[LeetCode]copy-list-with-random-pointer

题目：A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.Return a deep copy of the list.

题目理解:深拷贝一个带有random指针的链表，因为存在random指针故不能简单直接拷贝，要想让random指针指向正确的位置，必须链表中所有的节点都被复制好。也不能单独复制到一个新的链表，否则random指针也不能指向正确位置。

故考虑复制每个节点到原链表每个节点后面，然后将random指针指向正确位置后，把复制的节点剥离下来，形成新的链表即可。

代码：

 1 public class Solution {
 2     public RandomListNode copyRandomList(RandomListNode head) {
 3         if(head == null) return head;
 4 
 5         RandomListNode tmpList = head;
 6         while(tmpList != null){
 7             RandomListNode node = new RandomListNode(tmpList.label);
 8             node.next = tmpList.next;
 9             tmpList.next = node;
10 
11             tmpList = tmpList.next.next;
12         }
13 
14         tmpList = head;
15         while(tmpList != null){
16             if(tmpList.random != null)//这里的判断尤为重要，丢失的话会产生空指针的错误
17                 tmpList.next.random = tmpList.random.next;
18             tmpList = tmpList.next.next;
19         }
20         RandomListNode pHead = new RandomListNode(0);
21         pHead.next = head;
22         RandomListNode newlist = pHead;
23 
24         tmpList = head;
25         while (tmpList != null){
26             pHead.next = tmpList.next;
27             tmpList.next = pHead.next.next;
28             pHead = pHead.next;
29             tmpList = tmpList.next;
30         }
31         return newlist.next;
32     }
33 }

重点：指针操作要注意可能产生空指针的错误。