Question

我的任务是在一个程序中使用ADT List的基于引用的实现和ADT Stack的基于数组的实现，该程序让用户输入一串小写字母。我是通过字符串并将每个字母存储在列表和堆栈中，然后使用堆栈和列表内容来确定字符串是否是回文。我将以相反的顺序显示原始的字母序列，字母序列，最后显示是否是回文的陈述。出于某种原因，当我输入回文时，例如madamimadam，它输出它不是回文。我无法弄清楚为什么，请帮忙！这是我的方法代码：

import javax.swing.JOptionPane;

public class PalindromeTester
{    
    public static void main (String [] args)
    {    
        Character ch;
        boolean isPalindrome = true;
        LinkedList myList = new LinkedList();
        StackArrayBased myStack = new StackArrayBased();
        String response = JOptionPane.showInputDialog ("Please enter a string of lower-case letters" ) ;

        for ( int i = 0 ; i < response.length ( ) ; i++ )
        {
            ch = new Character ( response.charAt ( i ) ) ;
            myStack.push ( ch ) ;
            myList.add ( i + 1 , ch ) ;
        }

        System.out.println ( "The original sequence of characters is: " + response ) ;
        System.out.print ( "The sequence of letters backwards is: " ) ;

        int j = 1 ;
        while ( ! myStack.isEmpty ( ) )
        {
            System.out.print ( myStack.peek ( ) ) ;
            if ( ! myList.get ( j ).equals( myStack.pop (  ) ) ) ;
            isPalindrome = false ;
        }

        if ( isPalindrome )
            System.out.println ( "\nThe string is a palindrome." ) ;
        else
            System.out.println ( "\nThe string is not a palindrome." ) ;
    }
}

这是ADT Stack类：

public class StackArrayBased
{
    private static final int MAX_STACK = 15 ;
    private Object items [ ] ;
    private int top ;    

    public StackArrayBased ( )
    {
        items = new Object [ MAX_STACK ] ;
        top = -1 ;
    }

    public boolean isEmpty ( )
    {
        return top < 0 ;
    } 

    public boolean isFull ( )
    {
        return top == MAX_STACK - 1 ;
    }

    public void push ( Object newItem ) throws StackException
    {
        if ( ! isFull ( ) )
            items [ ++ top ] = newItem ;
        else
            throw new StackException ( "StackException on push: stack is full" ) ;
    }

    public void popAll ( )
    {
        items = new Object [ MAX_STACK ] ;
        top = -1 ;
    }

    public Object pop ( ) throws StackException
    {
        if ( ! isEmpty ( ) )
            return items [ top -- ] ;
        else
            throw new StackException ( "StackException on pop: stack is empty" ) ;
    }

    public Object peek ( ) throws StackException
    {
        if ( ! isEmpty ( ) )
            return items [ top ] ;
        else
            throw new StackException ( "StackException on peek: stack is empty" ) ;
    }
}

这是ADT列表：

public class LinkedList
{
    private Node head;
    private int numItems;    

    public LinkedList ( )
    {
        head = null ;
        numItems = 0 ;
    }

    public boolean isEmpty ( )
    {
        return numItems == 0 ;
    }

    public int size ( )
    {
        return numItems ;
    }

    private Node find ( int position )
    {
        Node curr = head ;
        for ( int i = 1 ; i < position ; i ++ )
            curr = curr.getNext ( ) ;

        return curr ;
    }

    public Object get ( int position )
    {
        if ( position >= 0 && position <= numItems )
        {
            Node curr = find ( position ) ;
            Object dataItem = curr.getItem ( ) ;
            return dataItem ;
        }
        else
        {
            System.out.println ( "Error in position value during get attempt." ) ;
            return null ;
        }
    }

    public void add ( int position, Object item )
    {
        if ( position >= 1 && position <= numItems + 1 )
        {
            if ( position == 1 )
            {
                Node newNode = new Node ( item, head ) ;
                head = newNode ;
            }
            else
            {
                Node prev = find ( position - 1 ) ;
                Node newNode = new Node ( item, prev.getNext ( ) ) ;
                prev.setNext ( newNode ) ;
            }

            numItems ++ ;
        }
        else
            System.out.println ( "Position is invalid on attempted add." ) ;
    }

    public void remove ( int position )
    {
        if ( position >= 1 && position <= numItems )
        {
            if ( position == 1 )
                head = head.getNext ( ) ;
            else
            {
                Node prev = find ( position - 1 ) ;
                Node curr = prev.getNext ( ) ;
                prev.setNext ( curr.getNext ( ) ) ;
            }

            numItems -- ;
        }
        else
            System.out.println ( "Position is invalid on attempted remove." ) ;
    }

    public void removeAll ( )
    {
        head = null ;
        numItems = 0 ;
    }
}

Answer 1

如果您想正确设置isPalindrome，那么您不应该在此循环中使用j做某事......？：

[...]
int j = 1 ;
while ( ! myStack.isEmpty ( ) )
{
  System.out.print ( myStack.peek ( ) ) ;
  if ( ! myList.get ( j ).equals( myStack.pop (  ) ) ) ;
        isPalindrome = false ;
}
[...]

Answer 2

在第二个循环中，您应该递增j。由于linkedlist索引可以为0，因此在添加时（第一个循环中）不应该执行i + 1索引。如果你使它成为一个基于0的索引，你应该在第二个循环之前将j初始化为0。

Answer 3

任务似乎很奇怪;如果你可以访问列表的最后一个元素（如大多数语言中的抽象列表所允许的那样），那么你可以只执行for i=[0,length) {if input[i]!=input[length-1-i], return false} return true

如果你只是要使用堆栈，你可以克隆并反转堆栈（例如，通过推入两个堆栈，然后将其中一个弹出一个新的堆栈，从而将其反转），并执行相同的操作for循环（逐个元素地遍历两个堆栈以查看它们是否相同）。

在上述两种线性时间算法（使用仅列表的算法或仅使用堆栈的算法）中，函数应该只有3行左右。

为什么我的程序没有检测到回文？

3 个答案: