【面试题24】复杂链表的复制

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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;
    }
}
*/
public class Solution {
     
    privatevoid copyNext(RandomListNode head){
        if(head==null){
            return;
        }
        while(head!=null){
            RandomListNode newNode=newRandomListNode(head.label);
            newNode.random=head.random;
            newNode.next=head.next;
            head.next=newNode;
            head=head.next.next;
        }
    }
     
    privatevoid copyRandom(RandomListNode head){
        if(head==null){
            return;
        }
        while(head!=null){
            if(head.next.random!=null){
                head.next.random=head.random.next;
            }           
            head=head.next.next;
        }
    }
     
    privateRandomListNode splitList(RandomListNode head){
        if(head==null){
            returnnull;
        }
        RandomListNode newHead=head.next;
        while(head!=null){
            RandomListNode temp=head.next;
            head.next=temp.next;
            head=head.next;
            if(temp.next!=null){
                temp.next=temp.next.next;
            }
        }
        returnnewHead;
    }
    publicRandomListNode Clone(RandomListNode phead)
    {
     if(phead==null){
         returnnull;
     }  
        copyNext(phead);
        copyRandom(phead);
        returnsplitList(phead);
    }
}

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