[leetcode]684. Redundant Connection

Analysis

突然想家。。。—— [每天刷题并不难。。。]

In this problem, a tree is an undirected graph that is connected and has no cycles.
The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, …, N), with one additional edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.
The resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] with u < v, that represents an undirected edge connecting nodes u and v.
Return an edge that can be removed so that the resulting graph is a tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array. The answer edge [u, v] should be in the same format, with u < v.

并查集，但是好像用DFS和BFS也可以解决~

Implement

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        vector<int> res;
        vector<int> root(2010, -1);
        for(auto e:edges){
            int x = find(root, e[0]);
            int y = find(root, e[1]);
            if(x == y){
                res.push_back(e[0]);
                res.push_back(e[1]);
                break;
            }
            root[x] = y;
        }
        return res;
    }
    int find(vector<int>& root, int i){
        while(root[i] != -1)
            i = root[i];
        return i;
    }
};