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  1. #1
    trinity is offline Member
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    Default Palindrome program

    Hi there , I have just written a palindrome program to check if a specific String is or not a palindrome . My program seems fare to complicated for such a simple task. Is there a more conventional way of doing this?
    Thankkkkkks
    Java Code:
    import tcdIO.*;
    
    public class Palyndrome {
    	public static void main(String args[]) {
    		String string;
    		Terminal t = new Terminal();
    		string = t.readString("Enter the word you want to test\n>");
    		
    		int counter = string.length() - 1;
    		int remainder;
    		int palyndrome = 0;
    		remainder = string.length() % 2;
    
    		for (int i = 0; i < string.length() / 2 + remainder; i++) {
    			while (counter >= string.length() / 2 + remainder) {
    				if (string.charAt(i) != string.charAt(counter)) {
    					palyndrome++;
    				}
    				counter--;
    
    			}
    
    		}
    		
    		
    		if(palyndrome==0){
    			t.println(""+string+" is a palyndrome");
    		}else{
    				t.println(""+string+" is  not a palyndrome");}
    
    
    	}
    
    }

  2. #2
    KevinWorkman's Avatar
    KevinWorkman is offline Crazy Cat Lady
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    Default

    I would just have two indexes- start (initialized to zero) and end (initialized to the length of the string minus one). In a loop, check that the char at index start and the char at index end are equal. If they aren't, the String is not a palindrome. Increment start and decrement end before looping again. When start is greater than or equal to end, then the String is a palindrome.

    But that's just how I do it by hand or in my head, without any code. If you do it a different way (think about how you decide whether some text is a palindrome or not without any code, just in your head), then that's how your program should work.
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  3. #3
    dlorde is offline Senior Member
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    Default

    Would this be cheating?
    Java Code:
    boolean palindrome = text.equals(new StringBuilder(text).reverse().toString());

  4. #4
    Maximus-EVG is offline Member
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    Default

    You could push the string which you're testing onto a stack data structure, char by char, and after its done, pop the stack and form another string with it.
    Then compare the two strings.

    Read up on Stack datastructure here:
    Stack (Java 2 Platform SE 5.0)

    (not as nice as dlorde's solution, but at least with this youre doing "work" :D. Actually Im pretty sure the .reverse() method does the same thing as I wrote).

  5. #5
    ra4king's Avatar
    ra4king is offline Senior Member
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    Default

    Hehe you misspelled Palindrome ;)

    Anyway, here's my 4-lines of code to check palindrome:
    Java Code:
    public boolean isPalindrome(String s) {
        for(int a = 0; a < s.length()/2; a++)
            if(s.charAt(a) != s.charAt(s.length()-a-1))
                return false;
        return true;
    }
    ;)

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