547. Friend Circles | LeetCode 解题思路

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.

题目解读

朋友圈即同学关系图中的连通分量，题目要求找到图中连通分量的个数，因此需要使用深度优先搜索算法（DFS）找出连通分量。声明一个bool数组visited记录每个点是否被遍历，题目中的图用二维数组M存储，所以visited的大小即为M的行数，对每个点进行DFS，查找一遍代表该点的行，每次找到一个与其相连的点（即矩阵该行中数值为1的列数），则对该点所在行再进行DFS，若在同一连通分量内，则visited为TRUE，dfs函数最终返回0，如果不在同一连通分量内（即dfs返回值大于0），则连通分量数增加1）。

class Solution {public:    int dfs(vector<vector<int>>& M, bool *visited, int ind){        if(visited[ind] == 1) return 0;        else{            visited[ind] = 1;            int ccnum = 1;            for(int j = 0; j < M.size(); j++){                if(ind != j && M[ind][j] == 1){                    ccnum += dfs(M, visited, j);                }            }            return ccnum;        }    }    int findCircleNum(vector<vector<int>>& M) {        int size = M.size();        if(size == 0) return 0;        else{            bool visited[size];            memset(visited, 0, size);            int cnum = 0;            for(int i = 0; i < size; i++){                if(dfs(M, visited, i) > 0)                cnum++;            }            return cnum;        }    }};