LeetCode 547: Friend Circles (并查集好题)

547. Friend Circles
     Medium

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.

Constraints:

1 <= N <= 200
M[i][i] == 1
M[i][j] == M[j][i]

解法1：并查集
我一开始把这题当成找岛屿个数那题了。但这题不一样，因为主对角线上面的点表示每个人，其他的点表示人与人之间的关系，两种点不一样，不能混作一团。
我的解法是用并查集，先把ans=n。然后沿着主对角线开始，每行往右找，如果是1，就和主对角线的那个人join，join过程中asn–。最后返回ans。

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

        for (int i = 0; i < n; ++i) {
                father[i] = i;
        }

        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (M[i][j] == 1) {
                    join(i, j);
                }
            }
        }
        
        return ans;
        
   }
private:
    vector<int> father;
    int ans = 0;
    int find(int x) {
        if (father[x] == x) {
            return x;
        }
        return father[x] = find(father[x]);
    }
    
    void join(int x, int y) {
        int fatherX = find(x);
        int fatherY = find(y);
        if (fatherX != fatherY) {
            father[fatherX] = fatherY;
            ans--;
        }
    }
    
};