leetcode Friend Circles

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 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:

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.

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;
    }
};