I am trying to reverse order of giv开发者_开发百科en 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
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.
Here was my solution:
def f(n,a=None):
if n > 0:
f(n-1, input())
if a is not None:
print a
f(input())
It bugs me a little bit that I had to use the check on a
, but then I realized this question is a bit artificial anyway. You don't need to know the number of items in the list in order to reverse it. Of course the None
is in case 0
appears in the inputs.
Demo:
$ <<< '5
7
4
1
0
8' python Desktop/foo.py
8
0
1
4
7
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