547. Friend Circles

最新推荐文章于 2020-02-21 08:56:52 发布

转载最新推荐文章于 2020-02-21 08:56:52 发布 · 148 阅读

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本文介绍了一种通过深度优先搜索（DFS）与并查集（Union Find）算法来解决社交网络中学生群体划分的问题。输入为一个表示学生间直接友谊关系的矩阵，输出则是社交圈的总数。文中提供了两种实现方案：一种利用DFS遍历每个节点，另一种采用并查集进行分组合并。

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 aredirect 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

DFS

class Solution {
public:
    int findCircleNum(vector<vector<int>>& M) {
        if (M.empty()) return 0;
        int n = M.size();
        vector<bool> visited(n, false);
        int groups = 0;
        for (int i = 0; i < visited.size(); i++) {
            groups += !visited[i] ? dfs(i, M, visited), 1 : 0;
        }
        return groups;
    }

private:
    void dfs(int i, vector<vector<int>>& M, vector<bool>& visited) {
        visited[i] = true;
        for (int j = 0; j < visited.size(); j++) {
            if (i != j && M[i][j] && !visited[j]) {
                dfs(j, M, visited);
            }
        }
    }
};

UnionFind

class Solution {
public:
    int findCircleNum(vector<vector<int>>& M) {
        if (M.empty()) return 0;
        int n = M.size();

        vector<int> leads(n, 0);
        for (int i = 0; i < n; i++) { leads[i] = i; }   // initialize leads for every kid as themselves

        int groups = n;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {   // avoid recalculate M[i][j], M[j][i]
                if (M[i][j]) {
                    int lead1 = find(i, leads);
                    int lead2 = find(j, leads);
                    if (lead1 != lead2) {       // if 2 group belongs 2 different leads, merge 2 group to 1
                        leads[lead1] = lead2;
                        groups--;
                    }
                }
            }
        }
        return groups;
    }

private:
    int find(int x, vector<int>& parents) {
        return parents[x] == x ? x : find(parents[x], parents);
    }
};