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){
 System.out.println(isPalindrome("noon"));
 System.out.println(isPalindrome("Madam I'm Adam"));
 System.out.println(isPalindrome("A man, a plan, a canal, Panama"));
 System.out.println(isPalindrome("A Toyota"));
 System.out.println(isPalindrome("Not a Palindrome"));
 System.out.println(isPalindrome("asdfghfdsa"));
}

public static boolean isPalindrome(String in){
 if(in.equals(" ") || in.length() == 1 ) return true;
 in= in.toUpperCase();
 if(Character.isLetter(in.charAt(0))
}

public static boolean isPalindromeHelper(String in){
 if(in.equals("") || in.length()==1){
  return true;
  }
 }
}

Can anyone supply a solution to my problem?

Answer 1

Here I am pasting code for you:

But, I would strongly suggest you to know how it works,

from your question , you are totally unreadable.

Try understanding this code. Read the comments from code

import java.util.Scanner;
public class Palindromes
{

    public static boolean isPal(String s)
    {
        if(s.length() == 0 || s.length() == 1)
            // if length =0 OR 1 then it is
            return true; 
        if(s.charAt(0) == s.charAt(s.length()-1))
            // check for first and last char of String:
            // if they are same then do the same thing for a substring
            // with first and last char removed. and carry on this
            // until you string completes or condition fails
            return isPal(s.substring(1, s.length()-1));

        // if its not the case than string is not.
        return false;
    }

    public static void main(String[]args)
    {
        Scanner sc = new Scanner(System.in);
        System.out.println("type a word to check if its a palindrome or not");
        String x = sc.nextLine();
        if(isPal(x))
            System.out.println(x + " is a palindrome");
        else
            System.out.println(x + " is not a palindrome");
    }
}

Answer 2

Well:

• It's not clear why you've got two methods with the same signature. What are they meant to accomplish?
• In the first method, why are you testing for testing for a single space or any single character?
• You might want to consider generalizing your termination condition to "if the length is less than two"
• Consider how you want to recurse. One option:
  • Check that the first letter is equal to the last letter. If not, return false
  • Now take a substring to effectively remove the first and last letters, and recurse
• Is this meant to be an exercise in recursion? That's certainly one way of doing it, but it's far from the only way.

I'm not going to spell it out any more clearly than that for the moment, because I suspect this is homework - indeed some may consider the help above as too much (I'm certainly slightly hesitant myself). If you have any problems with the above hints, update your question to show how far you've got.

Answer 3

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.

Answer 4

I think, recursion isn't the best way to solve this problem, but one recursive way I see here is shown below:

String str = prepareString(originalString); //make upper case, remove some characters 
isPalindrome(str);

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

private boolean isPalindrome(String str, int i) {
       if (i > str.length / 2) {
      return true;
   }
   if (!str.charAt(i).equals(str.charAt(str.length() - 1 - i))) {
      return false;
   }
   return isPalindrome(str, i+1);
}

Answer 5

public class palin
{ 
    static boolean isPalin(String s, int i, int j)
    {
        boolean b=true;
        if(s.charAt(i)==s.charAt(j))
        {
            if(i<=j)
                isPalin(s,(i+1),(j-1));
        }
        else
        {
            b=false;
        }
        return b;
    }
    public static void main()
    {
        String s1="madam";
        if(isPalin(s1, 0, s1.length()-1)==true)
            System.out.println(s1+" is palindrome");
        else
            System.out.println(s1+" is not palindrome");
    }
}

Answer 6

Some of the codes are string heavy. Instead of creating substring which creates new object, we can just pass on indexes in recursive calls like below:

private static boolean isPalindrome(String str, int left, int right) {
    if(left >= right) {
        return true;
    }
    else {
        if(str.charAt(left) == str.charAt(right)) {

            return isPalindrome(str, ++left, --right);
        }
        else {
            return false;
        }
    }
}


 public static void main(String []args){
    String str = "abcdcbb"; 
    System.out.println(isPalindrome(str, 0, str.length()-1));
 }

Answer 7

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 8

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 9

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 10

String source = "liril";
StringBuffer sb = new StringBuffer(source);
String r = sb.reverse().toString();
if (source.equals(r)) {
    System.out.println("Palindrome ...");
} else {
    System.out.println("Not a palindrome...");
}

Answer 11

public class chkPalindrome{

public static String isPalindrome(String pal){

if(pal.length() == 1){

return pal;
}
else{

String tmp= "";

tmp = tmp + pal.charAt(pal.length()-1)+isPalindrome(pal.substring(0,pal.length()-1));

return tmp;
}


}
     public static void main(String []args){

         chkPalindrome hwObj = new chkPalindrome();
         String palind = "MADAM";

       String retVal= hwObj.isPalindrome(palind);
      if(retVal.equals(palind))
       System.out.println(palind+" is Palindrome");
       else
       System.out.println(palind+" is Not Palindrome");
     }
}

Answer 12

Here is a recursive method that will ignore specified characters:

public static boolean isPal(String rest, String ignore) {
    int rLen = rest.length();
    if (rLen < 2)
        return true;
    char first = rest.charAt(0)
    char last = rest.charAt(rLen-1);
    boolean skip = ignore.indexOf(first) != -1 || ignore.indexOf(last) != -1;
    return skip || first == last && isPal(rest.substring(1, rLen-1), ignore);
}

Use it like this:

isPal("Madam I'm Adam".toLowerCase(), " ,'");
isPal("A man, a plan, a canal, Panama".toLowerCase(), " ,'");

It does not make sense to include case insensitivity in the recursive method since it only needs to be done once, unless you are not allowed to use the .toLowerCase() method.

