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]
class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
if matrix == []:
return []
n = len(matrix)
m = len(matrix[0])
result = []
if n<=m:
for i in range(n//2):
a = i
b = i
while a != i+1 or b != i:
result.append(matrix[a][b])
if a == i and b != m-1-i:
b += 1
elif b == m-1-i and a != n-1-i:
a += 1
elif a == n-1-i and b != i:
b -= 1
elif b == i and a != i:
a -= 1
result.append(matrix[a][b])
if n%2 == 1:
for item in range(0+n//2, m-n//2):
result.append(matrix[n//2][item])
else:
for i in range(m//2):
a = i
b = i
while a != i+1 or b != i:
result.append(matrix[a][b])
if a == i and b != m-1-i:
b += 1
elif b == m-1-i and a != n-1-i:
a += 1
elif a == n-1-i and b != i:
b -= 1
elif b == i and a != i:
a -= 1
result.append(matrix[a][b])
if m%2 == 1:
for item in range(0+m//2, n-m//2):
result.append(matrix[item][m//2])
return result
