<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) {

        vector<int> ans;
        
        int rowsize = matrix.size();
        int colsize = matrix[0].size();

        int top = 0;
        int left = 0;
        int right = colsize-1;
        int bottom = rowsize-1;

        while(top <= bottom && left <= right){
        for(int i = left; i <= right; i++){

            ans.push_back(matrix[top][i]);
        }
        top++;

        for(int i = top; i <= bottom; i++){

            ans.push_back(matrix[i][right]);
        }
        right--;

        if(top <= bottom){
        for(int i = right; i >= left; i--){

            ans.push_back(matrix[bottom][i]);
        }
        bottom--;
        }

        if(left <= right){
        for(int i = bottom; i >= top; i--){

            ans.push_back(matrix[i][left]);
        }
        left++;
        }
    }

    return ans;
}
};
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