Question

I am trying to reverse order of given numbers in python.The problem says that the first line of the standard input contains one integer t (t<1001) which is the number of numbers.In each of the next t lines there is one integer n (n<1001).

Now I am trying to solve this using recursion trick:

def f(n,a):
   if n > 0:
      a = input()
      f(n-1,a)
   print a    
f(input(),0)

But this is not working properly since the last number is output twice.How to fix this ?

My obvious solution for this is:

n=input()
a=[1]*n
for i in range(n):a[i]=input()
a=a[::-1]
for i in range(n):print a[i]

If anybody is aware of any other smart solution for the same please enlighten me too.
Thanks

Answer 1

You probably don’t want to print in the ‘n == 0‘ case.
Try indenting the ‘print‘ inside the ‘if‘.

I imagine your ‘a‘ variable is getting re-used in the ‘n == 0‘ case, causing your problem.

Also, is there a particular reason you don’t just read the input into an list, then reverse that list? [I see you know how to do that from your edit] If this is just to experiment, then more power to you. Lists in python have a reverse() function, though (:

FWIW, here’s a slight variation on your solution I tried out:

def f(n):
    if n == 0:
        return
    a = input()
    f(n-1)
    print a

f(input())

Also, there’s a fun function in Python called reversed(), which returns an iterator.

x = [1,2,3]
for i in reversed(x):
    print i    # prints 3, 2, 1

And lastly, not exactly related, but since you seem to be learning recursion… A handy way to debug your problem might be to put this as the first line of your f() function:

print '>' * n, n, a

Maybe then it will be clearer why you have the double-print problem.