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添加 problem 684 题解
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# [684. Redundant Connection](https://leetcode.com/problems/redundant-connection/)
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## 题目:
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In this problem, a tree is an **undirected** graph that is connected and has no cycles.
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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.
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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`.
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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`.
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**Example 1:**
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Input: [[1,2], [1,3], [2,3]]
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Output: [2,3]
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Explanation: The given undirected graph will be like this:
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1
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/ \
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2 - 3
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**Example 2:**
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Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
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Output: [1,4]
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Explanation: The given undirected graph will be like this:
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5 - 1 - 2
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| |
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4 - 3
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**Note:**
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- The size of the input 2D-array will be between 3 and 1000.
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- Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.
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**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.
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## 题目大意
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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。
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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。
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注意:
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- 输入的二维数组大小在 3 到 1000。
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- 二维数组中的整数在 1 到 N 之间,其中 N 是输入数组的大小。
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## 解题思路
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- 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。
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- 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。

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