Friend Circles

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 jthstudents 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.

思路：刚开始以为这题跟Number of Island 一样，后来测试之后发现不是一个东西。比如

[

1 0 0 1

0 1 1 0

0 1 1 0

1 0 1 1

]

如果number of island 返回4，然而这题返回1.也就是说，number island是探索孤立的岛屿，而这题是朋友圈，比如1，3认识，3跟2认识，2跟1认识，他们所有人的都在一个朋友圈里面，相互传染了，这题就是为了union find设计的。我参考了princeton的weighted-quick-union算法，2ms解决。memory beat 93%. 注意union的时候，应该是parent[rootP] = rootQ, 而不是p,q.

class Solution {
    
    public class UnionFind {
        private int[] parent;
        private int[] size;
        private int count;
        
        public UnionFind(int n) {
            count = n;
            parent = new int[n];
            size = new int[n];
            for(int i=0; i<n; i++){
                parent[i] = i;
                size[i] = 1;
            }
        }
        
        public int findRoot(int i){
            while(i!=parent[i]){
                i=parent[i];
            }
            return i;
        }
        
        public boolean connected(int p, int q){
            return findRoot(p) == findRoot(q);
        }
        
        public int count(){
            return count;
        }
        
        public void union(int p, int q){
            int rootP = findRoot(p);
            int rootQ = findRoot(q);
            if(rootP == rootQ) return;
            
            if(size[rootP] < size[rootQ]){
                parent[rootP] = rootQ;
                size[rootQ] += size[rootP];
            } else {
                // size[rootP] >= size[rootQ];
                parent[rootQ] = rootP;
                size[rootP] += size[rootQ];
            }
            count--;
        }
    }
    
    public int findCircleNum(int[][] M) {
        if(M == null || M.length == 0) return 0;
        int n = M.length;
        UnionFind unionFind = new UnionFind(n);
        for(int i=0; i<M.length; i++){
            for(int j=0; j<M[0].length; j++){
                if(M[i][j] == 1){
                    unionFind.union(i,j);
                }
            }
        }
        return unionFind.count;
    }
}