LeetCode #684 Redundant Connection

题目

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

Note:

1. The size of the input 2D-array will be between 3 and 1000.
2. Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.

解题思路

这道题用并查集可以很容易地解决。
遍历每一个 edge ，如果第二个顶点所在的树的根结点与第一个结点的根结点不相同，说明在当前构建的树中还不存在路径把两个顶点连接起来，则把当前 edge 的第二个顶点所在的树合并到第一个顶点的树中，使这两个顶点具有相同的根结点，这样就表示找到了一条路径可以连接当前 edge 的两个顶点；
如果第二个顶点所在的树的根结点与第一个结点的根结点相同，说明在前面的构建树的过程中，已经找到路径连接这两个顶点了，因此当前的这个 edge 就是冗余的，可以删掉。把当前的 edge 作为结果返回即可。
（注意一旦找到了冗余的 edge ，则这个 edge 肯定就是 the answer that occurs last in the given 2D-array，用题目的例子细细推导一下就可以知道）

C++代码实现

class Solution {public:    vector<int> findRedundantConnection(vector<vector<int>>& edges) {        int N = edges.size();        unordered_map<int, int> parents;        vector<int> rst;        for (int i = 1; i <= N; ++i) { parents[i] = i; }        for (int i = 0; i < N; ++i) {            int firstP = findParent(edges[i][0], parents);            int secondP = findParent(edges[i][1], parents);            if (firstP == secondP) {                rst.push_back(edges[i][0]);                rst.push_back(edges[i][1]);                break;            } else {                parents[secondP] = firstP;            }        }        return rst;    }    int findParent(int son, unordered_map<int, int>& parents) {        if (son == parents[son]) { return son; }        parents[son] = findParent(parents[son], parents);        return parents[son];    }};