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

类似200. Number of Islands, 但是lc考察的是题意的转化, 而不是光会个框架

考察: 对邻居的理解, 什么才是真正的邻居, 对邻居的遍历

dfs
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;
}
Bfs
public int findCircleNum(int[][] M) {
    int count = 0;
    for (int i=0; i<M.length; i++)
        if (M[i][i] == 1) { count++; BFS(i, M); }
    return count;
}

public void BFS(int student, int[][] M) {
    Queue<Integer> queue = new LinkedList<>();
    queue.add(student);
    while (queue.size() > 0) {
        int queueSize = queue.size();
        for (int i=0;i<queueSize;i++) {
            int j = queue.poll();
            M[j][j] = 2; // marks as visited
            for (int k=0;k<M[0].length;k++) 
                if (M[j][k] == 1 && M[k][k] == 1) queue.add(k);
        }
    }
}

　UF 面试常考

Union find
public class Solution {
   
        class UF {
            private int[] id;
            private int count;
            public UF(int N) {
                count = N;
                id = new int[N];
                for (int i = 0; i < N; i++) {
                    id[i] = i;
                }
            }
            public int count() {
                return count;
            }
            public int find(int p) {
                while (p != id[p]) p = id[p];
                return p;
            }
            public void union(int p, int q) {
                int pRoot = find(p);
                int qRoot = find(q);
                if (pRoot == qRoot) return;
                id[pRoot] = qRoot;
                count--;
            } 
        }
        
    public int findCircleNum(int[][] M) {
        int n = M.length;
        UF uf = new UF(n);
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                if (M[i][j] == 1) uf.union(i, j);
            }
        }
        return uf.count();
    }
}