题解 | #螺旋矩阵#
螺旋矩阵
https://www.nowcoder.com/practice/7edf70f2d29c4b599693dc3aaeea1d31
#coding:utf-8
#
# 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
#
#
# @param matrix int整型二维数组
# @return int整型一维数组
#
class Solution:
def spiralOrder(self , matrix ):
# write code here
if matrix == []:
return []
res = []
left = 0
top = 0
bottom = len(matrix) - 1
right = len(matrix[0]) - 1
while len(res) < (len(matrix) * len(matrix[0])):
for j in range(left, right + 1):
res.append(matrix[top][j])
top += 1
for i in range(top, bottom + 1):
res.append(matrix[i][right])
right -= 1
for j in range(right, left - 1, -1):
if top - 1 == bottom:
break
res.append(matrix[bottom][j])
bottom -= 1
for i in range(bottom, top - 1, -1):
if left - 1 == right:
break
res.append(matrix[i][left])
left += 1
return res
设置上下左右四个边界,后面两次需要增加上下和左右是否相等的判断(否则会重复)。开头增加特殊情况判断。

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