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The problem:
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
/ \_/
The implementation pattern used in this problem is very classic in cloning a graph. The key idea: use a hash map to store and retrieve the copy of a node. <original, copy> The basic idea is to use the invariant: 1. we dequeue a node from the queue, then we scan the node‘s neighbors. we check if the neighbor node has already been traversaled by using the "containKeys()" method. 1.1 iff the node is in the hashmap, we would not make a copy for it. 1.2 iff the node is not in the hashmap, we make a copy for this neighbor node. Put the <neighbor_node, copy> into the hashmap and enqueue the neighbor node. if (!map.containsKey(cur.neighbors.get(i))) {//this node has not been traversaled copy = new UndirectedGraphNode(cur.neighbors.get(i).label); map.put(cur.neighbors.get(i), copy); queue.offer(cur.neighbors.get(i)); } Note1: we only enqueue nodes from neighbors list, which means all neighbors list of remaining nodes in the queue(until the dequeue) has not been copyed yet. Thus, for each current node‘s neghbor list, we need to fully copy it. for (int i = 0; i < cur.neighbors.size(); i++) { if (!map.containsKey(cur.neighbors.get(i))) { ... } map.get(cur).neighbors.add(map.get(cur.neighbors.get(i))); } Note2: to use the invariant, we need to make necessary preparation for it (dummy layer). we should suppose there is dummy layer copy the starting node, and then put it into hashmap and enqueue the starting node.
My solution:
public class Solution { public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) { if (node == null) return null; HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode> (); Queue<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode> (); UndirectedGraphNode cur; UndirectedGraphNode copy = new UndirectedGraphNode(node.label); queue.offer(node); map.put(node, copy); while(!queue.isEmpty()) { cur = queue.poll(); for (int i = 0; i < cur.neighbors.size(); i++) { if (!map.containsKey(cur.neighbors.get(i))) {//this node has not been traversaled copy = new UndirectedGraphNode(cur.neighbors.get(i).label); map.put(cur.neighbors.get(i), copy); queue.offer(cur.neighbors.get(i)); } map.get(cur).neighbors.add(map.get(cur.neighbors.get(i))); } } return map.get(node); } }
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原文地址：http://www.cnblogs.com/airwindow/p/4220170.html
