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
代码语言:javascript复制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。
代码语言:javascript复制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);
}
}
}
};