<p>You are given an array of integers&nbsp;<code>nums</code>, there is a sliding window of size <code>k</code> which is moving from the very left of the array to the very right. You can only see the <code>k</code> numbers in the window. Each time the sliding window moves right by one position.</p>

<p>Return <em>the max sliding window</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,3,-1,-3,5,3,6,7], k = 3
<strong>Output:</strong> [3,3,5,5,6,7]
<strong>Explanation:</strong> 
Window position                Max
---------------               -----
[1  3  -1] -3  5  3  6  7       <strong>3</strong>
 1 [3  -1  -3] 5  3  6  7       <strong>3</strong>
 1  3 [-1  -3  5] 3  6  7      <strong> 5</strong>
 1  3  -1 [-3  5  3] 6  7       <strong>5</strong>
 1  3  -1  -3 [5  3  6] 7       <strong>6</strong>
 1  3  -1  -3  5 [3  6  7]      <strong>7</strong>
</pre>

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

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

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

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

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> ans;
        stack<int> s1, s2, mx1, mx2;
        for (int i = 0; i < k; i++){
            s2.push(nums[i]);
            if (mx2.empty()){
                mx2.push(nums[i]);
            }
            else {
                mx2.push(max(mx2.top(), nums[i]));
            }
        }
        ans.push_back(mx2.top());
        for (int i = k, n = nums.size(); i < n; i++){
            if (s1.empty()){
                for (int i = 0; i < k; i++){
                    int ph = s2.top();
                    s2.pop();
                    mx2.pop();
                    s1.push(ph);
                    mx1.push((mx1.empty() ? ph: max(ph, mx1.top())));
                }
            }
            s1.pop();
            mx1.pop();
            s2.push(nums[i]);
            mx2.push((mx2.empty() ? nums[i]: max(nums[i], mx2.top())));
            int mx;
            if (mx2.empty()) mx = mx1.top();
            else if (mx1.empty()) mx = mx2.top();
            else mx = max(mx1.top(), mx2.top());
            ans.push_back(mx);
        }
        return ans;
    }
};
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