Creating a recursive method for Palindrome

Question

I am trying to create a Palindrome program using recursion within Java but I am stuck, this is what I have so far:

 public static void main (String[] args){
         


        

Answer 1

Try this:

package javaapplicationtest;

public class Main {

    public static void main(String[] args) {

        String source = "mango";
        boolean isPalindrome = true;

        //looping through the string and checking char by char from reverse
        for(int loop = 0; loop < source.length(); loop++){          
            if( source.charAt(loop) != source.charAt(source.length()-loop-1)){
                isPalindrome = false;
                break;
            }
        }

         if(isPalindrome == false){
             System.out.println("Not a palindrome");
         }
         else
             System.out.println("Pailndrome");

    }

}

Answer 2

for you to achieve that, you not only need to know how recursion works but you also need to understand the String method. here is a sample code that I used to achieve it: -

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


    Scanner sc=new Scanner(System.in);
    System.out.println("Enter a string");
    String input=sc.next();
    System.out.println("is "+ input + "a palindrome : " +  isPalindrome(input));


  }

  public static  boolean isPalindrome(String s)
  {
    int low=0;
    int high=s.length()-1;
    while(low<high)
    {
      if(s.charAt(low)!=s.charAt(high))
      return false;
      isPalindrome(s.substring(low++,high--));
    }

    return true;
  }
}

Answer 3

Here is the code for palindrome check without creating many strings

public static boolean isPalindrome(String str){
    return isPalindrome(str,0,str.length()-1);
}

public static boolean isPalindrome(String str, int start, int end){
    if(start >= end)
        return true;
    else
        return (str.charAt(start) == str.charAt(end)) && isPalindrome(str, start+1, end-1);
}

Answer 4

Here is my go at it:

public class Test {

    public static boolean isPalindrome(String s) {
        return s.length() <= 1 ||
            (s.charAt(0) == s.charAt(s.length() - 1) &&
             isPalindrome(s.substring(1, s.length() - 1)));
    }


    public static boolean isPalindromeForgiving(String s) {
        return isPalindrome(s.toLowerCase().replaceAll("[\\s\\pP]", ""));
    }


    public static void main(String[] args) {

        // True (odd length)
        System.out.println(isPalindrome("asdfghgfdsa"));

        // True (even length)
        System.out.println(isPalindrome("asdfggfdsa"));

        // False
        System.out.println(isPalindrome("not palindrome"));

        // True (but very forgiving :)
        System.out.println(isPalindromeForgiving("madam I'm Adam"));
    }
}

Answer 5

public static boolean isPalindrome(String str)
{
    int len = str.length();
    int i, j;
    j = len - 1;
    for (i = 0; i <= (len - 1)/2; i++)
    {
      if (str.charAt(i) != str.charAt(j))
      return false;
      j--;
    }
    return true;
} 

Answer 6

public static boolean isPalindrome(String in){
   if(in.equals(" ") || in.length() < 2 ) return true;
   if(in.charAt(0).equalsIgnoreCase(in.charAt(in.length-1))
      return isPalindrome(in.substring(1,in.length-2));
   else
      return false;
 }

Maybe you need something like this. Not tested, I'm not sure about string indexes, but it's a start point.