CODEWITHSUNDEEP
javaarrays
ktouchie
ktouchie

Reputation: 391

Count occurrences of each digit in number using arrays

I need to write a program using arrays, which takes a number and returns the number of incidents of each digit within that number. I think I may be overcomplicating things here.

import java.util.*;
class Exercice7 {

  public static void main(String [] args) {
    Scanner sc = new Scanner(System.in);

    System.out.println("Veuillez saisir un nombre naturel:");   // Get number

    int n = sc.nextInt();                                       // Store number as n

    String str = Integer.toString(n);                           // Store n as string

    int length = str.length();                                    // Store string length as length

    int arr[] = new int[length];                                // Declare array with as many elements as n has digits

    int digit[] = {0,1,2,3,4,5,6,7,8,9};                        // Declare array with the digits to look for

    int count = 0;                                                // Number of occurences of each digit

    for (int i=(length-1); i>=0; i--) {                         // Fill array with digits from number input
        while (n>0) {
            arr[i]= n%10;
            n = n/10;
        }
    }

    for (int j=0; j<10; j++) {
        count = 0;
        for (int i=0; i<length; i++) {
            if (arr[i]==digit[j]) {
                count++;
            }
        }
        if (count>0) {
        System.out.println(digit[j] + " occurs " + count + " times.");
        }
    }
  }
}

This code only returns the number of 0s and 1s and it's wrong anyway. Could someone push me in the right direction?

Upvotes: 5

Views: 6678

Answers (3)

ktouchie
ktouchie

Reputation: 391

Thanks, deem, for your answer. I didn't completely understand what you meant, but it helped me get on the right track:

import java.util.*;
class Exercice7 {

public static void main(String [] args) {
    Scanner sc = new Scanner(System.in);

    System.out.println("Veuillez saisir un nombre naturel:"); //* Get number */
    int num = sc.nextInt(); //* Store number as n */
    String str = Integer.toString(num); //* Store n as string *//

    char digit[] = {'0','1','2','3','4','5','6','7','8','9'};
    int count = 0;

    for (int i=0; i<10; i++) {
        for (int j=0; j<(str.length()); j++) {
            if (str.charAt(j) == digit[i]) {
                count++;
            }
        }
        if (count>0) {
            System.out.println(digit[i] + " apparait " + count + " fois.");
            count = 0;
        }            
    }
}
}

It may not be the easiest way, but it works! Feel free to add more comments if you want to make improvements to the code.

Upvotes: 0

nafas
nafas

Reputation: 5423

you can try converting to String and read each char

int counts[] = {0,0,0,0,0,0,0,0,0,0};  
int myNumber=123222;//example
String string=""+myNumber; //converting integer to String

for(int i=0;i<string.length();i++){
 try{
   int n=Integer.parseInt(string.charAt(i)+"")
   counts[n]=counts[n]+1;
 }catch(Exception e){}

}

then to print it out:

for (int i=0; i<counts.length; i++) {
    System.out.println(i + " occurs " + counts[i] + " times.");
}

Upvotes: 0

deem
deem

Reputation: 1252

Declare array with ten elements ([0..9]) - there you will have occurences of each digit in your number. Simply using counts[3] will get you number of occurences of digit 3.

Then you just iterate over string number and read next char as integer and increase counter. This way you have only one loop. For example, having 3 in your number, you use counts[3]++.

Upvotes: 4

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