如何编写代码指示用户输入的文本中的第一个位置违反回文属性？

Question

Say I would input “Able were you ere you saw Elba”. This looks like a palindrome until you see the first “e” in “were”, so the output for this text would be : This is not a palindrome Mismatch detected at : Ablewe

下面是驱动程序

public static void main(String[] args) {

        Scanner input = new Scanner(System.in);
        String line;

        do {

            // prompt user for input String and store it in line
            System.out.print("Enter an expression (or hit Enter to exit) : ");
            line = input.nextLine();

            // convert input String to upper case
            line = line.toUpperCase();

            // if user hits Enter or simply types in spaces and hits enter, break out of loop
            if (line.trim().isEmpty()) {
                break;
            }

            // call isPalindrom
            // if it returns true, display one message, else display another
            if (Palindrome.isPalindrome(line)) {
                System.out.println("Your expression is a palindrome.");
            } else {
                System.out.println("Your expression is not a palindrome.");
            }

        } while (line.length() != 0);

        System.out.println("You didn't enter an expression.  Exiting application ...");
    }
}

Below is the class that has the isPalindrome method

 public static boolean isPalindrome(String input){
        
        // create Queue and Stack of Characters
        // to store the input String
        Queue<Character> q = new LinkedList<>();
        Stack<Character> s = new Stack();
        
        // temporarily store the individual Characters
        // in input String before they're pushed onto
        // Stack and added to Queue
        Character letter;
        
        // keep track of differences between Characters
        // in Stack and Queue
        int mismatches = 0;
        
        // loop through input String
        for (int i = 0; i < input.length(); i++){
            // get current Character
            letter = input.charAt(i);
            
            // if the current Character is an alphabetic character
            if (Character.isLetter(letter)){
                // push it onto the Stack
                s.push(letter);
                // add it to the Queue
                q.add(letter);
            }
        }
        
        // while the Queue isn't empty
        while (!q.isEmpty()){
            // remove a Character from the Queue
            // pop a Character from the Stack
            // if they're not equal, increment mismatches
            if (!q.remove().equals(s.pop()))
                // Stack will produce the input String backwards
                // Queue will produce the input String forwards
                mismatches++;
        }
        
        // return true if there are no mismatches, else return false
        return (mismatches == 0);
    }
}

I need the output to show where the mismatch occurs

Answer 1

通过用空字符串替换所有非字母来创建一个字符串（下面给出的代码中的allLetters），即input.replaceAll("[^\p{L}]", "")，其中^\p{L}指定一个non-letter character。
使用变量（下面给出的代码中的index）来跟踪已处理字母的位置。当检测到不匹配时，打印 allLetters.substring(0, index + 1) 并中断循环。

代码：

int index = 0;
String allLetters = input.replaceAll("[^\p{L}]", "");
// while the Queue isn't empty
while (!q.isEmpty()) {
    // remove a Character from the Queue
    // pop a Character from the Stack
    // if they're not equal, increment mismatches
    if (!q.remove().equals(s.pop())) {
        // Stack will produce the input String backwards
        // Queue will produce the input String forwards
        mismatches++;
        System.out.println("Mismatch detected at: " + allLetters.substring(0, index + 1));
        break;
    }
    index++;
}

样本运行:

Enter an expression (or hit Enter to exit) : Able were you ere you saw Elba
Mismatch detected at: ABLEWE
Your expression is not a palindrome.
Enter an expression (or hit Enter to exit) :

注意： 目前，您正在将转换为大写的用户输入传递给函数 isPalindrome，因此输出字符串 (ABLEWE在上面显示的示例运行中）是大写的。更改大小写的更好位置是 isPalindrome 本身，如下所示：

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.Stack;

public class Palindrome {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        String line;

        do {

            // prompt user for input String and store it in line
            System.out.print("Enter an expression (or hit Enter to exit) : ");
            line = input.nextLine();

            // if user hits Enter or simply types in spaces and hits enter, break out of
            // loop
            if (line.trim().isEmpty()) {
                break;
            }

            // call isPalindrom
            // if it returns true, display one message, else display another
            if (Palindrome.isPalindrome(line)) {
                System.out.println("Your expression is a palindrome.");
            } else {
                System.out.println("Your expression is not a palindrome.");
            }

        } while (line.length() != 0);

        System.out.println("You didn't enter an expression.  Exiting application ...");
    }

    public static boolean isPalindrome(String input) {
        // Create a string by replacing all non-letters by empty string
        String allLetters = input.replaceAll("[^\p{L}]", "");

        // convert input String to upper case
        input = input.toUpperCase();

        // create Queue and Stack of Characters
        // to store the input String
        Queue<Character> q = new LinkedList<>();
        Stack<Character> s = new Stack();

        // temporarily store the individual Characters
        // in input String before they're pushed onto
        // Stack and added to Queue
        Character letter;

        // keep track of differences between Characters
        // in Stack and Queue
        int mismatches = 0;

        // loop through input String
        for (int i = 0; i < input.length(); i++) {
            // get current Character
            letter = input.charAt(i);

            // if the current Character is an alphabetic character
            if (Character.isLetter(letter)) {
                // push it onto the Stack
                s.push(letter);
                // add it to the Queue
                q.add(letter);
            }
        }

        int index = 0;
        // while the Queue isn't empty
        while (!q.isEmpty()) {
            // remove a Character from the Queue
            // pop a Character from the Stack
            // if they're not equal, increment mismatches
            if (!q.remove().equals(s.pop())) {
                // Stack will produce the input String backwards
                // Queue will produce the input String forwards
                mismatches++;
                System.out.println("Mismatch detected at: " + allLetters.substring(0, index + 1));
                break;
            }
            index++;
        }

        // return true if there are no mismatches, else return false
        return (mismatches == 0);
    }
}

样本运行:

Enter an expression (or hit Enter to exit) : Able were you ere you saw Elba
Mismatch detected at: Ablewe
Your expression is not a palindrome.
Enter an expression (or hit Enter to exit) :

如何编写代码指示用户输入的文本中的第一个位置违反回文属性？

How can I write code that indicates the first position in the text input by the user violates the palindrome property?

java

stack

queue

palindrome

如何编写代码指示用户输入的文本中的第一个位置违反回文 属性？

How can I write code that indicates the first position in the text input by the user violates the palindrome property?

java

stack

queue

palindrome

如何编写代码指示用户输入的文本中的第一个位置违反回文属性？