leetcode 684. 冗余连接

c++

class Solution {
public:
    class UnionSet {
    public:
        int *father, n;
        UnionSet(int n) : n(n) {
            father = new int[n + 1]; 
            for (int i = 0; i <= n; i++) {
                father[i] = i;
            }
        }
        int find(int x) {
            return father[x] = (father[x] == x ? x : find(father[x]));
        }
        void merge(int a, int b) {
            father[find(a)] = find(b);
            return ;
        }
    };
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        UnionSet u(edges.size());
        vector<int> ans;
        for (auto edge : edges) {
            if (u.find(edge[0]) != u.find(edge[1])) u.merge(edge[0], edge[1]);
            else ans = edge;
        }
        return ans;
    }
};

js

class UnionSet {
    constructor(n) {
        this.father = new Array(n);
        for (var i = 0; i < n; i++) this.father[i] = i;
    }
    find(x) {
        return this.father[x] = (this.father[x] == x ? x : this.find(this.father[x]));
    }
    merge(a, b) {
        this.father[this.find(a)] = this.find(b); 
    } 
};
var findRedundantConnection = function(edges) {
    var u = new UnionSet(edges.length);
    for (var edge of edges) {
        if (u.find(edge[0]) == u.find(edge[1])) return edge;
        u.merge(edge[0], edge[1]);
    }
    return [];
};