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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
/ \_/
题目大意是给定一个无向图，可能有环，实现一个方法，deep clone这个图。
我的做法是采用BFS，从一个点开始，遍历它的邻居，然后加入队列，当队列非空，循环遍历，因为label是唯一的，采用map保存新生成的node，用label作为key。另外使用visited数组保存是否加入过队列，以免重复遍历。
Talk is cheap>>
public UndirectedGraphNode cloneGraph(UndirectedGraphNode root) { HashSet<Integer> visited = new HashSet<>(); if (root==null) return null; List<UndirectedGraphNode> queue = new ArrayList<>(); HashMap<Integer,UndirectedGraphNode> map = new HashMap<>(); queue.add(root); while (!queue.isEmpty()) { UndirectedGraphNode node = queue.get(0); if (map.get(node.label)==null) { map.put(node.label, new UndirectedGraphNode(node.label)); } UndirectedGraphNode tmp = map.get(node.label); queue.remove(0); for (int i = 0; i < node.neighbors.size(); i++) { int key = node.neighbors.get(i).label; if (map.get(key)==null){ map.put(key,new UndirectedGraphNode(key)); } tmp.neighbors.add(map.get(key)); visited.add(node.label); if (!visited.contains(key)){ queue.add(node.neighbors.get(i)); visited.add(key); } } } return map.get(root.label); }
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原文地址：http://www.cnblogs.com/aboutblank/p/4379648.html
