More efficient line of code in Java

Question

I work with strings in my programs for many times.

Is there a way to do this line of Java code more efficient:

String str2 = str.replaceAll("\\s+", " ").trim();

I don't really think so (other than hidden away in a commons library) and by the way, you're only removing spaces in the middle, not tabs, linefeeds, etc, I think.... — Davio
– Davio, Commented Feb 19, 2015 at 11:56
Trim first and then replace like String str2 = str.trim().replaceAll("\\s+", " "). because sometimes, larger trailing spaces would be trimmed in one shot! — Maheswaran Ravisankar
– Maheswaran Ravisankar, Commented Feb 19, 2015 at 11:59
Where the strings are comming from? If they are the result of a database query, I would try to let the database do the work (at least as much as possible). — SubOptimal
– SubOptimal, Commented Feb 19, 2015 at 12:04

Jurgen · Accepted Answer · 2015-02-20 16:04:04Z

1

You could try using a pre compiled pattern:

private Pattern p = Pattern.compile( "\\s+" );

And then use it like so:

str2 = p.matcher( str.trim() ).replaceAll( " " );

A more complex version that doesn't require trimming:

private Pattern p = Pattern.compile( "^\\s+|\\s+(?= )|\\s+$" );

str2 = p.matcher( str ).replaceAll( "" );  // no space

edited Feb 20, 2015 at 16:04

answered Feb 19, 2015 at 12:18

Jurgen

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4 Comments

oventarb Over a year ago

Thank you. But more complex version will not work there is any any tabs or line feeds in the middle of the string.

Jurgen Over a year ago

Sorry, you can try (?=\\s) instead of (?= ), but I think the simpler version would be better. You can also try moving the trim to after the replaceAll and see how that works - performance wise.

oventarb Over a year ago

(?=\\s) will change to tabulation or line feed or something else. But I want change to space, not any other whitespace.

Jurgen Over a year ago

(?=\\s) isn't for the replacement character, its part of the pattern being looked for. (?= ) means look ahead for a space, while (?=\\s) means look ahead for any white-space character.

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