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]

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.

題解：

這題用到了並查集的演算法

reference：https://blog.csdn.net/dm_vincent/article/details/7655764

從例子來看，是從給出的起點終點的陣列edges中，找出第一個形成環的元素。

思路：

題目中，每個數字代表一個點，定義一個起點與終點的陣列begin，begin有2001個元素，begin的下標代表終點，begin的元素代表原始起點（最開始的起點），題目可以化簡為找出第一個edge，這個edge的起點和終點都有同樣的一個原始起點，即形成環。

初始化為每個起點的終點都指向自身為原始起點。然後遍歷edges陣列，更新新遍歷的點，更新原始起點，直至找到環為止。

以example 1為例，

begin 的下標為 0 1 2 3 ... ...

初始化後元素為 0 1 2 3

[1 2] 進入後的為 0 1 1 3，2的原始起點更新為1

[1 3] 進入後的為 0 1 1 1，3的原始起點更新為1

[2 3] 進入後，發現2與3的原始起點一致，均為1，發現環。

reference:

https://blog.csdn.net/dong_beijing/article/details/78094443

class Solution {
    int[] roots;
    public int[] findRedundantConnection(int[][] edges) {
        roots = new int[edges.length + 1];
        for(int i = 0; i < roots.length; i++) {
            roots[i] = i;
        }
        for(int[] edge : edges) {
            int root1 = find(edge[0]);
            int root2 = find(edge[1]);
            if(root1 == root2)
                return edge;
            roots[root2] = root1;
        }
        int[] ans = new int[2];
        return ans;
    }
    
    public int find(int i) {
        while(i != roots[i]) {
            i = roots[i];
        }
        return i;
    }
}