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 #

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 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
}