Givenn 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
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1 --- 2 4Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], return 2.
Example 2:
0 4
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1 --- 2 --- 3Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [3, 4]], return 1.
Note:
You can assume that no duplicate edges will appear inedges. Since all edges are undirected,[0, 1] is the same as[1, 0]and thus will not appear together inedges.
Solution:
class Solution {
public int countComponents(int n, int[][] edges) {
if (n == 0 || edges == null || edges.length == 0)
return 0;
if (n == 1)
return n;
Map<Integer, ArrayList<Integer>> graph = new HashMap<>();
for (int i = 0; i < n; i++) {
graph.put(i, new ArrayList<>());
}
for (int[] edge : edges) {
graph.get(edge[0]).add(edge[1]);
graph.get(edge[1]).add(edge[0]);
}
// BFS using queue
Set<Integer> visited = new HashSet<>();
Queue<Integer> queue = new LinkedList<>();
int count = 0;
for (int i = 0; i < n; i++) {
if (!visited.contains(i)) {
count++;
queue.offer(i);
while (!queue.isEmpty()) {
int cur = queue.poll();
visited.add(cur);
for(int next : graph.get(cur)) {
if (!visited.contains(next)) {
queue.offer(next);
}
}
}
}
}
return count;
}
}