323. Number of Connected Components in an Undirected Graph
Problem:
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.
Solutions:
public class Solution {
public int countComponents(int n, int[][] edges) {
HashMap<Integer, List<Integer>> adj = new HashMap<Integer, List<Integer>>();
init(adj, edges, n);
boolean[] visited = new boolean[n];
int count = 0;
for (int i = 0; i < n; i ++) {
if (visited[i] == false) {
count ++;
dfs(adj, i, visited);
}
}
return count;
}
private void dfs(HashMap<Integer, List<Integer>> adj, int index, boolean[] visited) {
visited[index] = true;
for (Integer j:adj.get(index)) {
if (visited[j] == false) {
dfs(adj, j, visited);
}
}
}
private void init(HashMap<Integer, List<Integer>> adj, int[][] edges, int n) {
for (int i = 0; i < n; i ++) {
adj.put(i, new LinkedList<Integer>());
}
for (int i = 0; i < edges.length; i ++) {
adj.get(edges[i][0]).add(edges[i][1]);
adj.get(edges[i][1]).add(edges[i][0]);
}
}
}
public class Solution {
public int countComponents(int n, int[][] edges) {
int[] root = new int[n];
for (int i = 0; i < n; i ++) {
root[i] = i;
}
int count = n;
for (int i = 0; i < edges.length; i ++) {
int r1 = getRoot(root, edges[i][0]);
int r2 = getRoot(root, edges[i][1]);
if (r1 != r2) {
root[r1] = r2;
count --;
}
}
return count;
}
private int getRoot(int[] root, int i) {
while (root[i] != i) {
root[i] = root[root[i]];
i = root[i];
}
return i;
}
}