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。
输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出：[1,2,3,6,9,8,7,4,5]
观察给出的例子，根据上，右，下，左的顺序依次输出最外面的一层，然后再利用递归，打印第二层，直到矩阵的最内层。
image.png
class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
result = list()
rows = len(matrix)
cols = len(matrix[0])
self.spiralOrderOuterLayer(matrix, (0, 0), (rows - 1, cols - 1), result)
return result
def spiralOrderOuterLayer(self, matrix, top_left, bottom_right, current_sequence):
top_left_row, top_left_col = top_left
bottom_right_row, bottom_right_col = bottom_right
if top_left_row < bottom_right_row and top_left_col < bottom_right_col:
# print the outer layer in the orders of top, right, bottom, left
for col in range(top_left_col, bottom_right_col):
current_sequence.append(matrix[top_left_row][col])
for row in range(top_left_row, bottom_right_row):
current_sequence.append(matrix[row][bottom_right_col])
for col in range(bottom_right_col, top_left_col, -1):
current_sequence.append(matrix[bottom_right_row][col])
for row in range(bottom_right_row, top_left_row, -1):
current_sequence.append(matrix[row][top_left_col])
# recursively print for the rest matrix
self.spiralOrderOuterLayer(
matrix,
(top_left_row + 1, top_left_col + 1),
(bottom_right_row - 1, bottom_right_col - 1),
current_sequence
)
elif top_left_col == bottom_right_col:
# if the matrix is just one column
for row in range(top_left_row, bottom_right_row + 1):
current_sequence.append(matrix[row][top_left_col])
elif top_left_row == bottom_right_row:
# if the matrix is just one row
for col in range(top_left_col, bottom_right_col + 1):
current_sequence.append(matrix[top_left_row][col])
