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 M[j][i] = 1.

题解

找有几组小朋友，每组或直接或间接认识。

解法1

并查集，开始假设每个人都没有朋友，组的个数为n，之后每次合并减1

public class Solution {    public int findCircleNum(int[][] M) {        int n = M.length;        //建立并查集        int[] root = new int[n];        for(int i = 0; i < n; i++){            root[i] = i;        }        int count = n;        for(int i = 0; i < n; i++){            for(int j = 0; j < i; j++){                int ti = i, tj = j;                if(M[ti][tj] == 1){                    while(root[ti] != ti)   ti = root[ti];                    while(root[tj] != tj)   tj = root[tj];                    if(ti == tj)    continue;                    //合并                    root[ti] = tj;                    count--;                }            }        }        return count;    }}

解法2

DFS

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