sumDigits challenge

Hi,

http://codingbat.com/prob/p163932

I am working on above challenge
Given a non-negative int n, return the sum of its digits recursively (no loops). Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost digit (126 / 10 is 12).

sumDigits(126) → 9
sumDigits(49) → 13
sumDigits(12) → 3

i was not sure how to find the sum of digits recursively without loop which i worked earlier. please advise
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gudii9Asked:
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d-glitchCommented:
Do you remember how to break up an integer into the LAST digit and the REST using % and / .

Recursion works by letting a function call itself multiple times, each time with a smaller or simpler argument.

The pseudo code would be something like:
function sumDigits( n)
sum = 0
if arg = 0 
     return sum
else
     sum = sum + LAST( arg) + sumDigits( REST( arg))

Open in new window


RECURSION is probably the single most important and elegant topic in computer science.
You should really read up on it if you don't understand the concept.  Challenges are not going to be enough.
gudii9Author Commented:
Do you remember how to break up an integer into the LAST digit and the REST using % and / .
yes
RECURSION is probably the single most important and elegant topic in computer science.
You should really read up on it if you don't understand the concept.  Challenges are not going to be enough.
where should i read up on it? Any best material,book, site?
d-glitchCommented:
Here are a couple of references:
     http://introcs.cs.princeton.edu/java/23recursion/
     http://arxiv.org/pdf/cs/9301113


My pseudo code needs some mods:
function sumDigits( n)
if n = 1   
     return 1
else   
     return  LAST + sumDigits( REST)

Open in new window

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rrzCommented:
My pseudo code is a little different from d-glitch
if input is 0 then return 0
if not then return last + sumDigits(rest)
gudii9Author Commented:
I will try
gudii9Author Commented:
public int sumDigits(int n) {
  if (n < 10) 
{return n;
}
  return (n % 10) + sumDigits(n/10);
}

Open in new window


above passes all test. any improvements or alternate approaches?
rrzCommented:
 if (n < 10)
{return n;
Brilliant! I like your base case much better.
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