leetcode Copy List with Random Pointer

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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.

题目大意：对一个单向链表进行深度拷贝，单向链表的每个结点还有一个random指针，或者为NULL，或者指向链表中任意一个结点。

解题思路：

将新结点嵌入对应的结点后面如old1->new1->old2->new2.....oldn->newn，然后新节点的random指针指向的就是旧结点random指针指向的next

最后恢复原链表，构建新的链表 old->next = old->next->next; new->next = new->next->next;

代码实现如下：

/** * Definition for singly-linked list with a random pointer. * struct RandomListNode { *     int label; *     RandomListNode *next, *random; *     RandomListNode(int x) : label(x), next(NULL), random(NULL) {} * }; */class Solution {public:    RandomListNode *copyRandomList(RandomListNode *head)    {        if(head == NULL)            return NULL;        RandomListNode *iter = head;        RandomListNode *newiter, *olditer;                //将链表变成old1->new1->old2->new2.....oldn->newn        while(iter != NULL)        {            olditer = iter;            iter = iter->next;            newiter = new RandomListNode(olditer->label);            olditer->next = newiter;            newiter->next = iter;        }        //将新节点的random指针指向新节点的位置        olditer = head;        newiter = head->next;        while(true)        {            if(olditer->random == NULL)            {                newiter->random = NULL;            }            else            {                newiter->random = olditer->random->next;            }            olditer = olditer->next->next;            if(olditer == NULL)                break;            newiter = olditer->next;        }        //恢复原来的链表        olditer = head;        newiter = head->next;        iter = newiter;                while(true)        {            olditer->next = newiter->next;            olditer = newiter->next;            if(olditer == NULL)                break;            newiter->next = olditer->next;            newiter = newiter->next;        }                return iter;    }};

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