LeetCode 547. Friend Circles

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#547 朋友圈

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本文介绍了一种算法，用于计算由直接和间接朋友关系构成的朋友圈数量。通过递归搜索每个学生的所有朋友并对其进行标记，最终确定班级中独立朋友圈的总数。

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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.

题目要求找出班级里面朋友环的个数。可从第一个同学开始算，找出他的所有直接朋友，通过直接朋友找到间接朋友，并把所有这些被找到的人进行标记，表示他们为一个朋友环。而后对未被标记的人进行相同的操作，可以得到朋友环的总数。

class Solution {

public:

    void find(vector<vector<int>>& M,vector<int>& num,int i){

        num[i]=1;

        for(int j=0;j<M.size();j++){

            if(j!=i&&M[i][j]==1&&num[j]!=1)

                find(M,num,j);

        }

        return;

    }

    int findCircleNum(vector<vector<int>>& M) {

        if(M.size()==0)

            return 0;

        int count=0;

        vector<int> num(M.size(),0);

        for(int i=0;i<M.size();i++){

            if(num[i]==0){

                count++;

                find(M,num,i);

            }

        }

        return count;

    }

};