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/

 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     }

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