在java中使用链表制作dropout堆栈的麻烦 [英] trouble with making a dropout stack with linked lists in java

问题描述

所以在我的作业中，我必须用 Java 实现一个 dropout 堆栈.drop-out 堆栈在各个方面都像堆栈，除了如果堆栈大小为 n，那么当 n+1 元素被压入时，第一个元素会丢失.就我而言，我设置了 n=5.我的代码运行良好，但是当我在第 5 个元素之后添加更多元素时，底部的元素没有被删除.它只是像普通堆栈一样将新的堆栈堆叠在顶部.请帮助我了解如何解决此问题.这是我的堆栈实现代码:

So in my assignment, I have to implement a dropout stack in Java. A drop-out stack behaves like a stack in every respect except that if the stack size is n, then when the n+1 element is pushed, the first element is lost. In my case, I have set n=5. My code is running fine, but when I add more elements after the 5th element, the element at the bottom is not being removed. It just keeps stacking the new ones on top like a normal stack. Please help me understand how to fix this. Here is my code for the stack implementation:

/**
 * Represents a linked implementation of a stack.
 *
 * @author Java Foundations 
 * @version 4.0
 */
    public class DropOutStack<T> implements StackADT<T>
    {
    private int count; //number of elements in the stack
    private LinearNode<T> top; 

    /*Declares the maximum number of elements in the stack*/
    private final int n = 5;//max size
    private LinearNode<T> prev;
    private LinearNode<T> curr;

    /**
     * Creates an empty stack.
     */
    public DropOutStack()
    {
        count = 0;
        top = null;
    }

    /**
     * Adds the specified element to the top of this stack.
     * @param element element to be pushed on stack
     */
    public void push(T element)
    {
        LinearNode<T> temp = new LinearNode<T>(element);

        /*Verifies that the number of elements in the stack is
         * less than n. If yes, adds the new element to the stack*/
        if (count < n) {
            temp.setNext(top);
            top = temp;
            count++;
        }
        /*Verifies if the number of elements in the stack is greater
         * than or equal to n or not, and that the n is not equal to one.
         * If yes, removes the first element from the stack and adds
         * the new element to the stack*/
        else if(count>=n && n!=1) {
            prev = top;
            curr = top.getNext();

            while(curr != null) {
                prev = prev.getNext();
                curr = curr.getNext();
            }
            prev.setNext(null);
            count--;

            push(element);
        }
        else //if n=1
        {
            top.setElement(element);
        }
    }

    /**
     * Removes the element at the top of this stack and returns a
     * reference to it. 
     * @return element from top of stack
     * @throws EmptyCollectionException if the stack is empty
     */
    public T pop() throws EmptyCollectionException
    {
        if (isEmpty())
            throw new EmptyCollectionException("stack");

        T result = top.getElement();
        top = top.getNext();
        count--;

        return result;
    }

    /**
     * Returns a reference to the element at the top of this stack.
     * The element is not removed from the stack.  
     * @return element on top of stack
     * @throws EmptyCollectionException if the stack is empty  
     */
    public T peek() throws EmptyCollectionException
    {

        if (isEmpty())
            throw new EmptyCollectionException("stack");
        T result = top.getElement();

        return result;

    }

    /**
     * Returns true if this stack is empty and false otherwise. 
     * @return true if stack is empty
     */
    public boolean isEmpty()
    {
        return (count ==0);

    }

    /**
     * Returns the number of elements in this stack.
     * @return number of elements in the stack
     */
    public int size()
    {
        return count;
    }

    /**
     * Returns a string representation of this stack. 
     * @return string representation of the stack
     */
    public String toString()
    {
        String result = "";
        LinearNode<T> current = top;
        while (current != null) {
            result = current.getElement() + "\n" + result;
            current = current.getNext();
        }
        return result;
    }
    }

推荐答案

问题是您在 else if (count >= n && n !=1) 块.

当前你在 curr == null 时终止，但如果 curr 是 null 那么 prev 是最后一个列表中的节点，谁的下一个值已经是null (curr).

Currently you terminate when curr == null, but if curr is null then prev is the last node in the list, who's next value is already null (curr).

要解决此问题，请将循环条件更改为 curr.getNext() != null.由于您总是想推送元素，因此在最后执行并简化逻辑，前提是 n(最大大小 [可怜的名称])应至少为 1.

To resolve this, change the loop condition to curr.getNext() != null. Since you always want to push the element, just do that at the end and simplify the logic, given the precondition that n (the max size [poor name]) should be at least 1.

/**
 * @pre. n >= 1
 */
public void push(T element) {
    if (count >= n) {
        assert top != null;
        prev = null;
        curr = top;
        while (curr.getNext() != null) {
            prev = curr;
            curr = curr.getNext();
        }
        if (prev != null) {
            prev.setNext(null);
        } else {
            top = null;
        }
        count--;
    }
    LinearNode<T> temp = new LinearNode<>(element);
    temp.setNext(top);
    top = temp;
    count++;
}

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