[Leetcode] 323. Number of Connected Components in an Undirected Graph 解题报告

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:

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

.
思路：

一道难度适中的Union-Find题目。这类题目也有套路：首先为每个顶点初始化一个单独的集合；然后遍历，每次遇到一个边，就把边的两个顶点所属的集合进行合并，同时总的连通图数量减1。注意判断两个顶点是否属于同一个集合的经典代码，个人感觉应该背下来^_^。

这道题目用BFS和DFS也可以求解，但代码要相对复杂一些。

代码：

class Solution {
public:
int countComponents(int n, vector<pair<int, int>>& edges) {
vector<int> parents(n);
for (int i = 0; i < n; ++i) {
parents[i] = i;
}
int ret = n;
for (int i = 0; i < edges.size(); ++i) {
int par1 = edges[i].first, par2 = edges[i].second;
while (par1 != parents[par1]) {
par1 = parents[par1];
}
while (par2 != parents[par2]) {
par2 = parents[par2];
}
if (par1 != par2) {
parents[par2] = par1;
--ret;
}
}
return ret;
}
};