<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:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        n=len(matrix)
        m=len(matrix[0])
        lst=[]
        top,left=0,0
        bottom=n-1
        right=m-1
        while(top<=bottom and left<=right):
            for i in range(left,right+1):
                lst.append(matrix[top][i])
            top+=1
            for i in range(top,bottom+1):
                lst.append(matrix[i][right])
            right-=1
            if(top<=bottom):
                for i in range(right,left-1,-1):
                    lst.append(matrix[bottom][i])
                bottom-=1
            if(left<=right):
                for i in range(bottom,top-1,-1):
                    lst.append(matrix[i][left])
                left+=1

        return lst

16100