的数字 +// // 1~rank-1 : 第二个信息的max + int p2 = rank > 1 ? (st.max(rank - 1) + 1) : 1; + dp[i] = Math.max(p1, p2); + ans = Math.max(ans, dp[i]); + if (arr[i] > arr[i - 1]) { + cur++; + } else { + cur = 1; + } + // 我的当前值是rank + // 之前有没有还是rank的记录！ + if (st.get(rank) < cur) { + st.update(rank, cur); + } + } + return ans; + } + + public static int rank(int[] sorted, int num) { + int l = 0; + int r = sorted.length - 1; + int m = 0; + int ans = -1; + while (l <= r) { + m = (l + r) / 2; + if (sorted[m] >= num) { + ans = m; + r = m - 1; + } else { + l = m + 1; + } + } + return ans + 1; + } + + public static class SegmentTree { + private int n; + private int[] max; + private int[] update; + + public SegmentTree(int maxSize) { + n = maxSize + 1; + max = new int[n << 2]; + update = new int[n << 2]; + Arrays.fill(update, -1); + } + + public int get(int index) { + return max(index, index, 1, n, 1); + } + + public void update(int index, int c) { + update(index, index, c, 1, n, 1); + } + + public int max(int right) { + return max(1, right, 1, n, 1); + } + + private void pushUp(int rt) { + max[rt] = Math.max(max[rt << 1], max[rt << 1 | 1]); + } + + private void pushDown(int rt, int ln, int rn) { + if (update[rt] != -1) { + update[rt << 1] = update[rt]; + max[rt << 1] = update[rt]; + update[rt << 1 | 1] = update[rt]; + max[rt << 1 | 1] = update[rt]; + update[rt] = -1; + } + } + + private void update(int L, int R, int C, int l, int r, int rt) { + if (L <= l && r <= R) { + max[rt] = C; + update[rt] = C; + return; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + if (L <= mid) { + update(L, R, C, l, mid, rt << 1); + } + if (R > mid) { + update(L, R, C, mid + 1, r, rt << 1 | 1); + } + pushUp(rt); + } + + private int max(int L, int R, int l, int r, int rt) { + if (L <= l && r <= R) { + return max[rt]; + } + int mid = (l + r) >> 1; + pushDown(rt, mid - l + 1, r - mid); + int ans = 0; + if (L <= mid) { + ans = Math.max(ans, max(L, R, l, mid, rt << 1)); + } + if (R > mid) { + ans = Math.max(ans, max(L, R, mid + 1, r, rt << 1 | 1)); + } + return ans; + } + + } + + // 为了验证 + public static int[] randomArray(int len, int v) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * v) - (int) (Math.random() * v); + } + return arr; + } + + // 为了验证 + public static void main(String[] args) { + int n = 100; + int v = 20; + int testTime = 5000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int m = (int) (Math.random() * n); + int[] arr = randomArray(m, v); + int ans1 = maxLen1(arr); + int ans2 = maxLen2(arr); + if (ans1 != ans2) { + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_05_4_week/Code04_ABCSameNumber.java b/算法周更班/class_2022_05_4_week/Code04_ABCSameNumber.java new file mode 100644 index 0000000..65d61f4 --- /dev/null +++ b/算法周更班/class_2022_05_4_week/Code04_ABCSameNumber.java @@ -0,0 +1,169 @@ +package class_2022_05_4_week; + +// 来自京东 +// 4.2笔试 +// 给定一个长度为3N的数组，其中最多含有0、1、2三种值 +// 你可以把任何一个连续区间上的数组，全变成0、1、2中的一种 +// 目的是让0、1、2三种数字的个数都是N +// 返回最小的变化次数 +public class Code04_ABCSameNumber { + + // 暴力方法 + // 为了验证不会超过2次 + public static int minTimes1(int[] arr) { + int[] set = new int[arr.length]; + for (int i = 0; i < arr.length; i++) { + set[i] = arr[i]; + } + return process1(set, 0, arr); + } + + public static int process1(int[] set, int time, int[] origin) { + int[] cnt = new int[3]; + for (int num : set) { + cnt[num]++; + } + if (cnt[0] == cnt[1] && cnt[0] == cnt[2]) { + return time; + } else { + if (time == 2) { + return 3; + } + int ans = Integer.MAX_VALUE; + for (int L = 0; L < set.length; L++) { + for (int R = L; R < set.length; R++) { + set(set, L, R, 0); + ans = Math.min(ans, process1(set, time + 1, origin)); + set(set, L, R, 1); + ans = Math.min(ans, process1(set, time + 1, origin)); + set(set, L, R, 2); + ans = Math.min(ans, process1(set, time + 1, origin)); + rollback(set, L, R, origin); + } + } + return ans; + } + } + + public static void set(int[] set, int L, int R, int v) { + for (int i = L; i <= R; i++) { + set[i] = v; + } + } + + public static void rollback(int[] set, int L, int R, int[] origin) { + for (int i = L; i <= R; i++) { + set[i] = origin[i]; + } + } + + // 正式方法 + // 时间复杂度O(N) + public static int minTimes2(int[] arr) { + int[] cnt = new int[3]; + for (int num : arr) { + cnt[num]++; + } + if (cnt[0] == cnt[1] && cnt[0] == cnt[2]) { + return 0; + } + int n = arr.length; + int m = n / 3; + if ((cnt[0] < m && cnt[1] < m) || (cnt[0] < m && cnt[2] < m) || (cnt[1] < m && cnt[2] < m)) { + return 2; + } else { // 只有一种数的个数是小于m的 + return once(arr, cnt, m) ? 1 : 2; + } + } + + // 只有一种数是少于N/3 + public static boolean once(int[] arr, int[] cnt, int m) { + int lessV = cnt[0] < m ? 0 : (cnt[1] < m ? 1 : 2); + int lessT = lessV == 0 ? cnt[0] : (lessV == 1 ? cnt[1] : cnt[2]); + if (cnt[0] > m && modify(arr, 0, cnt[0], lessV, lessT)) { + return true; + } + if (cnt[1] > m && modify(arr, 1, cnt[1], lessV, lessT)) { + return true; + } + if (cnt[2] > m && modify(arr, 2, cnt[2], lessV, lessT)) { + return true; + } + return false; + } + + // 0 -> 10个 + // 1 -> 10个 + // 2 -> 10个 + // ========== + // 0 -> 7个 + // 2 -> 12个 1 -> 11个 + // 多的数 2 + // 少的数 0 + public static boolean modify(int[] arr, + int more, int moreT, + int less, int lessT) { + int[] cnt = new int[3]; + cnt[less] = lessT; + cnt[more] = moreT; + // 目标 + int aim = arr.length / 3; + int L = 0; + int R = 0; + while (R < arr.length || cnt[more] <= aim) { + // cnt[more] 窗口之外，多的数有几个？ + if (cnt[more] > aim) { + // R++ 窗口右边界，右移 + cnt[arr[R++]]--; + } else if (cnt[more] < aim) { + cnt[arr[L++]]++; + } else { // 在窗口之外，多的数，够了！ + // 少的数，和，另一种数other，能不能平均！都是10个！ + if (cnt[less] + R - L < aim) { + cnt[arr[R++]]--; + } else if (cnt[less] + R - L > aim) { + cnt[arr[L++]]++; + } else { + return true; + } + } + } + return false; + } + + // 为了验证 + public static int[] randomArray(int len) { + int[] arr = new int[len]; + for (int i = 0; i < len; i++) { + arr[i] = (int) (Math.random() * 3); + } + return arr; + } + + // 为了验证 + public static void main(String[] args) { + // 数组长度一定是3的整数倍，且 <= 3*n + // 如下代码是验证操作次数一定不大于2次 + // 24个，8个 + int n = 8; + int testTime = 2000; + System.out.println("测试开始"); + for (int i = 0; i < testTime; i++) { + int m = ((int) (Math.random() * n) + 1) * 3; + int[] arr = randomArray(m); + int ans1 = minTimes1(arr); + int ans2 = minTimes2(arr); + if (ans1 != ans2) { + for (int num : arr) { + System.out.print(num + " "); + } + System.out.println(); + System.out.println(ans1); + System.out.println(ans2); + break; + } + } + System.out.println("测试结束"); + } + +} diff --git a/算法周更班/class_2022_06_1_week/Code01_WhereWillTheBallFall.java b/算法周更班/class_2022_06_1_week/Code01_WhereWillTheBallFall.java new file mode 100644 index 0000000..6727152 --- /dev/null +++ b/算法周更班/class_2022_06_1_week/Code01_WhereWillTheBallFall.java @@ -0,0 +1,46 @@ +package class_2022_06_1_week; + +// 最好打开链接看图 +// 用一个大小为 m x n 的二维网格 grid 表示一个箱子 +// 你有 n 颗球。