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Recursively find sum of digits including negative numbers

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1

1

I'm working on a few challenges using recursion in Java (a new concept to me). I'm currently working on a recursive method that gives the sum of digits given a long parameter (n). Intended output:

• 257 = 14


• -257 = -14

I've tried editing my base case to include the numbers only between 1 & 10, then decrement the sum for sumOfDigits(), but obviously you cannot use an if, else if, or else statement in a recursive method. I am stuck! Can someone steer me towards the solution? Here's my method:

 public static long sumOfDigits(long n)

 long sum = n %10;
 if ( n < 10)
 return sum;
 else 
 return sum += sumOfDigits(n%10);
 

java recursion

share|improve this question

edited Mar 9 at 4:17

cdlane

19.9k21245

asked Mar 9 at 1:05

Sam MatthewsSam Matthews

208

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why can you not use if, else if and else statements in a recursive method?
  
  – OptimusCrime
  Mar 9 at 1:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You are correct, thanks for your comment! You made me do some more digging and I solved my own problem.
  
  – Sam Matthews
  Mar 9 at 1:23
  

  
  
  
  

add a comment | 

1

1

I'm working on a few challenges using recursion in Java (a new concept to me). I'm currently working on a recursive method that gives the sum of digits given a long parameter (n). Intended output:

• 257 = 14


• -257 = -14

I've tried editing my base case to include the numbers only between 1 & 10, then decrement the sum for sumOfDigits(), but obviously you cannot use an if, else if, or else statement in a recursive method. I am stuck! Can someone steer me towards the solution? Here's my method:

 public static long sumOfDigits(long n)

 long sum = n %10;
 if ( n < 10)
 return sum;
 else 
 return sum += sumOfDigits(n%10);
 

java recursion

share|improve this question

edited Mar 9 at 4:17

cdlane

19.9k21245

asked Mar 9 at 1:05

Sam MatthewsSam Matthews

208

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why can you not use if, else if and else statements in a recursive method?
  
  – OptimusCrime
  Mar 9 at 1:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You are correct, thanks for your comment! You made me do some more digging and I solved my own problem.
  
  – Sam Matthews
  Mar 9 at 1:23
  

  
  
  
  

add a comment | 

I'm working on a few challenges using recursion in Java (a new concept to me). I'm currently working on a recursive method that gives the sum of digits given a long parameter (n). Intended output:

• 257 = 14


• -257 = -14

I've tried editing my base case to include the numbers only between 1 & 10, then decrement the sum for sumOfDigits(), but obviously you cannot use an if, else if, or else statement in a recursive method. I am stuck! Can someone steer me towards the solution? Here's my method:

 public static long sumOfDigits(long n)

 long sum = n %10;
 if ( n < 10)
 return sum;
 else 
 return sum += sumOfDigits(n%10);
 

java recursion

share|improve this question

edited Mar 9 at 4:17

cdlane

19.9k21245

asked Mar 9 at 1:05

Sam MatthewsSam Matthews

208

 public static long sumOfDigits(long n)

 long sum = n %10;
 if ( n < 10)
 return sum;
 else 
 return sum += sumOfDigits(n%10);
 

share|improve this question

edited Mar 9 at 4:17

cdlane

19.9k21245

asked Mar 9 at 1:05

Sam MatthewsSam Matthews

208

share|improve this question

edited Mar 9 at 4:17

cdlane

19.9k21245

asked Mar 9 at 1:05

Sam MatthewsSam Matthews

208

edited Mar 9 at 4:17

cdlane

19.9k21245

edited Mar 9 at 4:17

cdlane

19.9k21245

asked Mar 9 at 1:05

Sam MatthewsSam Matthews

208

asked Mar 9 at 1:05

Sam MatthewsSam Matthews

208

3 Answers
3

active

oldest

votes

1

For anyone wondering, I managed to solve this myself!
Solution:

public static long sumOfDigits(long n)

 long sum = n %10;

 if (n == 0) return 0;

