Medium
Given n nodes labeled from “0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to find the number of connected components in an undirected graph.

Example 1:

 0          3
 |          |
 1 --- 2    4

Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], return 2.

Example 2:

 0           4
 |           |
 1 --- 2 --- 3

Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [3, 4]], return 1.

Note:
You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as[1, 0] and thus will not appear together in edges.

这道题主要是为了复习一下Union Find这个数据结构，三部分写好Union Find: int[] father, int find(), void union().

class Solution {
    public int[] father;
    public int countComponents(int n, int[][] edges) {
        if (edges == null || edges.length == 0 || edges[0].length == 0){
            return n;
        } 
        father = new int[n];
        int res = n;
        for (int i = 0; i < n; i++){
            father[i] = i;
        }
        for (int i = 0; i < edges.length; i++){
            int u = edges[i][0];
            int v = edges[i][1];
            if (find(u) != find(v)){
                union(u, v);
                res--;
            }
        }
        return res;
    }
    
    private int find(int a){
        if (father[a] == a){
            return a;
        } 
        return find(father[a]);
    }
    
    private void union(int a, int b){
        int root_a = find(a);
        int root_b = find(b);
        if (root_a != root_b){
            father[root_a] = root_b;
        }
    }
}