leetcode547. Friend Circles

547. Friend Circles

1、原题

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 directfriends 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: 2Explanation: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: 1Explanation: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.

2、解析及思路

这道题其实比较简单，但是我在做时却遇到了问题；主要是审题不清造成的。这道题是说当M[i][j]为1时，i与j则为fiend，同时这种friend的关系还是可以传递的。然后就给了我们相应的二维数组来表示他们的关系。我的失误是一开始将其理解成了二维数组中有多少个独立的1圈，让我的思路一下子就偏了。

具体思路就直接在代码注释里写了，这里就不多啰嗦啦。

3、代码实现

public class Solution {    public int findCircleNum(int[][] M) {        //标记数组，用于标记该数是否已经遍历过了；int[] circleArray = new int[M.length];//初始计数int circleNum = 0;//遍历一遍数组for (int i = 0; i < M.length; i++) {    //当该数没有被访问过时，计数+1，进入逻辑if (circleArray[i] == 0) {circleNum++;//标记circleArray[i] = 1;//入栈Stack<Integer> s = new Stack<Integer>();s.push(i);//深搜while(!s.isEmpty()) {int current = s.pop();for (int j = 0; j < M.length; j++) {if (M[current][j] == 1 && circleArray[j] == 0) {circleArray[j] = 1;s.push(j);}}}}}return circleNum;}}