class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
res=[]
m=len(matrix)
if m==0:
return res
else:
n=len(matrix[0])
if n==0:
return res
count=(m+1)//2
k=0
su=m*n
cc=0
while k<count and cc<su:
for i in range(k,n-k):
res.append(matrix[k][i])
cc+=1
for i in range(k+1,m-1-k):
res.append(matrix[i][n-1-k])
cc+=1
if k!=m-1-k:
for i in range(k,n-k):
res.append(matrix[m-1-k][n-1-i])
cc+=1
if k!=n-1-k:
for i in range(k+1,m-1-k):
res.append(matrix[m-1-i][k])
cc+=1
k+=1
return res
Leetcode 54. 螺旋矩阵（python）
