msb-public-java
/
LeetCode
mirror of https://github.com/yuanguangxin/LeetCode
1
2
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Merge remote-tracking branch 'origin/master'

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# Conflicts:
#	.idea/workspace.xml
#	README.md
pull/6/head
yuanguangxin 5 years ago
parent 67c2adc01a 522f0c76d8
commit 7cfdb1282a
15 changed files with 458 additions and 150 deletions
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277
.idea/workspace.xml
Unescape View File

@ -2,10 +2,7 @@
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9
README.md
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@ -9,6 +9,8 @@
### 链表操作
* [q2_两数相加](/src/链表操作/q2_两数相加)
* [q19_删除链表的倒数第N个节点](/src/链表操作/q19_删除链表的倒数第N个节点)
* [q61_旋转链表](/src/链表操作/q61_旋转链表)
* [q138_复制带随机指针的链表](/src/链表操作/q138_复制带随机指针的链表)
* [q206_反转链表](/src/链表操作/q206_反转链表)
@ -52,10 +54,13 @@
* [q54_螺旋矩阵](/src/数组操作/q54_螺旋矩阵)
* [q73_矩阵置零](/src/数组操作/q73_矩阵置零)
* [q945_使数组唯一的最小增量](/src/数组操作/q945_使数组唯一的最小增量)
### 栈相关
* [q20_有效的括号](/src/栈相关/q20_有效的括号)
* [q32_最长有效括号](/src/栈相关/q32_最长有效括号)
* [q155_最小栈](/src/栈相关/q155_最小栈)
* [q224_基本计算器](/src/栈相关/q224_基本计算器)
* [q316_去除重复字母](/src/栈相关/q316_去除重复字母)
@ -68,6 +73,7 @@
* [q21_合并两个有序链表](/src/递归/q21_合并两个有序链表)
* [q101_对称二叉树](/src/递归/q101_对称二叉树)
* [q104_二叉树的最大深度](/src/递归/q104_二叉树的最大深度)
* [q226_翻转二叉树](/src/递归/q226_翻转二叉树)
* [q236_二叉树的最近公共祖先](/src/递归/q236_二叉树的最近公共祖先)
@ -75,11 +81,14 @@
* [q23_合并K个排序链表](/src/分治法/q23_合并K个排序链表)
* [q33_搜索旋转排序数组](/src/分治法/q33_搜索旋转排序数组)
* [q34_在排序数组中查找元素的第一个和最后一个位置](/src/分治法/q34_在排序数组中查找元素的第一个和最后一个位置)
### 动态规划
* [q5_最长回文子串](/src/动态规划/q5_最长回文子串)
* [q53_最大子序和](/src/动态规划/q53_最大子序和)
* [q64_最小路径和](/src/动态规划/q64_最小路径和)
* [q70_爬楼梯](/src/动态规划/q70_爬楼梯)
* [q118_杨辉三角](/src/动态规划/q118_杨辉三角)
* [q300_最长上升子序列](/src/动态规划/q300_最长上升子序列)
* [q746_使用最小花费爬楼梯](/src/动态规划/q746_使用最小花费爬楼梯)

1
Rocket.md
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@ -601,7 +601,6 @@ TCP是一个双向通信协议，通信双方都有能力发送信息，并接
因为当Server端收到Client端的SYN连接请求报文后，可以直接发送SYN+ACK报文。其中ACK报文是用来应答的，SYN报文是用来同步的。但是关闭连接时，当Server端收到FIN报文时，很可能并不会立即关闭SOCKET，所以只能先回复一个ACK报文，告诉Client端，"你发的FIN报文我收到了"。只有等到我Server端所有的报文都发送完了，我才能发送FIN报文，因此不能一起发送。故需要四步握手。
## 数据结构与算法
### 排序算法

60
src/分治法/q34_在排序数组中查找元素的第一个和最后一个位置/Solution.java
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@ -0,0 +1,60 @@
package 分治法.q34_在排序数组中查找元素的第一个和最后一个位置;
/**
* 二分法 o(log(n))
*/
public class Solution {
public int[] searchRange(int[] nums, int target) {
if (nums == null || nums.length < 1) {
return new int[]{-1, -1};
}
int midIndex = find(0, nums.length - 1, nums, target);
int[] rs = new int[2];
rs[0] = midIndex;
rs[1] = midIndex;
if (midIndex == -1) {
return rs;
}
while (nums[rs[0]] == target && rs[0] > 0) {
int temp = find(0, rs[0] - 1, nums, target);
if (temp == -1) {
break;
} else {
rs[0] = temp;
}
}
while (nums[rs[1]] == target && rs[1] < nums.length - 1) {
int temp = find(rs[1] + 1, nums.length - 1, nums, target);
if (temp == -1) {
break;
} else {
rs[1] = temp;
}
}
return rs;
}
public int find(int beginIndex, int endIndex, int[] nums, int target) {
if (beginIndex == endIndex) {
if (nums[beginIndex] == target) {
return beginIndex;
} else {
return -1;
}
}
int mid = (endIndex - beginIndex) / 2 + beginIndex;
if (nums[mid] > target) {
return find(beginIndex, mid, nums, target);
} else if (nums[mid] < target) {
return find(mid + 1, endIndex, nums, target);
} else {
return mid;
}
}
public static void main(String[] args) {
new Solution().searchRange(new int[]{2, 2}, 2);
}
}

