547. Friend Circles

最新推荐文章于 2019-05-09 21:25:08 发布

BigFatSheep

最新推荐文章于 2019-05-09 21:25:08 发布

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分类专栏： Leetcode Medium

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本文介绍了一种使用深度优先搜索（DFS）算法来计算给定学生群体中朋友圈数量的方法。通过遍历二维数组，该算法能够有效地识别并计算出独立的朋友圈数目。

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

Problem URL

Solution

给一个二维数组，表示人与人之间的朋友关系，找到所有的朋友圈数量。使用DFS，每次找到所有圈子里的人，记录下来。

For this problem, we could use DFS(depth first search). We could use an array to record this people is visited or not. Then using a for loop to DFS each people. If this people has been visited, it has been in another friend circle.

Code

class Solution {
    public int findCircleNum(int[][] M) {
        int[] visited = new int[M.length];
        int count = 0;
        for (int i = 0; i < M.length; i++){
            if (visited[i] == 0){
                dfs(M, visited, i);
                count++;
            }
        }
        return count;
    }
    
    public void dfs(int[][] M, int[] visited, int i){
        for (int j = 0; j < M.length; j++){
            if (visited[j] == 0 && M[i][j] == 1){
                visited[j] = 1;
                dfs(M, visited, j);
            }
        }
    }
}