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package 第02期.mca_02;
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// 每周有营养的大厂算法面试题
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// 课时111~课时112
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// 来自美团
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// 最大子段和是
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// 一个经典问题，即对于一个数组找出其和最大的子数组。
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// 现在允许你在求解该问题之前翻转这个数組的连续一段
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// 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)，
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// 则翻转后该数组的最大子段和最大能达到多少？
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public class Code03_MaxSumOnReverseArray {
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public static int maxSumReverse1(int[] arr) {
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int ans = Integer.MIN_VALUE;
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for (int L = 0; L < arr.length; L++) {
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for (int R = L; R < arr.length; R++) {
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reverse(arr, L, R);
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ans = Math.max(ans, maxSum(arr));
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reverse(arr, L, R);
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}
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}
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return ans;
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}
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public static void reverse(int[] arr, int L, int R) {
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while (L < R) {
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int tmp = arr[L];
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arr[L++] = arr[R];
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arr[R--] = tmp;
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}
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}
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public static int maxSum(int[] arr) {
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int pre = arr[0];
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int max = arr[0];
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for (int i = 1; i < arr.length; i++) {
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pre = Math.max(arr[i], arr[i] + pre);
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max = Math.max(max, pre);
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}
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return max;
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}
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public static int maxSumReverse2(int[] arr) {
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int n = arr.length;
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int[] prefix = new int[n];
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prefix[n - 1] = arr[n - 1];
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for (int i = n - 2; i >= 0; i--) {
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prefix[i] = arr[i] + Math.max(0, prefix[i + 1]);
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}
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int ans = prefix[0];
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int suffix = arr[0];
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int maxSuffix = suffix;
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for (int i = 1; i < n; i++) {
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ans = Math.max(ans, maxSuffix + prefix[i]);
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suffix = arr[i] + Math.max(0, suffix);
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maxSuffix = Math.max(maxSuffix, suffix);
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}
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ans = Math.max(ans, maxSuffix);
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return ans;
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}
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// 为了测试
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public static int[] randomArray(int n, int v) {
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int[] arr = new int[n];
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for (int i = 0; i < n; i++) {
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arr[i] = (int) (Math.random() * v) - (int) (Math.random() * v);
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}
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return arr;
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}
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// 为了测试
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public static void main(String[] args) {
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int len = 50;
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int value = 20;
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int testTime = 20000;
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System.out.println("测试开始");
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for (int i = 0; i < testTime; i++) {
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int n = (int) (Math.random() * len) + 1;
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int[] arr = randomArray(n, value);
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int ans1 = maxSumReverse1(arr);
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int ans2 = maxSumReverse2(arr);
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if (ans1 != ans2) {
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System.out.println("出错了!");
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for (int num : arr) {
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System.out.print(num + " ");
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}
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System.out.println();
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System.out.println(ans1);
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System.out.println(ans2);
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break;
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}
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}
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System.out.println("测试结束");
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}
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}
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