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LeetCode题解(0286):墙与门(Python)


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 ( N × D ): D为门的数量

O ( 1 )

180ms (24.58%)

Ans 2 (Python)

Ans 3 (Python)

解法一:

class Solution:
def wallsAndGates(self, rooms: List[List[int]]) -> None:
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 rooms[xx][yy] > 0:
res.append((xx, yy))
return res

if not rooms or not rooms[0]:
return

s1, s2 = len(rooms), len(rooms[0])

doors = []
for i1 in range(s1):
for i2 in range(s2):
if rooms[i1][i2] == 0:
doors.append((i1, i2))

for door in doors:
queue = collections.deque([door])
while queue:
(i1, i2) = queue.popleft()
distance = rooms[i1][i2] + 1
for (j1, j2) in get_near(i1, i2):
if rooms[j1][j2] > distance:
rooms[j1][j2] = distance
queue.append((j1, j2))



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