### 题目

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.

Note:

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

### 思路

• 第一种 “Union Find”。$O(n^2)$，其中n代表学生的数量。
• 第二种 “DFS”挖出整棵树。$O(n^2\log_{}{n})$，其中n代表学生的数量。

### DFS解法

#### 代码

class Solution {
private boolean[] table = new boolean[0];
private int[][] local = new int[0][0];
private void init(int[][] m) {
table = new boolean[m.length];
local = m;
}
public int findCircleNum(int[][] M) {
init(M);
int count = 0;
for (int i = 0; i <table.length; i++) {
if (!table[i]) {
count++;
dfs(i);
}
}
return count;
}
private void dfs(int student) {
for (int i = 0; i < local.length; i++) {
if ((local[student][i] == 1) && (table[i] == false)) {
table[i] = true;
dfs(i);
}
}
}
}