难题部分Copy List with Random Pointer-优快云博客

本文探讨如何深拷贝一个包含随机指针的单链表，并提供了两种实现方案：一种使用HashMap来跟踪旧节点与新节点的关系；另一种通过在原链表中插入拷贝节点的方式，避免使用额外的空间。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

别的都不说，这部分题对我来说，首先就是看不懂题意

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.

/**
* Definition for singly-linked list with a random pointer.
* class RandomListNode {
* int label;
* RandomListNode next, random;
* RandomListNode(int x) { this.label = x; }
* };
*/

1 label和random，next关系，做hashmap的时候怎么放key 和value？

2 初一看题查了deep copy的说明，但是还是给想复杂了，人家就是让你copy而已，把random信息也给copy过去，而不是reorder这个list。这样就好理解ref link里面题解里“如果要copy一个带有random pointer的list，主要的问题就是有可能这个random指向的位置还没有被copy到，所以解决方法都是多次扫描list。”的意义

ref http://www.cnblogs.com/springfor/p/3864457.html

3 HashMap 里面存的是指向node的指针还是node本身呢？

4 第二种作法，不利用hashmap，关键是看懂node.next.random = node.random.next

那些破人，连个“滚”的眼神都懒得甩

soln 1

    public RandomListNode copyRandomList(RandomListNode head) {
        //http://www.cnblogs.com/springfor/p/3864457.html
        if(head==null) return null;
        
        //soln 1 HashMap 遍历2次list，所以时间复杂度是O(2n)=O(n)，然后使用了HashMap，所以空间复杂度是O(n)。
        
        RandomListNode newhd = new RandomListNode(head.label);
        RandomListNode oldhd = head;
        RandomListNode oldp = head.next;
        RandomListNode newp = newhd;
        HashMap<RandomListNode, RandomListNode> hm = new HashMap<RandomListNode, RandomListNode>();
        hm.put(oldhd, newhd); //Map<K,V> , which is key，which is value？
        
        while(oldp!=null){
            RandomListNode newnode = new RandomListNode(oldp.label);
            newp.next = newnode;
            newp = newp.next;
            hm.put(oldp,newp);
            oldp = oldp.next;
            
        }
        
        oldp = head;
  
  
        newp = newhd;
        
        while(oldp!=null){
            newp.random = hm.get(oldp.random);  
            newp = newp.next;
            oldp = oldp.next;
        }
        
        return newhd;

soln 2

//soln 2 a) replica old nodes twice, b) replica node.next(which is node's replica).random = node.random.next(which is random's replica) , c) separate node and its replica
        RandomListNode node = head;
        while(node!=null){
            RandomListNode newnode = new RandomListNode(node.label);
            newnode.next = node.next;
            node.next = newnode;   // a) replica old nodes twice
            node =newnode.next;
        }
        
        node = head;
        while(node!=null){
            if(node.random!=null)
                node.next.random =  node.random.next; // b) replica node.next(which is node's replica).random = node.random.next(which is random's replica)
            node = node.next.next;
        }
        
        node = head;
        RandomListNode newhd = head.next;
        while(node!=null){
            RandomListNode rep = node.next;
            node.next = rep.next;
            if(rep.next!=null)
                rep.next = rep.next.next;
            node = node.next;
        }
        
        return newhd;
    }
}