第十一周LeetCode

题目
Redundant Connection
难度 Medium

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:
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.

实现思路

可以利用UnionSet的思想实现，相当于把结点分组，两个结点连通则分到同一组。初始时每个点自成一组，当加入一条（i，j）边时，把i和j放到同一组。假设最开始只有一条（0，1）的边，那么把0和1分到同一组，再加入一条（1，2）的边，因为0与1相连，因此0和2也有一条边，把0，1，2分到一组。使用组号标记他们是哪一组，此处组号都设为2，先把0分到组1，再把组1分到组2。当加入新的边（0，2）时，由于他们是同一组，说明没有这条边之前他们已经连通，因此这条边是冗余的。
实现时建立一个UnionSet的数组，每个值初始化为下标号。当加入一条边时就将查看边的两个结点的分组，如果分组一样就是冗余，否则将他们分到同一组。

实现代码

//分组int find(vector<int>& unionSet, int w) {        while (unionSet[w] != w) {           w = unionSet[w];         }        return w;    }    vector<int> findRedundantConnection(vector<vector<int>>& edges) {        vector<int> unionSet(2001,0);        for (int i = 1; i < 2001; i++) {            unionSet[i] = i;        }        for (int i = 0; i < edges.size(); i++) {            if (find(unionSet, edges[i][0]) != find(unionSet,edges[i][1])) {                    unionSet[find(unionSet, edges[i][0])] = find(unionSet, edges[i][1]);            } else {                return edges[i];            }        }        return {};    }