I'm working to familiarize myself with regex in Java, and it's a bit of a learning curve for me.

So the solution to the problem is as follows:

public static String validate (String ip){
        return ip.replaceAll("(?<=^|\\.)0 (?!\\.|$)", "");
    }

I just need clarification on why this regex solution works.

I get that the "?" specifies zero or one instances, the "^" represents the beginning of a string, the "\." is the escape character for a period, the "$" represents the end of the line, and the "" at the end represents deletion of a character, but I don't understand the totality of the regex. If someone could just quick walk me through what it all means together, that would be greatly appreciated. Thank you!

CodePudding user response：

In my opinion, this regex is a tricky one, because it uses a pattern not very common : lookahead and lookbehind patterns.

I get that the "?" specifies zero or one instances

That is true, but only when it follows a character to search for.

Here, it follows no character, as it is the first character of a group opening. It means it is a special construct. There's a dedicated javadoc subsection for it (see [1]).

In your examples, we can find two different constructs:

1. (?<=^|\\.)
   • This is cited in [1] as (?<=X) X, via zero-width positive lookbehind
   • What is this "positive lookbehind" stuff ? Source [2] defines it as Asserts that what immediately precedes the current position in the string is X
   • In your case, we ask to verify that 0 is either (|) the first character of the input text (^), or is just after a .
2. (?!\\.|$)
   • [1] define it as (?!X) X, via zero-width negative lookahead
   • [2] explains: Asserts that what immediately follows the current position in the string is not X
   • In your context, it ensures that we don't match trailing zero of a number (either the last zero or a zero that is just before a dot).

Sources:

[1] Oracle Pattern javadoc, Special constructs

[2] Regex tutorial : look around