使用Java的回文测试器,忽略空格和标点符号

Question

我有程序编写,直到它必须忽略和线程中的标点和空格,我想知道是否有人可以帮我编码？我一直在尝试的东西似乎没有起作用.这是我到目前为止:

import java.util.Scanner;

public class PalindromeTester
{
public static void main (String[] args)

{

    String str, another = "y";

    int left, right;

    char charLeft, charRight;


    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 (left < right)
        {
            charLeft = str.charAt(left);
            charRight = str.charAt(right);


            if (charLeft == charRight)
            {
                left++;
                right--;
            }

            else if (charLeft == ',' || charLeft == '.' ||
 charLeft == '-' || charLeft == ':' ||
 charLeft == ';' || charLeft == ' ')

                left++;

            else if (charRight == ',' || charRight == '.' ||
 charRight == '-' || charRight == ':' ||
 charRight == ';' || charRight == ' ')
                right--;
            else

                break;

        }

    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

只是为了澄清Jim Garrison所说的,你需要的正则表达式如下

String m = "Madam, I'm'',.,.''   Adam";
m = m.toLowerCase().replaceAll("\\W", "");

这将只留下字母和数字,并删除空格和标点符号,即m将成为"madamimadam",你可以对该字符串进行常规的回文测试.

您可以在此处详细了解正则表达式

Answer 2

用于确定单词是否是回文结构的该代码可以更加简化.查找更新的代码

String word;
int z;
int y = 0;
int i = 0;

char letter;

Scanner input = new Scanner(System.in);

System.out.print("Enter a word: ");
word = input.nextLine();

word = word.replaceAll("\\s+", "");
word = word.toLowerCase();

z = word.length()-1;
while (i <= z){

    if ((letter = word.charAt(i)) == (letter = word.charAt(z-i))){
        y += 1;
    }
    i += 1;
}

if (y == (z+1)){
    System.out.println("The word IS a palindrome");
}
else{
    System.out.println("The word is NOT a palindrome");
}

}
}

Answer 3

这看起来像一个很老的帖子,但我想我偶然发现了一个更简单的回文测试解决方案.这将检查第一个和最后一个字符并向内移动,并在字符不匹配时立即退出程序.

public class CharTest {
    public static void main(String[] args) {
             //converts string to lowercase and replaces everything except numbers
             // and alphabets
        String s = "Niagara. O roar again!".toLowerCase().replaceAll("\\W", "");
        int j=0;
        int k = s.length() - 1;
        while(j < s.length() / 2) { //loops until half the length of the string if 
                                        //even and floor value if odd.
            if (s.charAt(j++) != s.charAt(k--)){//check for first and last chars                                                                                                
                                              //and  go inwards. if char do not match print 'Not a Palindrome' and exit 
                System.out.println("Not a Palindrome");
            System.exit(0);}
        }
        System.out.println("Palindrome");  //if every chars match print "Palindrome"
    }
}