学步园

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
         / \
         \_/

解题思路：

	1. 用一个Queue来存储待处理的节点

	2. 遍历的时候对于不存在的节点先要创建出来，所有节点通过Map<int,node> 的方式实现快速存取

/**
 * Definition for undirected graph.
 * class UndirectedGraphNode {
 *     int label;
 *     ArrayList<UndirectedGraphNode> neighbors;
 *     UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); }
 * };
 */
public class Solution {
    /**
     * @param node: A undirected graph node
     * @return: A undirected graph node
     */
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if(node == null)
            return null;
            
        Map<Integer,UndirectedGraphNode> map = new HashMap<Integer,UndirectedGraphNode>();
        Set<Integer> pass = new HashSet<Integer>();
        Queue<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
        queue.offer(node);
        pass.add(node.label);
        map.put(node.label,new UndirectedGraphNode(node.label));
        
        UndirectedGraphNode t = null;
        while((t = queue.poll()) !=null){
            UndirectedGraphNode t1 = map.get(t.label);
            ArrayList<UndirectedGraphNode> neighbors = new ArrayList<UndirectedGraphNode>();
            for(UndirectedGraphNode n : t.neighbors){
                if(map.get(n.label)==null){
                    map.put(n.label,new UndirectedGraphNode(n.label));
                    queue.offer(n);
                }
                neighbors.add(map.get(n.label));
	        }
	        t1.neighbors = neighbors;
        }
             
        return map.get(node.label);
    }
}