How to count the length of integer using recursion

Question

I've been trying to find how do I do check the length of integer using recursion. For example: 100, it will say that there's 3 number. 12, it will say that there's 2 number.

What I've saw online is all about the summation of the last digit. I couldn't source for help on the length of an integer.

can someone help me out on this? I've tried to use lens to count, but my output is always 0.

def num(x):
   if x == 0:
       return 0
   else:
       return num(x / 10)
print len(str(num(100)))

I've also tried this

def num(x):
       if x == 0:
           return 0
       else:
           return num(x / 10) + len(str(x % 10))
    print num(100)

Kasravnd · Accepted Answer · 2015-10-08 06:47:12Z

1

You need to add 1 with num(N/10) if N is not 0 :

>>> def num(N):
...   if N==0:
...      return 0
...   return 1+num(N/10)
... 
>>> num(100)
3

This code will add ones in each calling the function with N/10 until it get to 0, so at the end based on what time your number can bi divide by 10 you'll have 1's which have been sum together, and at last for 0 they eill be add with 0.

For example for 100 your function will do the following :

1+(100/10)
      ^
1+1+(10/10)
       ^
 1+1+1(1/10)
        ^
  1+1+1+0

edited Oct 8, 2015 at 6:47

answered Oct 8, 2015 at 6:31

Kasravnd

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7 Comments

stack Over a year ago

Hi, Can you explain to me why is there a 1+ needed?

stack Over a year ago

meaning,if let say I've this number (12345) n/10 will basically check it will loop how many time, right? & each time when it check, it will + 1, thus the output will be 5. Am I right to say that?

Kasravnd Over a year ago

@stack Exactly! for better demonstration check the edit again ;)

stack Over a year ago

thank you so much. Do you know if I will to use a while loop in it, how it should be?

Kasravnd Over a year ago

@stack Welcome, that would be a simple task you can just try it by your self! ;) as a hint you need to add ones till N gets to 0.

|

Sharon Dwilif K · Accepted Answer · 2015-10-08 06:32:15Z

1

Are you looking for this -

def numlen(i):
    if i:
        return numlen(i//10) + 1
    return 0

answered Oct 8, 2015 at 6:32

Sharon Dwilif K

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