Leetcode 323. 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.

题目要我们找到一个graph 里 互相连接的component 的个数

1、使用 adjacent list 来表示一个图

2、遍历每个 节点 从1到N

3、DFS (深度优先搜索)， 用一个boolean array 来记录访问的节点

public class Solution {
    public int countComponents(int n, int[][] edges) {
        if (n <= 1) return n;
        
        List> adj_list = new ArrayList<>();
        
        for (int i = 0; i < n; i++) {
            adj_list.add(new ArrayList());
        }
        
        for (int[] edge : edges) {
            adj_list.get(edge[0]).add(edge[1]);
            adj_list.get(edge[1]).add(edge[0]);
        }
        
        boolean[] visited = new boolean[n];
        
        int count = 0;
        
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                count++;
                dfs(visited, i, adj_list);
            }
        }
        
        return count;
    }
    
    private void dfs(boolean[] visited, int index, List> adj_list) {
        visited[index] = true;
        
        for (int i : adj_list.get(index)) {
            if (!visited[i]) dfs(visited, i, adj_list);
        }
    }
}