/Python/Python sum of natural numbers recursion

Python sum of natural numbers recursion

Python recursive sum of digit

def sum_digits(num: int) -> int:
	
	#base case when your "positive" integer get to 0
    if num == 0: 
        return 0
    #base case when your "negative" integer get is between -10 and 0
    if num > -10 and num < 0:
        return num
	
	# recursion for positive integer
    elif num > 0:
        return (num % 10) + sum_digits(num//10)
	
	# recursion for negative integer
    elif num < 0:
        return -(abs(num) % 10) + sum_digits(-(abs(num)//10))
      
sum_digits(123) # returns 6
sum_digits(-123) # returns -6

Python sum of natural numbers recursion

# Python program to find the sum of natural using recursive function

def recur_sum(n):
   if n <= 1:
       return n
   else:
       return n + recur_sum(n-1)

# change this value for a different result
num = 16

if num < 0:
   print("Enter a positive number")
else:
   print("The sum is",recur_sum(num))


#Output - The sum is 210

Source: programiz.com

Similar codes

0 vote231 views2 answers
Python list all youtube channel videos
Python
0 vote222 views2 answers
Python dict for k v
Python
0 vote209 views1 answer
Length of list without len function
Python
8 votes226 views2 answers
Linear search in python
Python
9 votes215 views3 answers
TypeError: can only concatenate str (not "int") to str
Python
-1 vote213 views2 answers
Template strings in python
Python
0 vote246 views1 answer
Django save object in view
Python Django
12 votes217 views3 answers
Setwd python
Python
23 votes205 views1 answer
Python year month from date
Python
0 vote224 views1 answer
Show pythonpath
Python