From 6041ac97e141b896c435c4181b0ddb5bf6c411b6 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=B7=A6=E7=A8=8B=E4=BA=91?= Date: Mon, 28 Dec 2020 12:47:30 +0800 Subject: [PATCH] add new problem --- src/class12/Code10_SplitSumClosed.java | 106 +++++++++++++++ .../Code11_SplitSumClosedSizeHalf.java | 122 ++++++++++++++++++ 2 files changed, 228 insertions(+) create mode 100644 src/class12/Code10_SplitSumClosed.java create mode 100644 src/class12/Code11_SplitSumClosedSizeHalf.java diff --git a/src/class12/Code10_SplitSumClosed.java b/src/class12/Code10_SplitSumClosed.java new file mode 100644 index 0000000..d12b567 --- /dev/null +++ b/src/class12/Code10_SplitSumClosed.java @@ -0,0 +1,106 @@ +package class12; + +import java.util.TreeSet; + +/* + * 给定一个整型数组arr，请把arr中所有的数分成两个集合，尽量让两个集合的累加和接近 + * 返回最接近的情况下，较小集合的累加和（较大集合的累加和一定是所有数累加和减去较小集合的累加和） + * 为了方便起见，假设arr中没有负数，其实也可以有 + * 但是处理起来会比较麻烦，而且有没有负数都不影响算法流程的理解 + * */ +public class Code10_SplitSumClosed { + + public static int right(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int sum = 0; + for (int num : arr) { + sum += num; + } + TreeSet ans = new TreeSet<>(); + process(arr, 0, 0, 0, ans, sum >> 1); + return ans.last(); + } + + public static void process(int[] arr, int i, int sum, int picks, TreeSet ans, int limit) { + if (i == arr.length) { + if (sum <= limit) { + ans.add(sum); + } + } else { + process(arr, i + 1, sum, picks, ans, limit); + process(arr, i + 1, sum + arr[i], picks + 1, ans, limit); + } + } + + public static int dp(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; + boolean[][] dp = new boolean[N][sum + 1]; + for (int i = 0; i < N; i++) { + dp[i][0] = true; + } + if (arr[0] <= sum) { + 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]]; + } + } + } + for (int j = sum; j >= 1; j--) { + if (dp[N - 1][j]) { + return j; + } + } + return 0; + } + + 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/src/class12/Code11_SplitSumClosedSizeHalf.java b/src/class12/Code11_SplitSumClosedSizeHalf.java new file mode 100644 index 0000000..bf2c28b --- /dev/null +++ b/src/class12/Code11_SplitSumClosedSizeHalf.java @@ -0,0 +1,122 @@ +package class12; + +import java.util.TreeSet; + +/* + * 给定一个整型数组arr，请把arr中所有的数分成两个集合 + * 如果arr长度为偶数，两个集合包含数的个数要一样多 + * 如果arr长度为奇数，两个集合包含数的个数必须只差一个 + * 请尽量让两个集合的累加和接近 + * 返回最接近的情况下，较小集合的累加和（较大集合的累加和一定是所有数累加和减去较小集合的累加和） + * 为了方便起见，假设arr中没有负数，其实也可以有 + * 但是处理起来会比较麻烦，而且有没有负数都不影响算法流程的理解 + * */ +public class Code11_SplitSumClosedSizeHalf { + + public static int right(int[] arr) { + if (arr == null || arr.length < 2) { + return 0; + } + int sum = 0; + for (int num : arr) { + sum += num; + } + TreeSet ans = new TreeSet<>(); + process(arr, 0, 0, 0, ans, sum >> 1); + return ans.last(); + } + + public static void process(int[] arr, int i, int sum, int picks, TreeSet ans, int limit) { + if (i == arr.length) { + if ((arr.length & 1) == 0) { + if (picks == (arr.length >> 1) && sum <= limit) { + ans.add(sum); + } + } else { + if ((picks == (arr.length >> 1) || picks == (arr.length >> 1) + 1) && sum <= limit) { + ans.add(sum); + } + } + } else { + process(arr, i + 1, sum, picks, ans, limit); + process(arr, i + 1, sum + arr[i], picks + 1, ans, limit); + } + } + + public static int dp(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]); + } + + 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"); + } + +} \ No newline at end of file