题目：原题链接（中等）
标签：广度优先搜索、深度优先搜索
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N M ) | O ( N × M ) | 364ms (29.88%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class Solution:
def pondSizes(self, land: List[List[int]]) -> List[int]:
if not land or not land[0]:
return []
s1, s2 = len(land), len(land[0])
def is_valid(x, y):
return 0 <= x < s1 and 0 <= y < s2
def neighbours(x1, y1):
res = []
for x2, y2 in [(x1 + 1, y1 - 1), (x1 + 1, y1), (x1 + 1, y1 + 1),
(x1, y1 - 1), (x1, y1 + 1),
(x1 - 1, y1 - 1), (x1 - 1, y1), (x1 - 1, y1 + 1)]:
if is_valid(x2, y2):
res.append((x2, y2))
return res
ans = []
visited = set()
for i1 in range(s1):
for i2 in range(s2):
if land[i1][i2] == 0 and (i1, i2) not in visited:
pools = {(i1, i2)}
now = 1
waiting = collections.deque([(i1, i2)])
while waiting:
j1, j2 = waiting.popleft()
for k1, k2 in neighbours(j1, j2):
if land[k1][k2] == 0 and (k1, k2) not in visited and (k1, k2) not in pools:
waiting.append((k1, k2))
now += 1
pools.add((k1, k2))
ans.append(now)
visited |= pools
ans.sort()
return ans
