复杂链表的复制

问题描写叙述：

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.

解题思路：

将1->2->3->4->NULL先复制为1->1->2->2->3->3->4->4->NULL。这样就会有例如以下关系式：

pCur->next->random = pCur->random->next;

/**
 * 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 *pCur = head, *tmp, *new_head;
        while(pCur != NULL)/*复制结点*/
        {
            tmp =  new RandomListNode(pCur->label);
            tmp->next = pCur->next;
            pCur->next = tmp;
            pCur = tmp->next;
        }
        pCur = head;
        /* 注意调用空指针的next指针。会报错*/
        while(pCur != NULL)/*复制随意指针*/
        {
            if (pCur->random != NULL)
                pCur->next->random = pCur->random->next;
            else
                pCur->next->random = NULL;
            pCur = pCur->next->next;
        }
        pCur = head;
        tmp = new_head = head->next;
        while(tmp != NULL)/* 将链表拆分为两个链表 */
        {
            pCur->next = tmp->next;
            if(pCur->next == NULL)
                break;
            else
                tmp->next = pCur->next->next;
            pCur = pCur->next;
            tmp = tmp->next;
        }
        return new_head;
    }
};