How to delete elements of a circular list until there is only one element left using python? [closed]

Question

How would I go about iterating through a list from 1-100, where I delete every other element starting with the first element, and repeat that step until there in only one element left in the list. Would I have to use a circular linked list, or can it be done just using loops and conditional statements?

Answer 1

This deletes every other element over and over until just one is left

>>> L = range(100)       # for Python3, use L = list(range(100))
>>> while len(L) > 1:
...     del L[::2]
... 
>>> L
[63]

I'm not sure what the "circular list" means, but maybe this modification is needed

>>> L = range(100)
>>> while len(L) > 1:
...     del L[len(L)%2::2]
... 
>>> L
[99]

The len(L)%2 means to del L[1::2] if the length of L is odd

Or if you like to see what's going on:

>>> L = range(100)
>>> while len(L) > 1:
...     del L[len(L)%2::2]
...     L
... 
[1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99]
[3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99]
[3, 11, 19, 27, 35, 43, 51, 59, 67, 75, 83, 91, 99]
[3, 19, 35, 51, 67, 83, 99]
[3, 35, 67, 99]
[35, 99]
[99]

Answer 2

How about using Python's handy slice syntax:

while len(best_list_ever) > 1:
    best_list_ever = best_list_ever[1::2]

The expression best_list_ever[1::2] is a list of every other element in the original list.

Edit: I'm actually pretty confused about the circular constraint thing, but if it's accurately documented by ysakamoto then maybe look to gnibbler's answer.