<p>Given an <code>m x n</code> <code>matrix</code>, return <em>all elements of the</em> <code>matrix</code> <em>in spiral order</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/13/spiral1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> matrix = [[1,2,3],[4,5,6],[7,8,9]]
<strong>Output:</strong> [1,2,3,6,9,8,7,4,5]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/13/spiral.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
<strong>Output:</strong> [1,2,3,4,8,12,11,10,9,5,6,7]
</pre>

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

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10</code></li>
	<li><code>-100 &lt;= matrix[i][j] &lt;= 100</code></li>
</ul>

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int m=matrix.size();
        int n=matrix[0].size();
        int left=0,up=0,bottom=m-1,right=n-1;
        vector<int>res;
        while(left<=right && up<=bottom){
           for(int i=left;i<=right;i++)
           {
            res.push_back(matrix[up][i]);
           }
           up++;

           for(int i=up;i<=bottom;i++)
           {
            res.push_back(matrix[i][right]);
           }
           right--;

           if(up<=bottom)
           {
            for(int i=right;i>=left;i--)
            {
                res.push_back(matrix[bottom][i]);
            }
            bottom--;
           }

            if(left<=right)
           {
            for(int i=bottom;i>=up;i--)
            {
                res.push_back(matrix[i][left]);
            }
            left++;
           }
        }
        return res;
    }
};
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