Clone Graph

Clone Graph

问题：

Clone an undirected graph. Each node in the graph contains a

label

and a list of its

neighbors

.

思路：

　　dfs 或者 bfs

我的代码1：（dfs）

public class Solution {
public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
if(node == null) return null;
HashMap<UndirectedGraphNode,UndirectedGraphNode> hm = new HashMap<UndirectedGraphNode,UndirectedGraphNode>();
Queue<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
queue.offer(node);
UndirectedGraphNode rst = new UndirectedGraphNode(node.label);
hm.put(node, rst);
while(!queue.isEmpty())
{
UndirectedGraphNode first = queue.poll();
UndirectedGraphNode ugn = hm.get(first);
for(UndirectedGraphNode tmp : first.neighbors)
{
if(!hm.containsKey(tmp))
{
UndirectedGraphNode newNode = new UndirectedGraphNode(tmp.label);
hm.put(tmp,newNode);
queue.offer(tmp);
}
ugn.neighbors.add(hm.get(tmp));
}
}
return rst;
}
}

View Code
学习之处：

无论是DFS和BFS写的过程中都出现了死循环的问题，纠结而言是因为对于节点1 有邻居2,3 HashMap里面便有了（1，**）（2，**），（3，**），对于节点2 有邻居2 则我们不需要进一步的访问（进一步的访问是指对于BFS加入到Queue中，对于DFS是下一步的递归），仅仅需要加入到Neighbor中就可以了。