def unique_in_order(string):
s = []
for key, group in itertools.groupby(string):
s.append(key)
return s
print(unique_in_order('AAAABBBCCDAABBB'))
print(unique_in_order('ABBCcAD'))
print(unique_in_order())
kankan def fun65(tar):
return for each in range(len(tar)) if tar != tar]
print(fun65('AAAABBBCCDAABBB'))
print(fun65('ABBCcAD'))
print(fun65())
print(fun65("")) 这道题太简单了,我可以直接写写。def unique_in_order(s):
l = []
for t in s:
if t not in l:
l.append(t)
return l
def unique_in_order(string):
l=]
for i in string:
if l[-1]!=i:l.append(i)
return l 想用迭代器的,没弄出来{:10_285:}def unique_in_order(string:str):
answer = []
iter_string = iter(string)
for each in iter_string:
answer.append(each)
temp = next(iter_string)
while each == temp:
each = next(iter_string)
return answer import itertools
def uniq_func(st):
list_k =[]
for key, group in itertools.groupby(st):
list_k.append(key)
st =input("请输入字符串:")
print(uniq_func(st)) 1 让我康康楼主的答案 看看 1
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