[LintCode] Find the Connected Component in the Undirected Graph

Find the Connected Component in the Undirected Graph

Find the number connected component in the undirected graph. Each node in the graph contains a label and a list of its neighbors. (a connected component (or just component) of an undirected graph is a subgraph in which any two vertices are connected to each other by paths, and which is connected to no additional vertices in the supergraph.)

 
Example

Given graph:

A------B  C
      |  | 
      |  |
      |  |
      |  |
      D   E

Return {A,B,D}, {C,E}. Since there are two connected component which is {A,B,D}, {C,E}

DFS:

 1 /**
 2  * Definition for Undirected graph.
 3  * struct UndirectedGraphNode {
 4  *     int label;
 5  *     vector<UndirectedGraphNode *> neighbors;
 6  *     UndirectedGraphNode(int x) : label(x) {};
 7  * };
 8  */
 9 class Solution {
10 public:
11     /**
12      * @param nodes a array of Undirected graph node
13      * @return a connected set of a Undirected graph
14      */
15     void dfs(vector<UndirectedGraphNode*> &nodes, vector<int> &path,
16             unordered_set<UndirectedGraphNode*> &visit, UndirectedGraphNode* n) {
17         visit.insert(n);
18         path.push_back(n->label);
19         for (auto &nn : n->neighbors) if (visit.find(nn) == visit.end()) {
20             dfs(nodes, path, visit, nn);
21         }
22     }
23     vector<vector<int>> connectedSet(vector<UndirectedGraphNode*>& nodes) {
24         // Write your code here
25         unordered_set<UndirectedGraphNode*> visit;
26         vector<vector<int>> res;
27         vector<int> path;
28         for (auto &n : nodes) {
29             if (visit.find(n) == visit.end()) {
30                 path.clear();
31                 dfs(nodes, path, visit, n);
32                 sort(path.begin(), path.end());
33                 res.push_back(path);
34             } 
35         }
36         return res;
37     }
38 };

BFS:

 1 /**
 2  * Definition for Undirected graph.
 3  * struct UndirectedGraphNode {
 4  *     int label;
 5  *     vector<UndirectedGraphNode *> neighbors;
 6  *     UndirectedGraphNode(int x) : label(x) {};
 7  * };
 8  */
 9 class Solution {
10 public:
11     /**
12      * @param nodes a array of Undirected graph node
13      * @return a connected set of a Undirected graph
14      */
15     vector<vector<int>> connectedSet(vector<UndirectedGraphNode*>& nodes) {
16         // Write your code here
17         unordered_set<UndirectedGraphNode*> visit;
18         vector<vector<int>> res;
19         vector<int> path;
20         queue<UndirectedGraphNode*> que;
21         for (auto &n : nodes) {
22             if (visit.find(n) == visit.end()) {
23                 path.clear();
24                 visit.insert(n);
25                 for (que.push(n); !que.empty(); que.pop()) {
26                     auto u = que.front();
27                     path.push_back(u->label);
28                     for (auto nn : u->neighbors) if (visit.find(nn) == visit.end()) {
29                         visit.insert(nn);
30                         que.push(nn);
31                     }
32                 }
33                 sort(path.begin(), path.end());
34                 res.push_back(path);
35             } 
36         }
37         return res;
38     }
39 };
原文地址:https://www.cnblogs.com/easonliu/p/4606255.html