[LeetCode 题解] Spiral Matrix

前言

题目链接

题目描述

[1,2,3,6,9,8,7,4,5]

题意

[1, 2, 3, 6, 9, 8, 7, 4, 5]

思路

解法

class Solution(object):
def bfs(self, matrix, i, j, last = 0):
"""
:type matrix: List[List[int]]
:type i: int current row index
:type j: int current column index
:rtype: void
"""
if self.visited[i][j] == 0:
self.visited[i][j] = 1
self.res.append(matrix[i][j])
retry = 0
while retry < 4:
if last % 4 == 0:
if j + 1 <= self.n - 1 and self.visited[i][j + 1] == 0:
self.bfs(matrix, i, j + 1, last)
elif last % 4 == 1:
if i + 1 <= self.m - 1 and self.visited[i + 1][j] == 0:
self.bfs(matrix, i + 1, j, last)
elif last % 4 == 2:
if j - 1 >= 0 and self.visited[i][j - 1] == 0:
self.bfs(matrix, i, j - 1, last)
elif last % 4 == 3:
if i - 1 >= 0 and self.visited[i - 1][j] == 0:
self.bfs(matrix, i - 1, j, last)
last += 1
retry += 1

def spiralOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
self.visited = []
self.res = []
self.m = len(matrix)
if self.m == 0:
# the matrix is empty
return self.res
self.n = len(matrix[0])
for i in range(self.m):
line = []
for j in range(self.n):
line.append(0)
self.visited.append(line[:])
self.bfs(matrix, 0, 0)
return self.res

声明