# 利用并查集
class Solution:
def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
n = len(edges)
parent = list(range(n+1))
def find(index): # 找到当前节点index的父节点
if parent[index]==index:
return index
else:
return find(find(parent[index]))
def union(index1, index2): # 将两个节点连在同一个根节点上
parent[find(index1)] = find(index2)
for index1, index2 in edges: # 判断两个节点是不是同一个根节点
if find(index1) == find(index2):
return [index1, index2]
else:
union(index1, index2)
参考:
https://leetcode-cn.com/problems/redundant-connection/solution/684-rong-yu-lian-jie-bing-cha-ji-ji-chu-eartc/
https://leetcode-cn.com/problems/redundant-connection/solution/rong-yu-lian-jie-by-leetcode-solution-pks2/