class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
res = 0 # 全部腐烂需要的时间
ls = [] # 存放腐烂的橘子
dx = [1, -1, 0, 0] # 位置坐标
dy = [0, 0, 1, -1]
if len(grid) == 0: # 如果没有数据,返回-1
return -1
# 找到腐烂的橘子,将坐标放入ls中
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 2:
ls.append([i,j])
while ls:##bfs遍历
newls = []
for x in ls:
# 从一个腐烂的橘子开始,遍历其上下左右的橘子并将其腐烂
x0 = x[0]
y0 = x[1]
for k in range(4):
x = x0+dx[k]
y = y0+dy[k]
if 0<=x<len(grid) and 0<=y<len(grid[0]) and grid[x][y] == 1:
grid[x][y] = 2
newls.append([x,y])
# 如果newls为空,则表明该橘子的上下左右无法继续腐烂
if not newls:
break
ls = newls[:]
res += 1
# 判断原数组中还有没有新鲜橘子,如果有,返回-1
for row in grid:
for i in row:
if i == 1:
return -1
return res
腐烂的橘子------BFS练习
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转载自blog.csdn.net/qzcrystal/article/details/104649562
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