Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to find the number of connected components in an undirected graph.
Example 1:
0 3
| |
1 --- 2 4
Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], return 2.
Example 2:
0 4
| |
1 --- 2 --- 3
Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [3, 4]], return 1.
Note:
You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.
给定n个标记为0到n-1的节点和无向边列表（每个边都是一对节点），编写一个函数来查找无向图中连接的组件的数量。
例1：
0 3
| |
1 --- 2 4
假设n=5，edges = [[0, 1], [1, 2], [3, 4]]，返回2。
例2：
0 4
| |
1 --- 2 --- 3
如果n=5，edges = [[0, 1], [1, 2], [2, 3], [3, 4]],返回1。
注：
可以假定边中不会出现重复的边。因为所有边都是无向的，[0，1]与[1，0]相同，因此不会出现在边中。
Solution:
1 class Solution {
2 func countComponents(_ n:Int,_ edges:inout [[Int]]) -> Int {
3 var res:Int = n
4 var root:[Int] = [Int](repeating:0,count:n)
5 for i in 0..<n
6 {
7 root[i] = i
8 }
9 for a in edges
10 {
11 var x:Int = find(&root, a[0])
12 var y:Int = find(&root, a[1])
13 if x != y
14 {
15 res -= 1
16 root[y] = x
17 }
18 }
19 return res
20 }
21
22 func find(_ root:inout [Int],_ i:Int) -> Int
23 {
24 var i = i
25 while(root[i] != i)
26 {
27 i = root[i]
28 }
29 return i
30 }
31 }
点击：Playground测试
1 let n1:Int = 5 2 var edge1:[[Int]] = [[0, 1], [1, 2], [3, 4]] 3 print(Solution().countComponents(n1,&edge1)) 4 //Print 2 5 let n2:Int = 5 6 var edge2:[[Int]] = [[0, 1], [1, 2], [2, 3], [3, 4]] 7 print(Solution().countComponents(n2,&edge2)) 8 //Print 1
原文地址：https://www.cnblogs.com/strengthen/p/10706799.html