问题
I'm doing some python online tutorials, and I got stuck at an exercise:A palindrome is a word which is spelled the same forwards as backwards. For example, the word
racecar
is a palindrome: the first and last letters are the same (r), the second and second-last letters are the same (a), etc. Write a function isPalindrome(S) which takes a string S as input, and returns True if the string is a palindrome, and False otherwise. These is the code I wrote :
def isPalindrome(S):
if S[0] == S[-1]
return print("True")
elif S[0] == S[-1] and S[1] == S[-2] :
return print("True")
else:
return print("False")
But, if the word is for example ,,sarcas,, , the output is incorect. So I need a fix to my code so it works for any word.
回答1:
A one line solution but O(n) and memory expensive is :
def isPalindrome(word) : return word == word[::-1]
A O(n/2) solution that uses the same amount of memory is:
def palindrome(word):
for i in range(len(word)//2):
if word[i] != word[-1-i]:
return False
return True
This is the trick @LennartRegebro mentioned
回答2:
def is_palindrome(text):
text = text.replace(' ', '')
text = text.casefold()
if list(text) == list(reversed(text)):
return True
else:
return False
回答3:
Try this
word='malayalam'
print(word==word[::-1])
回答4:
Here is my solution.
S = input("Input a word: ")
def isPalindrome(S):
for i in range(0, len(S)):
if S[0 + i] == S[len(S) - 1]:
return "True"
else:
return "False"
print(isPalindrome(S))
回答5:
Here's my solution:
def isPalindrome(S):
l = len(S)-1
for i in range(0,l):
if S[i]!=S[l-i]:
return False
return True
回答6:
Another way of doing it using recursion is:
def isPalindrome(word):
if len(word) <= 1: return True
return (word[0] == word[-1]) and isPalindrome(word[1:-1])
回答7:
this is my solution S = input("Input a word: ")
def isPalindrome(S):
for i in range(0, len(S)):
if S[0 + i] == S[len(S) - 1]:
return "True"
else:
return "False"
print(isPalindrome(S))
来源:https://stackoverflow.com/questions/11758791/python-3-2-palindrome