Check for Palindrome string in Java

Question

I was wondering if you can give me advice on this solution to assess if an user input string is a palindrome or not. How can I optimize this solution?

import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.IOException;

public class Palindrome {
public static void main(String[] args)
    throws IOException {

    // user enters string
    System.out.println ("Please enter string");

    InputStreamReader input = new InputStreamReader (System.in);
    BufferedReader reader = new BufferedReader (input);
    String userinput = "";
    userinput = reader.readLine();

    // test if string is valid
        if (!userinput.matches("[A-Za-z]+")){
            System.out.println ("Invalid Input");
            System.exit(1);
        }

        else if (userinput.length() > 10){
            System.out.println ("String too long");
            System.exit(2);
        }


    String stringReverse = "";
    for (int i = userinput.length()-1; i >= 0; i--){
        stringReverse += String.valueOf(userinput.charAt(i));
    }

    if (userinput.equals(stringReverse)){
        System.out.println ("String is a Palindrome");
        System.exit(0);
    }

    else if (userinput != stringReverse) {
        System.out.println ("String is not a Palindrome");
        System.exit(0);
    }

    }
}

Answer 1

When programming a solution like this, in order to simplify the problem create a separate function for each component. So, I would break it up into three different parts:

String getUserInput() {
    // Get input string and return it...
}

boolean isValidInput(String input) {
    // Return true or false based off of the validity of the input string.
}

boolean isPalindrome(String input) {
    int left = 0;
    int right = input.length();
    while(left < right) {
        if(input.charAt(left++) != input.charAt(--right)) return false;
    }
    return true;
}

---

The reason for returning a boolean rather than throwing an exception is that a try catch can be expensive especially if there is a good chance of an error being thrown. Now, a boolean means later that there may be some sort of conditional branch, but that is less expensive than if there is a high chance of an error being thrown. Also, there is more usability with returning a boolean, because now you can decide if you want to throw an exception or just print true or false.

---

Since the question was more geared towards the optimization of the palindrome detection algorithm, I provided a solution that I have found to work the best for when detecting palindromic inputs.

The solution cuts the number of steps down to a quarter of your solution. This is done by assuming that if the first half is the same as the second half then the String must be palindrome. So, we start from both ends and work our way to the center (essentially when left reaches or passes right). If any where along the way we find a point in which is not equal, then we know that we do not have a palindromic String. Therein, we can exit returning false. If we successfully exit the loop, then we know we were given a palindromic String. Therein, we return false.

Answer 2

You're implementation can be improved which in turn can save some memory space and execution time. Space is saved by not using another string in stringReverse and time is saved by iterating from 0 - (strlen()-1)/2 instead of 0 - strlen().

Hope this helps!

import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.IOException;

public class Palindrome {
public static void main(String[] args) throws IOException {

    // user enters string
    System.out.println ("Please enter string");

    InputStreamReader input = new InputStreamReader (System.in);
    BufferedReader reader = new BufferedReader (input);
    String userinput = "";
    userinput = reader.readLine();

    // test if string is valid
        if (!userinput.matches("[A-Za-z]+")){
            System.out.println ("Invalid Input");
            System.exit(1);
        }

        else if (userinput.length() > 10){
            System.out.println ("String too long");
            System.exit(2);
        }

       //Modified this part
       int n = userinput.length();
       for (int i = 0; i <= (n - 1) / 2; i++){
         if(userinput.charAt(i) != userinput.charAt(n-1-i)) {
            System.out.println("String is not a palindrome");
            System.exit(0);
         }
       }

       System.out.println("String is a palindrome");

    }
}

Answer 3

Another way of doing it may be to create 2 indexes, the first starting from 0 and the second starting at the end of the string. Loop upon equality until they cross each other or they are not equal.

edit: Ok I am no Java specialist but the code could be something like

int a = 0;
int b = userInput.length()-1;

while (a<=b && userInput.charAt(a) == userInput.charAt(b)){
    a++;
    b--;
}

if (a>=b) {
    System.out.println("palyndrome");
} else {
    System.out.println("not palyndrome");
}

Answer 4

Do not use System.exit() (somewhere else than in a chatch block inside main) this terminates the JVM and this is not what you want in most cases.

  // user enters string
  // test if string is valid

This type of comment tells you that the code below sould be placed in a method with a name derived from the comment. eg:

String userEntersString(){
  // ...
}

void testIfStringIsValid(String string){

}

then your main would read like this:

public static void main(String[] argd){
   String string userEntersString();
   testIfStringIsValid(string);
   testIfStringIsPalindrome(string);
   System.out.println ("String is a Palindrome");
}

To handle the validation you should have your validation method throwing Exceptions:

void testIfStringIsValid(String string) throws Exception {
    if (!userinput.matches("[A-Za-z]+")){
        throw new Exception("Invalid Input");
    }
   //...
}

then you main would change like this:

public static void main(String[] argd){
   String string userEntersString();
   try{
       testIfStringIsValid(string);
       testIfStringIsPalindrome(string);
       System.out.println ("String is a Palindrome");
   } (catch Exception e) {
       System.err.println(e.getMessage());
       // if you really need to communicate the programms result:
       System.exit(1);
   }
}