题目：原题链接（中等）
标签：广度优先搜索、图
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 ) | 532ms (61.37%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class Solution:
def maxDistance(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
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)]
visited = set()
queue = collections.deque()
for i in range(m):
for j in range(n):
if grid[i][j] == 1:
queue.append((i, j))
visited.add((i, j))
if len(visited) == m * n:
return -1
step = 0
while queue:
for _ in range(len(queue)):
i1, j1 = queue.popleft()
for i2, j2 in _get_neighbors(i1, j1):
if (i2, j2) not in visited:
queue.append((i2, j2))
visited.add((i2, j2))
if not queue:
return step
else:
step += 1
return -1
