Palindrome Checker using Stack in Java (Always returning false)

Question

I'm trying my hand at creating my own Stack class in Java. I'm not sure where my code is wrong, but if I enter an actual palindrome into my program console it always returns false. Where is my problem happening?

Here is my palindrome client file: (my stack.java class is accessible in the same program folder.)

public static void main(String args[]) {

//declare both stacks
Stack<String> fwd = new Stack<String>();
Stack<String> rev = new Stack<String>();

//instance variables
String st1;
boolean palindrome = true;

Scanner in = new Scanner(System.in);

System.out.println("Enter a String and I will check if it is a palindrome: \n");

st1 = in.nextLine();

// Read the data in forward order into the Stack
for (int i =0; i < st1.length(); i++) {
  fwd.push(Character.toString(st1.charAt(i)));
}
// Read the same data in reverse order into another Stack
for (int j = st1.length()-1; j >= 0; j--) {
  rev.push(Character.toString(st1.charAt(j)));
}
System.out.println("The string you entered was: ");
System.out.println(fwd.display());
System.out.println(rev.display());
System.out.println();
System.out.println("Checking to see if " + st1 + " is a palindrome.");
System.out.println("/**********************************************");

//check fwd and rev against each other
while (!fwd.isEmpty() && !rev.isEmpty()) {  // make sure stack in not empty
  for (int i = 0; i < st1.length(); i++) {  // go through each element in the stack

     if (fwd.pop() == rev.pop()) //check if fwd pop and rev pop are the same
        palindrome = true;
     else
        palindrome = false;

  }//end for lop
}// end while loop

System.out.println(palindrome);


}// end main

}// end Palindrome

Here is my Stack.java file:

public class Stack<E> implements StackInterface<E> {

   //variables
   private ArrayList<E> data;
   private E element;


   //constructor
   public Stack() {

      data = new ArrayList<E>();
   }


   //stack methods
   public void push(E element) { //push new element into the stack

      data.add(element);
   }

   public E pop() { //pop the element from the top

      if (data.isEmpty()) //if stack is empty, throw exception
         throw new EmptyStackException("The stack is empty.");
      else //else, remove and return the element that is on top of the stack
         return data.remove(data.size()-1); 
   }

   public E peek() { //peek at the element on top of the stack without removing it

      if (data.isEmpty()) //if stack is empty, throw exception
         throw new EmptyStackException("The stack is empty.");
      else //else, return the element that is on top of the class
         return data.get(data.size()-1);   
   }

   public String display() { //display the elements in the stack in the form of a String

      if (data.isEmpty()) //if stack is empty, throw exception
         throw new EmptyStackException("The stack is empty");
      else //else, return elements as a String
         return data.toString();
   }

   public boolean isEmpty() { //check to see if the stack is empty

      if (data.size() == 0)
         return true;
      else
         return false;

   }

   public int size() { //retrurn the number of elements in the stack

      return data.size();

   }

}// end Stack class

possible duplicate of [How do I compare strings in Java?](http://stackoverflow.com/questions/513832/how-do-i-compare-strings-in-java) — chrylis -cautiouslyoptimistic-, Oct 22 '13 at 08:11
That moment when you think "today there will be no `==` vs `equals` questions!". — Maroun, Oct 22 '13 at 08:12
This is not the solution, but if you add to (!fwd.isEmpty() && !rev.isEmpty()) && palindrome , you save a little bit of runtime. Also after palindrome = false; you can call break, because you already know its no palindrom. — Akkusativobjekt, Oct 22 '13 at 08:15
@Maroun Still familiarizing myself with the java language. Will not make that silly mistake again, now that I understand the difference! — Maggie S., Oct 22 '13 at 08:15

Alexis C. · Accepted Answer · 2013-10-22T10:53:11.173

5

if (fwd.pop() == rev.pop()) should be if (fwd.pop().equals(rev.pop())) since you're comparing content of two strings.

You should also break your while loop if you detect that it's not a palindrome, otherwise "test" would return true.

if (fwd.pop().equals(rev.pop())) //check if fwd pop and rev pop are the same
                palindrome = true;
             else {
                palindrome = false;
                break;
                 }
}

Edit :

As said in commment the for loop inside the while is actually useless.

Another option would be to create two stacks of Characters and now you can use == to compare them :

Stack<Character> fwd = new Stack<>();
Stack<Character> rev = new Stack<>();

/**/
for (int i =0; i < st1.length(); i++) {
    fwd.push(st1.charAt(i));
}
for (int j = st1.length()-1; j >= 0; j--) {
    rev.push(st1.charAt(j));
}
while (!fwd.isEmpty() && !rev.isEmpty()) {  // make sure stack in not empty
        if (fwd.pop()==rev.pop()) //check if fwd pop and rev pop are the same
           palindrome = true;
        else {
           palindrome = false;
           break; //here break the while loop it's not a palindrom
        }
}// end while loop

edited Oct 22 '13 at 10:53

answered Oct 22 '13 at 08:10

Alexis C.

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1

just to add information, the for loop inside the while is not needed. – RamonBoza Oct 22 '13 at 08:13
Thank you that worked! Still familiarizing myself with java in the form of fun little test projects like these. :) – Maggie S. Oct 22 '13 at 08:14
@RamonBoza Thank you, I see what you mean and I took it out now. – Maggie S. Oct 22 '13 at 08:20
Aha. I see. I tried making a char stack initially but when I got a compile error for making it like this: 'code'Stack fwd = new Stack<>(); I didn't try any further. Now I know how to define a Character Stack, thank you! – Maggie S. Oct 22 '13 at 08:23
@MagdaleneB. Notify the break statement. It's very important, otherwise words like "test" would say that is a palindrome. – Alexis C. Oct 22 '13 at 08:24
@ZouZou What do you mean by notify? What is that? – Maggie S. Oct 22 '13 at 08:26
@MagdaleneB. Pay attention to the `break;` instruction. It actually makes your programm correct. Check my edit =) – Alexis C. Oct 22 '13 at 08:27
I see now, for the Character Stack version. So is the break statement still useful/necessary if I were keep it as a String Stack? Probably not since it's not going through each character one by one right? – Maggie S. Oct 22 '13 at 08:29
1

@MagdaleneB. You can keep your String stack too it's correct. But the main idea of checking a word to be a palindrom is to check character by character. Instead of put the character in a String object you could directly but the character itself in the stack, but your version is ok too =) – Alexis C. Oct 22 '13 at 08:31

Palindrome Checker using Stack in Java (Always returning false)

1 Answers1