LeetCode_copy-list-with-random-pointer

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.

思路1：用HashMap把原来节点和CoPY节点一起放到HashMap中，然会，根据原来的节点的随机指针指向的键所对应的值就是该节点在map中对应值的随机值，说起来比较绕，直接看代码吧；

思路2：直接操作链表，先复制链表插在原来链表中，更新复制节点的随机数，千万注意这里，当原来的随机数是null的时候，就不用操作复制节点了，然后再把复制的节点进行分离。

import java.util.*;
/**
 * Definition for singly-linked list with a random pointer.
 * class RandomListNode {
 *     int label;
 *     RandomListNode next, random;
 *     RandomListNode(int x) { this.label = x; }
 * };
 */
public class Solution {
    //用HashMap
    public RandomListNode copyRandomList(RandomListNode head) {
        if(head==null)
            return null;
        HashMap<RandomListNode,RandomListNode> map=new HashMap<>();
        RandomListNode nHead=new RandomListNode(head.label);
        RandomListNode preNode=nHead;
        map.put(head,preNode);
        RandomListNode cur=head.next;
        while(cur!=null){
            RandomListNode CopyNode=new RandomListNode(cur.label);
            preNode.next=CopyNode;
            map.put(cur,CopyNode);
            preNode=CopyNode;
            cur=cur.next;
        }
        for(Map.Entry<RandomListNode,RandomListNode> m:map.entrySet()){
            m.getValue().random=map.get(m.getKey().random);
        }
        return nHead;
    }
    
    //直接用链表
    public RandomListNode copyRandomList(RandomListNode head) {
        if(head==null)
            return null;
        RandomListNode cur=head;
        while(cur!=null){
            RandomListNode copyNode=new RandomListNode(cur.label);
            copyNode.next=cur.next;
            cur.next=copyNode;
            cur=cur.next.next;
        }
        cur=head;
        while(cur!=null){
            if(cur.random!=null)//千万注意这一步，当random为null的时候 不能执行不然会空指针
                cur.next.random=cur.random.next;
            cur=cur.next.next;
        }
        cur=head.next;
        while(cur.next!=null){
            cur.next=cur.next.next;
            cur=cur.next;
        }
        return head.next;
    }
}