不考虑空格、标点符号和大小写的回文程序

Question

我正在尝试创建一个回文程序，在确定字符串是否为回文时不考虑空格、标点符号以及大小写。

我如何更改此代码以执行我之前所说的操作？

package palindrome;

import java.util.Scanner;

public class Palindrome {

   public static void main (String[] args)
 {
  String str, another = "y";
  int left, right;
  Scanner scan = new Scanner (System.in);

  while (another.equalsIgnoreCase("y")) // allows y or Y
  {
     System.out.println ("Enter a potential palindrome:");
     str = scan.nextLine();



     left = 0;
     right = str.length() - 1;

     while (str.charAt(left) == str.charAt(right) && left < right)
     {
        left++;
        right--;
     }

     System.out.println();

     if (left < right)
        System.out.println ("That string is NOT a palindrome.");
     else
        System.out.println ("That string IS a palindrome.");

     System.out.println();
     System.out.print ("Test another palindrome (y/n)? ");
     another = scan.nextLine();
  }

} }

Answer 1

您当前的代码查看变量 str 并检查从左到右和从右到左读取的字符是否相同（也就是说，毕竟，回文是什么）。

目前，它对用户输入的原始字符串执行此操作。要更改此设置，请在内部 while 循环之前对变量 str 进行一些过滤。我将让您弄清楚到底要过滤什么/如何过滤，但请查看 String 类，如 indexOf()、replace() 和 substring()。

Answer 2

The heart of your program is in following loop

while (str.charAt(left) == str.charAt(right) && left < right)
    {
        left++;
        right--;
    }


what you are doing here is to compare the characters without honoring the condition of ignoring space and punctuation here. Your logic should be such that while picking the characters for comparison (either from left or right) you will need to skip that character and move for the next character.
To make picture clearer see the following :

Input string :
inStr = Ma'lyalam

Step 1:
Since you have to ignore the case do following
inStr = inStr.toLowerCase();

Step 2:

 1. int left =0 , right = 8
 2. char chLeft, chRight
 3. chLeft = m , chRight = m 
 4. chLeft == chRight -> true, increment left and decrement right
 5. chLeft = a , chRight = a
 6. chLeft == chRight -> true, increment left and decrement right
 7. chLeft = ' , chRight = l -> Since chLeft = ' skip it, increment left so chLeft = l. Now continue like above

so the code should look like

    boolean isPlaindrome = false;
    str = str.toLowerCase();
    char chLeft, chRight;
    while (left < right)
    {
    chLeft = str.charAt(left);
    chRight = str.charAt(right)
    if (chLeft == ''' || chLeft == ' ')
     {
       left++
       continue;
     }
     else if (chRight == ''' || chRight == ' ')
     {
       right--;
       continue;
     }

     if (chLeft == chRight)
    {
     left++; right--;
    }
    else
    {
    break;
    }
    }

    if (left == right)
     isPlaindrome = true;