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）

/*
public class RandomListNode {
    int label;
    RandomListNode next = null;
    RandomListNode random = null;

    RandomListNode(int label) {
        this.label = label;
    }
}
*/
import java.util.*;
public class Solution {
    public RandomListNode Clone(RandomListNode pHead)
    {
        if(pHead == null)   return null;
        cloneOneByOne(pHead);
        cloneRandom(pHead);
        return split(pHead);
    }
    
    public void cloneOneByOne(RandomListNode pHead)
    {
        while(pHead != null) {
            RandomListNode newNode = new RandomListNode(pHead.label);
            newNode.next = pHead.next;
            pHead.next = newNode;
            pHead = newNode.next;
        }
    }
    
    public void cloneRandom(RandomListNode pHead)
    {
        while(pHead != null) {
            if(pHead.random != null)
                pHead.next.random = pHead.random.next;
            pHead = pHead.next.next;
        }
    }
    
    public RandomListNode split(RandomListNode pHead)
    {
        RandomListNode tmp = pHead.next;
        RandomListNode root1 = pHead, root2 = pHead.next;
        
        while(pHead.next.next != null) {
            pHead.next = pHead.next.next;
            pHead = pHead.next;
            tmp.next = tmp.next.next;
            tmp = tmp.next;
        }
        
        pHead.next = null;
        return root2;
    }
}