Friend Circles问题及解法

问题描述：

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.

示例:

Input: [[1,1,0], [1,1,0], [0,0,1]]Output: 2Explanation: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.

Input: [[1,1,0], [1,1,1], [0,1,1]]Output: 1Explanation: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:

1. N is in range [1,200].
2. M[i][i] = 1 for all students.
3. If M[i][j] = 1, then M[j][i] = 1.

问题分析：

矩阵M是个对称矩阵，根据Union Find数据结构可以求解此类问题。

过程详见代码：

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;int group = n;for (int i = 0; i < n; i++){for (int j = i + 1; j < n; j++){if (M[i][j]){int lead1 = find(i,leads);int lead2 = find(j, leads);if (lead2 != lead1){leads[lead2] = lead1;group--;}}}}return group;}private:int find(int x, vector<int>& parents){return parents[x] == x ? x : find(parents[x], parents);}};