# Clean C++ BFS solution

• ``````class Solution {
public:
int countComponents(int n, vector<pair<int, int>>& edges) {
int ret = 0;
vector<bool> visited(n, false);
unordered_map<int, vector<int>> graph = make_graph(edges);
for ( int i = 0; i < n; i++ ) {
if (!visited[i]) {
bfs(i, graph, visited);
ret++;
}
}
return ret;
}
private:
unordered_map<int, vector<int>> make_graph(vector<pair<int, int>>& edges) {
unordered_map<int, vector<int>> ret;
for ( int i = 0; i < edges.size(); i++ ) {
int key = edges[i].first;
int neigh = edges[i].second;
ret[key].push_back(neigh);
ret[neigh].push_back(key);
}
return ret;
}

void bfs(int node, unordered_map<int, vector<int>>& graph, vector<bool>& visited) {
visited[node] = true;
queue<int> Q;
Q.push(node);
while(!Q.empty()) {
int key = Q.front();
Q.pop();
for ( int i = 0; i < graph[key].size(); i++ ) {
int cur = graph[key][i];
if (!visited[cur]) {
visited[cur] = true;
Q.push(cur);
}
}
}
}
};
``````

Explanation:
Store edges into a hash table for easy finding. Becasue it is an undirected graph, therefore, storing a edge in both node lists. For example, (1,0). In the hash table, `key: 1` has element `0`; and `key: 0` has element `1`.
And rest of code follows the basic bfs structure.

One bfs() can find all connected nodes. And the times of bfs() called is the final answers.

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