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.

Example 1:

Input: 
[[1,1,0],
 [1,1,0],
 [0,0,1]]
Output: 2
Explanation:The 0th and 1st students are direct friends, so they are in a friend circle. 
The 2nd student himself is in a friend circle. So return 2.

Example 2:

Input: 
[[1,1,0],
 [1,1,1],
 [0,1,1]]
Output: 1
Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends, 
so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.

Note:

1. N is in range [1,200].

2. M[i][i] = 1 for all students.

3. If M[i][j] = 1, then M[j][i] = 1.

一开始理解错了题意，当成了类似于找岛屿数那道题了，其实是不对的，这道题给的矩阵，其实是和找连通图的数量那道题是一样的，比如在0行这里，找到除了0列以外所有是1的纵坐标，然后放到queue里，循环

时间复杂度 o(n^2), space o(n)

class Solution {
    public int findCircleNum(int[][] M) {
        if(M == null || M.length == 0|| M[0].length== 0)
            return 0;
        
        int size = M.length;
        
        boolean[] visited = new boolean[size];
        Queue<Integer> queue = new LinkedList<>();
        int res = 0;
        for(int i = 0; i < size; i++){
            if(!visited[i]){
                queue.offer(i);
                visited[i] = true;
                while(!queue.isEmpty()){
                    int cur = queue.poll();
                    
                    for(int j = 0; j < size; j++){
                        if(!visited[j] && M[cur][j] == 1){
                            queue.offer(j);
                            visited[j] = true;
                        }
                    }
                }
                
                res++;
            }
            
            
        }
        
        return res;
    }
}