delete repeated letters in a word

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spidey12345 - 08 Feb 2007 14:08 GMT

so If i enter the letter blbabh

the word will acutally become: blah

// repeatble letters in the key

char checkkey[] = pkey.toCharArray();

int pkeysize = checkkey.length;
    int counter = 0;
    for(int i = 1; i< pkeysize; i++)
    {
        for(int j = 0; j< i; j++)
        if(checkkey[i] ==checkkey[j])
        {
        checkkey[i] = '0';
        }
    }

// what i did here was just assign all repeated b's to 0 so i would
have
// bl0a0h

    for(int i = 0; i<pkeysize; i++)
    {
        System.out.println(checkkey[i]);
        if(checkkey[i] == '0')
            counter++;
    }

// it works up to this point, i get bl0a0h

// some thing is wrong with the aglorithm here....

    System.out.println(counter);
    int nsize = pkeysize-counter;
    char nkey[] = new char[pkeysize];
    for(int i = 0; i< pkeysize; i++)
    {
        int j = 0;

        if(checkkey[i] != '0')
        {
            nkey[j] = checkkey[i];
            j++;
        }
    }

    for(int i = 0; i< nsize; i++)
    System.out.print(nkey[i]);

// but when i try to print out the nkey, it gives me bunch of sh.t

Reply to this Message

Andrew Thompson - 08 Feb 2007 15:17 GMT

> so If i enter the letter blbabh

...

> // but when i try to print out the nkey, it gives me bunch of sh.t

Did you have a question, or were you just
'keeping us informed' of your progress?

Andrew T.

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spidey12345 - 08 Feb 2007 17:35 GMT

> > so If i enter the letter blbabh
> ...
[quoted text clipped - 4 lines]
>
> Andrew T.

yeah i have a question, i am not sure what to do when i reach
"bl0a0h", read my comments the algorithm

so i am not sure why it's not reading correctly in that part of the
code

when i say if(checkkey[i] != '0') , and going through i iterations,
shouldn't it only set the ones that is not equal to 0 to the indexes,
why is it still
setting 0 to the answer

so when i print out the new key, it gives me bla0 rather than blah for
some reasons

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Oliver Wong - 08 Feb 2007 17:48 GMT

> so If i enter the letter blbabh
>
> the word will acutally become: blah

[code snipped]

First thing that comes to mind for me is to put the characters in a
LinkedHashSet, and then pull them out again in the correct order.

- Oliver

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spidey12345 - 08 Feb 2007 18:58 GMT

> > the word will acutally become: blah
>
[quoted text clipped - 4 lines]
>
> - Oliver

Does java have any standard libary for this, i am new to java, so i
delt linkedhashset would be something i like to use.

Reply to this Message

Oliver Wong - 08 Feb 2007 19:06 GMT

>> enter the letter blbabh
>>
[quoted text clipped - 9 lines]
> Does java have any standard libary for this, i am new to java, so i
> delt linkedhashset would be something i like to use.

Yup: http://java.sun.com/javase/6/docs/api/java/util/LinkedHashSet.html

- Oliver

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spidey12345 - 09 Feb 2007 04:33 GMT

> >> "spidey12345" <swq_sha...@hotmail.com> wrote in message
>
[quoted text clipped - 18 lines]
>
> - Show quoted text -

ok, fine, i am using hashset:)

but which function in there allow me to delete repeated values or 0,
that is not character

Reply to this Message

Lew - 09 Feb 2007 07:04 GMT

> ok, fine, i am using hashset:)
>
> but which function in there allow me to delete repeated values or 0,
> that is not character

add()

Read the documentation for Set:
<http://java.sun.com/javase/6/docs/api/java/util/Set.html>

You will note the very first phrase of the explanation for Set: "A collection
that contains no duplicate elements."

The "0" was an artifact of your first approach and not part of the original
problem definition.

- Lew

Reply to this Message

Michael Rauscher - 09 Feb 2007 12:10 GMT

> ok, fine, i am using hashset:)
>
> but which function in there allow me to delete repeated values or 0,
> that is not character

Why do you use an array? You don't even need a HashSet if speed doesn't
matter.

Untested (I even didn't tried to compile it) code:

public String removeDuplicateLetters( String word ) {
if ( word == null )
return null;

StringBuilder builder = new StringBuilder();
for ( int i = 0, n = word.length(); i < n; i++ ) {
String sub = word.substring(i,i+1);
if ( builder.indexOf(sub) == -1 )
builder.append(sub);
}

return builder.toString();
}

If you want to use a HashSet due to performance reasons the only thing
you'd have to replace is the if-condition and of course, you'd have to
add the element to the Set.

Bye
Michael

Reply to this Message

Daniel Pitts - 12 Feb 2007 21:22 GMT

> > ok, fine, i am using hashset:)
>
[quoted text clipped - 27 lines]
> Bye
> Michael

This might be a little better:
Its known to compile and run, for one thing.
public class Test2 {
public static void main(String[] args) {
final String s = "My string has a lot of duplicate letters";
final StringBuilder builder = new StringBuilder(s);
for (int i = 0; i < builder.length(); ++i) {
final String charStr =
Character.toString(builder.charAt(i));
int index = builder.indexOf(charStr, i+1);
while (index > i) {
builder.deleteCharAt(index);
index = builder.indexOf(charStr, index);
}
}
System.out.println(builder.toString());
}
}

Reply to this Message

Bart Rider - 12 Feb 2007 17:15 GMT

> so If i enter the letter blbabh
>
[quoted text clipped - 36 lines]
> {
> int j = 0;

Move this statement out of the loop, than it will behave
like you want.

>         if(checkkey[i] != '0')
>         {
[quoted text clipped - 5 lines]
>     for(int i = 0; i< nsize; i++)
>     System.out.print(nkey[i]);

Hint: form a string out of the new char-array and print this.

> // but when i try to print out the nkey, it gives me bunch of sh.t

Think about it: what happens if your word contains the character
'0'? Maybe you should consider using '\0' instead of '0'.

Best regards,
Bart

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delete repeated letters in a word

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