### 题目

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:

Input: n = 5 and edges = [[0, 1], [1, 2], [3, 4]]

0          3
|          |
1 --- 2    4

Output: 2


Example 2:

Input: n = 5 and edges = [[0, 1], [1, 2], [2, 3], [3, 4]]

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

Output:  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.

### union-find

union-find的核心原理就是： #### 代码

class Solution {
public int countComponents(int n, int[][] edges) {
if (n <= 0 || edges == null) { return 0; }
for (int i = 0; i < n; i++) { broad[i] = i; }
for (int[] edge : edges) {
union(edge,edge);
}
int count = 0;
for (int i = 0; i < n; i++) {
if (broad[i] == i) { count++; }
}
return count;
}
//union-find
private void union(int a, int b) {
int rootA = find(a);
int rootB = find(b);
}
private int find(int a) { 