[LeetCode] 547. Friend Circles

题目链接: https://leetcode.com/problems/friend-circles/description/

Description

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

解题思路

题意中朋友圈定义为包括所有朋友还有朋友的朋友、朋友的朋友的朋友……不管直接间接都属于同个朋友圈，问给出的关系矩阵中共有几个朋友圈。对于这道题，最直接想到的就是DFS了，而且这道题不需要知道哪些人属于哪个朋友圈，所以只用一个数组来判断是否访问过就可以了，具体实现也很简单，不过多赘述。

Code

class Solution {
public:
    int findCircleNum(vector<vector<int>>& M) {
        vector<bool> visited(M.size(), false);

        int count = 0;
        for (int i = 0; i < M.size(); ++i) {
            if (!visited[i]) {
                visited[i] = true;
                count++;
                dfs(i, visited, M);
            }
        }

        return count;
    }

    void dfs(int pos, vector<bool>& visited, vector<vector<int>>& M) {
        for (int i = 0; i < M.size(); ++i) {
            if (M[pos][i] == 1 && !visited[i]) {
                visited[i] = true;
                dfs(i, visited, M);
            }
        }
    }
};