<p>Given an integer array <code>nums</code>, you need to find one <b>continuous subarray</b> such that if you only sort this subarray in non-decreasing order, then the whole array will be sorted in non-decreasing order.</p>

<p>Return <em>the shortest such subarray and output its length</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4]
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Can you solve it in <code>O(n)</code> time complexity?
class Solution {
public:
 Solution(){
         ios_base::sync_with_stdio(false);
         cin.tie(NULL);
         cout.tie(NULL);
    }
    int findUnsortedSubarray(vector<int>& nums) {
        int n=nums.size();
        if (n<2){return 0;}
        int i=0,j=n-1;
        while(i<n-1){
            if(nums[i]>=nums[i+1]){
                break;
            }
            i++;
        }
        while(j>0){
            if(nums[j]<nums[j-1]){
                break;
            }
            j--;
        }
        int minEle=nums[i],maxEle=nums[j];
        for (int k=i;k<=j;k++){
            minEle=min(minEle,nums[k]);
            maxEle=max(maxEle,nums[k]);
        }
        for(int k=i-1;k>=0;k--){
            if(nums[k]>minEle) i=k;
        }
        for(int k=j+1;k<n;k++){
            if(nums[k]<maxEle) j=k;
        }
        if (j<=i) return 0;
        return j-i+1;
    }
};
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