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 }