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.
算法1：我们在构建新链表的节点时，保存原始链表的next指针映射关系，并把指针做如下变化(蓝色为原始链表节点，紫红色为新链表节点)：
然后在上图的基础上进行如下两步
1、构建新链表的random指针：比如new1->random = new1->random->random->next, new2->random = NULL, new3-random = NULL, new4->random = new4->random->random->next
2、恢复原始链表：根据最开始保存的原始链表next指针映射关系恢复原始链表
该算法时间空间复杂度均为O(N)
算法2：该算法更为巧妙，不用保存原始链表的映射关系，构建新节点时，指针做如下变化，即把新节点插入到相应的旧节点后面：
同理分两步
1、构建新节点random指针：new1->random = old1->random->next, new2-random = NULL, new3-random = NULL, new4->random = old4->random->next
2、恢复原始链表以及构建新链表：例如old1->next = old1->next->next, new1->next = new1->next->next
该算法时间复杂度O(N)，空间复杂度O(1)
算法2的代码：
public RandomListNode copyRandomList(RandomListNode head) { if(head == null) return head; RandomListNode cur = head; while(cur != null){ RandomListNode sec = new RandomListNode(cur.label); sec.next = cur.next; cur.next = sec; cur = cur.next.next; } cur = head; while(cur != null){ if(cur.random != null) cur.next.random = cur.random.next; cur = cur.next.next; } cur = head; RandomListNode res = head.next; RandomListNode sec = res; while(sec.next != null){ cur.next = cur.next.next; cur = cur.next; sec.next = sec.next.next; sec = sec.next; } cur.next = cur.next.next; return res; } --------------------- 作者：liuchongee 来源： 原文：https://blog.csdn.net/liuchonge/article/details/74858192 版权声明：本文为博主原创文章，转载请附上博文链接！图示
reference：
https://blog.csdn.net/liuchonge/article/details/74858192
http://www.cnblogs.com/TenosDoIt/p/3387000.html
转载于:https://www.cnblogs.com/hygeia/p/10057625.html
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