如何有效地检查列表的前半部分是否与另一部分相等？

Question

I need to check if a Python list consists of two equal halves. For example, this list does:

[6, 2, 0, 2, 3, 2, 6, 2, 0, 2, 3, 2]

and this doesn't

[6, 2, 0, 2, 4, 6]

I've tried this check: len(lst) % 2 == 0 and lst[:len(lst)//2] == lst[len(lst)//2:] but it seems to be too slow for bigger lists. Any other solutions?

Answer 1

It's possible to do the check without creating the two sublists. It might be faster for really big lists.

n = len(lst)//2
all(lst[i]==lst[i+n] for i in range(n))

If you want to also check that your list is of even length, you could also add

len(lst)%2==0

as a condition.