Spiral Matrix
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0054.Spiral%20Matrix/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0054.Spiral%20Matrix/images/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>
Array; Matrix; Simulation
class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        m, n = len(matrix), len(matrix[0])
        dirs = (0, 1, 0, -1, 0)
        vis = [[False] * n for _ in range(m)]
        i = j = k = 0
        ans = []
        for _ in range(m * n):
            ans.append(matrix[i][j])
            vis[i][j] = True
            x, y = i + dirs[k], j + dirs[k + 1]
            if x < 0 or x >= m or y < 0 or y >= n or vis[x][y]:
                k = (k + 1) % 4
            i += dirs[k]
            j += dirs[k + 1]
        return ans
