[Leetcode][python]Spiral Matrix/Spiral Matrix II/螺旋矩阵/螺旋矩阵 II
[Leetcode][python]Spiral Matrix/Spiral Matrix II/螺旋矩阵/螺旋矩阵 II
蛮三刀酱
发布于 2019-03-26 16:43:54
发布于 2019-03-26 16:43:54
Spiral Matrix
题目大意
将一个矩阵中的内容螺旋输出。 注意点：
矩阵不一定是正方形 例子：
输入: matrix = [ [1, 2, 3], [4, 5, 6], [7, 8, 9] ] 输出: [1, 2, 3, 6, 9, 8, 7, 4, 5]
解题思路
看代码就可以理解 注意：
while True： 然后里面自行判断结束
代码
class Solution(object):
def spiralOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
if not matrix:
return []
up = 0
left = 0
down = len(matrix)-1
right = len(matrix[0])-1
direct = 0 # 0: go right 1: go down 2: go left 3: go up
res = []
while True:
if direct == 0:
for i in range(left, right+1):
res.append(matrix[up][i])
up += 1
if direct == 1:
for i in range(up, down+1):
res.append(matrix[i][right])
right -= 1
if direct == 2:
for i in range(right, left-1, -1):
res.append(matrix[down][i])
down -= 1
if direct == 3:
for i in range(down, up-1, -1):
res.append(matrix[i][left])
left += 1
if up > down or left > right:
return res
direct = (direct+1) % 4Spiral Matrix II
题目大意
将一个正方形矩阵螺旋着填满递增的数字。
解题思路
螺旋填满数字
代码
class Solution(object):
def generateMatrix(self, n):
"""
:type n: int
:rtype: List[List[int]]
"""
if n == 0:
return []
matrix = [[0 for i in range(n)] for j in range(n)] # 创建矩阵
up = 0
down = len(matrix)-1
left = 0
right = len(matrix[0])-1
direct = 0
count = 0
while True:
if direct == 0:
for i in range(left, right+1):
count += 1; matrix[up][i] = count
up += 1
if direct == 1:
for i in range(up, down+1):
count += 1; matrix[i][right] = count
right -= 1
if direct == 2:
for i in range(right, left-1, -1):
count += 1; matrix[down][i] = count
down -= 1
if direct == 3:
for i in range(down, up-1, -1):
count += 1; matrix[i][left] = count
left += 1
if count == n*n:
return matrix
direct = (direct+1) % 4总结
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原始发表：2017年10月10日，如有侵权请联系 cloudcommunity@tencent.com 删除
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