leetcode第十六周解题总结--图

547. Friend Circles

There are N students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a direct friend of B, and B is a direct friend of C, then A is an indirect friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.

Given a N*N matrix M representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jth students are direct friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.

题意解析： 求一个无向图中的连通分量。

解题思路： 从一个节点开始搜索，若还存在无法访问的节点，则加一个连通分量，并从该点继续进行搜索，一直进行，直到所有节点都被访问。

广度优先搜索BFS

class Solution {
public:
int findCircleNum(vector<vector<int>>& M) {
int n = M.size();
vector<int> a(n);
int count = 0;

stack<int> g;
g.push(0);
a[0] = 1;
while(g.size()!=0) {
int i = g.top();
g.pop();
for(int j = 0; j < n; j++) {
if(M[i][j] == 1 && a[j] != 1) {
a[j] = 1;
g.push(j);
}
}

if(g.size() == 0){
count ++;
for(int p = 0; p < n; p ++) {
if(a[p] == 0) {
g.push(p);
a[p] = 1;
break;
}
}
}
}
return count;
}
};

深度优先搜索DFS

class Solution {
public:
int findCircleNum(vector<vector<int>>& M) {
int n = M.size();
vector<int> a(n);
int count = 0;

for(int i = 0; i < n; i++) {
if(a[i] == 0) {
dfs(i,M,a);
count ++;
}

}

return count;

}

void dfs(int i,vector<vector<int>>& M, vector<int>& a){
a[i] = 1;
for(int j = 0; j < a.size(); j++) {
if(M[i][j] == 1 && a[j] == 0){
dfs(j,M,a);
}
}
}

};