Friend Circles

最新推荐文章于 2020-11-18 21:50:03 发布

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本文解析了LeetCode上一道关于朋友关系的问题，利用深度优先搜索（DFS）算法来找出学生之间的朋友圈数量。通过遍历矩阵并使用递归方法确定直接或间接的朋友关系。

题目来源LeetCode
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.

这道题其实是图论问题，求图的连通分量，通过DFS来遍历朋友，代码如下：

void Findfriend(vector<vector<int>>& M,bool judge[],int m)
     {
         judge[m] = true;
         for (int i = 0; i < M.size();i++)
         {
             if (judge[i] == false &&  M[m][i]==1)  Findfriend(M,judge, i);
         }
     }

class Solution {
public:
     int findCircleNum(vector<vector<int>>& M)
     {
         int row = M.size();
         if (row == 0)return 0;         
         bool judge[row+1];
         memset(judge,false,sizeof(judge));
         int cir_num = 0;
         for (int i = 0; i < row;i++)
         {
             if (judge[i]==false)
             {
                 Findfriend(M, judge, i);
                 cir_num++;
             }
         }
         return cir_num;
     }
};