CODEWITHSUNDEEP
javaarraysstring
mra419
mra419

Reputation: 413

How to get last number from string like 12345678910111213 (numbers from 1 to 13 without any space in between)

I have a String output like 12345678910111213 from a method (numbers from 1 to 13 without any space in between) I want to get the last number from this series, In this case i need number 13 as output in order to use that with another method.

Can anyone help me in writing java code for the above logic?

Thanks in Adavance!!

Upvotes: 0

Views: 543

Answers (6)

DarkDust
DarkDust

Reputation: 92326

If you known the input is always well-formed with 1, 2, ..., 9, 10, 11, 12, …, 99 concatenated you can derive the last value simply by looking at the string length:

public static void main(String[] args) {
        System.out.println(getLastNumber("123")); // 3
        System.out.println(getLastNumber("12345678910111213")); // 13
}

public static int getLastNumber(String input) {
        if (input == null) {
                return 0;
        } else if (input.length() > 9) {
                return 9 + ((input.length() - 9) / 2);
        } else {
                return input.length();
        }
}

For values larger than 99, the algorithm would need to be expanded (there surely is a generic solution to this problem that deals with arbitrarily large numbers…).

Upvotes: 0

Nim
Nim

Reputation: 33655

Here is an outline of an algorithm, the assumption is that the last number you want is a valid number up to and including 13.

  1. Read a character, if it's 1, then (potentially could be 10/11/12/13), go to 2.
  2. Read next character, if it's not 0|1|2|3, then track the value of character in 1 as the last_value, then go to 4, else go to 3
  3. Track the value of the characters collected in 1 & 2 as the last_value (so for example, characters 1 & 1, would end up saving 11 in last_value.
  4. If there are more characters, go to 1, else return the last_value.

Upvotes: 0

Szabolcs Albert
Szabolcs Albert

Reputation: 1

Do you always want to read the number using the last 2 characters? If not how can you know how many characters you want in your number?

int desiredNumberLength = 2;
String str = "12345678910111213";
int number = Integer.parseInt( str.substring( str.length - desiredNumberLength, str.length ) );
if ( number > 13 ) number = number % 10;

Upvotes: 0

nafas
nafas

Reputation: 5423

just split the last two numbers if greater than 13 then return last char otherwise return last two characters. e.g.: 

public int getLastNum(String numbers) throws Exception{
  if(numbers.length()==0) return 0;
  if(numbers.length()==1) return Integer.parseInt(numbers);
  int num=Integer.parseInt(numbers.subString(numbers.length-2,numbers.length));
  if(num>13) return num%10;
  else return num;
}

Upvotes: 0

nomis
nomis

Reputation: 2715

If the numbers will always be ordered integers starting from 1 then you can try the following (highly inefficient) approach

public int getLast(String in) {
    int current = 0;
    while (in.length() != 0) {
        current++;
        in = in.replaceFirst(String.valueOf(current), "");
    }
    return current;
}

Upvotes: 2

Ofer Lando
Ofer Lando

Reputation: 824

I believe the simple answer is: you can't.

What you MIGHT want to do, if you have control on the string structure, is to pad each number with a zero (or zeros, if relevant), and then you can split the string into pairs (or trios, again, if relevant), and get the last group.

For example, instead of 12345678910111213 you would have "010203040506070809010111213", and then you can break it into "01 02 03 04 05 06 07 08 09 10 11 12 13" - which is then easy to get your "13" out of.

Upvotes: 0

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