323. Number of Connected Components in an Undirected Graph

问题描述：

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:

Input: n = 5 and edges = [[0, 1], [1, 2], [3, 4]]

     0          3
     |          |
     1 --- 2    4 

Output: 2

Example 2:

Input: n = 5 and edges = [[0, 1], [1, 2], [2, 3], [3, 4]]

     0           4
     |           |
     1 --- 2 --- 3

Output:  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.

 

解题思路：

建立邻接链表

开始准备dfs该图

首先判断改顶点是否被访问过

若被访问过，则跳过。

若没有被访问过，则从这个顶点开始dfs，然后遍历，对遍历到的每一个点，设其为被访问过。

 

需要注意的是：要将每个顶点都加入其对应的邻接连表

for(auto e : edges){
            adj_list[e.first].push_back(e.second);
            adj_list[e.second].push_back(e.first);
        }

 

代码：

class Solution {
public:
    int countComponents(int n, vector<pair<int, int>>& edges) {
        vector<vector<int>> adj_list(n);
        for(auto e : edges){
            adj_list[e.first].push_back(e.second);
            adj_list[e.second].push_back(e.first);
        }
        vector<bool> visited(n, false);
        int ret = 0;
        for(int i = 0; i < n; i++){
            if(visited[i])
                continue;
            visited[i] = true;
            dfs(adj_list, i, visited);
            ret++;
        }
        return ret;
    }
private:
    void dfs(vector<vector<int>> &adj, int node, vector<bool> &visited){
        for(int i = 0; i < adj[node].size(); i++){
            if(visited[adj[node][i]])
                continue;
            visited[adj[node][i]] = true;
            dfs(adj, adj[node][i], visited);
        }
    }
};