Python: Null Output Instead of Correct Value

Question

I'm trying to write a function to add the digits of a given number repeatedly until I have a single digit.

So addDigits(345) = 12, 1+2 = 3, so this should return 3

Here's my attempt:

class Solution(object):
 def addDigits(self, num):
 """
 :type num: int
 :rtype: int
 """
 if (num >= 0 and num < 10):
 return num
 else:
 total = 0
 while (num >= 10):
 total += (num % 10)
 num /= 10
 total += num
 self.addDigits(total)

On input 10, I'm getting null back and I have no idea why. I've traced through the code and it looks right to me...

• thefourtheye: You need to return the result of recursive call. return self.addDigits(total)

Answer 1

The old technique of "casting out nines" gives a quick answer

def sumOfDigits(x):
 if x == 0: return 0 
 y = x % 9
 return 9 if y == 0 else y

An example shows why this works:

433 = 4*(99+1) +3*(9+1) +2 = 4*99 + 3*9 + (4+3+3)

reducing modulo 9 gives (4+3+3) (mod 9)

As Andrea's comment below points out, this gives the wrong answer when the sum of the digits is congruent to 9.

• saulspatz: @Andrea Corbellini My mistake. Let me see if I can correct it.

Answer 2

Here is one approach that splits your number into individual digits, and continuously sums them until you end up with a single digit.

total = 12348736
while len(str(total)) > 1:
 total = sum(map(int,str(total)))
print total

EDIT: Just to explain this further:

• str(12348736): turns your number into a string
• map(int, '12348736'): turns your string (list of characters) into a list of integers (applies int to every character)
• sum([1, 2, 3, 4, 8, 7, 3, 6]): adds up all the digits in the list

• Dan D.: Omit the list: sum(map(int, str(total))).