Friend Circles(dfs)——LeetCode进阶路

原题链接https://leetcode.com/problems/friend-circles/

题目描述

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:

N is in range [1,200].
M[i][i] = 1 for all students.
If M[i][j] = 1, then M[j][i] = 1.

思路分析

遍历小伙伴的朋友圈，并进行标记，依次遍历为标记过的小伙伴，并对遍历的环的个数计数。
其实是一道经典的并查集，但是想到蓝桥杯dfs比较多，笔者依然坚强的用了dfs……

贴一下大神的并查集博客，当年的入门利器，爱看武林外传的阿猿不会太差~
并查集入门篇https://www.cnblogs.com/xzxl/p/7226557.html
并查集进阶篇 http://www.cnblogs.com/xzxl/p/7341536.html

AC解

class Solution {
    public int findCircleNum(int[][] M) {
        if(M == null || M.length == 0 || M[0].length == 0)
        {
            return 0;
        }

        boolean[] flag = new boolean[M.length];
        int result = 0;

        for(int i=0;i<M.length;i++)
        {
            if(!flag[i])
            {
                result ++;
                dfs(M,i,flag);
            }
        }

        return result;
    }
    public void dfs(int[][] M,int i,boolean[] flag)
    {
        flag[i] = true;

        for(int j=0;j<M.length;j++)
        {
            if(M[i][j] == 1 && !flag[j])
            {
                dfs(M,j,flag);
            }
        }
    }
}