<p>You are given an <code>n x n</code> 2D <code>matrix</code> representing an image, rotate the image by <strong>90</strong> degrees (clockwise).</p>

<p>You have to rotate the image <a href="https://en.wikipedia.org/wiki/In-place_algorithm" target="_blank"><strong>in-place</strong></a>, which means you have to modify the input 2D matrix directly. <strong>DO NOT</strong> allocate another 2D matrix and do the rotation.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/28/mat1.jpg" style="width: 500px; height: 188px;" />
<pre>
<strong>Input:</strong> matrix = [[1,2,3],[4,5,6],[7,8,9]]
<strong>Output:</strong> [[7,4,1],[8,5,2],[9,6,3]]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/28/mat2.jpg" style="width: 500px; height: 201px;" />
<pre>
<strong>Input:</strong> matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
<strong>Output:</strong> [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]
</pre>

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

<ul>
	<li><code>n == matrix.length == matrix[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 20</code></li>
	<li><code>-1000 &lt;= matrix[i][j] &lt;= 1000</code></li>
</ul>

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        
        int n = matrix.size();

        for(int i=0;i<n;i++)
        {
            for(int j=i+1; j<n;j++)
            {
                swap(matrix[i][j],matrix[j][i]);
            }

            reverse(matrix[i].begin(),matrix[i].end());
        }

        
        
    }
};
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