Copy List with Random Pointer

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)

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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;
    }
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作者：liuchongee 
来源：CSDN 
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