Given an m x n matrix, return all elements of the matrix in spiral order.
\n
Example 1:
\n![\"\"](\"https://assets.leetcode.com/uploads/2020/11/13/spiral1.jpg\")
\n\nInput: matrix = [[1,2,3],[4,5,6],[7,8,9]]\nOutput: [1,2,3,6,9,8,7,4,5]\n\n\n
Example 2:
\n![\"\"](\"https://assets.leetcode.com/uploads/2020/11/13/spiral.jpg\")
\n\nInput: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]\nOutput: [1,2,3,4,8,12,11,10,9,5,6,7]\n\n\n
\n
Constraints:
\n\nm == matrix.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100给你一个 m 行 n 列的矩阵 matrix ，请按照 顺时针螺旋顺序 ，返回矩阵中的所有元素。
\n\n
示例 1：
\n\n\n输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]\n输出：[1,2,3,6,9,8,7,4,5]\n\n\n
示例 2：
\n\n\n输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]\n输出：[1,2,3,4,8,12,11,10,9,5,6,7]\n\n\n
\n\n
提示：
\n\nm == matrix.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100iout of
[i, j], you'd be shifting in the same column. Similarly, by changing values for
j, you'd be shifting in the same row.\r\nAlso, keep track of the end of a boundary so that you can move inwards and then keep repeating. It's always best to run the simulation on edge cases like a single column or a single row to see if anything breaks or not." ], "solution": null, "status": null, "sampleTestCase": "[[1,2,3],[4,5,6],[7,8,9]]", "metaData": "{\r\n \"name\": \"spiralOrder\",\r\n \"params\": [\r\n {\r\n \"name\": \"matrix\",\r\n \"type\": \"integer[][]\"\r\n }\r\n ],\r\n \"return\": {\r\n \"type\": \"list
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