> For the complete documentation index, see [llms.txt](https://dfine.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://dfine.gitbook.io/leetcode/581.shortest_unsorted_continuous_subarray.md).

# 581.Shortest Unsorted Continuous Subarray

**581.Shortest Unsorted Continuous Subarray**

难度:Easy

> 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 <=.
```

找出中间不是按递增顺序的子串，然后找出其中最大最小值，依次往前比较，确定需要排序的最短长度。

```
class Solution {
public:
    int findUnsortedSubarray(vector<int>& nums) {
     int left=0;
     int right=nums.size()-1;
     
     while(left<right)
     {
         
         if(nums[left]>nums[left+1] && nums[right-1]>nums[right])
             break;
         if(nums[left]<=nums[left+1]) left++;
         if(nums[right-1]<=nums[right])right--;
     }
     if(left >= right) return 0;
     int res=0;
     int Nmax=INT_MIN;
     int Nmin=INT_MAX;
      for(int i=left;i<=right;i++)
      {
          Nmax=max(Nmax,nums[i]);
          Nmin=min(Nmin,nums[i]);
      }
      for(int i=left-1;i>=0;i--)
          if(nums[i]>Nmin) res++;
          else break;
       for(int j=right+1;j<nums.size();j++)
           if(nums[j]<Nmax) res++;
           else break;
        //cout<<Nmax<< " "<< Nmin<<endl; 
        //cout<<left << " "<< right<<" " <<res<<endl; 
        return res+right-left+1;
    }
};
```

> 执行用时 :52 ms, 在所有 C++ 提交中击败了59.45%的用户\
> 内存消耗 :10.5 MB, 在所有 C++ 提交中击败了89.21%的用户


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