箱子的顶部和底部都是开着的。 +// 箱子中的每个单元格都有一个对角线挡板，跨过单元格的两个角， +// 可以将球导向左侧或者右侧。 +// 将球导向右侧的挡板跨过左上角和右下角，在网格中用 1 表示。 +// 将球导向左侧的挡板跨过右上角和左下角，在网格中用 -1 表示。 +// 在箱子每一列的顶端各放一颗球。每颗球都可能卡在箱子里或从底部掉出来。 +// 如果球恰好卡在两块挡板之间的 "V" 形图案，或者被一块挡导向到箱子的任意一侧边上，就会卡住。 +// 返回一个大小为 n 的数组 answer ， +// 其中 answer[i] 是球放在顶部的第 i 列后从底部掉出来的那一列对应的下标， +// 如果球卡在盒子里，则返回 -1 +// 本题测试链接 : https://leetcode.com/problems/where-will-the-ball-fall/ +public class Code01_WhereWillTheBallFall { + + public static int[] findBall(int[][] grid) { + int n = grid.length; + int m = grid[0].length; + int[] ans = new int[m]; + for (int col = 0; col < m; col++) { + // (0,0) (0,1) (0,2) + int i = 0; + int j = col; + while (i < n) { + // (i,j) 左上 -> 右下的格子 grid[i][j] == 1 + // (i+1, j+1) + // (i,j) 右上 -> 左下的格子 grid[i][j] == -1 + // (i+1, j-1) + int jnext = j + grid[i][j]; + if (jnext < 0 || jnext == m || grid[i][j] != grid[i][jnext]) { + ans[col] = -1; + break; + } + i++; + j = jnext; + } + if (i == n) { + ans[col] = j; + } + } + return ans; + } + +} diff --git a/算法周更班/class_2022_06_1_week/Code02_UniqueSubstringsInWraparoundString.java b/算法周更班/class_2022_06_1_week/Code02_UniqueSubstringsInWraparoundString.java new file mode 100644 index 0000000..7e0b6a9 --- /dev/null +++ b/算法周更班/class_2022_06_1_week/Code02_UniqueSubstringsInWraparoundString.java @@ -0,0 +1,34 @@ +package class_2022_06_1_week; + +// 把字符串 s 看作 "abcdefghijklmnopqrstuvwxyz" 的无限环绕字符串， +// 所以 s 看起来是这样的： +// ...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd.... +// 现在给定另一个字符串 p 。返回 s 中 不同 的 p 的 非空子串 的数量 +// 测试链接 : https://leetcode.com/problems/unique-substrings-in-wraparound-string/ +public class Code02_UniqueSubstringsInWraparoundString { + + public int findSubstringInWraproundString(String s) { + char[] str = s.toCharArray(); + int n = str.length; + int ans = 0; + int len = 1; + // 256 0~255 + int[] max = new int[256]; + max[str[0]]++; + for (int i = 1; i < n; i++) { + char cur = str[i]; + char pre = str[i - 1]; + if ((pre == 'z' && cur == 'a') || pre + 1 == cur) { + len++; + } else { + len = 1; + } + max[cur] = Math.max(max[cur], len); + } + for (int i = 0; i < 256; i++) { + ans += max[i]; + } + return ans; + } + +} diff --git a/算法周更班/class_2022_06_1_week/Code03_NumberOfAtoms.java b/算法周更班/class_2022_06_1_week/Code03_NumberOfAtoms.java new file mode 100644 index 0000000..b41256f --- /dev/null +++ b/算法周更班/class_2022_06_1_week/Code03_NumberOfAtoms.java @@ -0,0 +1,104 @@ +package class_2022_06_1_week; + +import java.util.TreeMap; + +// 给你一个字符串化学式 formula ，返回 每种原子的数量 。 +// 原子总是以一个大写字母开始，接着跟随 0 个或任意个小写字母，表示原子的名字。 +// 如果数量大于 1，原子后会跟着数字表示原子的数量。如果数量等于 1 则不会跟数字。 +// 例如，"H2O" 和 "H2O2" 是可行的，但 "H1O2" 这个表达是不可行的。 +// 两个化学式连在一起可以构成新的化学式。 +// 例如 "H2O2He3Mg4" 也是化学式。 +// 由括号括起的化学式并佐以数字（可选择性添加）也是化学式。 +// 例如 "(H2O2)" 和 "(H2O2)3" 是化学式。 +// 返回所有原子的数量，格式为：第一个（按字典序）原子的名字，跟着它的数量（如果数量大于 1）， +// 然后是第二个原子的名字（按字典序），跟着它的数量（如果数量大于 1），以此类推。 +// 示例 1： +// 输入：formula = "H2O" +// 输出："H2O" +// 解释：原子的数量是 {'H': 2, 'O': 1}。 +// 示例 2： +// 输入：formula = "Mg(OH)2" +// 输出："H2MgO2" +// 解释：原子的数量是 {'H': 2, 'Mg': 1, 'O': 2}。 +// 示例 3： +// 输入：formula = "K4(ON(SO3)2)2" +// 输出："K4N2O14S4" +// 解释：原子的数量是 {'K': 4, 'N': 2, 'O': 14, 'S': 4}。 +// 测试链接 : https://leetcode.com/problems/number-of-atoms/ +public class Code03_NumberOfAtoms { + + public static String countOfAtoms(String str) { + char[] s = str.toCharArray(); + Info info = process(s, 0); + StringBuilder builder = new StringBuilder(); + for (String key : info.cntMap.keySet()) { + builder.append(key); + int cnt = info.cntMap.get(key); + if (cnt > 1) { + builder.append(cnt); + } + } + return builder.toString(); + } + + public static class Info { + public TreeMap cntMap; + public int end; + + public Info(TreeMap c, int e) { + cntMap = c; + end = e; + } + } + + public static Info process(char[] s, int i) { + TreeMap cntMap = new TreeMap<>(); + int cnt = 0; + StringBuilder builder = new StringBuilder(); + Info info = null; + while (i < s.length && s[i] != ')') { + if (s[i] >= 'A' && s[i] <= 'Z' || s[i] == '(') { + if (builder.length() != 0 || info != null) { + cnt = cnt == 0 ? 1 : cnt; + if (builder.length() != 0) { + String key = builder.toString(); + cntMap.put(key, cntMap.getOrDefault(key, 0) + cnt); + builder.delete(0, builder.length()); + } else { + for (String key : info.cntMap.keySet()) { + cntMap.put(key, cntMap.getOrDefault(key, 0) + info.cntMap.get(key) * cnt); + } + info = null; + } + cnt = 0; + } + if (s[i] == '(') { + info = process(s, i + 1); + i = info.end + 1; + } else { + builder.append(s[i++]); + } + } else if (s[i] >= 'a' && s[i] <= 'z') { + builder.append(s[i++]); + } else { + cnt = cnt * 10 + s[i++] - '0'; + } + } + if (builder.length() != 0 || info != null) { + cnt = cnt == 0 ? 1 : cnt; + if (builder.length() != 0) { + String key = builder.toString(); + cntMap.put(key, cntMap.getOrDefault(key, 0) + cnt); + builder.delete(0, builder.length()); + } else { + for (String key : info.cntMap.keySet()) { + cntMap.put(key, cntMap.getOrDefault(key, 0) + info.cntMap.get(key) * cnt); + } + info = null; + } + cnt = 0; + } + return new Info(cntMap, i); + } + +} diff --git a/算法周更班/class_2022_06_1_week/Code04_SubstringWithLargestVariance.java b/算法周更班/class_2022_06_1_week/Code04_SubstringWithLargestVariance.java new file mode 100644 index 0000000..86d8d59 --- /dev/null +++ b/算法周更班/class_2022_06_1_week/Code04_SubstringWithLargestVariance.java @@ -0,0 +1,98 @@ +package class_2022_06_1_week; + +// 字符串的 波动 定义为子字符串中出现次数 最多 的字符次数与出现次数 最少 的字符次数之差。 +// 给你一个字符串 s ，它只包含小写英文字母。请你返回 s 里所有 子字符串的 最大波动 值。 +// 子字符串 是一个字符串的一段连续字符序列。 +// 注意：必须同时有，最多字符和最少字符的字符串才是有效的 +// 测试链接 : https://leetcode.cn/problems/substring-with-largest-variance/ +public class Code04_SubstringWithLargestVariance { + + public static int largestVariance1(String s) { + if (s == null || s.length() == 0) { + return 0; + } + int n = s.length(); + // a b a c b b a + // 0 1 0 2 1 1 0 + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = s.charAt(i) - 'a'; + } + int ans = 0; + // 26 * 26 * n O(N) + for (int more = 0; more < 26; more++) { + for (int less = 0; less < 26; less++) { + if (more != less) { + int continuousA = 0; + boolean appearB = false; + int max = 0; + // 从左到右遍历， + for (int i = 0; i < n; i++) { + if (arr[i] != more && arr[i] != less) { + continue; + } + if (arr[i] == more) { // 当前字符是more + continuousA++; + if (appearB) { + max++; + } + } else { // 当前字符是B + max = Math.max(max, continuousA) - 1; + continuousA = 0; + appearB = true; + } + ans = Math.max(ans, max); + } + } + } + } + return