 if ( n >= 1 && n < 10)
 return sum;
 else if (n < 0)
 return sum - sumOfDigits(-n %10);
 else 
 return sum + sumOfDigits(n%10);
 

share|improve this answer

edited Mar 9 at 3:03

answered Mar 9 at 1:24

Sam MatthewsSam Matthews

208

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This works by coincidence, not by design. It simply returns twice the last digit. But since the last digit of test number 257 equals the sum of the first two digits, it appears to work. Change the test number from 257 to 256 and instead of getting 13, you'll get 12 -- the last digit doubled. Who upvoted this?
  
  – cdlane
  Mar 9 at 4:00
  
  

  
  
  
  

add a comment | 

1

I believe the solution is simpler than folks are trying to make it:

public static long sumOfDigits(long n) 

 long digit = n % 10;

 if (n == digit) 
 return n;
 

 return digit + sumOfDigits(n / 10);

The sum in the return statement is independent of sign as you're either adding positives to positive or negatives to negatives.

The tricky part is the digit extraction and test -- it depends on Java's remainder operator (%) preserving the sign of the number on the left. In some languages (e.g. Python) this is a modulus operator that always returns a positive result. If you're working with a language with modulus, which doesn't preserve sign, you'll need a workaround for this (e.g. the abs() and/or sign() functions).

share|improve this answer

edited Mar 9 at 5:04

answered Mar 9 at 4:11

cdlanecdlane

19.9k21245

add a comment | 

0

You can do it like this:

public class SumOfDigits 

 public static void main(String[] args) 
 System.out.println("257 -> " + sumOfDigits(257));
 System.out.println("-257 -> " + sumOfDigits(-257));
 System.out.println("0 -> " + sumOfDigits(0));
 System.out.println("1 -> " + sumOfDigits(1));
 

 public static long sumOfDigits(long n) 
 if (n < 0) 
 return sumOfDigitsOfPositive(Math.abs(n)) * -1;
 
 else 
 return sumOfDigitsOfPositive(n);
 
 

 // Parameter n must be positive
 private static long sumOfDigitsOfPositive(long n) 
 if (n < 10) 
 return n;
 
 else 
 long lastDigit = n % 10;
 return lastDigit + sumOfDigitsOfPositive(n / 10);
 
 

share|improve this answer

answered Mar 9 at 3:04

Prasad KarunagodaPrasad Karunagoda

1,7532814

add a comment | 

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1

For anyone wondering, I managed to solve this myself!
Solution:

public static long sumOfDigits(long n)

 long sum = n %10;

 if (n == 0) return 0;

 if ( n >= 1 && n < 10)
 return sum;
 else if (n < 0)
 return sum - sumOfDigits(-n %10);
 else 
 return sum + sumOfDigits(n%10);
 

share|improve this answer

edited Mar 9 at 3:03

answered Mar 9 at 1:24

Sam MatthewsSam Matthews

208

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This works by coincidence, not by design. It simply returns twice the last digit. But since the last digit of test number 257 equals the sum of the first two digits, it appears to work. Change the test number from 257 to 256 and instead of getting 13, you'll get 12 -- the last digit doubled. Who upvoted this?
  
  – cdlane
  Mar 9 at 4:00
  
  

  
  
  
  

add a comment | 

1

For anyone wondering, I managed to solve this myself!
Solution:

public static long sumOfDigits(long n)

 long sum = n %10;

 if (n == 0) return 0;

 if ( n >= 1 && n < 10)
 return sum;
 else if (n < 0)
 return sum - sumOfDigits(-n %10);
 else 
 return sum + sumOfDigits(n%10);
 

share|improve this answer

edited Mar 9 at 3:03

answered Mar 9 at 1:24

Sam MatthewsSam Matthews

208

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This works by coincidence, not by design. It simply returns twice the last digit. But since the last digit of test number 257 equals the sum of the first two digits, it appears to work. Change the test number from 257 to 256 and instead of getting 13, you'll get 12 -- the last digit doubled. Who upvoted this?
  
