Leetcode 547: 省份数量 Number of Provinces

中文描述：

有 n 个城市，其中一些彼此相连，另一些没有相连。如果城市 a 与城市 b 直接相连，且城市 b 与城市 c 直接相连，那么城市 a 与城市 c 间接相连。
省份 是一组直接或间接相连的城市，组内不含其他没有相连的城市。
给你一个 n x n 的矩阵 isConnected ，其中 isConnected[i][j] = 1 表示第 i 个城市和第 j 个城市直接相连，而 isConnected[i][j] = 0 表示二者不直接相连。
返回矩阵中 省份 的数量。

题目描述：

There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c.
A province is a group of directly or indirectly connected cities and no other cities outside of the group.
You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the ith city and the jth city are directly connected, and isConnected[i][j] = 0 otherwise.
Return the total number of provinces.

方法1:
Time complexity: $O(N^2)$
Space complexity: $O(N)$
并查集：
标准并查集模版，初始时，每个城市都属于不同的连通分量。遍历矩阵如果两个城市之间有相连关系，则它们属于同一个连通分量，对它们进行合并。最后连通块的个数即为省份的总数。

class Solution {
    public int findCircleNum(int[][] isConnected) {
        int n = isConnected.length;
        UnionFind uf = new UnionFind(n);
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                if(isConnected[i][j] == 1){
                    uf.union(i, j);
                }
            }
        }
        
        return uf.getCount();
    }
}

class UnionFind{
    int[] father;
    int count;
    UnionFind(int n){
        this.father = new int[n+1];
        this.count = 0;
        for(int i = 1; i <=n; i++){
            father[i] = i;
            count++;
        }
    }
    
    public int getCount(){
        return this.count;
    }
    
    public int find(int x){
        if(x != father[x]){
            father[x] = find(father[x]);
        }
        return father[x];
    }
    
    public void union(int x, int y){
        int fx = find(x);
        int fy = find(y);
        if(fx != fy){
            father[fx] = fy;
            count--;
        }
    }
}