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.

Please clarify on "nearest position". Is it the first occurance in the file? — Andreas Dolk, Oct 01 '12 at 21:23
yes, i meant the first occurance. The position just means the number that the indexOf function in java would give as a number. — DVPeer, Oct 01 '12 at 21:28
You can find what you are looking for here: http://stackoverflow.com/questions/8938498/get-the-index-of-a-pattern-in-a-string-using-regex — Andrew Backes, Oct 01 '12 at 21:29
Are you looking for a sequence of *speicific* digits or just a sequence of *any* digits? — Code-Apprentice, Oct 01 '12 at 21:35

score 4 · Answer 1 · edited May 23 '17 at 11:43

4

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)

edited May 23 '17 at 11:43

Community

1
1

answered Oct 01 '12 at 21:28

La bla bla

8,558
13
60
109

for very large string it likely will fail with OOME – jdevelop Oct 01 '12 at 21:28
2

why not use REGEX \\d{6} instead? – gtgaxiola Oct 01 '12 at 21:30
Well, I don't know how long is string is, but I think it's a good option. Thanks for pointing it out though. – La bla bla Oct 01 '12 at 21:30
@gtgaxiola, good question. I didn't remember it.. Edited. thanks. – La bla bla Oct 01 '12 at 21:31
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. – DVPeer Oct 01 '12 at 21:44

Konrad Reiche · Answer 2 · 2012-10-01T21:36:40.603

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.

edited Oct 01 '12 at 21:36

answered Oct 01 '12 at 21:24

Konrad Reiche

27,743
15
106
143

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. – DVPeer Oct 01 '12 at 21:30
@user1637234 you could use a regular expression like Lablabla has posted. – Luiggi Mendoza Oct 01 '12 at 21:30
2

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. – Luiggi Mendoza Oct 01 '12 at 21:34
@gtgaxiola yeah, one of my String matching favorite algorithms when I can't rely on RegEx. – Luiggi Mendoza Oct 01 '12 at 21:38

Johan Sjöberg · Accepted Answer · 2012-10-01T21:36:52.450

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"
  }
}

edited Oct 01 '12 at 21:36

answered Oct 01 '12 at 21:25

Johan Sjöberg

47,929
21
130
148

Thank you, I will definitely try this. – DVPeer Oct 01 '12 at 21:42

Java - How to search a string for 6 random numbers

3 Answers3