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 - 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.

解題思路
利用一個數組，下標對應頂點的編號，值對應著與之相連的頂點。如果沒有頂點與之相連，則置為0 。如果頂點已經與其他頂點相連，則通過通過陣列找到這一組相連頂點中的那個度為1的點，即數值為0的點，將其數值設定為該邊另一個端點的編號。
每加入一條邊，都要找到最後那個度為1的點（起點和終點都要找）。如果起點和終點對應的是同一個度為1的點，那麼說明加入這條邊就會產生環，這條邊可以去掉。否則，將起點對應的度為1的點的值設定為終點的編號。

程式碼如下：

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        vector<int> p(2000, 0);
        vector<int> result;
        for(auto v : edges) {
            int n1 = v[0];
            int n2 = v[1];
            while(p[n1] != 0) {
                n1 = p[n1];
            }
            while(p[n2] != 0) {
                n2 = p[n2];
            }
            if(n1 == n2) 
                result = v;
            else
                p[n1] = n2;
        }
        return result;
    }
};