How to check for palindrome using Python logic [closed]

Answer 1

the "algorithmic" way:

import math

def isPalindrome(inputString):
    if inputString == None:
        return False

    strLength = len(inputString)
    for i in range(math.floor(strLength)):
        if inputString[i] != inputString[strLength - 1 - i]:
            return False
    return True

Answer 2

An alternative to the rather unintuitive [::-1] syntax is this:

>>> test = "abcba"
>>> test == ''.join(reversed(test))
True

The reversed function returns a reversed sequence of the characters in test.

''.join() joins those characters together again with nothing in between.

Answer 3

The real easy way to do that it is

word = str(raw_input(""))
is_palindrome = word.find(word[::-1])
if is_palindrome == 0:
    print True
else:
    print False

And if/else here just for fancy looks. The question about palindrome was on Amazon's interview for QA

Answer 4

#compare 1st half with reversed second half
# i.e. 'abba' -> 'ab' == 'ba'[::-1]

def is_palindrome( s ):
   return True if len( s ) < 2 else s[ :len( s ) // 2 ] == s[ -( len( s ) // 2 ):][::-1]

Answer 5

def is_palindrome(string):
   return string == ''.join([letter for letter in reversed(string)])

Answer 6

There is also a functional way:

def is_palindrome(word):
  if len(word) == 1: return True
  if word[0] != word[-1]: return False
  return is_palindrome(word[1:-1])