LeetCode 547. Friend Circles

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 Mj = 1.

大致题意

Mi = 1 表示i和j是直接朋友，如果A和B是朋友，B和C是朋友，那么A和C是间接朋友。
根据所给关系划分出共有多少个朋友圈。

题解

题目类型：DFS

直接对每个人迭代，一个人最多拥有一个朋友圈，直接dfs验证其是否拥有朋友圈，有的话从节点中删除（不再查找此人）

代码如下

class Solution {
public:
    void dfs(vector<int> & v, vector<vector<int>>& M, int line) {
        for (int i = 0; i < M[line].size(); i++) {
            if (M[line][i] == 1 && v[i] == 0) {
                v[i] = 1;
                dfs(v, M, i);
            }
        }
    }
    int findCircleNum(vector<vector<int>>& M) {
        vector<int> v(M.size(), 0);
        int count = 0;
        for (int i = 0; i < M.size(); i++) {
            if (v[i] == 0) {
                
                dfs(v, M, i);
                count++;
            }
        }
        return count;
    }
};