我目前正在使用LinkedStacks，我想知道为什么toString和pop方法不起作用？我使用的方法是本书给出的默认方法，即Pop和toString方法，其余的我工作并且运行良好。推送方法完美地添加元素。窥视顶部元素而不改变List和Size，返回我使用push方法的次数。 pop方法奇怪地只工作一次然后发出错误。注意本书在Stacks部分给出toString方法的例子似乎也不起作用。我很乐意接受任何指示，但请知道我只是一个初学者而且我正在学习。这是该类的代码：
public class LinkedStack<T> implements Stack<T> {
private int count;
private LinearNode<T> top;
//-----------------------------------------------------------------
// Creates an empty stack using the default capacity.
//-----------------------------------------------------------------
public LinkedStack()
{
count = 0;
top = null;
}
@Override
public boolean IsEmpty()
{
if(top == null)
{
System.out.println("Stack is empty");
}
return top == null;
}
@Override
public void Push(T element)
{
top = new LinearNode(element, top);
System.out.println(element);
count++;
}
@Override
public T Pop()
{
T result;
System.out.println("Lets pop the top element!");
if (count == 0)
{
System.out.println("Pop operation failed. "+ "The stack is empty.");
}
result = top.getElement();
top = top.getNext();
count--;
System.out.println("The element that we have poped is :" + result);
return result;
}
Override
public String toString()
{
String result = "<top of stack>\n";
LinearNode current = top;
while (current != null)
{
result += current.getElement() + "\n";
current = current.getNext();
}
return result + "<bottom of stack>";
}
@Override
public T Peek() {
System.out.println("Lets peek the top element!");
if(count == 0)
{
System.out.println("Peek failed stack is empty");
}
System.out.println("The element that we have peeked is: " + top.getElement());
return top.getElement();
}
@Override
public int Size() {
System.out.println("The size of the list now is: " + count);
return count;
}
}
public class LSmain {
public static void main(String[]args)
{
LinkedStack<Integer> main = new LinkedStack<>();
main.Push(1);
main.Push(2);
main.Push(3);
main.Size();
main.Peek();
main.Pop();
main.Pop();
main.Size();
main.toString();
}
}
1
Exception in thread "main" java.lang.NullPointerException
2
3
The size of the list now is: 3
Lets peek the top element!
The element that we have peeked is: 3
Lets pop the top element!
The element that we have poped is: 3
Lets pop the top element!
at LinkNod.LinkedStack.Pop(LinkedStack.java:64)
at LinkNod.LSmain.main(LSmain.java:22)
Java Result: 1
BUILD SUCCESSFUL (total time: 0 seconds)
run:
1
2
3
The size of the list now is: 3
Lets peek the top element!
The element that we have peeked is: 3
Lets pop the top element!
The element that we have poped is: 3
The size of the list now is: 2
/ 这实际上会创建一个新节点并存储插入参数中的值 错误是我不能给同一个列表赋予最高分配，因为如果我这样做，List将只由一个顶部的Node组成。就像我过去的代码所说的那样，我只是说顶砖会与自己相等。所以我从类中理解的是创建一个新的对象类型LinearNode，然后存储元素，然后top将等于新Node的值。由于这种情况，Pop会工作，因为会有更多节点而不是一个节点。关于toString方法的额外注意事项就是返回;在java中有时会显示值，大多数时候它并不意味着你需要添加一个System.out.println（）;在调用方法或方法时在驱动程序中。 /
@Override
public void Push(T element)
{
LinearNode<T> current = new LinearNode<>(element);
current.setNext(top);
top = current;
count++;
}
public class LSmain {
public static void main(String[]args)
{
LinkedStack<Integer> list = new LinkedStack<>();
System.out.println("Let's make a List!");
System.out.println("Push 3 times.");
System.out.println("Check the size.");
System.out.println("Peek the top element.");
System.out.println("Pop three times.");
System.out.println("The size now should be zero!" + "\n");
list.Push(1);
list.Push(2);
list.Push(3);
System.out.println(list.toString());
list.Size();
list.Peek();
list.Pop();
list.Pop();
list.Pop();
list.Size();
}
run:
Let's make a List!
Push 3 times.
Check the size.
Peek the top element.
Pop three times.
The size now should be zero!
<top of stack-->[3][2][1]<--bottom of stack>
Let's check the size of the list!
The size of the list is: '3'
Lets peek the top element!
The element that we have peeked is: [3]
Lets pop the top element!
The element that we have poped is: '3'
Lets pop the top element!
The element that we have poped is: '2'
Lets pop the top element!
The element that we have poped is: '1'
The size of the list is...Woah.
The list size is now: '0'
Push more elements!
BUILD SUCCESSFUL (total time: 3 seconds)
感谢您的帮助！ PS：我忘了将方法声明更改为驼峰案例。
/ 当我尝试通过创建新的Node对象来纠正push方法时，出现了不正确的参数。 /
包LinkNod;
public class LinearNode<T> {
private LinearNode<T> next; //se guarda la referencia del Nodo
private T element; //Lista vacia
public LinearNode()
{
next = null;
element = null;
}
//-----------------------------------------------------------------
// Creates a node storing the specified element.
//-----------------------------------------------------------------
public LinearNode (T elem)
{
next = null;
element = elem;
}
//-----------------------------------------------------------------
// Returns the node that follows this one.
//-----------------------------------------------------------------
public LinearNode<T> getNext()
{
return next;
}
//-----------------------------------------------------------------
// Sets the node that follows this one.
//-----------------------------------------------------------------
public void setNext (LinearNode<T> node)
{
next = node;
}
//-----------------------------------------------------------------
// Returns the element stored in this node.
//-----------------------------------------------------------------
public T getElement()//asigna valor
{
return element;
}
public void setElement(T elem)
{
element = elem;
}
}
答案 0 :(得分：0)
这看起来像是问题：
if (count == 0)
{
System.out.println("Pop operation failed. "+ "The stack is empty.");
}
result = top.getElement(); //NPE here perhaps
top = top.getNext();
count--;
System.out.println("The element that we have poped is :" + result);
return result;
当您触发count == 0时，您不会从函数返回，而是继续处理弹出操作。
答案 1 :(得分：0)
Pop不起作用，因为您没有更新LinearNode toString不起作用，看起来没问题，但我会使用StringBuilder和.append()方法。