684. Redundant Connection -----------M ~~~

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. 

A:

因为u<v,  很自然想到union 的思路。  

可是， merge的时候，我们还需要 再map back 回来。 因为不能简单地用指针都指向同一个地址。 （会给后续的修改 造成missing ends)

为了简单快捷，我这里用了for loop

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        int N = edges.size();
        vector<int> V(N+1, INT_MAX); // node value as index, V[i] is their smallest parent
        for (auto edge : edges) {
            int u = edge[0];
            int v = edge[1];
            int smallest = min(u, min(V[u], V[v]) );
            
            if(V[u] != INT_MAX && V[u] == V[v]) // if already same parent
                return vector<int>{u,v};
            
            int preUsmallest = V[u];
            int preVsmallest = V[v];
            for(int i =1;i<=N;i++){
                if(i==u || i==v || (V[i] != INT_MAX && (V[i] == preUsmallest || V[i] == preVsmallest) )){// if parent 
                    V[i] = smallest;
                }
            }
        }
        return vector<int>();// shall not hit
    }
};