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.

Analysis:

From here: https://discuss.leetcode.com/topic/32752/easiest-2ms-java-solution

This is 1D version of Number of Islands II. For more explanations, check out this 2D Solution.

1. n points = n islands = n trees = n roots.
2. With each edge added, check which island is e[0] or e[1] belonging to.
3. If e[0] and e[1] are in same islands, do nothing.
4. Otherwise, union two islands, and reduce islands count by 1.
5. Bonus: path compression can reduce time by 50%.

Hope it helps!

Solution:

 1 public class Solution {
 2     public int countComponents(int n, int[][] edges) {
 3         int[] roots = new int[n];
 4         for (int i=0;i<n;i++) roots[i] = i;
 5         int num = n;
 6
 7         int len = edges.length;
 8         for (int i=0;i<len;i++){
 9             int root1 = findRoot(roots,edges[i][0]);
10             int root2 = findRoot(roots,edges[i][1]);
11
12             if (root1!=root2){
13                 roots[root1] = root2;
14                 num--;
15             }
16         }
17
18         return num;
19     }
20
21     public int findRoot(int[] roots, int node){
22         while (roots[node]!=node){
23             roots[node] = roots[roots[node]];
24             node = roots[node];
25         }
26         return node;
27     }
28 }