  – cdlane
  Mar 9 at 4:00
  
  

  
  
  
  

add a comment | 

For anyone wondering, I managed to solve this myself!
Solution:

public static long sumOfDigits(long n)

 long sum = n %10;

 if (n == 0) return 0;

 if ( n >= 1 && n < 10)
 return sum;
 else if (n < 0)
 return sum - sumOfDigits(-n %10);
 else 
 return sum + sumOfDigits(n%10);
 

share|improve this answer

edited Mar 9 at 3:03

answered Mar 9 at 1:24

Sam MatthewsSam Matthews

208

public static long sumOfDigits(long n)

 long sum = n %10;

 if (n == 0) return 0;

 if ( n >= 1 && n < 10)
 return sum;
 else if (n < 0)
 return sum - sumOfDigits(-n %10);
 else 
 return sum + sumOfDigits(n%10);
 

share|improve this answer

edited Mar 9 at 3:03

answered Mar 9 at 1:24

Sam MatthewsSam Matthews

208

answered Mar 9 at 1:24

Sam MatthewsSam Matthews

208

answered Mar 9 at 1:24

Sam MatthewsSam Matthews

208

1

I believe the solution is simpler than folks are trying to make it:

public static long sumOfDigits(long n) 

 long digit = n % 10;

 if (n == digit) 
 return n;
 

 return digit + sumOfDigits(n / 10);

The sum in the return statement is independent of sign as you're either adding positives to positive or negatives to negatives.

The tricky part is the digit extraction and test -- it depends on Java's remainder operator (%) preserving the sign of the number on the left. In some languages (e.g. Python) this is a modulus operator that always returns a positive result. If you're working with a language with modulus, which doesn't preserve sign, you'll need a workaround for this (e.g. the abs() and/or sign() functions).

share|improve this answer

edited Mar 9 at 5:04

answered Mar 9 at 4:11

cdlanecdlane

19.9k21245

add a comment | 

1

I believe the solution is simpler than folks are trying to make it:

public static long sumOfDigits(long n) 

 long digit = n % 10;

 if (n == digit) 
 return n;
 

 return digit + sumOfDigits(n / 10);

The sum in the return statement is independent of sign as you're either adding positives to positive or negatives to negatives.

The tricky part is the digit extraction and test -- it depends on Java's remainder operator (%) preserving the sign of the number on the left. In some languages (e.g. Python) this is a modulus operator that always returns a positive result. If you're working with a language with modulus, which doesn't preserve sign, you'll need a workaround for this (e.g. the abs() and/or sign() functions).

share|improve this answer

edited Mar 9 at 5:04

answered Mar 9 at 4:11

cdlanecdlane

19.9k21245

add a comment | 

I believe the solution is simpler than folks are trying to make it:

public static long sumOfDigits(long n) 

 long digit = n % 10;

 if (n == digit) 
 return n;
 

 return digit + sumOfDigits(n / 10);

The sum in the return statement is independent of sign as you're either adding positives to positive or negatives to negatives.

The tricky part is the digit extraction and test -- it depends on Java's remainder operator (%) preserving the sign of the number on the left. In some languages (e.g. Python) this is a modulus operator that always returns a positive result. If you're working with a language with modulus, which doesn't preserve sign, you'll need a workaround for this (e.g. the abs() and/or sign() functions).

share|improve this answer

edited Mar 9 at 5:04

answered Mar 9 at 4:11

cdlanecdlane

19.9k21245

public static long sumOfDigits(long n) 

 long digit = n % 10;

 if (n == digit) 
 return n;
 

 return digit + sumOfDigits(n / 10);

share|improve this answer

edited Mar 9 at 5:04

answered Mar 9 at 4:11

cdlanecdlane

19.9k21245

answered Mar 9 at 4:11

cdlanecdlane

19.9k21245

answered Mar 9 at 4:11

cdlanecdlane

19.9k21245

0

You can do it like this:

public class SumOfDigits 

 public static void main(String[] args) 
 System.out.println("257 -> " + sumOfDigits(257));
 System.out.println("-257 -> " + sumOfDigits(-257));
 System.out.println("0 -> " + sumOfDigits(0));
 System.out.println("1 -> " + sumOfDigits(1));
 

 public static long sumOfDigits(long n) 
 if (n < 0) 
 return sumOfDigitsOfPositive(Math.abs(n)) * -1;
 
 else 
 return sumOfDigitsOfPositive(n);
 
