leetCode-Shortest Unsorted Continuous Subarray

Description:
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:

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

Solution:

public class Solution {
    public int findUnsortedSubarray(int[] nums) {
        int n = nums.length;
        int[] temp = nums.clone();
        Arrays.sort(temp);

        int start = 0;
        while (start < n  && nums[start] == temp[start]) start++;

        int end = n - 1;
        while (end > start  && nums[end] == temp[end]) end--;

        return end - start + 1;
    }
}

Better Solution:

//low和high分别存储nums中从左边和右边遍历不满足有序的下标,找到low和high，然后找出low和high之间的最小值和最大值，从low向左遍历nums,将low定位到比low和high之间最小值还小的小标,从high向右遍历nums，将high定位到比low和high之间最大值还大的小标。将high-low-1即为无序长度
class Solution {
    public int findUnsortedSubarray(int[] nums) {
        int low = 0, high = nums.length-1, max = Integer.MIN_VALUE, min = Integer.MAX_VALUE;
        while(low<high&&nums[low+1]>=nums[low]) low++;
        while(high>low&&nums[high-1]<=nums[high]) high--;

        if(low>=high) return 0;

        for(int i= low;i<=high;i++){
            max =Math.max(max, nums[i]);
            min = Math.min(min, nums[i]);
        }

        while(low>=0&&nums[low]>min) low--;
        while(high<=nums.length-1&&nums[high]<max) high++;

        return high- low -1;
    }