题目：原题链接（中等）
标签：广度优先搜索
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( M × N ) | O ( M × N ) | 64ms (45.32%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
def _is_valid(x, y):
return 0 <= x < m and 0 <= y < n
def _get_neighbors(x1, y1):
return [(x2, y2) for (x2, y2) in [(x1 - 1, y1), (x1 + 1, y1), (x1, y1 - 1), (x1, y1 + 1)]
if _is_valid(x2, y2)]
m, n = len(grid), len(grid[0])
queue = collections.deque()
waiting = set()
for i in range(m):
for j in range(n):
if grid[i][j] == 2:
queue.append((i, j))
elif grid[i][j] == 1:
waiting.add((i, j))
if not queue:
if not waiting:
return 0
else:
return -1
time = 0
while queue:
for _ in range(len(queue)):
i1, j1 = queue.popleft()
for (i2, j2) in _get_neighbors(i1, j1):
if (i2, j2) in waiting:
waiting.remove((i2, j2))
queue.append((i2, j2))
time += 1
if waiting:
return -1
return time - 1
