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.
题目含义：查找一个班级里面有几个独立的朋友圈
思路：
声明一个visited，用于记录遍历过的结点。每次dfs找到一个原矩阵为1的位置（除了左上到右下的对角线），就把这个位置的列数变成行数再dfs，如果是在同一个圈里，最终会绕回已经遍历过的行，visited为true，return 0；如果不是同一个圈，则增加1。（mat[i][j]==1这个判断相当于i的邻接点，深度优先遍历）
1 private int friendCircle(int[][] M,int row,boolean[] visited) 2 { 3 if (visited[row]) return 0; 4 visited[row] = true; 5 int count=1; 6 for (int j=0;j<M.length;j++) 7 { 8 if (j!=row && M[row][j] == 1) 9 { 10 count += friendCircle(M,j,visited); 11 } 12 } 13 return count; 14 } 15 16 public int findCircleNum(int[][] M) { 17 if (M.length==0) return 0; 18 int count=0; 19 boolean[] visited = new boolean[M.length]; 20 for (int i=0;i<M.length;i++) 21 { 22 if (friendCircle(M,i,visited)>0) count++; 23 } 24 return count; 25 }