https://leetcode.com/problems/redundant-connection/description/

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.

顯然是個(gè)并查集的題。
直接初始化并查集，每碰到一條邊合并集合即可。

class Solution {
public:
    int fa[1005];
    
    bool unionTest(int u, int v) {
        while (fa[u] != u) u = fa[u];
        while (fa[v] != v) v = fa[v];
        
        if (u != v) {
            fa[v] = u;
            return false;
        } else {
            return true;
        }
    }
    
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        int n = edges.size();
        int ansu, ansv;
        
        ansu = ansv = 0;
        for (int i = 1; i <= n; i++) {
            fa[i] = i;
        }
        
        for (int i = 0; i < n; i++) {
            int u = edges[i][0];
            int v = edges[i][1];
            
            if (unionTest(u, v)) {
                ansu = u;
                ansv = v;
            }
        }
        
        vector<int> ans;
        ans.push_back(ansu);
        ans.push_back(ansv);
        return ans;
    }
};