1，問題描述

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.

數據結構：

1 /**
2  * Definition for singly-linked list with a random pointer.
3  * class RandomListNode {
4  *     int label;
5  *     RandomListNode next, random;
 
6  *     RandomListNode(int x) { this.label = x; }
7  * };
8  */

2，邊界條件：root==null情況可以處理，所以沒有邊界條件，不需要特殊處理

3，解題思路：先不管random指針，把鏈表節點依次復制，不過不是生成一個新的鏈表，而是把新節點放在原節點的next。然後再根據random指針把新節點的random指針指向原節點的next，即新節點。最後把大鏈表分離，恢復舊鏈表，把新節點生成一個新鏈表，即為舊鏈表的deep copy。

4，代碼實現

 1 public RandomListNode copyRandomList(RandomListNode head) {
 
 2     if (head == null) {
 3         return null;
 4     }
 5     RandomListNode cur = head;
 6     while (cur != null) {
 7         RandomListNode node = new RandomListNode(cur.label);
 8         node.next = cur.next;
 9         cur.next = node;
10         cur = node.next;
11     }
12 
13     cur = head;
 
14     while (cur != null && cur.next != null) {
15         if (cur.random != null) {
16             cur.next.random = cur.random.next;
17         }
18         cur = cur.next.next;
19     }
20 
21     RandomListNode dummy = new RandomListNode(-1);
22     // RandomListNode copyCur = head.next;//這種寫法在leetcode不通過，在lintcode上面能通過
23     // dummy.next = copyCur;
24     // cur = head.next.next;
25     RandomListNode copyCur = dummy;//這裏是註意下，比上面寫法簡潔，可以處理head=null的情況
26     cur = head;
27     while (cur != null && cur.next != null) {
28         copyCur.next = cur.next;
29         cur.next = cur.next.next;
30         cur = cur.next;
31         copyCur = copyCur.next;
32     }
33     return dummy.next;
34 }

5，時間復雜度：O（n），空間復雜度：O（1）

6，api：無

leetcode--138. Copy List with Random Pointer