Leetcode 994. Rotting Oranges
Posted SnailTyan
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文章作者:Tyan
博客:noahsnail.com | CSDN | 简书
1. Description
2. Solution
**解析:**Version 1,先统计新鲜水果数量,如果不存在,则时间为0
,同时记录腐败水果的位置。按照广度优先搜索,记录下一轮腐败水果的位置,同时时间加1
,新鲜水果数量减1
,递归执行,直至不存在腐败的水果位置或者新鲜水果为0
。如果此时仍存在新鲜水果,则返回-1
,否则,返回时间。
- Version 1
class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
m = len(grid)
n = len(grid[0])
fresh = 0
rotten = []
for i in range(m):
for j in range(n):
if grid[i][j] == 2:
rotten.append([i, j])
elif grid[i][j] == 1:
fresh += 1
if fresh == 0:
return 0
time = 0
while rotten and fresh:
temp = []
for x, y in rotten:
if x > 0 and grid[x-1][y] == 1:
grid[x-1][y] = 2
temp.append([x-1, y])
fresh -= 1
if y > 0 and grid[x][y-1] == 1:
grid[x][y-1] = 2
temp.append([x, y-1])
fresh -= 1
if x < m - 1 and grid[x+1][y] == 1:
grid[x+1][y] = 2
temp.append([x+1, y])
fresh -= 1
if y < n - 1 and grid[x][y+1] == 1:
grid[x][y+1] = 2
temp.append([x, y+1])
fresh -= 1
time += 1
rotten = temp
if fresh == 0:
return time
else:
return -1
Reference
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