Return to Revisions

Your code looks great!

Here, maybe another option that we might exercise would be to possibly do the entire task with an expression, maybe something similar to these:

^(.+?\/).+\/(.+)$
(.+?\/).+\/(.+)

Our first capturing group is non-greedy, collects our desired path1 for both inputs, followed by a greedy .+ that'd continue upto the last slash, and our path2 and path3 are in this group: (.+), and our desired output can be called using 1円2円.

####Escaping might be unnecessary, just following based on the demo .

###Test

import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "^(.+?\\/).+\\/(.+)$";
final String string = "path1/path2/../path3\n"
 + "path1/./path2";
final String subst = "1ドル2ドル";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
final String result = matcher.replaceAll(subst);
System.out.println(result);

###Demo

console.log(`path1/path2/../path3
path1/./path2`.replace(/^(.+?\/).+\/(.+)$/gm, `1ドル2ドル`));

###Performance

const repeat = 1000000;
const start = Date.now();
for (var i = repeat; i >= 0; i--) {
 const regex = '/^(.+?/).+/(.+)$/gm';
 const str = `path1/path2/../path3`;
 const subst = `1ドル2ドル`;
 var match = str.replace(regex, subst);
}
const end = Date.now() - start;
console.log("YAAAY! \"" + match + "\" is a match 💚💚💚 ");
console.log(end / 1000 + " is the runtime of " + repeat + " times benchmark test. 😳 ");

RegEx Circuit

jex.im visualizes regular expressions:

enter image description here