# 螺旋矩阵
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。
示例 1：
输入：matrix = [[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：matrix = [[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]
提示：
m == matrix.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100
## template
```python
class Solution(object):
def spiralOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
if matrix is None or len(matrix) == 0:
return matrix
m, n = len(matrix), len(matrix[0])
return self.get_spiralOrder(matrix, 0, m - 1, 0, n - 1)
def get_spiralOrder(self, matrix, r_start, r_end, c_start, c_end):
if r_start > r_end or c_start > c_end:
return []
elif r_start == r_end:
return matrix[r_start][c_start:c_end + 1]
elif c_start == c_end:
return [matrix[j][c_end] for j in range(r_start, r_end + 1)]
curr = matrix[r_start][c_start:c_end + 1] + [matrix[j][c_end] for j in range(r_start + 1, r_end)] +\
matrix[r_end][c_start:c_end + 1][::-1] +\
[matrix[j][c_start] for j in reversed(range(r_start + 1, r_end))]
res = curr + self.get_spiralOrder(matrix, r_start + 1, r_end - 1, c_start + 1, c_end - 1)
return res
if __name__ == '__main__':
s = Solution()
print (s.spiralOrder([[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]))
```
## 答案
```python
```
## 选项
### A
```python
```
### B
```python
```
### C
```python
```