LeetCode-581. Shortest Unsorted Continuous Subarray (JAVA)最短未排序数组

581. Shortest Unsorted Continuous Subarray

Given an integer array, you need to find one continuous subarray that if you only sort this subarray in ascending order, then the whole array will be sorted in ascending order, too.

You need to find the shortest such subarray and output its length.

Example 1:

Input: [2, 6, 4, 8, 10, 9, 15]Output: 5Explanation: You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order.

Note:

1. Then length of the input array is in range [1, 10,000].
2. The input array may contain duplicates, so ascending order here means <=.

先用一个数组temp保存nums，然后对tmp排序，然后用两个变量low和high去找两个数组出现不同之处的第一个位置和最后一个位置，最后返回high-low+1就是要找的数组长度

// O(nlogn) Java sortpublic int findUnsortedSubarray(int[] nums) {// 深拷贝int[] tmp = nums.clone();Arrays.sort(tmp);int n = nums.length;int low = 0;int high = n - 1;while (low < n && nums[low] == tmp[low]) {low++;}// 注意是大于等于lowwhile (high >= low && nums[high] == tmp[high]) {high--;}return high - low + 1;}