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 #.

1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
2. Second node is labeled as 1. Connect node 1 to node 2.
3. 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
         / \
         \_/

[java]  view plain  copy

1. /** 
2.  * Definition for undirected graph. 
3.  * class UndirectedGraphNode { 
4.  *     int label; 
5.  *     List<UndirectedGraphNode> neighbors; 
6.  *     UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); } 
7.  * }; 
8.  */  
9. public class Solution {  
10.     public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {  
11.         Map<UndirectedGraphNode,UndirectedGraphNode> map =   
12.                 new HashMap<UndirectedGraphNode,UndirectedGraphNode>();  
13.         if(node==null) return null;  
14.         Queue<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();  
15.         queue.add(node);  
16.         UndirectedGraphNode root = new UndirectedGraphNode(node.label);  
17.         map.put(node, root);  
18.         while(!queue.isEmpty()){  
19.             UndirectedGraphNode cur = queue.poll();       
20.             for(UndirectedGraphNode child:cur.neighbors){  
21.                 if(!map.containsKey(child)){  
22.                     UndirectedGraphNode p = new UndirectedGraphNode(child.label);  
23.                     map.put(child, p);  
24.                     queue.add(child);  
25.                 }  
26.                 map.get(cur).neighbors.add(map.get(child));  
27.             }  
28.         }  
29.         return root;  
30.     }  
31. }

原文链接http://blog.youkuaiyun.com/crazy__chen/article/details/46560567