Question

Question on User-defined methods.

WAP using a method IsPalindrome(String str) to check if the string is palindrome or not.
The function returns true if it is palindrome else will return false.

Programming language: Java.

Good luck, :)
​

Answer 1

import java.util.*;

public class PalindromeString

{

public static void main(String args [ ] )

{

Scanner scan = new Scanner(System.in);

int length;

String reverseString = '' '' ;

System.out.println(''Enter a word: '');

String inputString = scan.nextLine();

length = inputString.length();

for (int i =length-1; i >=0; i --)

{

reverseString = reverseString + inputString.charAt(i);

}

if (reverseString.equalsIgnoreCase(inputString))

System.out.println(inputString + '' is palindrome'');

else

System.out.println(inputString + '' is not palindrome'');

scan.close();

}

}

Output :-

Enter a word:

Radar

Radar is Palindrome

Palindrome :

It a number/digit which remains same even after reversing the digits/letters of it. Ex: 111 , Madam.

Answer 2

Solution:

The given problem is solved using Java.

import java.util.*;

public class CheckPalindrome{

   public static void main(String s[]){

       System.out.print("Enter a string: ");

       String str=(new Scanner(System.in)).nextLine();

       System.out.println(">> isPalindrome(\""+str+"\") = "+isPalindrome(str));

   }

   static boolean isPalindrome(String str){

       return str.equalsIgnoreCase(new StringBuilder(str).reverse().toString());

   }

}

Sample I/O:

Enter a string: LOL

>> isPalindrome("LOL") = true

See attachment for output.