LeetCode: Clone Graph

LeetCode: Clone Graph

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

OJ's undirected graph serialization:

Nodes are labeled uniquely.

We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.

As an example, consider the serialized graph {0,1,2#1,2#2,2}.

The graph has a total of three nodes, and therefore contains three parts as separated by #.

First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
Second node is labeled as 1. Connect node 1 to node 2.
Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.

Visually, the graph looks like the following:

       1
      / 
     /   
    0 --- 2
         / 
         \_/

地址：https://oj.leetcode.com/problems/clone-graph/
算法：把整个图复制下来，基本就是利用map来完成的，复制的过程采用递归完成。代码：

 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     UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
12         if(!node){
13             return NULL;
14         }
15         map<int,UndirectedGraphNode *> hash_table;
16         set<int> isTraved;
17         return subClone(node,hash_table,isTraved);
18     }
19     UndirectedGraphNode *subClone(UndirectedGraphNode *node, map<int,UndirectedGraphNode*> &hash_table, set<int> &isTraved){
20         if(!node){
21             return NULL;
22         }
23         int label = node->label;
24         if(isTraved.find(label) != isTraved.end()){
25             return hash_table[label];
26         }
27         if(hash_table.find(label) == hash_table.end()){
28             hash_table[label] = new UndirectedGraphNode(label);
29         }
30         isTraved.insert(label);
31         UndirectedGraphNode *p = hash_table[label];
32         vector<UndirectedGraphNode *>::iterator it = node->neighbors.begin();
33         for(; it != node->neighbors.end(); ++it){
34             p->neighbors.push_back(subClone(*it,hash_table,isTraved));
35         }
36         return p;
37     }
38 };