<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:

    int findUnsortedSubarray(vector<int>& nums) {
        int left = nums.size(), right = 0;
        vector<int> st;
        for (int i = 0 ; i < nums.size() ; i++) {
            while (st.size() && nums[st.back()] > nums[i]) {
                st.pop_back();
            }
            st.push_back(i);
        }
        for (int i = 0 ; i < st.size() ; i ++) {
            if (st[i] != i) {
                left = i;
                break;
            }
        }
        if (left == nums.size()) return 0;
        st.clear();
        for (int i = nums.size() - 1 ; i >= 0 ; i--) {
            while (st.size() && nums[st.back()] < nums[i]) {
                st.pop_back();
            }
            st.push_back(i);
        }
        for (int i = 0 ; i < st.size() ; i ++) {
            if (nums.size() - st[i] - 1 != i) {
                right = i;
                break;
            }
        }
        //cout << left << " " << right;
        return nums.size() - left - right;
    }
};
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