1. 题目

问有几个连通网络

2. 解题

2.1 BFS 广度优先

参考图的数据结构

class Solution {
public:int findCircleNum(vector<vector<int>>& M) {int n = M.size(), groups = 0, i;bool visited[n] = {false};for(i = 0; i < n; ++i){if(!visited[i]){bfs(M, visited, n, i);++groups;}}return groups;}void bfs(vector<vector<int>>& M, bool *visited, int &n, int idx) {queue<int> q;q.push(idx);visited[idx] = true;int i, j;while(!q.empty()){i = q.front();q.pop();for(j = 0; j < n; ++j){if(M[i][j] && !visited[j]){visited[j] = true;q.push(j);}}}}
};

2.2 DFS 深度优先

参考图的DFS，还可以不用递归的写法，见前面链接。

class Solution {
public:int findCircleNum(vector<vector<int>>& M) {int n = M.size(), groups = 0, i;bool visited[n] = {false};for(i = 0; i < n; ++i){if(!visited[i]){dfs(M, visited, n, i);++groups;}}return groups;}void dfs(vector<vector<int>>& M, bool *visited, int &n, int idx) {visited[idx] = true;for(int j = 0; j < n; ++j){if(M[idx][j] && !visited[j]){dfs(M, visited, n, j);}}}
};