• 博客园logo
  • 会员
  • 众包
  • 新闻
  • 博问
  • 闪存
  • 赞助商
  • HarmonyOS
  • Chat2DB
    • 搜索
      所有博客
    • 搜索
      当前博客
  • 写随笔 我的博客 短消息 简洁模式
    用户头像
    我的博客 我的园子 账号设置 会员中心 简洁模式 ... 退出登录
    注册 登录
neverlandly
博客园    首页    新随笔    联系   管理    订阅  订阅

Leetcode: Number of Connected Components in an Undirected Graph

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.

Use Union Find to figure out the number of connected components

 1 public class Solution {
 2     public int countComponents(int n, int[][] edges) {
 3         unionFind uf = new unionFind(n);
 4         for (int[] edge : edges) {
 5             if (!uf.isConnected(edge[0], edge[1])) {
 6                 uf.union(edge[0], edge[1]);
 7             }
 8         }
 9         return uf.findCount();
10     }
11     
12     public class unionFind{
13             int[] ids;
14             int count;
15             
16             public unionFind(int num) {
17                 this.ids = new int[num];
18                 for (int i=0; i<num; i++) {
19                     ids[i] = i;
20                 }
21                 this.count = num;
22             }
23             
24             public int find(int i) {
25                 return ids[i];
26             }
27             
28             public void union(int i1, int i2) {
29                 int id1 = find(i1);
30                 int id2 = find(i2);
31                 if (id1 != id2) {
32                     for (int i=0; i<ids.length; i++) {
33                         if (ids[i] == id2) {
34                             ids[i] = id1;
35                         }
36                     }
37                     count--;
38                 }
39             }
40             
41             public boolean isConnected(int i1, int i2) {
42                 return find(i1)==find(i2);
43             }
44             
45             public int findCount() {
46                 return count;
47             }
48         }
49 }

 

posted @ 2015-12-30 13:04  neverlandly  阅读(2903)  评论(0)    收藏  举报
刷新页面返回顶部
博客园  ©  2004-2025
浙公网安备 33010602011771号 浙ICP备2021040463号-3