ans; + } + + public static int largestVariance2(String s) { + if (s == null || s.length() == 0) { + return 0; + } + int n = s.length(); + // a b a c b b a + // 0 1 0 2 1 1 0 + int[] arr = new int[n]; + for (int i = 0; i < n; i++) { + arr[i] = s.charAt(i) - 'a'; + } + // dp[a][b] = more a less b max + // dp[b][a] = more b less a max + int[][] dp = new int[26][26]; + // continuous[a][b] more a less b 连续出现a的次数 + // continuous[b][a] more b less a 连续出现b的次数 + int[][] continuous = new int[26][26]; + // appear[a][b] more a less b b有没有出现过 + // appear[b][a] more b less a a有没有出现过 + boolean[][] appear = new boolean[26][26]; + int ans = 0; + // 26 * N + for (int i : arr) { + for (int j = 0; j < 26; j++) { + if (j != i) { + // i,j + // more i less j 三个变量 连续出现i，j有没有出现过，i-j max + // more j less i 三个变量 连续出现j，i有没有出现过，j-i max + ++continuous[i][j]; + if (appear[i][j]) { + ++dp[i][j]; + } + if (!appear[j][i]) { + appear[j][i] = true; + dp[j][i] = continuous[j][i] - 1; + } else { + dp[j][i] = Math.max(dp[j][i], continuous[j][i]) - 1; + } + continuous[j][i] = 0; + ans = Math.max(ans, Math.max(dp[j][i], dp[i][j])); + } + } + } + return ans; + } + +} diff --git a/算法周更班/class_2022_06_2_week/Code01_MostStonesRemovedWithSameRowOrColumn.java b/算法周更班/class_2022_06_2_week/Code01_MostStonesRemovedWithSameRowOrColumn.java new file mode 100644 index 0000000..fd951c1 --- /dev/null +++ b/算法周更班/class_2022_06_2_week/Code01_MostStonesRemovedWithSameRowOrColumn.java @@ -0,0 +1,87 @@ +package class_2022_06_2_week; + +import java.util.HashMap; + +// n块石头放置在二维平面中的一些整数坐标点上 +// 每个坐标点上最多只能有一块石头 +// 如果一块石头的 同行或者同列 上有其他石头存在，那么就可以移除这块石头。 +// 给你一个长度为 n 的数组 stones ， +// 其中 stones[i] = [xi, yi] 表示第 i 块石头的位置， +// 返回 可以移除的石子 的最大数量。 +// 测试链接 : https://leetcode.com/problems/most-stones-removed-with-same-row-or-column/ +public class Code01_MostStonesRemovedWithSameRowOrColumn { + + public static int removeStones(int[][] stones) { + int n = stones.length; + HashMap rowPre = new HashMap(); + HashMap colPre = new HashMap(); + UnionFind uf = new UnionFind(n); + for (int i = 0; i < n; i++) { + int x = stones[i][0]; + int y = stones[i][1]; + if (!rowPre.containsKey(x)) { + rowPre.put(x, i); + } else { + uf.union(i, rowPre.get(x)); + } + if (!colPre.containsKey(y)) { + colPre.put(y, i); + } else { + uf.union(i, colPre.get(y)); + } + } + return n - uf.sets(); + } + + public static class UnionFind { + + public int[] father; + public int[] size; + public int[] help; + public int sets; + + public UnionFind(int n) { + father = new int[n]; + size = new int[n]; + help = new int[n]; + for (int i = 0; i < n; i++) { + father[i] = i; + size[i] = 1; + } + sets = n; + } + + private int find(int i) { + int hi = 0; + while (i != father[i]) { + help[hi++] = i; + i = father[i]; + } + while (hi != 0) { + father[help[--hi]] = i; + } + return i; + } + + public void union(int i, int j) { + int fi = find(i); + int fj = find(j); + if (fi != fj) { + if (size[fi] >= size[fj]) { + father[fj] = fi; + size[fi] += size[fj]; + } else { + father[fi] = fj; + size[fj] += size[fi]; + } + sets--; + } + } + + public int sets() { + return sets; + } + + } + +} diff --git a/算法周更班/class_2022_06_2_week/Code02_Solution.HEIC b/算法周更班/class_2022_06_2_week/Code02_Solution.HEIC new file mode 100644 index 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