题目：原题链接（困难）
标签：广度优先搜索、深度优先搜索、图
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( N 2 × M 2 ) O(N^2×M^2) O(N2×M2) | O ( N × M ) O(N×M) O(N×M) | 956ms (39.78%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
class Solution:
def shortestDistance(self, grid: List[List[int]]) -> int:
def is_valid(x, y):
return 0 <= x < s1 and 0 <= y < s2
def get_near(x, y):
res = []
for xx, yy in [(x - 1, y), (x + 1, y), (x, y - 1), (x, y + 1)]:
if is_valid(xx, yy) and grid[xx][yy] == 0:
res.append((xx, yy))
return res
if not grid or not grid[0]:
return 0
s1, s2 = len(grid), len(grid[0])
buildings = []
for i1 in range(s1):
for i2 in range(s2):
if grid[i1][i2] == 1:
buildings.append((i1, i2))
dp1 = [[0] * s2 for _ in range(s1)]
dp2 = [[0] * s2 for _ in range(s1)]
for person in buildings:
visited = {person}
queue = collections.deque([person])
distance = 0
while queue:
distance += 1
for _ in range(len(queue)):
(i1, i2) = queue.popleft()
for (j1, j2) in get_near(i1, i2):
if (j1, j2) not in visited:
visited.add((j1, j2))
queue.append((j1, j2))
dp1[j1][j2] += distance
dp2[j1][j2] += 1
# for row in dp1:
# print(row)
ans = float("inf")
for i1 in range(s1):
for i2 in range(s2):
if grid[i1][i2] == 0 and dp2[i1][i2] == len(buildings):
ans = min(ans, dp1[i1][i2])
return ans if ans != float("inf") else -1
