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对于手头的练习，您的代码似乎有些复杂。这是一个更简单的版本，可以做相同的功能（我认为）。aba。
当我提交代码时，有一个自动系统检查您的算法，该系统告诉我的算法存在一些问题，我找不到问题。这是我功能的代码：def question3(str):#define function with input string i=0#define counter which will run through the whole string from the begining j=0#define counter which will run through the whole string without the char we deleted from the begining of the string k=0#define counter which will run through the string without the deleted char from the end of the string answer=False #define answer while i<len(str):#run a loop through the string k=len(str)-1 while j<len(str):# run a loop through the string without the deleted char if i==j:# if j is in the place of the deleted chart,j skip to the next char j=j+1 continue if k==i:# if k is in the place of the deleted chart,k skip to the next char k=k-1 continue if str[j]==str[k]:#check if the chart in the j's place equal to the char in the k's place answer=True else: answer=False#if the answer is false,we don't need to check the rest of the string because it's already not almost a polyndrom break#exit the inner loop j=j+1#j moves to the next chart k=k-1# k moves to the next chart if answer==True:# if we found that the string is almost a polyndrom without a specific char, #we don't need to check the rest of the string without the rest of the chars and we can exit the outer loop break# exit the loop i=i+1# move to the next chart j=0#nullify the counter that runs through the whole string from the beginning without a specific char print(answer) return;
要检查python中的alindrinmes，最简单的方法是（来自this Answer
）：
def check_for_palindrome(s):
return s == s[::-1]
您正在检查至少一个可能的新字符串是一个回文：
def is_almost_palindrome_v1a(s):
"""
'True' if there is at least one option that is palindrome.
"""
for i in range(len(s)):
new_s = s[:i] + s[i+1:]
is_palindrome = check_for_palindrome(new_s)
print(new_s, is_palindrome)
if is_palindrome:
return True
return False
可以使用
any()：
可以缩短这一点def is_almost_palindrome_v1b(s):
return any(
check_for_palindrome(s[:i] + s[i+1:])
for i in range(len(s)))
您正在检查所有可能的新字符串都是alindromesdef is_almost_palindrome_v2a(s):
"""
'False' if there is at least one option that is NOT a palindrome.
"""
for i in range(len(s)):
new_s = s[:i] + s[i+1:]
is_palindrome = check_for_palindrome(new_s)
print(new_s, is_palindrome)
if not is_palindrome:
return False
return True
可以使用
all()可以缩短这一点
def is_almost_palindrome_v2b(s):
return all(
check_for_palindrome(s[:i] + s[i+1:])
for i in range(len(s)))
def question3(str):#define function with input string
strCopy = "" # define an empty string to copy a string without a char we want to
delete
i = 0 # define counter which will run through the original string
j = 0 # define first help counter
answer = False # define answer
while i < len(str): # run a loop through the original string
while j < len(
str): # a loop where the original string will be copied to a new string without the deleted char
if i == j: # check if the current char is the char we want to delete,if yes,skip to the next char
j = j + 1
continue
strCopy = strCopy + str[j] # add the char in the j'th place from the original string to the new string
j = j + 1#moving to the next char
if strCopy==strCopy[::-1]:#check if the string without the deleted char is a palindrome,if yes,so the original string is almost a palindrome
answer=True
break
i = i + 1#moving to the next char
strCopy = ""#nullify the string without the deleted char
j = 0#nullify j
print(answer) # print the result(True/False)
return;
def check_palindrom(s):
right = len(s)-1
count = 0
fl = len(s) // 2
for i in range(fl):
if s[i] != s[right]:
count += 1
if count > 1:
return "String is not palindrom"
right -= 1
return "String is almost palindrom" if count == 1 else "String is palindrom"
s="abcdcbeea"
print(check_palindrom(s))