palindrome checker 4 methods.py

# -*- coding: utf-8 -*-
"""
Created on Sun Apr 8 18:10:50 2018

@author: Administrator
"""
#check if a string is a palindrome
#reversing a string is extremely easy
#at first i was thinking about using pop function
#however, for an empty string, pop function is still functional
#it would not stop itself
#it would show errors of popping from empty list
#so i have to use the classic way, slicing
#each time we slice the final letter
#to remove the unnecessary marks, we have to use regular expression

importre

deff(n):
#this is the base case, when string is empty
#we just return empty
ifn=='':
returnn
else:
returnn[-1]+f(n[:-1])

defcheck(n):
#this part is the regular expression to get only characters
#and format all of em into lower cases
c1=re.findall('[a-zA-Z0-9]',n.lower())
c2=re.findall('[a-zA-Z0-9]',(f(n)).lower())
ifc1==c2:
returnTrue
else:
returnFalse



#alternatively, we can use deque
#we pop from both sides to see if they are equal
#if not return false
#note that the length of string could be an odd number
#we wanna make sure the pop should take action while length of deque is larger than 1

fromcollectionsimportdeque

defg(n):

c=re.findall('[a-zA-Z0-9]',n.lower())
de=deque(c)
whilelen(de) >=1:
ifde.pop()!=de.popleft():
returnFalse
returnTrue


#or we can use non recursive slicing
defh(n):
c=re.findall('[a-zA-Z0-9]',n.lower())
ifc[::-1]==c:
returnTrue
else:
returnFalse

#or creating a new list
#using loop to append new list from old list s popped item
defi(n):
c=re.findall('[a-zA-Z0-9]',n.lower())
d=[]
foriinrange(len(c)):
d.append(c.pop())
c=re.findall('[a-zA-Z0-9]',n.lower())

ifd==c:
returnTrue
else:
returnFalse



print(check('Evil is a deed as I live!'))
print(g('Evil is a deed as I live!'))
print(h('Evil is a deed as I live!'))
print(i('Evil is a deed as I live!'))
#for time and space complexity, python non recursive slicing wins