力扣 684. 冗余连接 python AC

并查集

class Solution:
    def findRedundantConnection(self, edges):
        n = len(edges)
        bcj = {i: i for i in range(1, n + 1)}
        ans = None

        def find(x):
            if x != bcj[x]:
                bcj[x] = find(bcj[x])
            return bcj[x]

        def union(x, y):
            a, b = find(x), find(y)
            if a != b:
                bcj[a] = b
                return True
            return False

        for i, j in edges:
            if not union(i, j):
                ans = [i, j]
        return ans