Python：如何使用递归反转整数

Question

I am given a problem where I have to reverse given digits using recursion only, but the problem is when I try to print out the integer reversed, for example if the integer was 1234, it would print the reversed numbers partially once at a time, eg: 21, 321, 4321. How can I fix this so that it can print 4321 in one go?

def reverse_digits(n):
    if n < 10:
        return n
    else:
        reverse = str(n % 10) + str(reverse_digits(n // 10))
        print(reverse)
        return reverse

Answer 1

Just don't have the function print anything at all, only returns the value, and you can print it when you call it:

def reverse_digits(n):
    if n < 10:
        return n
    else:
        reverse = str(n % 10) + str(reverse_digits(n // 10))
        return reverse

print(reverse_digits(1234))

Output:

4321

If you still want the function to print, you can print each digit separately in the same line (using end='' in the print function):

def reverse_digits(n):
    if n < 10:
        print(n)  #  print the very last digit and add a newline
        return n
    else:
        print(n % 10, end='')  #  print the last digit so far and stay on the same line
        reverse = str(n % 10) + str(reverse_digits(n // 10))
        return reverse

reverse_digits(1234)

Output:

4321

Answer 2

Remove the print from within the function:

def reverse_digits(n):
    if n < 10:
        return n
    else:
        reverse = str(n % 10) + str(reverse_digits(n // 10))
        return reverse
print(reverse_digits(1234))