Find the last match with Java regex matcher

Question

I'm trying to get the last result of a match without having to cycle through .find()

Here's my code:

String in = "num 123 num 1 num 698 num 19238 num 2134";
Pattern p = Pattern.compile("num '([0-9]+) ");
Matcher m = p.matcher(in);

if (m.find()) {
     in = m.group(1);
}

That will give me the first result. How do I find the LAST match without cycling through a potentionally huge list?

Answer 1

You could prepend .* to your regex, which will greedily consume all characters up to the last match:

import java.util.regex.*;

class Test {
  public static void main (String[] args) {
    String in = "num 123 num 1 num 698 num 19238 num 2134";
    Pattern p = Pattern.compile(".*num ([0-9]+)");
    Matcher m = p.matcher(in);
    if(m.find()) {
      System.out.println(m.group(1));
    }
  }
}

Prints:

2134

You could also reverse the string as well as change your regex to match the reverse instead:

import java.util.regex.*;

class Test {
  public static void main (String[] args) {
    String in = "num 123 num 1 num 698 num 19238 num 2134";
    Pattern p = Pattern.compile("([0-9]+) mun");
    Matcher m = p.matcher(new StringBuilder(in).reverse());
    if(m.find()) {
      System.out.println(new StringBuilder(m.group(1)).reverse());
    }
  }
}

But neither solution is better than just looping through all matches using while (m.find()), IMO.