java replaceLast() [duplicate]

by

It could (of course) be done with regex:

public class Test {

    public static String replaceLast(String text, String regex, String replacement) {
        return text.replaceFirst("(?s)"+regex+"(?!.*?"+regex+")", replacement);
    }

    public static void main(String[] args) {
        System.out.println(replaceLast("foo AB bar AB done", "AB", "--"));
    }
}

although a bit cpu-cycle-hungry with the look-aheads, but that will only be an issue when working with very large strings (and many occurrences of the regex being searched for).

A short explanation (in case of the regex being AB):

(?s)     # enable dot-all option
A        # match the character 'A'
B        # match the character 'B'
(?!      # start negative look ahead
  .*?    #   match any character and repeat it zero or more times, reluctantly
  A      #   match the character 'A'
  B      #   match the character 'B'
)        # end negative look ahead

EDIT

Sorry to wake up an old post. But this is only for non-overlapping instances.
For example .replaceLast("aaabbb", "bb", "xx"); returns "aaaxxb", not "aaabxx"

True, that could be fixed as follows:

public class Test {

    public static String replaceLast(String text, String regex, String replacement) {
        return text.replaceFirst("(?s)(.*)" + regex, "$1" + replacement);
    }

    public static void main(String[] args) {
        System.out.println(replaceLast("aaabbb", "bb", "xx"));
    }
}

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