螺旋矩阵 #

给定一个包含  m x n  个元素的矩阵（m 行，n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。 示例  1: ``` 输入: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] 输出: [1,2,3,6,9,8,7,4,5] ``` 示例  2: ``` 输入: [ [1, 2, 3, 4], [5, 6, 7, 8], [9,10,11,12] ] 输出: [1,2,3,4,8,12,11,10,9,5,6,7] ```

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order. Example 1: ``` Input: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] Output: [1,2,3,6,9,8,7,4,5] ``` Example 2: ``` Input: [ [1, 2, 3, 4], [5, 6, 7, 8], [9,10,11,12] ] Output: [1,2,3,4,8,12,11,10,9,5,6,7] ```

```c ```

```c++ ```

```c# ```

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```python ```

```python class Solution: def spiralOrder(self, matrix: List[List[int]]) -> List[int]: if not matrix: return [] # 第一行，按顺序 result = matrix.pop(0) # 最右侧 for i, l in enumerate(matrix[:-1]): if l: result.append(matrix[i].pop()) # 最后一行 if matrix: result.extend(matrix.pop()[::-1]) # 最左侧 for i, l in enumerate(matrix[::-1]): if l: result.append(matrix[::-1][i].pop(0)) # 递归，合并结果 result.extend(self.spiralOrder(matrix)) return result ```

```ruby ```

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```swift ```