No.547. Friend Circles

最新推荐文章于 2024-05-29 15:55:53 发布

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最新推荐文章于 2024-05-29 15:55:53 发布

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文章标签： LeetCode DFS 算法

本文链接：https://blog.youkuaiyun.com/qiaosen20/article/details/73695109

本文介绍了一种使用深度优先搜索（DFS）算法来计算学生间直接和间接朋友关系组成的社交圈数量的方法。通过给出的矩阵表示学生之间的友谊关系，利用DFS遍历每个未访问的学生节点，寻找所有相连的朋友并标记已访问，以此来确定社交圈的数量。

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一、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.

二、主要思想

运用dfs遍历一个点所能连接到的所有点，如果遇到一个没有访问过的点，计数器加一，同时将与该点所连接的所有点都找到，并标记为访问过的点。

三、代码实现

class Solution {
private:
    vector<bool> visit;
    void dfs(vector<vector<int>>& M,int num){
        if(visit[num]==false){
            visit[num]=true;
            for(int i=0;i<M.size();i++){
                if(M[num][i]==1)
                    dfs(M,i);
            }
        }
    }
public:
    int findCircleNum(vector<vector<int>>& M) {
        visit.resize(M.size(),false);
        int count=0;
        for(int i=0;i<M.size();i++){
            if(visit[i]==false){
                count++;
                dfs(M,i);
            }
        }
        return count;
    }
};