Java - How to search a string for 6 random numbers

Question

I have a (large) string file in my application which contains a series of random character [a-Z] and [0-9] but also ";","/", "?", ":" and "@". I would like my application to tell me the nearest position where 6 digits are displayed consecutively ( like "105487" or "558463").

What would be the best way to achieve this? Thank you for looking into this.

Hothouse

Please clarify on "nearest position". Is it the first occurance in the file? – Slotter

Does the position mean line number and column in the file? – Crazed

yes, i meant the first occurance. The position just means the number that the indexOf function in java would give as a number. – Hothouse

You can find what you are looking for here: #8938998 – Plumbism

Are you looking for a sequence of speicific digits or just a sequence of any digits? – Necrolatry

I am looking for a sequence of any digits. – Hothouse

Answer 1

An effective approach would be to iterate the characters of the string and test if each is a digit. Upon finding a match continue to look for the rest of the sequence. Something like

int nDigits=0, i = 0;
CharacterIterator it = new StringCharacterIterator("very long string123456");
for (char ch=it.first(); ch != CharacterIterator.DONE; ch=it.next()) {
  i++;
  nDigits = (ch.isDigit() ? nDigits++ : 0);
  if (nDigits == 5) {
      // DONE. Position is "i"
  }
}

Usurer

Thank you, I will definitely try this. – Hothouse

Answer 2

You could use regex.

String regex = "(\\d{6})";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(YOU STRING HERE);
// Check all occurrences
while (matcher.find()) {
    System.out.print("Start index: " + matcher.start());
    System.out.print(" End index: " + matcher.end());
    System.out.println(" Found: " + matcher.group());
}

This would do the job.

(Code sample from here)

Resistless

for very large string it likely will fail with OOME – Staggs

why not use REGEX \\d{6} instead? – Adolescence

Well, I don't know how long is string is, but I think it's a good option. Thanks for pointing it out though. – Resistless

@gtgaxiola, good question. I didn't remember it.. Edited. thanks. – Resistless

Thank you. I will also try this approach, but I think that the answer that Johan Sjöberg posted would be best suited for my application. – Hothouse

Answer 3

Use Character.isDigit while iterating the string's characters and then count a number up until you have found 6 consecutive numbers or reset it if the sequence breaks. Keep track of the index and you can simply calculate the nearest position by subtraction.

This is not very efficient, but I think O(n) is sufficient if the strings are not too large. For optimization take a look on what Luiggi Mendoza has suggested in the comments.

Headwaters

Thank you for your answer. I thought this would be the only solution, but i was hoping that java had a more efficient way of doing this. I will look into this. – Hothouse

@user1637234 you could use a regular expression like Lablabla has posted. – Tubman

You can optimize this: if you find a digit in position X, then check the character in position X + 5, if that character is not a a digit, then start searching from here. If it's a digit, then go back (X+4, X+3...) until you find a non-digit and start again from the least position. – Tubman

@Adolescence yeah, one of my String matching favorite algorithms when I can't rely on RegEx. – Tubman