LeetCode|Friend Circles

最新推荐文章于 2025-01-02 19:21:53 发布

DouMiaoO_Oo

最新推荐文章于 2025-01-02 19:21:53 发布

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分类专栏： LeetCode 文章标签：并查集

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博客围绕求解学生朋友圈数量问题展开，给定表示学生朋友关系的N*N矩阵M，若M[i][j]=1则第i和第j个学生为直接朋友。通过并查集（Union - find Sets）的思路来计算所有学生中的朋友圈总数，并给出了示例及相关注意事项。

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Friend Circles

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：并查集 (Union-find Sets)
参考博客：http://www.cnblogs.com/cyjb/p/UnionFindSets.html


class Solution {
public:
    int findCircleNum(vector<vector<int>>& M) {
        if(M.size() == 0) return 0;
        parent = vector<int>(M.size());
        rank = vector<int>(M.size(), 0);
        // make_set
        for(int i = 0; i < M.size(); ++i) parent[i] = i;
        for(int i = 0; i < M.size(); ++i){
            for(int j = 0; j < M.size(); ++j){
                if(M[i][j]) union_set(i, j);
            }
        } int res = 0;
        for(int i = 0; i < M.size(); ++i) {
            if(i == find(i))
                ++res;
        } return res;
    }
private:
    int find(int x){
        if(x != parent[x]) parent[x] = find(parent[x]);
        return parent[x];
    }
    void union_set(int x, int y){
        if((x = find(x)) == (y = find(y))) return;
        if(rank[x] > rank[y]) {
            parent[y] = x;
        }
        else{
            parent[x] = y;
            if(rank[y] == rank[x]){
                ++rank[y];
            }
        }
    }
    vector<int> parent, rank;
};