Clone Graph

题目地址：
https://leetcode.com/problems/clone-graph/#/description

题目：

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
         / \
         \_/

解题思路：
这道题就是用bfs的思想，用一个map来存储新的node，用queue来存储原来的node然后做bfs。

代码：

public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
    if(node == null){
        return null;
    }

    Map<Integer, UndirectedGraphNode> map = new HashMap<>();
    UndirectedGraphNode tempNode = new UndirectedGraphNode(node.label);
    map.put(node.label, tempNode);
    Queue<UndirectedGraphNode> q = new LinkedList<>();
    q.offer(node);
    while(!q.isEmpty()){
        UndirectedGraphNode pollNode = q.poll();
        UndirectedGraphNode newNode = map.get(pollNode.label);
        for(UndirectedGraphNode n : pollNode.neighbors){
            if(map.containsKey(n.label)){
                UndirectedGraphNode neighbor = map.get(n.label);
                newNode.neighbors.add(neighbor);
            }
            else{
                q.offer(n);
                UndirectedGraphNode temp = new UndirectedGraphNode(n.label);
                newNode.neighbors.add(temp);
                map.put(n.label, temp);
            }
        }
    }
    return map.get(node.label);
}