Removing every other character in a string using Java regex

You are indeed very close to the answer: just make matching the second char optional.

String s = "1a2b3c4d5";
System.out.println(s.replaceAll(".(.)?", "$1"));
// prints "abcd"

This works because:

• Regex is greedy by default, it will take the second character if it's there
  • When the input is of odd length, the second char won't be there at the last replacement, but you'd still match one char (i.e. last char in input)
• You can still use backreferences in substitution even if the group fails to match
  • It will substitute in the empty string, not "null"
  • This is different from Matcher.group(int), which returns null for failed groups

References

• regular-expressions.info/Optional

A closer look at the first part

Let's take a closer look at the first part of the homework:

String s = "1a2b3c4d5";
System.out.println(s.replaceAll("(.).", "$1"));
// prints "12345"

Here you didn't have to use ? for the second char, but it "works" because even though you didn't match the last char, you didn't have to! The last char can remain unmatched, unreplaced, due to the problem specification.

Now suppose that we want to delete chars at index 1,3,5..., and put the chars at index 0,2,4... in brackets.

String s = "1a2b3c4d5";
System.out.println(s.replaceAll("(.).", "($1)"));
// prints "(1)(2)(3)(4)5"

A-ha!! Now you're experiencing the exact same problem with odd-length input! You couldn't match the last char with your regex, because your regex needs two chars, but there's only one char at the end for odd-length input!

The solution, again, is to make matching the second char optional:

String s = "1a2b3c4d5";
System.out.println(s.replaceAll("(.).?", "($1)"));
// prints "(1)(2)(3)(4)(5)"