Finding number position

[hugo84]

Hi i have a problem here im reading 2 int values f and digit from a user and i want to count the position of the number he input in a reverse order starting from 1. i Kept getting -1 from this. I think the problem is because of INT mismatch with String?Or can i use indexOfSubList? how can i implement it

public static int pos(int f, int digit){
      
         StringBuffer phrase = new StringBuffer(f);
         phrase.reverse();
         String x=Integer.toString(digit);
         int firstoccur = phrase.indexOf(x); // find where the Digit x first occurs
        
         return firstoccur;

[holestary]

public static int pos(int f, int digit){
      
         StringBuffer phrase = new StringBuffer(f);
         phrase.reverse();
         String x=Integer.toString(digit);
         int firstoccur = phrase.indexOf(x); // find where the Digit x first occurs
        
         return firstoccur;

i dont think it makes f a string..

[holestary]

        String phrase = Integer.toString(f);
        String reverse = new StringBuffer(phrase).reverse().toString();

u can reverse like this..

[hugo84]

        String phrase = Integer.toString(f);
        String reverse = new StringBuffer(phrase).reverse().toString();

u can reverse like this..

Hmmm.Weird. its not reversing the order from the results of my output.

       public static int position(int f, int digit){
         String phrase = Integer.toString(f);
         String reverse = new StringBuffer(phrase).reverse().toString();
         String x=Integer.toString(digit);
         int firstoccur = phrase.indexOf(x); // find where the Digit x first occurs from the right
        
         return firstoccur;

[holestary]

public class position {

    public static int pos(int f, int digit){
          
        String phrase = Integer.toString(f);
        String reverse = new StringBuffer(phrase).reverse().toString();
        System.out.println(reverse);
        String x=Integer.toString(digit);
        int firstoccur = reverse.indexOf(x); // find where the Digit x first occurs
        System.out.print(firstoccur+1);  //because index starts 0
        return firstoccur;
        }
       public static void main(String[] args) {
     
        pos(123414,3);       
        
    }

}

i think u understand what i mean with codes..
output:

414321
4

[hugo84]

thanks man learnt alot from you.

[jcaccia]

Hi i have a problem here im reading 2 int values f and digit from a user and i want to count the position of the number he input in a reverse order starting from 1. i Kept getting -1 from this. I think the problem is because of INT mismatch with String?Or can i use indexOfSubList? how can i implement it

public static int pos(int f, int digit){
      
         StringBuffer phrase = new StringBuffer(f);
         phrase.reverse();
         String x=Integer.toString(digit);
         int firstoccur = phrase.indexOf(x); // find where the Digit x first occurs
        
         return firstoccur;

Let's see if I understood the question: given an integer number and a single digit you want to know the first position of that digit in the number from right to left, correct?
I think this gives you the answer (-1 means the digit is not in the number):

private static int pos (int f, int digit) {
	String s = Integer.toString(f);
	int p = s.lastIndexOf('0' + digit);
	return (p < 0 ? p : s.length() - p);
}

StringBuffer phrase = new StringBuffer(f); does not convert f to String; it creates a StringBuffer with a capacity to hold f characters.

[nuzzle]

thanks man learnt alot from you.

This is an alternative (if you by reverse order mean from the right to the left),

public static int pos(int f, int digit) {
   if (f==0 && digit==0) return 1; // handle special case
   int pos = 1;
   while (f != 0) {
       if (f%10 == digit) return pos; // check rightmost digit of f
       f = f/10; // skip rightmost digit of f
       pos++;
   }
   return 0;
}

Digits from f are extracted from the right to the left. If none matches digit, 0 is returned.

Apart from being fast this code has the advantage of working with negative f.

[jcaccia]

Apart from being fast this code has the advantage of working with negative f.

When f is negative this code always returns 0. The test in the if should be:

if (Math.abs(f)%10 == digit) ...

[nuzzle]

When f is negative this code always returns 0. The test in the if should be:

if (Math.abs(f)%10 == digit) ...

Okay, you're right. I don't know why I claimed that. I think it's because one common use of this code is to count the digits in a number and that works for negative numbers too.

So this code is just a fast alternative. If you want it to work with negative f you can put in this line in the function before the loop starts,

if (f<0) f = -f;