<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 m=matrix.size(), n=matrix[0].size(), rs=0, cs=0, re=m-1, ce=n-1, k=m*n;

        while(k>0) {
            // left
            int c=cs;

            while(k>0 && c<=ce) {
                ans.push_back(matrix[rs][c]);
                k--;
                c++;
            }
            rs++;

            // down
            int r=rs;

            while(k>0 && r<=re) {
                ans.push_back(matrix[r][ce]);
                k--;
                r++;
            }
            ce--;

            // right
            c=ce;

            while(k>0 && c>=cs) {
                ans.push_back(matrix[re][c]);
                k--;
                c--;
            }
            re--;

            // up
            r=re;

            while(k>0 && r>=rs) {
                ans.push_back(matrix[r][cs]);
                k--;
                r--;
            }
            cs++;
        }

        return ans;
    }
};
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