Clone Graph
这里使用BFS来解本题,BFS需要使用queue来保存neighbors
但这里有个问题,在clone一个节点时我们需要clone它的neighbors,而邻居节点有的已经存在,有的未存在,如何进行区分?
这里我们使用Map来进行区分,Map的key值为原来的node,value为新clone的node,当发现一个node未在map中时说明这个node还未被clone,
http://oj.leetcode.com/problems/clone-graph/
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) { 2 if(node==null) 3 return null; 4 UndirectedGraphNode copyroot=new UndirectedGraphNode(node.label); 5 UndirectedGraphNode copyfollow=copyroot; 6 UndirectedGraphNode follow=node; 7 8 Queue<UndirectedGraphNode> queue=new LinkedList(); 9 HashMap<UndirectedGraphNode,UndirectedGraphNode> copyHash=new HashMap<UndirectedGraphNode,UndirectedGraphNode>(); 10 queue.add(follow); 11 copyHash.put(follow, copyfollow); 12 while(queue.isEmpty()==false) 13 { 14 UndirectedGraphNode p=queue.poll(); 15 copyfollow=copyHash.get(p); 16 if(p.neighbors!=null) 17 copyfollow.neighbors=new ArrayList<UndirectedGraphNode>(); 18 for(UndirectedGraphNode elem:p.neighbors) 19 { 20 if(copyHash.containsKey(elem)) 21 { 22 copyfollow.neighbors.add(copyHash.get(elem)); 23 } 24 else 25 { 26 queue.add(elem); 27 UndirectedGraphNode temp=new UndirectedGraphNode(elem.label); 28 temp.neighbors=null; 29 copyfollow.neighbors.add(temp); 30 copyHash.put(elem, temp); 31 } 32 } 33 } 34 return copyroot; 35 }