26
src/动态规划/q64_最小路径和/Solution.java
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@ -0,0 +1,26 @@
package 动态规划.q64_最小路径和;
/**
* 动态规划 dp(j)=grid(i,j)+min(dp(j),dp(j+1)) o(m*n)
*/
public class Solution {
public int minPathSum(int[][] grid) {
int[] dp = new int[grid[0].length];
for (int i = grid.length - 1; i >= 0; i--) {
for (int j = grid[0].length - 1; j >= 0; j--) {
if (i == grid.length - 1 && j != grid[0].length - 1) {
dp[j] = grid[i][j] + dp[j + 1];
} else if (j == grid[0].length - 1 && i != grid.length - 1) {
dp[j] = grid[i][j] + dp[j];
} else if (j != grid[0].length - 1 && i != grid.length - 1) {
dp[j] = grid[i][j] + Math.min(dp[j], dp[j + 1]);
} else {
dp[j] = grid[i][j];
}
}
}
return dp[0];
}
}

20
src/动态规划/q70_爬楼梯/Solution.java
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@ -0,0 +1,20 @@
package 动态规划.q70_爬楼梯;
/**
* 动态规划 dp[i]表示到达第i阶的方法总数dp[i]=dp[i−1]+dp[i−2] o(n)
*/
public class Solution {
public int climbStairs(int n) {
if (n == 1) {
return 1;
}
int[] dp = new int[n + 1];
dp[1] = 1;
dp[2] = 2;
for (int i = 3; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
}

28
src/数组操作/q945_使数组唯一的最小增量/Solution.java
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@ -0,0 +1,28 @@
package 数组操作.q945_使数组唯一的最小增量;
import java.util.Arrays;
/**
* 先排序再遍历一次 o(n*log(n))
*/
public class Solution {
public int minIncrementForUnique(int[] A) {
if (A == null || A.length == 0 || A.length == 1) {
return 0;
}
int rs = 0;
Arrays.sort(A);
int t = A[0];
for (int i = 1; i < A.length; i++) {
if (A[i] <= t) {
rs = rs + t - A[i] + 1;
A[i] = t + 1;
}
t = A[i];
}
return rs;
}
}

43
src/栈相关/q155_最小栈/MinStack.java
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@ -0,0 +1,43 @@
package 栈相关.q155_最小栈;
import java.util.Stack;
/**
* 不使用辅助栈，每次push两个元素
*/
public class MinStack {
private Stack<Integer> stack;
public MinStack() {
stack = new Stack<>();
}
public void push(int x) {
if (stack.isEmpty()) {
stack.push(x);
stack.push(x);
} else {
int tmp = stack.peek();
stack.push(x);
if (tmp < x) {
stack.push(tmp);
} else {
stack.push(x);
}
}
}
public void pop() {
stack.pop();
stack.pop();
}
public int top() {
return stack.get(stack.size() - 2);
}
public int getMin() {
return stack.peek();
}
}

38
src/栈相关/q32_最长有效括号/Solution.java
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@ -0,0 +1,38 @@
package 栈相关.q32_最长有效括号;
import java.util.Stack;
/**
* 利用索引栈 o(n)
*/
public class Solution {
public int longestValidParentheses(String s) {
if (s == null || s.length() < 2) {
return 0;
}
int maxLen = 0;
Stack<Integer> stack = new Stack<>();
stack.push(-1);
for (int i = 0; i < s.length(); i++) {
char temp = s.charAt(i);
if (temp == '(') {
stack.push(i);
} else {
stack.pop();
if (stack.empty()) {
stack.push(i);
} else {
maxLen = Math.max(maxLen, i - stack.peek());
}
}
}
return maxLen;
}
public static void main(String[] args) {
System.out.println(new Solution().longestValidParentheses(")()())"));
}
}

17
src/递归/q104_二叉树的最大深度/Solution.java
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@ -0,0 +1,17 @@
package 递归.q104_二叉树的最大深度;
/**
* 递归 o(n)
*/
public class Solution {
public int maxDepth(TreeNode root) {
if (root == null) {
return 0;
} else {
int leftHeight = maxDepth(root.left);
int rightHeight = maxDepth(root.right);
return Math.max(leftHeight, rightHeight) + 1;
}
}
}

11
src/递归/q104_二叉树的最大深度/TreeNode.java
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@ -0,0 +1,11 @@
package 递归.q104_二叉树的最大深度;
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}

12
src/链表操作/q19_删除链表的倒数第N个节点/ListNode.java
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@ -0,0 +1,12 @@
package 链表操作.q19_删除链表的倒数第N个节点;
public class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
}
}

25
src/链表操作/q19_删除链表的倒数第N个节点/Solution.java
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@ -0,0 +1,25 @@
package 链表操作.q19_删除链表的倒数第N个节点;
/**
* 利用两个指针 o(n)
*/
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode first = dummy;
ListNode second = dummy;
for (int i = 1; i <= n + 1; i++) {
first = first.next;
}
while (first != null) {
first = first.next;
second = second.next;
}
second.next = second.next.next;
return dummy.next;
}
}

10
src/链表操作/q61_旋转链表/ListNode.java
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@ -0,0 +1,10 @@
package 链表操作.q61_旋转链表;
public class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
}
}

31
src/链表操作/q61_旋转链表/Solution.java
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@ -0,0 +1,31 @@
package 链表操作.q61_旋转链表;
/**
* 先连接成环再找断点 o(n)
*/
public class Solution {
public ListNode rotateRight(ListNode head, int k) {
if (head == null) {
return null;
}
if (head.next == null) {
return head;
}
ListNode oldTail = head;
int n;
for (n = 1; oldTail.next != null; n++) {
oldTail = oldTail.next;
}
oldTail.next = head;
ListNode newTail = head;
for (int i = 0; i < n - k % n - 1; i++) {
newTail = newTail.next;
}
ListNode newHead = newTail.next;
newTail.next = null;
return newHead;
}
}
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