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]
很蠢得用了模拟法。。
class Solution:
def spiralOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
if len(matrix)==0:
return []
m,n = len(matrix),len(matrix[0])
flag = [[True for i in range(n)]for j in range(m)]
i,j,dir = 0,0,0
res = [matrix[0][0]]
flag[0][0] = False
while True:
if dir==0: #go right
if j<n-1 and flag[i][j+1]:
res.append(matrix[i][j+1])
j += 1
flag[i][j] = False
else:
dir = 1
if i==m-1 or flag[i+1][j] is False:
break
else:
res.append(matrix[i+1][j])
i += 1
flag[i][j] = False
continue
if dir==1: #go down
if i<m-1 and flag[i+1][j]:
res.append(matrix[i+1][j])
i += 1
flag[i][j] = False
else:
dir = 2
if j==0 or flag[i][j-1] is False:
break
else:
res.append(matrix[i][j-1])
j -= 1
flag[i][j] = False
continue
if dir==2: #go left
if j>0 and flag[i][j-1]:
res.append(matrix[i][j-1])
j -= 1
flag[i][j] = False
else:
dir = 3
if i==0 or flag[i-1][j] is False:
break
else:
res.append(matrix[i-1][j])
i -= 1
flag[i][j] = False
continue
if dir==3: #go up
if i>0 and flag[i-1][j]:
res.append(matrix[i-1][j])
i -= 1
flag[i][j] = False
else:
dir = 0
if j==n-1 or flag[i][j+1] is False:
break
else:
res.append(matrix[i][j+1])
j += 1
flag[i][j] = False
continue
return res