java - How to format string to a particular pattern -

let's have input string , need format this:

### ### ############........etc 

so has have first 3 chars, space, 3 chars, space , rest. there third-party library or jdk class able that?

i trying use regular expressions

system.out.println(inputstring.replaceall(".{3}", "$0 ")); 

but it's not working because result is

### ### ### ### ### etc. 

you this:

system.out.println(inputstring.replacefirst("(.{3})(.{0,3})", "$1 $2 ")); 

explanation:

just $0 entire matched string, $1 , $2 are, respectively, first , second matched things in brackets.

i modified {3} {0,3} strings 6 characters or shorter work (it add trailing space when string between 4 , 6 characters, can removed .trim() (which have unwanted other effects) or more complex).

hopefully no explanation required rest, since it's similar code, feel free ask if you're unsure. java regex reference.

example:

system.out.println("12345678901234567890".replacefirst("(.{3})(.{0,3})", "$1 $2 ")); system.out.println("12".replacefirst("(.{3})(.{0,3})", "$1 $2 ")); system.out.println("12345".replacefirst("(.{3})(.{0,3})", "$1 $2 ")); 

prints:

123 456 78901234567890 12 123 45