Question

我创建了一个链表，以及一些方法。我的任务是创建一个名单。在main中，指示用户输入名称，然后程序查看名称是否位于列表中。它确实编译，但我唯一无法弄清楚的是如何打印找不到名称。谢谢您的帮助！这是我的代码

package linkedlists;
import java.util.*;

public class Link {

public String Name;
public Link next;

public Link(String Name){
    this.Name = Name;
}

public void display(){
    System.out.println("Name in List " + Name);
}

@Override
public String toString(){
    return Name;
}

public static void main(String[] args) {

    Scanner input = new Scanner (System.in);


    LinkList theLinkedList = new LinkList();
    theLinkedList.insertFirstLink("James");
    theLinkedList.insertFirstLink("John");
    theLinkedList.insertFirstLink("Michael");
    theLinkedList.insertFirstLink("Peter");
    theLinkedList.insertFirstLink("Allison");
    theLinkedList.insertFirstLink("Daniel");
    theLinkedList.insertFirstLink("George");
    theLinkedList.insertFirstLink("Simon");
    theLinkedList.insertFirstLink("Jason");
    theLinkedList.insertFirstLink("Mark");

    System.out.print("Please enter a Name to search for");
    System.out.println();

    String name;
    name = input.nextLine();

    //System.out.println(theLinkedList.find(name).Name);

     if (theLinkedList.find(name).Name == null){
        System.out.println("Not Found");
        System.out.println();
        theLinkedList.display();
    }
    else {
        System.out.println(theLinkedList.find(name).Name + " was found.");
        System.out.println();
        theLinkedList.display();

    }






    }

}

这是LinkList类 class LinkList { public Link firstLink;

LinkList(){

    firstLink = null;

}

public boolean isEmpty(){
    return (firstLink == null);

}

public void insertFirstLink(String Name){
    Link newLink = new Link(Name);
    newLink.next = firstLink;
    firstLink = newLink;
}

public Link removeFirst(){
    Link linkReference = firstLink;
    if (!isEmpty()){
        firstLink = firstLink.next;

    } 
    else {
      System.out.println("Empty Linked List");

    }
    return linkReference;
}

public void display(){

    Link theLink = firstLink;

    while (theLink != null){
        theLink.display();
        System.out.println("Next Name in List: " + theLink.next);
        theLink = theLink.next;
        System.out.println();             
    }

}

public Link find(String Name){
    Link theLink = firstLink;

    if(!isEmpty()){

        while(!(theLink.Name.equals(Name))){

            if(theLink.next == null){
                return null;

            } else {
                theLink = theLink.next;

            }

        }

    } else {

        System.out.println("Empty LinkedList");

    }

    return theLink;

}

public Link removeLink(String Name){
    Link currentLink = firstLink;
    Link previousLink = firstLink;

    while(currentLink.Name != Name){
        if (currentLink.next == null){
            return null;
        }
        else{
            previousLink = currentLink;
            currentLink = currentLink.next;

        }

    }
    if (currentLink == firstLink){
        firstLink = firstLink.next;
    }
    else { 
        System.out.println(" Found a match!");
        System.out.println("Current Link: " + currentLink);
        System.out.println("First Link: " + firstLink);

        previousLink.next = currentLink.next;

    }
    return currentLink;
    }

}

Answer 1

遍历LinkedList的模式通常保持不变。

检查列表是否为空 - ＆gt;可以立即返回
存储对第一个链接的引用
虽然还有任何链接，但请继续，否则请转到6
做一些逻辑，当你遇到某种情况时回来
移动链接，（返回4）
循环结束，返回函数的最终状态

代码

public Link find(String Name){

    if(isEmpty()) {
        System.out.println("Empty LinkedList");
        return null;
    } 

    Link theLink = firstLink;
    while(theLink != null) {
        if (theLink.Name.equals(Name)) {
            return theLink;
        }

        theLink = theLink.next;
    }

     // looped through the whole list, didn't find anything
    return null;
}

现在，这是一个迭代解决方案，但更先进的解决方案是以递归方式编写

public Link find(String Name) {
    if(isEmpty()) {
        System.out.println("Empty LinkedList");
        return null;
    } 
    return findRecur(firstLink, Name);
}

private Link findRecur(Link link, String Name){
    if (link == null) return null;
    if (link.Name.equals(Name)) return link;
    return findRecur(link.next, Name);
}

无论哪种方式，请致电

final Link found = theLinkedList.find(name);
if (found == null){
    System.out.println("Not Found");
} else {
    System.out.println(found.Name + " was found.");
}
System.out.println();
theLinkedList.display();

Answer 2

更换线 if（theLinkedList.find（name）.Name == null）{

与

if（null == theLinkedList.find（name））{

您的语句尝试获取与输入的名称相关的Link对象，但是当输入的名称不在列表中时，返回null而不是Link对象。只需检查查找函数中返回的值是否为null，以确保列表中不存在该名称即可。您的代码尝试对名称不存在时返回的null Link对象进行操作 - 这将导致您必须看到的空指针异常。

链接列表输出

2 个答案: