题目：原题链接（中等）
标签：并查集、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 2 ) O(N^2) O(N2) | O ( N ) O(N) O(N) | 64ms (52.11%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
class DSU:
def __init__(self, n):
self.array = [i for i in range(n)]
self.size = [1] * n
def find(self, i):
if self.array[i] != i:
self.array[i] = self.find(self.array[i])
return self.array[i]
def union(self, i, j):
i = self.find(i)
j = self.find(j)
if self.size[i] >= self.size[j]:
self.array[j] = i
self.size[i] += self.size[j]
else:
self.array[i] = j
self.size[j] += self.size[i]
class Solution:
def findCircleNum(self, M: List[List[int]]) -> int:
dsu = DSU(len(M))
for i in range(len(M)):
for j in range(i + 1, len(M)):
if M[i][j] == 1:
dsu.union(i, j)
return len(set(dsu.find(i) for i in range(len(M))))
