Can someone explain the regular expression part of String#replaceAll(..) method?

Question

//Its a question on replacement of duplicate characters

public class RemoveDuplicateChars {

    static String testcase1 = "DPMD Jayawardene";

    public static void main(String args[]){
        RemoveDuplicateChars testInstance= new RemoveDuplicateChars();
        String result = testInstance.remove(testcase1);
        System.out.println(result);

    }
    //write your code here
    public String remove(String str){
        return str.replaceAll("(.)(?=.*\\1)", "");//how this replacement working
    }       
}

Use [this](http://rick.measham.id.au/paste/explain.pl) – takendarkk Feb 03 '14 at 20:10 — takendarkk, Feb 03 '14 at 20:10

score 2 · Answer 1 · answered Feb 03 '14 at 20:09

2

As you can see from the name of the class - it removes characters that repeat in a string.

Breakdown:
(.) - stands for any character, the brackets are used for grouping, so we'll be able to reference it later on using \1

(?=) - lookahead

(?=.*\\1) - we're looking forward
.* consuming any number of characters and looking for our first character\1

If the regex is truthy, the referenced character will be replaced with the empty string.

See Fiddle

answered Feb 03 '14 at 20:09

Nir Alfasi

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Does the `(` in front of `?=` start its own capture group? – Sotirios Delimanolis Feb 03 '14 at 20:13
@SotiriosDelimanolis yes it does, but in this case it would be same as `(.)` – Алексей Feb 03 '14 at 20:19
Nope. `(?=XXX)` will look forward for "XXX". The opening and closing brackets her are used as opening and closing tokens for the "look ahead" – Nir Alfasi Feb 03 '14 at 20:19
@alfasin if you redo w/ Pattern and Matcher you would end up with two groups; group(0) that corresponds to `(.)` and group(1) that correspond to positive look ahead – Алексей Feb 03 '14 at 20:22
@user1631616 are you sure it works with all flavors (that support lookahead) ? – Nir Alfasi Feb 03 '14 at 20:27
@alfasin positive/negative look ahead/behind def works with `Pattern/Matcher` as i used it for one of my projects. If you rewrite what OP has w/ Pattern/Matcher - you can see that whenever it matches there are 2 grps hence `Does the ( in front of ?= start its own capture group?` the answer is yes :) – Алексей Feb 03 '14 at 20:34
@user1631616 you ignored my question. I understand that it works in Java, but I doubt it'll work with other flavor of regex. That said, I'll definitely try it sometime soon with Python, Ruby, and JS - thanks! – Nir Alfasi Feb 03 '14 at 20:36
@alfasin i guess i miss understood your question (my bad) – Алексей Feb 03 '14 at 20:37

score 2 · Answer 2 · answered Feb 03 '14 at 20:12

From java.util.Pattern:

(.) : Match any character in a capture group (basically a variable named \1)
(?= : Zero-width positive lookahead (make sure the rest of the string matches)
- .* any number of characters followed by
- \\1 the captured group

In other words, it matches any character that also appears later in the string (i.e. is a duplicate). In Java, this would be:

for(int i=0; i<str.length(); i++) { 
  char captured = str.charAt(i); // (.)
  if (str.substring(i+1).matches(".*" + captured))  { // (?=.*\1)
    // the char is a duplicate, replace it with "" 
  }
}

Can someone explain the regular expression part of String#replaceAll(..) method?

2 Answers2