leetcode -- Clone Graph

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

[解题思路]

图的遍历有两种方式，BFS和DFS

这里使用BFS来解本题，BFS需要使用queue来保存neighbors

但这里有个问题，在clone一个节点时我们需要clone它的neighbors，而邻居节点有的已经存在，有的未存在，如何进行区分？

这里我们使用Map来进行区分，Map的key值为原来的node，value为新clone的node，当发现一个node未在map中时说明这个node还未被clone，

将它clone后放入queue中处理neighbors。

使用Map的主要意义在于充当BFS中Visited数组，它也可以去环问题，例如A--B有条边，当处理完A的邻居node，然后处理B节点邻居node时发现A已经处理过了

处理就结束，不会出现死循环！

queue中放置的节点都是未处理neighbors的节点！！！！

 1 /**
 2  * Definition for undirected graph.
 3  * class UndirectedGraphNode {
 4  *     int label;
 5  *     ArrayList<UndirectedGraphNode> neighbors;
 6  *     UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); }
 7  * };
 8  */
 9 public class Solution {
10     public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
11         // Note: The Solution object is instantiated only once and is reused by each test case.
12         if(node == null){
13             return node;
14         }
15         UndirectedGraphNode result = new UndirectedGraphNode(node.label);
16         LinkedList<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
17         queue.add(node);
18         Map<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
19         map.put(node, result);
20         
21         while(!queue.isEmpty()){
22             UndirectedGraphNode nodeInQueue = queue.poll();
23             ArrayList<UndirectedGraphNode> neighbors = nodeInQueue.neighbors;
24             for(int i = 0; i < neighbors.size(); i++){
25                 UndirectedGraphNode n1 = neighbors.get(i);
26                 if(map.containsKey(n1)){
27                     map.get(nodeInQueue).neighbors.add(map.get(n1));
28                 } else {
29                     UndirectedGraphNode n1clone = new UndirectedGraphNode(n1.label);
30                     map.get(nodeInQueue).neighbors.add(n1clone);
31                     map.put(n1, n1clone);
32                     queue.add(n1);
33                 }
34             }
35             
36         }
37         return result;
38     }
39 }

ref:

http://leetcode.com/2012/05/clone-graph-part-i.html