Graph-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.

int findParent(vector<int>& parent, int k) {  
    if (parent[k] != k)   
        parent[k] = findParent(parent, parent[k]);  
    return parent[k];  
}  
  
vector<int> findRedundantConnection(vector<vector<int> >& edges) {  
    vector<int> parent;  
    for (int i = 0; i < 2001; i++)  // 初始化  
        parent.push_back(i);  
  
    int point1, point2;  
    for (int j = 0; j < edges.size(); j++) {  
        point1 = findParent(parent, edges[j][0]);  
        point2 = findParent(parent, edges[j][1]);  
        if (point1 == point2)  
            return edges[j];  
        parent[point2] = point1;  
    }  
    return vector<int>(0, 0);  
}