### 题目

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.

### 抽象成数学模型

[[1,2], [1,3]]

1
/ \
2   3

1
/ \
2 - 3


[[1,2], [2,3], [3,4]]

5   1 - 2
|
4 - 3

5   1 - 2
|   |
4 - 3


### Union Find

此时4个节点已经连通。都属于1个小组，根节点只有一个。
1 - 2
|
4 - 3

1 - 2
|   |
4 - 3


#### 代码

class Solution {
public int[] findRedundantConnection(int[][] edges) {
int size = 1000;
board = new int[size + 1];
for (int i = 1; i <= size; i++) board[i] = i;
for (int[] edge : edges) {
if (!union(edge, edge)) return edge;
}
return null;
}
private int[] board;
private boolean union(int a, int b) {
int rootA = find(a);
int rootB = find(b);
if (rootA == rootB) return false;
board[rootB] = rootA;
return true;
}
private int find(int a) {
if (board[a] == a) return a;
int prev = find(board[a]);
board[a] = prev; // path compression
return prev;
}
}


#### 结果 