[LeetCode][Java] Clone Graph

题目：

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

OJ's undirected graph serialization:

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 #.

1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
2. Second node is labeled as 1. Connect node 1 to node 2.
3. Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.

Visually, the graph looks like the following:

       1      / \     /   \    0 --- 2         / \         \_/

题意：

克隆无向图。图中的每一个节点包含一个label 和 一组neighbors.

本题中的无向图的定义为：

每个节点都是唯一标记的。

我们通过#来划分每一个节点， 通过,来划分节点标记和节点的每一个邻居。

举个例子，给定一个图{0,1,2#1,2#2,2}.

这个图中共3个节点，因此包含了通过#划分的三个部分.

   1.第一个节点被标记为 0.连接节点0和节点1 and 2.

   2.第二个节点被标记为 1.将节点1和 节点2相连接.

  3.第三个节点被标记为 2. 将节点2和节点2（它自身）想连接，这样就形成了自环.

这个图就如下面所示：

       1      / \     /   \    0 --- 2         / \         \_/

算法分析：

 * 共两种方法，DFS--递归, BFS--队列
  
 * 拷贝方法是用用HashMap，key存原始值，value存copy的值，用DFS或者BFS方法遍历帮助拷贝neighbors的值。
   
 * 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，

AC代码：

/** * Definition for undirected graph. * class UndirectedGraphNode  * { *     int label; *     List<UndirectedGraphNode> neighbors; *     UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); } * }; */

方法一：广度优先搜索BFS 利用队列

public class Solution {    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node)    {        if(node == null)            return null;                    HashMap<UndirectedGraphNode, UndirectedGraphNode> hm = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();        LinkedList<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();        UndirectedGraphNode head = new UndirectedGraphNode(node.label);        hm.put(node, head);//key存原始值，value存copy的值        queue.add(node);                while(!queue.isEmpty())        {            UndirectedGraphNode curnode = queue.poll();            for(UndirectedGraphNode aneighbor: curnode.neighbors)//check each neighbor            {                if(!hm.containsKey(aneighbor))//if not visited,then add to queue                {                    queue.add(aneighbor);                    UndirectedGraphNode newneighbor = new UndirectedGraphNode(aneighbor.label);                    hm.put(aneighbor, newneighbor);                }                hm.get(curnode).neighbors.add(hm.get(aneighbor));//head.neighbors.add(newneightbor)            }        }        return head;    }}

方法二：通过递归进行DFS深度优先遍历

public class Solution {    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node)     {            if(node == null)                return null;                            HashMap<UndirectedGraphNode, UndirectedGraphNode> hm = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();            UndirectedGraphNode head = new UndirectedGraphNode(node.label);//初始化话节点时没有邻接节点            hm.put(node, head);//key存原始值，value存copy的值                        DFS(hm, node);//DFS            return head;    }    public void DFS(HashMap<UndirectedGraphNode, UndirectedGraphNode> hm, UndirectedGraphNode node)    {        if(node == null)            return;                    for(UndirectedGraphNode aneighbor: node.neighbors)        {             if(!hm.containsKey(aneighbor))            {                UndirectedGraphNode newneighbor = new UndirectedGraphNode(aneighbor.label);                hm.put(aneighbor, newneighbor);                DFS(hm, aneighbor);//DFS            }            hm.get(node).neighbors.add(hm.get(aneighbor));//对该节点拷贝neighbors的值。        }    }}