复杂链表的复制

## 题目描述

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）

```cpp
/*
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;
}
RandomListNode* pNew = pHead;
while(pNew){
RandomListNode* tmp = new RandomListNode(pNew->label);
tmp->next = pNew->next;
pNew->next = tmp;
pNew = pNew->next->next;
}
pNew = pHead;
while(pNew){
if(pNew->random){
pNew->next->random = pNew->random->next;
}
pNew = pNew->next->next;
}
RandomListNode* pResult = pHead->next;
pNew = pHead->next;
while(pNew){
pHead->next = pNew->next;
if(pNew->next){
pNew->next = pNew->next->next;
}
pHead = pHead->next;
pNew = pNew->next;
}
return pResult;
}
};
```

The issue names no repository file or test; the only entry point mentioned is the inline C++ Solution::Clone method. First clarify whether this is meant to become or update content, then identify the target location and validation criteria before starting; the issue currently does not define what done looks like.

Written by the indexing model from the issue text.