""" Moving Zeros To The End
алгоритм, который берет массив и перемещает все нули в конец, сохраняя порядок остальных элементов """
import time
def move_zeros(s):
list_new = []
list_0 = []
for s1 in s:
if s1 == 0:
list_0.append(0)
else:
list_new.append(s1)
for s0 in list_0:
list_new.append(s0)
return list_new
if __name__ == '__main__':
#move_zeros([1, 0, 1, 2, 0, 1, 3]) # returns [1, 1, 2, 1, 3, 0, 0]
start_time = time.time()
print(move_zeros([1, 0, 1, 2, 0, 1, 3]), "Время выполнения: ", time.time() - start_time)
Public
Type hints help humans and linters (like mypy) to understand what to expect "in" and "out" for a function. Not only it serves as a documentation for others (and you after some time, when the code is wiped from your "brain cache"), but also allows using automated tools to find type errors.
Simple rule: read a variable name as if you were English-speaking person without a context and guess what's inside - and if you have no clue, then you should give it a better name. Exceptions: i, j, x, y etc.
s and s1 are not good variable names. Should be list_mixed and element, for example.
Seems like things could be organized in a better way.
list_0 makes no sense because it always contains zeros. So instead of storing [0, 0, 0, 0, ...] you could just store number of zeros you found. With this knowledge, you could write the whole thing without actually using list_0 and list_new: count number of zeros in s, remove them from s, add required number of zeros to what's left from s. See code below.
def move_zeros(s):
num_items = len(s)
s_without_zeros = [item for item in s if item != 0]
num_zeros = num_items - len(s_without_zeros)
s_with_zeros_in_the_end = s_without_zeros + [0] * num_zeros
return s_with_zeros_in_the_end
This code is not really needed or may be simplified
list_new += list_0
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