leetcode.547.friendcircles.朋友圈.题解

Description:

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.

题解:

定义一个搜索过的数组，用来标记被搜索过的数字，不用重复遍历，然后深度优先，一直向下搜索，直到边界；

每搜索一次给count++，结果就是朋友圈的个数；

Code:

class  {
    public int findCircleNum(int[][] M) {
        if (M == null) {
            return 0;
        }
        int[] auth = new int[M.length];
        int count = 0;
        for (int i = 0; i < M.length; i++) {
            if (auth[i] == 0) {
                dfs(M,auth,i);
                count++;
            }
        }
        return count;
    }

    public void dfs(int[][] M,int[] auth,int i) {
        for (int j = 0; j < M.length; j++) {
            if (M[i][j] == 1 && auth[j] == 0) {
                auth[j] = 1;
                dfs(M,auth,j);
            }
        }
    }
}

leetcode.547.friendcircles.朋友圈.题解

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题解:

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