Redundant Connection问题及解法

问题描述：

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.

Example 1:

Input: [[1,2], [1,3], [2,3]]Output: [2,3]Explanation: The given undirected graph will be like this:  1 / \2 - 3

Example 2:

Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]Output: [1,4]Explanation: The given undirected graph will be like this:5 - 1 - 2    |   |    4 - 3

问题分析：

这是一类查找公共头结点的问题，如果两个相连的节点有公共头结点，那么就可能存在环。这里使用并查集求解。

过程详见代码：

class Solution {public:    vector<int> findRedundantConnection(vector<vector<int>>& edges) {        int n = edges.size();vector<int> index(n + 1);vector<int> res(2);for (int i = 0; i <= n; i++) index[i] = i;for (auto edge : edges){int id0 = index[edge[0]];int id1 = index[edge[1]];while (id0 != index[id0]){id0 = index[id0];}while (id1 != index[id1]){id1 = index[id1];}if (id0 != id1){int id = min(id0, id1);index[id0] = id;index[id1] = id;}else res = edge;}return res;    }};