 

 // Parameter n must be positive
 private static long sumOfDigitsOfPositive(long n) 
 if (n < 10) 
 return n;
 
 else 
 long lastDigit = n % 10;
 return lastDigit + sumOfDigitsOfPositive(n / 10);
 
 

share|improve this answer

answered Mar 9 at 3:04

Prasad KarunagodaPrasad Karunagoda

1,7532814

add a comment | 

0

You can do it like this:

public class SumOfDigits 

 public static void main(String[] args) 
 System.out.println("257 -> " + sumOfDigits(257));
 System.out.println("-257 -> " + sumOfDigits(-257));
 System.out.println("0 -> " + sumOfDigits(0));
 System.out.println("1 -> " + sumOfDigits(1));
 

 public static long sumOfDigits(long n) 
 if (n < 0) 
 return sumOfDigitsOfPositive(Math.abs(n)) * -1;
 
 else 
 return sumOfDigitsOfPositive(n);
 
 

 // Parameter n must be positive
 private static long sumOfDigitsOfPositive(long n) 
 if (n < 10) 
 return n;
 
 else 
 long lastDigit = n % 10;
 return lastDigit + sumOfDigitsOfPositive(n / 10);
 
 

share|improve this answer

answered Mar 9 at 3:04

Prasad KarunagodaPrasad Karunagoda

1,7532814

add a comment | 

You can do it like this:

public class SumOfDigits 

 public static void main(String[] args) 
 System.out.println("257 -> " + sumOfDigits(257));
 System.out.println("-257 -> " + sumOfDigits(-257));
 System.out.println("0 -> " + sumOfDigits(0));
 System.out.println("1 -> " + sumOfDigits(1));
 

 public static long sumOfDigits(long n) 
 if (n < 0) 
 return sumOfDigitsOfPositive(Math.abs(n)) * -1;
 
 else 
 return sumOfDigitsOfPositive(n);
 
 

 // Parameter n must be positive
 private static long sumOfDigitsOfPositive(long n) 
 if (n < 10) 
 return n;
 
 else 
 long lastDigit = n % 10;
 return lastDigit + sumOfDigitsOfPositive(n / 10);
 
 

share|improve this answer

answered Mar 9 at 3:04

Prasad KarunagodaPrasad Karunagoda

1,7532814

public class SumOfDigits 

 public static void main(String[] args) 
 System.out.println("257 -> " + sumOfDigits(257));
 System.out.println("-257 -> " + sumOfDigits(-257));
 System.out.println("0 -> " + sumOfDigits(0));
 System.out.println("1 -> " + sumOfDigits(1));
 

 public static long sumOfDigits(long n) 
 if (n < 0) 
 return sumOfDigitsOfPositive(Math.abs(n)) * -1;
 
 else 
 return sumOfDigitsOfPositive(n);
 
 

 // Parameter n must be positive
 private static long sumOfDigitsOfPositive(long n) 
 if (n < 10) 
 return n;
 
 else 
 long lastDigit = n % 10;
 return lastDigit + sumOfDigitsOfPositive(n / 10);
 
 

share|improve this answer

answered Mar 9 at 3:04

Prasad KarunagodaPrasad Karunagoda

1,7532814

answered Mar 9 at 3:04

Prasad KarunagodaPrasad Karunagoda

1,7532814

answered Mar 9 at 3:04

Prasad KarunagodaPrasad Karunagoda

1,7532814