求一个字符串在循环字符串中出现的总次数

Question

我目前正在学习Python，我被这个特定的问题困住了。图片

下面是我当前的代码:

word = input()
text = 0
wordch = 0
positions = 0
repeated = 0

while repeated != 2:
for i in range(0, len(tablet)):
if tablet[i] == word[wordch]:
text += 1
wordch += 1
if text == len(word):
positions += 1
text = 0
wordch = 0
elif repeated == 1 and text == len(word):
positions += 1
text = 0
wordch = 0
break
elif i == len(tablet)-1:
repeated += 1
break
elif tablet[i] != word[wordch]:
text == 0
wordch == 0

print(positions)

我希望一个代码，是真正的基本使用相同的概念，但请回答。谢谢你！

Answer 1

我尝试用另一种方法来解决这个问题。正如我们所知道的，如果我们试图从末尾以循环的方式创建子字符串，我们只能使用(len(fav_word)) - 1字母，因为如果我们取更多的字符，我们将从开始本身创建它们，而不需要循环。

因此，我只是通过将开始的(len(fav_word)) - 1附加到原始字符串中创建了一个新字符串，然后找到在新字符串中出现的所有fav_string。

def find_all(a_str, sub):
start = 0
while True:
start = a_str.find(sub, start)
if start == -1: return
yield start
start += 1
x = "cabccabcab"
fav = "abc"
y = x + x[0:len(fav)-1]
print(len(list(find_all(y, fav)))) # Output: 3
x = "ababa"
fav = "aba"
y = x + x[0:len(fav)-1]
print(len(list(find_all(y, fav)))) # Output: 2
x = "aaaaaa"
fav = "aa"
y = x + x[0:len(fav)-1]
print(len(list(find_all(y, fav)))) # Output: 6
x = "abaaba"
fav = "aaba"
y = x + x[0:len(fav)-1]
print(len(list(find_all(y, fav)))) # Output: 2

Answer 2

def find_str(g,find):
lg = len(g)
lf = len(find)
x=0
s=""
for index, i in enumerate(g):
if i == find[0]:
if index+lf <= lg:
s = "".join(g[index:index+lf])
if s == find:
x+=1
else:
rem = "".join(g[index:])
lr = len(rem)
for index,i in enumerate(g):
rem+=i
lr+=1
if lr == lf:
if rem == find:
x+=1
break
return x
print(find_str("abaaba","aaba"))

Answer 3

def split(word): 
return [char for char in word]          

x = "aaaaaa"
pattern = "aa"
mylist=split(x)
ok=True
occurrences=0
buffer=""
while ok:
char=mylist.pop(0)
buffer+=char
if buffer==pattern:
occurrences+=1
buffer=""
if len(mylist)==0:
ok=False
print(occurrences)
output:3