684. Redundant Connection

/*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 - 3Example 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 - 3Note:The size of the input 2D-array will be between 3 and 1000.Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.Update (2017-09-26):We have overhauled the problem description + test cases and specified clearly the graph is an undirected graph. For the directed graph follow up please see Redundant Connection II). We apologize for any inconvenience caused.采用并查集的方法来解决问题，对无向图所有的边一次加入不同的分组，查看是否存在环路，如果加入一条边形成环路即加入的这条边的两个点已经在同一个分组中了，返回该边即可。关于并查集的优化以及讲解请查看：http://blog.csdn.net/dm_vincent/article/details/7655764博主讲的十分清楚。动态连通性的几种操作：    查询节点属于的组        数组对应位置的值即为组号    判断两个节点是否属于同一个组        分别得到两个节点的组号，然后判断组号是否相等    连接两个节点，使之属于同一个组        分别得到两个节点的组号，组号相同时操作结束，不同时，将其中的一个节点的组号换成另一个节点的组号    获取组的数目        对构造树进行优化，来避免树的深度过大*/class Solution {public:    vector<int> findRedundantConnection(vector<vector<int>>& edges) {        vector<int> parent(2001);        for(int i=0;i<2001;i++) parent[i]=i;//分组，每组的根节点为i        for(auto edge : edges){            int l=findRoot(edge[0],parent),r=findRoot(edge[1],parent);            if(l == r) //新加入的边形成的环路 是否在同一个分组中                return edge;            parent[l] = r; //l组加入到r组中，成为r组根节点的一个分支 此处可以采用树的深度来优化        }        return vector<int> (0, 0);//不存在    }    int findRoot(int p,vector<int>& parent){ //寻找p节点对应分组的根节点        while(p != parent[p])            p = parent[p];        return p;    }};