[LeetCode] 547. Friend Circles

題目鏈接: 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.

解題思路

題意中朋友圈定義爲包括所有朋友還有朋友的朋友、朋友的朋友的朋友……不管直接間接都屬於同個朋友圈，問給出的關係矩陣中共有幾個朋友圈。對於這道題，最直接想到的就是DFS了，而且這道題不需要知道哪些人屬於哪個朋友圈，所以只用一個數組來判斷是否訪問過就可以了，具體實現也很簡單，不過多贅述。

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);
            }
        }
    }
};