25. 复杂链表的复制

题目：输入一个复杂链表（每个节点中有节点值，以及两个指针，一个指向下一个节点，另一个特殊指针指向任意一个节点），返回结果为复制后复杂链表的head。（注意，输出结果中请不要返回参数中的节点引用，否则判题程序会直接返回空）

/*

struct RandomListNode {
    int label;
    struct RandomListNode *next, *random;
    RandomListNode(int x) :
            label(x), next(NULL), random(NULL) {
    }
};
*/
class Solution {
public:
    RandomListNode* Clone(RandomListNode* pHead)
    {
        if(!pHead) 
            return NULL;
		//复制每个结点，并与原链表连接：原先是A B C D，变成 A A' B B' C C' D D'
        RandomListNode *pNode = pHead;
        while(pNode){
            RandomListNode *pCloned = new RandomListNode(pNode->label);
            pCloned->next = pNode->next;
            pCloned->random = NULL;
            pNode->next = pCloned;
            pNode = pCloned->next;
        }
		//
        pNode = pHead;
        while(pNode){
            RandomListNode *pCloned = pNode->next;
            if(pNode->random){               
                pCloned->random = pNode->random->next;
            }
            pNode = pCloned->next;
        }
        //拆分两个链表
        RandomListNode *pCloneHead = pHead->next;
        RandomListNode *pCloned;
        pNode = pHead;
        if(pNode){
            pCloned = pNode->next;
            pNode->next = pCloned->next;
            pNode = pNode->next;
        }
        while(pNode){
            pCloned->next = pNode->next;
            pCloned = pCloned->next;
            pNode->next = pCloned->next;
            pNode = pNode->next;
        }
        return pCloneHead;
    }
};