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Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.
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 #.
0. Connect node 0 to both nodes 1 and 2.1. Connect node 1 to node 2.2. Connect node 2 to node 2 (itself), thus forming a self-cycle.
Visually, the graph looks like the following:
1
/ / 0 --- 2
/ \_/
SOLUTION:
思路上讲这个题可以分两个部分1，遍历图 2，clone图。
遍历需要用到BFS，基本图的搜索都是BFS，那么也就是要用到queue，先讲node进入队列，然后pop的时候访问他的所有neighbors，把没访问过的neighbor放入队列中，依次pop重复以上操作，在这个过程中将入栈的neighbor存入hashmap中，以便以后确认入栈时是否曾经入队列过，保证只放没遍历过的点入队列。在这个题里，用一个指针＋ArrayList模拟一个队列也是没问题的。
clone时，就是要做deep copy。用newnode作为一个map里指向copynode的指针，再次遍历容器里的node，把原来点的neighbors在map里的映射（BFS时候会克隆到）加入新newnode的neighbors里面。
最后return map里的node的映射就可以了
public class Solution { /** * @param node: A undirected graph node * @return: A undirected graph node */ public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) { if (node == null){ return null; } //because I need the collection 2 times, so I can‘t use queue, then I choice ArrayList and a pointer(start) ArrayList<UndirectedGraphNode> nodes = new ArrayList<UndirectedGraphNode>(); HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>(); nodes.add(node); map.put(node, new UndirectedGraphNode(node.label)); int start = 0; //clone nodes while (start < nodes.size()){ UndirectedGraphNode cur = nodes.get(start++); for (int i = 0; i < cur.neighbors.size(); i++){ UndirectedGraphNode neighbor = cur.neighbors.get(i); if (!map.containsKey(neighbor)){ map.put(neighbor, new UndirectedGraphNode(neighbor.label)); nodes.add(neighbor); } } } // clone neighbors for (int i = 0; i < nodes.size(); i++){ UndirectedGraphNode newNode = map.get(nodes.get(i)); for (int j = 0; j < nodes.get(i).neighbors.size(); j++){ newNode.neighbors.add(map.get(nodes.get(i).neighbors.get(j))); } } return map.get(node); } }
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原文地址：http://www.cnblogs.com/tritritri/p/4919769.html