Answer 13

there's no code smaller than this:

public static boolean palindrome(String x){
    return (x.charAt(0) == x.charAt(x.length()-1)) && 
        (x.length()<4 || palindrome(x.substring(1, x.length()-1)));
}

if you want to check something:

public static boolean palindrome(String x){
    if(x==null || x.length()==0){
        throw new IllegalArgumentException("Not a valid string.");
    }
    return (x.charAt(0) == x.charAt(x.length()-1)) && 
        (x.length()<4 || palindrome(x.substring(1, x.length()-1)));
}

LOL B-]

Answer 14

public static boolean isPalindrome(String p)
    {
        if(p.length() == 0 || p.length() == 1)
            // if length =0 OR 1 then it is
            return true; 

         if(p.substring(0,1).equalsIgnoreCase(p.substring(p.length()-1))) 
            return isPalindrome(p.substring(1, p.length()-1));


        return false;
    }

This solution is not case sensitive. Hence, for example, if you have the following word : "adinida", then you will get true if you do "Adninida" or "adninida" or "adinidA", which is what we want.

I like @JigarJoshi answer, but the only problem with his approach is that it will give you false for words which contains caps.

Answer 15

Palindrome example:

static boolean isPalindrome(String sentence) {

    /*If the length of the string is 0 or 1(no more string to check), 
     *return true, as the base case. Then compare to see if the first 
     *and last letters are equal, by cutting off the first and last 
     *letters each time the function is recursively called.*/

    int length = sentence.length();

    if (length >= 1)
        return true;
    else {
        char first = Character.toLowerCase(sentence.charAt(0));
        char last = Character.toLowerCase(sentence.charAt(length-1));

        if (Character.isLetter(first) && Character.isLetter(last)) {
            if (first == last) {
                String shorter = sentence.substring(1, length-1);
                return isPalindrome(shorter);
            } else {
                return false;
            }
        } else if (!Character.isLetter(last)) {
            String shorter = sentence.substring(0, length-1);
            return isPalindrome(shorter);
        } else {
            String shorter = sentence.substring(1);
            return isPalindrome(shorter);
        }
    }
}

Called by:

System.out.println(r.isPalindrome("Madam, I'm Adam"));

Will print true if palindrome, will print false if not.

If the length of the string is 0 or 1(no more string to check), return true, as the base case. This base case will be referred to by function call right before this. Then compare to see if the first and last letters are equal, by cutting off the first and last letters each time the function is recursively called.

Answer 16

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 17

Here are three simple implementations, first the oneliner:

public static boolean oneLinerPalin(String str){
    return str.equals(new StringBuffer(str).reverse().toString());
}

This is ofcourse quite slow since it creates a stringbuffer and reverses it, and the whole string is always checked nomatter if it is a palindrome or not, so here is an implementation that only checks the required amount of chars and does it in place, so no extra stringBuffers:

public static boolean isPalindrome(String str){

    if(str.isEmpty()) return true;

    int last = str.length() - 1;        

    for(int i = 0; i <= last / 2;i++)
        if(str.charAt(i) != str.charAt(last - i))
            return false;

    return true;
}

And recursively:

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

private static boolean check (String str,int start,int stop){
    return stop - start < 2 ||
           str.charAt(start) == str.charAt(stop) &&
           check(str, start + 1, stop - 1);
}

Answer 18

public class PlaindromeNumbers {

int func1(int n)
{
    if(n==1)
        return 1;

    return n*func1(n-1);
}

static boolean check=false;
int func(int no)
{

    String a=""+no;

    String reverse = new StringBuffer(a).reverse().toString();

    if(a.equals(reverse))
    {

        if(!a.contains("0"))
        {
           System.out.println("hey");
            check=true;
            return Integer.parseInt(a);
        }

    }

      //  else
      //  {
    func(no++);
    if(check==true)
    {
        return 0;
    }
       return 0;   
   }
public static void main(String[] args) {
    // TODO code application logic here
    Scanner in=new Scanner(System.in);
    System.out.println("Enter testcase");
   int testcase=in.nextInt(); 
   while(testcase>0)
   {
     int a=in.nextInt();
     PlaindromeNumbers obj=new PlaindromeNumbers();
       System.out.println(obj.func(a));
       testcase--;
   }
}

}

Answer 19

/**
     * Function to check a String is palindrome or not
     * @param s input String
     * @return true if Palindrome
     */
    public boolean checkPalindrome(String s) {

        if (s.length() == 1 || s.isEmpty())
            return true;

        boolean palindrome = checkPalindrome(s.substring(1, s.length() - 1));

        return palindrome && s.charAt(0) == s.charAt(s.length() - 1);

    }

Answer 20

Simple Solution 2 Scenario --(Odd or Even length String)

Base condition& Algo recursive(ch, i, j)

1. i==j //even len

2. if i< j recurve call (ch, i +1,j-1)

3. else return ch[i] ==ch[j]// Extra base condition for old length

public class HelloWorld {


 static boolean ispalindrome(char ch[], int i, int j) {
  if (i == j) return true;

  if (i < j) {
   if (ch[i] != ch[j])
    return false;
   else
    return ispalindrome(ch, i + 1, j - 1);
  }
  if (ch[i] != ch[j])
   return false;
  else
   return true;

 }
 public static void main(String[] args) {
  System.out.println(ispalindrome("jatin".toCharArray(), 0, 4));
  System.out.println(ispalindrome("nitin".toCharArray(), 0, 4));
  System.out.println(ispalindrome("jatinn".toCharArray(), 0, 5));
  System.out.println(ispalindrome("nittin".toCharArray(), 0, 5));

 }
}

Answer 21

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;
  }
}