<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:
    void printList(list<int> l)
    {
        list<int> dummy = l;
        while(dummy.size() > 0)
        {
            cout<<dummy.front()<<" ";
            dummy.pop_front();
        }
    }
    vector<int> maxSlidingWindow(vector<int>& nums, int k) 
    {
        int n = nums.size();
        vector<int> ans(n-k+1,0);
        list<int> list;
        int i,j,l;
        i=0;
        j=0;
        l=0;
        while( i < n)
        {
            while(list.size() > 0 && nums[i] > list.back())
            {
                list.pop_back();
            }  
            list.push_back(nums[i]);
            if(i - j + 1 == k)
            {  
                ans[l++]=list.front();
                if(nums[j] == list.front())
                { 
                    list.pop_front();
                }
                j++;
            }
           i++;
        }  
        return ans;
    }
};
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