Lintcode105 Copy List with Random Pointer solution 题解

【题目描述】

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.

给出一个链表，每个节点包含一个额外增加的随机指针可以指向链表中的任何节点或空的节点。

返回一个深拷贝的链表。

【题目链接】

www.lintcode.com/en/problem/copy-list-with-random-pointer/

let data = {
    val:1,
    next:{
       val:2,
       next:{
          val:3,
          next:{
              val:4,
              next:null
          } 
       }
    }
};

data.rand = data.next.next;
data.next.rand = null;
data.next.next.rand = null;
data.next.next.next.rand = data;

const copyRandomList = function(head){
    if(head === null){
        return null;
    }
    let cur = head;
    let next = null;

    while(cur !== null){
        next = cur.next;
        cur.next = {};
        cur.next.next = next;
        cur = next;
    }

    cur = head;
    let curCopy = null;

    while(cur !== null){
        next = cur.next.next;
        curCopy = cur.next;
        curCopy.val = cur.val;
        curCopy.rand = (function(){
            if(cur.rand !== null){
                // cur.rand是正本，cur.rand.next是副本，所以返回的是副本指向
                return cur.rand.next;
            }

            return null;
        })();
        cur = next;
    }

    let res = head.next;
    cur = head;

    while(cur !== null){
        next = cur.next.next;
        curCopy = cur.next;
        cur.next = next;
        curCopy.next = (function(){
            if(next !== null){
                return next.next;
            }

            return null;
        })();
        cur =next;
    }

    return res;
};

let result = copyRandomList(data);
console.log(result);