0684. Redundant Connection

# 684. 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). We apologize for any inconvenience caused.

## 题目大意 #

• 输入的二维数组大小在 3 到 1000。
• 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。

## 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。

## 代码 #

``````
package leetcode

import (
"github.com/halfrost/LeetCode-Go/template"
)

func findRedundantConnection(edges [][]int) []int {
if len(edges) == 0 {
return []int{}
}
uf, res := template.UnionFind{}, []int{}
uf.Init(len(edges) + 1)
for i := 0; i < len(edges); i++ {
if uf.Find(edges[i][0]) != uf.Find(edges[i][1]) {
uf.Union(edges[i][0], edges[i][1])
} else {
res = append(res, edges[i][0])
res = append(res, edges[i][1])
}
}
return res
}

``````

Sep 6, 2020