Python Reverse Integer Using Recursion

July 26, 2023 Post a Comment

I am working on a problem that need to reverse a integer input without using list or string. But my program only return first and last digits. def reverseDisplay(number): if num

Solution 1:

This should work:

from math import log10
def rev(num):
    if num < 10:
        return num
    else:
        ones = num % 10
        rest = num // 10
        #print ones, rest, int(log10(rest) + 1), ones * 10 ** int(log10(rest) + 1)
        return ones * 10 ** int(log10(rest) + 1) + rev(rest)
print rev(9000), rev(1234), rev(1234567890123456789)

You could also reduce the number of times you call log10 and number of math operations by using a nested recursive function:

defrev(num):
    defrec(num, tens):
        if num < 10:
            return num        
        else:
            return num % 10 * tens + rec(num // 10, tens // 10)
    return rec(num, 10 ** int(log10(num)))

Solution 2:

It's running multiple times, but it only returns the left-most digit. It's pretty easy to see that's the case, if you consider under what circumstance the if will be true.

To make this work, you need to add in the digits that you skipped as you pass out of the recursion. The following works by simply tacking the passed digits to the left of the previous result. I'd rather do it without the string conversion, but I couldn't come up with an elegant way to do that...

defreverseDisplay(number):
   if number<10:
      return number 
   else:
      returnint(str(number%10) + str(reverseDisplay(number//10)))
defmain():
    number=int(input("Enter a number:"))
    print(number%10,end='')
    print(reverseDisplay(number))
main()

Solution 3:

Edit: This is a working solution.

def rev(x,prod=0):if x <10:returnprod+ x
    else:prod=prod*10+ x%10*10return rev(x /10,prod)

Result:

>>>rev(123)
321
>>>rev(12345)
54321
>>>rev(72823780029)
92008732827L
>>>rev(1)
1

Solution 4:

This should work : checked on Python3

# reverse a number
def reverseNum(n, rem=0):
   if n == 0:
      return (rem)//10else:
      return reverseNum(n//10, (rem+(n%10))*10)print(reverseNum(7165))
print(reverseNum(123456789))

Little modified version:

def reverseNum(n, rem=0):
    if n == 0:
        return rem
    elif n<10:
        return reverseNum(n//10, (rem+(n%10)))else:
        return reverseNum(n//10, (rem+(n%10))*10)

print(reverseNum(7165))
print(reverseNum(123456789))

Solution 5:

defreverseDisplay(number):
    if number<10:     
        return number                      #1 
    first_digit = number % 10#2
    the_rest = reverseDisplay(number//10)
    ex = 0while1:                               #3
        ex = ex + 1if number//(pow(10,ex)) == 0:
            break
    ex = ex - 1return first_digit*pow(10,ex) + the_rest #4

Here's how it works... (I've labelled the lines Im referring to here)

line 1: exit condition. But you know this
line 2: get the last digit. % means find the remainder of the division
line 3: if say we pass in 123. At this point we have first_digit=3 and the_rest=21. We want the result to be 321 = 300 + 21. So pretty much what we need is to know how many times we need to times first_digit by 10 for this to work
line 4: mmm delicious

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