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.

分析：UNION-FIND

 class Solution {
     public int[] findRedundantConnection(int[][] edges) {
         int[] parent = new int[edges.length + ];
         for (int i = ; i < parent.length; i++) parent[i] = i;

         for (int[] edge: edges){
             int f = edge[], t = edge[];
             if (find(parent, f) == find(parent, t)) return edge;
             else parent[find(parent, f)] = find(parent, t);
         }

         return new int[];
     }

     private int find(int[] parent, int f) {
         if (f != parent[f]) {
           parent[f] = find(parent, parent[f]);
         }
         return parent[f];
     }
 }