regular expression to replace two (or more) consecutive characters by only one?

Question

In java, which regular expression can be used to replace these, for example:

before: aaabbb after: ab

before: 14442345 after: 142345

thanks!

Answer 1

In perl

s/(.)\1+/$1/g;

Does the trick, I assume if java has perl compatible regexps it should work too.

Edit: Here is what it means

s {
    (.)  # match any charater ( and capture it )
    \1   # if it is followed by itself 
    +    # One or more times
}{$1}gx;  # And replace the whole things by the first captured character (with g modifier to replace all occurences)

Edit: As others have pointed out, the syntax in Java would become

original.replaceAll("(.)\\1+", "$1");

remember to escape the \1

Answer 2

 String a = "aaabbb";
 String b = a.replaceAll("(.)\\1+", "$1");
 System.out.println("'" + a + "' -> '" + b + "'");

Answer 3

"14442345".replaceAll("(.)\\1+", "$1");

Answer 4

originalString.replaceAll( "(.)\\1+", "$1" );

Answer 5

in TextEdit (assuming posix expressions) find: [a]+[b]+ replace with: ab

Answer 6

match pattern (in Java/languages where \ must be escaped):

(.)\\1+

or (in languages where you can use strings which don't treat \ as escape character)

(.)\1+

replacement:

$1

Answer 7

In Perl:

tr/a-z0-9//s;

Example:

$ perl -E'@a = (aaabbb, 14442345); for(@a) { tr/a-z0-9//s; say }'
ab
142345

If Java has no tr analog then:

s/(.)\1+/$1/sg; 
#NOTE: `s` modifier. It takes into account consecutive newlines.

Example:

$ perl -E'@a = (aaabbb, 14442345); for(@a) { s/(.)\1+/$1/sg; say }'
ab
142345