Leetcode 547. Friend Circles

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本文介绍了一种算法，用于计算学生之间的社交圈子数量。通过输入一个N×N的关系矩阵，该算法能够找出所有直接或间接的朋友关系，并计算出总的社交圈子数目。

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.

Union Find

public int findCircleNum(int[][] M) {
        int len = M.length;
        int[] root = new int[len];
        int cnt = 0;
        for (int i = 0; i < len; i++) {
            root[i] = i;
        }
        for (int i = 0; i < len; i++) {
            for (int j = i + 1; j < len; j++) {
                if (M[i][j] == 1) {
                    union(root, i, j);
                }
            }
        }
        for (int i = 0; i < len; i++) {
            if (root[i] == i) cnt++;
        }
        return cnt;
        
    }
    
    private void union(int[] root, int i, int j) {
        int _i = find(root, i);
        int _j = find(root, j);
        if (_i != _j) root[_i] = _j;
    }
    
    private int find(int[] root, int idx) {
        if (root[idx] == idx) return idx;
        return find(root, root[idx]);
    }

public void dfs(int[][] M, int[] visited, int i) {
        for (int j = 0; j < M.length; j++) {
            if (M[i][j] == 1 && visited[j] == 0) {
                visited[j] = 1;
                dfs(M, visited, j);
            }
        }
    }
    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 int findCircleNum(int[][] M) {
        int[] visited = new int[M.length];
        int count = 0;
        Queue < Integer > queue = new LinkedList < > ();
        for (int i = 0; i < M.length; i++) {
            if (visited[i] == 0) {
                queue.add(i);
                while (!queue.isEmpty()) {
                    int s = queue.remove();
                    visited[s] = 1;
                    for (int j = 0; j < M.length; j++) {
                        if (M[s][j] == 1 && visited[j] == 0)
                            queue.add(j);
                    }
                }
                count++;
            }
        }
        return count;
    }