leetcode Friend Circles

最新推荐文章于 2019-07-11 21:45:00 发布

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#c++ #算法 #leetcode #并查集

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本文介绍了一种使用并查集算法来解决寻找朋友圈数量的问题。通过构建并查集，将矩阵中的直接朋友关系进行合并，最终统计不同朋友圈的数量。

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Friend Circles

题目详情：

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 i_th and j_th 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 0_th and 1_st students are direct friends, so they are in a friend circle. 
The 2_nd 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 0_th and 1_st students are direct friends, the 1_st and 2_nd students are direct friends, 
so the 0_th and 2_nd 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.

解题方法：

这是一道经典的并查集问题。题目就是寻找朋友圈的个数。把它转化为并查集后就可以理解为寻找parent的个数。首先构造parent数组，M中的每个元素的parent'都是它本身。然后两个人如果是朋友，分别寻找这两个人的parent，如果不同，则其中一个人parent的parent被赋值为另一个人的parent。然后将每个人的parent插入到一个set中，set的大小即为朋友圈的个数，也就是不同的parent的个数。

代码详情：

class Solution {
public:
    int findCircleNum(vector<vector<int> >& M) {
        vector<int> parent;
        int n = M.size();
        for (int i = 0; i < n; i++) {
           parent.push_back(i);
       }
        for (int i = 0; i < n; i++) {
            for (int j = i+1; j < n; j++) {
               if ( M[i][j] == 1) {
                         int x = find(i, parent);
                         int y = find(j, parent);
                         if (x != y)
                    parent[x] = y;
               }
            }
        }
        set<int> result;
        for (int i = 0; i < n; i++) {
           //cout << i << " parent: " << parent[i] << endl;
            result.insert(find(i, parent));
        }
        return result.size();
    }
    int find(int tar, vector<int>& parent) {
        while (parent[tar] != tar) {
            tar = parent[tar];
        }
        return tar;
    }
};