添加 problem 684 题解

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# [684. Redundant Connection](https://leetcode.com/problems/redundant-connection/)
## 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](https://leetcode.com/problems/redundant-connection-ii/description/)**). We apologize for any inconvenience caused.
## 题目大意
在本问题中, 树指的是一个连通且无环的无向图。输入一个图该图由一个有着N个节点 (节点值不重复1, 2, ..., N) 的树及一条附加的边构成。附加的边的两个顶点包含在1到N中间这条附加的边不属于树中已存在的边。结果图是一个以边组成的二维数组。每一个边的元素是一对[u, v] 满足 u < v表示连接顶点u 和v的无向图的边
返回一条可以删去的边使得结果图是一个有着N个节点的树如果有多个答案则返回二维数组中最后出现的边答案边 [u, v] 应满足相同的格式 u < v
注意:
- 输入的二维数组大小在 3 1000
- 二维数组中的整数在 1 N 之间其中 N 是输入数组的大小
## 解题思路
- 给出一个连通无环无向图和一些连通的边要求在这些边中删除一条边以后图中的 N 个节点依旧是连通的如果有多条边输出最后一条
- 这一题可以用并查集直接秒杀依次扫描所有的边把边的两端点都合并 `union()` 到一起如果遇到一条边的两端点已经在一个集合里面了就说明是多余边删除最后输出这些边即可