leetcode -- 547. Friend Circles

最新推荐文章于 2020-11-24 14:57:32 发布

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最新推荐文章于 2020-11-24 14:57:32 发布

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分类专栏： leetCode 反复看文章标签：并查集

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本文介绍了一种使用并查集算法解决朋友圈数量问题的方法。在一个班级中，有些学生是朋友，有些则不是，但友谊具有传递性。通过构建一个N*N的矩阵来表示学生之间的直接友谊关系，我们可以确定班级中有多少个独立的朋友圈。文章详细解释了如何使用并查集进行查找和合并操作，以找出所有学生中朋友圈的总数。

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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.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/friend-circles
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

我的思路：

1、并查集做，并查集查找和合并

2、刚开始，每个学生自成朋友圈

3、两两学生一起，如果是diret friend，则合并他们，即他们的祖先相同

4、再遍历一遍father数组，找到原始祖先，就是朋友圈个数

class Solution {
    
public:
    int findCircleNum(vector<vector<int>>& M) {
        int n = M.size();  //班级人数
        vector<int> father(n);
        for(int i = 0; i < n; i++){
            father[i] = i;
        }
        
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                if(M[i][j] == 1){
                    Union(i,j,father);
                }
            }
        }
       int res = 0;
        for(int i = 0; i < n; i++){
            if(i == father[i]){
                res++;
            }
        }
        return res;
    }
    
    int findFather(int x,vector<int>& father){
        while(x != father[x]){
            x = father[x];
        }
        return x;
    }
    
    void Union(int a, int b,vector<int>& father){
        int fa = findFather(a,father);
        int fb = findFather(b,father);
        if(fa != fb){
            father[fa] = fb;
        }
    }
};