java链表数据插入在第N个位置.为什么函数insertN()不起作用? [英] java linked list data insertion at Nth position.Why The function insertN() is not working.?

问题描述

class Link{
    public  String val;
    public  Link nextlink;
    public Link(String val) {
        this.val = val;
    }
    public void linkNext(Link l){
        this.nextlink=l;
    }
    public void printlink()
    {
        System.out.print(" "+val);
   }
}
class linkList{
    Link first;
    public static int  count=0;
    public linkList(){
        first=null;
    }
    boolean isEmpty(){
        return first==null;
    }

    public void insertfirst(String str){

        Link next=new Link(str);
       count++;
       next.nextlink=first;
        first=next;
    }
    public void insertN(int pos,String str){
        if(pos<count){

         Link temp=first;
       count++;
       for(int i=0;i<pos-1;i++)
           temp=temp.nextlink;
       Link next=new Link(str);
     
       next=temp.nextlink;
         temp=next;
        }
    }
    public void printList(){
        Link currentLink=first;
      System.out.println("data is:");
        while(currentLink!=null){
            
            currentLink.printlink();
            currentLink=currentLink.nextlink;

    }
        System.out.println("  ");


}
}


public class LinkedList2 {
    public static void main(String[] args) {
        linkList li=new linkList();
        li.insertfirst("neeraj");
        li.insertfirst("rohit");
        li.insertfirst("lalit");
        li.insertfirst("meenu");
        li.printList();
        li.insertfirst("newly");
        li.insertN(9, "newlaay");
        li.printList();
        System.out.println(li.count);
    }

}

推荐答案

1. 为什么要public static int count=0;?
这意味着每个实例只有一个count，请删除static！

2.
将成员变量的访问权限更改为private.

3.
将方法linknext 更改为setNextLink，这是Java中的一个很好的约定.

4.
insertFirst必须使用方法setNextLink，因为成员varianble无法访问:

1. Why public static int count=0;?
That means there is a single count for every instance, remove the static!

2.
Change the access for the member variables to private.

3.
Change method linknext to setNextLink, this is a convention in Java that works well.

4.
insertFirst must use the method setNextLink, as the member varianble cannot be accessed:

public void insertFirst(String str) {
    Link next = new Link(str);
    count++;
    next.setNextLink(first);
    first=next;
}

5.
insertN将无法按计划工作.允许Link类为您完成工作:

5.
insertN will not work as planned. Allow the Link class to do the work for you :

class Link
    // stuff

    Link getLink (int position) {
        if (position <= 0 || this.nextlink == null) {
            return this.nextLink;
        }
        else
        {
            return this.nextlink.getLink(position - 1);
        }
    }
}

然后在位置后添加:

Then to add after a position:

public void insertAfter(int pos,String str){
    Link previous = this.first.getLink(pos);
    if (previous != null) {
        Link next = previous.getNextLink();
        Link insert = new Link(str);
        previous.setNextLink(insert);
        insert.setNextLink(next);
    }
}

简单且减少错误的代码.

Simple and less code to go wrong.

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