Java 如何修复删除节点并在之后转到最后一个节点?
我在从用户输入中删除一个节点并正确转到最后一个节点时遇到问题,因此它将准备在之后添加一个新节点。我正在将此代码重构为一个更大的实现。Java 如何修复删除节点并在之后转到最后一个节点?,java,generic-list,Java,Generic List,我在从用户输入中删除一个节点并正确转到最后一个节点时遇到问题,因此它将准备在之后添加一个新节点。我正在将此代码重构为一个更大的实现。但是,我无法删除该节点并转到之后的最后一个节点。这还使用用户输入来查找要删除的适当节点。这是一个类似类型的通用链接列表 import java.util.Scanner; import java.io.*; class MyGenericList <T extends Comparable<T>> { private class
但是,我无法删除该节点并转到之后的最后一个节点。这还使用用户输入来查找要删除的适当节点。这是一个类似类型的通用链接列表
import java.util.Scanner;
import java.io.*;
class MyGenericList <T extends Comparable<T>>
{
private class Node<T>
{
T value;
Node<T> next;
}
private Node<T> first = null;
int count = 0;
public void add(T element)
{
Node<T> newnode = new Node<T>();
newnode.value = element;
newnode.next = null;
if (first == null)
{
first = newnode;
}
else
{
Node<T> lastnode = gotolastnode(first);
lastnode.next = newnode;
}
count++;
}
public void remove(T element)
{
Node<T> nn = new Node<T>();
Node<T> cur = first.next;
Node<T> prev = first;
nn.value = element;
boolean deleted = false;
while(cur != null && deleted == false)
{
if(cur.equals(element)) //data cannot be resolved or is not a field
{
prev.next = cur.next;
this.count--;
deleted = true;
}
}
prev = gotolastnode(prev);
prev.next = nn;
}
public T get(int pos)
{
Node<T> Nodeptr = first;
int hopcount=0;
while (hopcount < count && hopcount<pos)
{ if(Nodeptr != null)
{
Nodeptr = Nodeptr.next;
}
hopcount++;
}
return Nodeptr.value;
}
private Node<T> gotolastnode(Node<T> nodepointer)
{
if (nodepointer== null )
{
return nodepointer;
}
else
{
if (nodepointer.next == null)
return nodepointer;
else
return gotolastnode( nodepointer.next);
}
}
}
class Employee implements Comparable<Employee>
{
String name;
int age;
@Override
public int compareTo(Employee arg0)
{
// TODO Auto-generated method stub
return 0;
// implement compareto method here.
}
Employee( String nm, int a)
{
name =nm;
age = a;
}
}
class City implements Comparable<City>
{
String name;
int population;
City( String nm, int p)
{
name =nm;
population = p;
}
@Override
public int compareTo(City arg0) {
// TODO Auto-generated method stub
return 0;
// implement compareto method here.
}
}
public class GenericLinkedList
{
public static void main(String[] args) throws IOException
{
MyGenericList<Employee> ml = new MyGenericList<>();
ml.add(new Employee("john", 32));
ml.add(new Employee("susan", 23));
ml.add(new Employee("dale", 45));
ml.add(new Employee("eric", 23));
Employee e1 = ml.get(0);
System.out.println( "Name " + e1.name + " Age "+ e1.age );
ml.remove(new Employee("john", 32));
System.out.println( "Name " + e1.name + " Age "+ e1.age );
ml.add(new Employee("jerry", 35));
Employee e2 = ml.get(2);
System.out.println( "Name " + e2.name + " Age "+ e2.age );
}
}
import java.util.Scanner;
导入java.io.*;
类MyGenericList
{
私有类节点
{
T值;
节点下一步;
}
private Node first=null;
整数计数=0;
公共无效添加(T元素)
{
Node newnode=新节点();
newnode.value=元素;
newnode.next=null;
if(first==null)
{
第一个=新节点;
}
其他的
{
Node lastnode=gotolastnode(第一个);
lastnode.next=newnode;
}
计数++;
}
公共无效删除(T元素)
{
节点nn=新节点();
节点cur=first.next;
节点prev=第一;
nn.value=元素;
布尔删除=假;
while(cur!=null&&deleted==false)
{
if(cur.equals(element))//数据无法解析或不是字段
{
上一个=当前下一个;
这是伯爵;
删除=真;
}
}
prev=gotolastnode(prev);
prev.next=nn;
}
公共T-get(内部位置)
{
节点Nodeptr=第一个;
int hopcount=0;
而(hopcountremove
方法的实现出现故障。请参阅下面的固定remove
方法。已添加注释以解释更改
该解决方案已通过测试,并经验证工作正常
public void remove(T element)
{
if(first == null) { // edge case - empty list
return;
}
else if(first.value.equals(element)) { // edge case - removing the first element
first = first.next;
this.count--;
return;
}
//Node<T> nn = new Node<T>(); // no need to create a new node, but rather remove an existing node.
Node<T> cur = first.next;
Node<T> prev = first;
//nn.value = element; //no need to create a new node and set its value attribute
boolean deleted = false;
while(cur != null && deleted == false)
{
if(cur.value.equals(element)) //data cannot be resolved or is not a field
{
prev.next = cur.next;
this.count--;
deleted = true;
}
else { // added missing advancement of the loop iterator - cur. prev must also be advanced
cur = cur.next;
prev = prev.next;
}
}
// This implementation adds the removed element to the end of the list, meaning
// it is not a remove method, but rather a move to the end implementation.
// In order to conform to what a remove method does, the last two code lines were commented out.
//prev = gotolastnode(prev);
//prev.next = nn;
}
感谢您的详细解释。为了正确使用remove,我将使用.remove(newemployee(“name”,int))?我一直在玩弄代码,它似乎只是在删除第一个员工。当我尝试删除任何其他员工时,代码不起作用。这是我的问题还是你有相同的问题?我只能删除第一个节点,但当我尝试任何其他节点时,它不起作用。@MorrisZieve:抱歉。更正的解决方案,请参阅更新上面的代码(必须对节点的值进行比较,而不是对节点本身进行比较。缺少prev迭代器的高级)。如果我想向类中添加另一个对象。例如,如果我想添加课程。我会在.compareTo()中返回其他对象吗在age之下并将其添加到公共布尔值?@MorrisZieve:对于其他类,必须同时实现
copareTo
和equals
。对于现有类中的其他属性,还必须更新compareTo
和equals
方法以说明新属性。作为补充说明,这是很高的我们强烈建议您也重写hashcode
方法,因为它与equals
方法耦合在一起。
class Employee implements Comparable<Employee>
{
String name;
int age;
@Override
public int compareTo(Employee arg0)
{
// sort first by name, then by age
if(name.equals(arg0.name)) {
return age - arg0.age;
}
return name.compareTo(arg0.name);
}
Employee( String nm, int a)
{
name =nm;
age = a;
}
@Override
public boolean equals(Object emp) {
boolean result = false;
if(emp != null && emp instanceof Employee) {
Employee e = (Employee)emp;
result = name.equals(e.name) && (age == e.age);
}
return result;
}
}