LeetCode 581. Shortest Unsorted Continuous Subarray

原题链接在这里：https://leetcode.com/problems/shortest-unsorted-continuous-subarray/

题目：

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: 5
Explanation: 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 <=.

题解：

If the input nums is not sorted, then we want to find out the boundary index. 

Then how, lets take start position as example. If the first part is sorted, then traversing right -> left, the updated minimum should be nums[i].

We want to find out the position when minimum != nums[i], and the most left one, then it should be start position of output. 

Vice versa for the ending position. 

Time Complexity: O(nums.length).

Space: O(1).

AC Java:

class Solution {
    public int findUnsortedSubarray(int[] nums) {
        if(nums == null || nums.length == 0){
            return 0;
        }
        
        int s = -1;
        int e = -2;
        int len = nums.length;
        int max = nums[0];
        int min = nums[len-1];
        for(int i = 0; i<len; i++){
            max = Math.max(max, nums[i]);
            if(max != nums[i]){
                e = i;
            }
            
            min = Math.min(min, nums[len-1-i]);
            if(min != nums[len-1-i]){
                s = len-1-i;
            }
        }
        
        return e-s+1;
    }
}