如何在Python中不使用reversed（）或[::-1]来反转字符串列表

Question

I made a post about this recently but didn't go into proper detail and I don't really know how commenting works on this website so I thought I'd create another thread with proper detail (if that's okay, sorry if it isn't)

Simply put, I'm writing code that prompts for and reads a string from a user (eg: 12345). This string could be infinitely long (eg: 123456789123456789, etc). After this string is entered, my code then takes each element from that string and puts it into a list (eg, 12345 is turned into ['1', '2', '3', '4', '5']).

What I need help with is reversing this list of strings without using reversed() or [::-1], and for it to work with a list of strings that could be infinitely long (eg, ['1', '2', '3', '4', '5'] turns into ['5', '4', '3', '2', '1']).

I know this is very basic, but I've spent quite a while trying to think of how to do this and for some reason my slow brain can't grasp a way how to. The best way for me to learn is to see it being done, with an explanation as to how it works (or I could look at the code and figure out the 'how' part by myself). I would be extremely appreciative for help on this, and thankyou in advance!

Answer 1

How about this

x = input('enter the values:')
x = list(x)

res = []

for i in range(len(x) -1, -1, -1):
    res.append(x[i])

print(res)

Answer 2

In addition to using reversed() or L[::-1] , you could use list.reverse() to reverse the elements of the list in-place :

>>> L = ['1', '2', '3', '4', '5']
>>> L.reverse()
>>> L
['5', '4', '3', '2', '1']

You can implement reversed() yourself with a for loop:

>>> L = ['1', '2', '3', '4', '5']
>>> R = []
>>> for i in range(len(L)-1, -1, -1):
...     R.append(L[i])
... 
>>> R
['5', '4', '3', '2', '1']

Answer 3

lst1 = [1,2,3,4,5,6]
reversed_list = []
length = len(lst1)
for r in range(0, length ):
    reversed_list.append( lst1[length - r - 1] )
print reversed_list

Answer 4

x="hey there"
l=len(x)
out=''
while l:
    out=out+x[l-1]
    l-=1
print out

This is what you are looking for. Start at end of sting on work backwards and append it to an out var