Leetcode -- 547. Friend Circles(dfs)

547. Friend Circles(dfs)
medium
https://leetcode.com/problems/friend-circles/#/description

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 directfriends 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.

思路：- define visited array for each person, mark all ppl in one cycle visited using dfs
           - traverse the he specific row, if get 1, traverse the corresponding row of that column #
             since 2's friend is considered indirect friend of 1
           - once finishing visit a 1 cycle, increment circle res

Complexity: Time O(N^2) -- n cycle  Space O(1)

public class Solution {
    public int findCircleNum(int[][] M) {
        boolean visited[] = new boolean[M.length];
        int cycle = 0;
        for(int i = 0; i < M.length; i ++){
            if(!visited[i]){
                cycle++;
                visited[i] = true;
                dfs(M, visited, i);//dfs to mark all ppl in the cycle visited
            }
        }
        return cycle;
    }
    public void dfs(int[][] M, boolean[] visited, int i){
        for(int j = 0; j < M[0].length; j++){//no return case here!!
            if(!visited[j] && M[i][j] == 1){
                visited[j] = true;
                dfs(M, visited, j);
            }
        }
    }
}