Help needed with String.split()

[rtaig]

Hi

I'd like to split a String with the following rule : Parse everything between a "( )" parenthesis besides nested parenthesis, meaning:

String s = "(example) (ex) (am (p (k) t) le)"

needs to be parsed into :
example
ex
am (p (k) t) le

I tried to work with split() and regular expressions but couldn't find the right one.

Thanks in advance:)

[JosAH]

Hi

I'd like to split a String with the following rule : Parse everything between a "( )" parenthesis besides nested parenthesis, meaning:

String s = "(example) (ex) (am (p (k) t) le)"

needs to be parsed into :
example
ex
am (p (k) t) le

I tried to work with split() and regular expressions but couldn't find the right one.

Regular expressions can't do that; in normal terms without mathematical proof: regular expressions can't count. You have to parse your String 'manually'.

kind regards,

Jos

[Fubarable]

Regular expressions can't do that; in normal terms without mathematical proof: regular expressions can't count. You have to parse your String 'manually'.

And he'll probably want to use a stack to do this, right?

[JosAH]

And he'll probably want to use a stack to do this, right?

Formally yes, practically a simple counter will do fine ;-)

kind regards,

Jos

[Fubarable]

Formally yes, practically a simple counter will do fine ;-)

As always, thanks!

[rtaig]

Thanks , I know that a full parsing of nested parenthesis can't be done with regular expressions but please note that my demand is much simpler and doesn't involve counting - I'm saying : "parse everything inside a parenthesis accept if it's inside other parenthesis - i don't care how many " - to be clear :

String s = "(I) (don't) (care)" should give : [I , don't , care];
String s = (I) (do (n'(t c) a) (re) should give: [I, do (n'(t c) a, re;

Is this equivalent hard as the general task ?

Thanks

[Junky]

my demand is much simpler and doesn't involve counting

Don't you love it when n00bs refute the excellent advice provided by much more knowledgable people.

So your problem doesn't involve counting huh, so how does it know how many brackets to ignore? Using your original example "(am (p (k)" satisfies your requirements.

[JavaHater]

@rtaig, this is toy example

                String s  = "(I) (don't) (care)";
                int c = 0;
                String joiner = "";
                for(int i =0 ;i <s.length(); i++){
                        if ( c!= 0 ){
                                joiner+=s.charAt(i);
                        }else if ( c==0 && joiner !="" ){
                                System.out.println ("joiner: " + joiner);
                                joiner="";
                        }
                        if ( s.charAt(i) == '(' ){
                                c++;
                        }else if ( s.charAt(i) == ')' ) {
                                c--;
                        }
                }
                System.out.println("joiner: " + joiner);

[mine0926]

Thanks , I know that a full parsing of nested parenthesis can't be done with regular expressions but please note that my demand is much simpler and doesn't involve counting - I'm saying : "parse everything inside a parenthesis accept if it's inside other parenthesis - i don't care how many " - to be clear :

String s = "(I) (don't) (care)" should give : [I , don't , care];
String s = (I) (do (n'(t c) a) (re) should give: [I, do (n'(t c) a, re;

Is this equivalent hard as the general task ?

Thanks

If you will not use a counting then the result will be
which is different from your expected output:
INPUT:
String s = (You) (are (no (t li) st) (ening)
POSSIBLE OUTPUT: [You, are not li st, ening];

INPUT:
String s = (You) (are (no (t li) st) (ening)
OUTPUT: [You, are (no(t li) st, ening];

[JosAH]

Thanks , I know that a full parsing of nested parenthesis can't be done with regular expressions but please note that my demand is much simpler and doesn't involve counting

It does involve counting; this morning I tried to come up with a simple proof that parsing balanced parentheses can't be done with regular languages but I failed. I always fall back to the pumping lemmas which show that the language of balanced (nested) parentheses is a context free language; but then again, I didn't have my coffee yet and I'm going to fix that right now ;-)

kind regards,

Jos