Recursive Function palindrome in Python [closed]

爷,独闯天下 提交于 2019-12-28 05:40:07

问题


I need help writing a recursive function which detects whether a string is a palindrome. But i can't use any loops it must be recursive. Can anyone help show me how this is done . Im using Python.


回答1:


def ispalindrome(word):
    if len(word) < 2: return True
    if word[0] != word[-1]: return False
    return ispalindrome(word[1:-1])

And here is the best one liner

def ispalindrome(word):
    return word == word[::-1]



回答2:


From a general algorithm perspective, the recursive function has 3 cases:

1) 0 items left. Item is a palindrome, by identity.

2) 1 item left. Item is a palindrome, by identity.

3) 2 or more items. Remove first and last item. Compare. If they are the same, call function on what's left of string. If first and last are not the same, item is not a palindrome.

The implementation of the function itself is left as an exercise to the reader :)




回答3:


If a string is zero or one letters long, it's a palindrome.

If a string has the first and last letters the same, and the remaining letters (I think it's a [1: -1] slice in Python, but my Python is a bit rusty) are a palindrome, it's a palindrome.

Now, write that as a palindrome function that takes a string. It will call itself.




回答4:


Here's another viewpoint

A palindromic string is

  1. Some letter, x.

  2. Some palindromic substrinng.

  3. The same letter, x, repeated.

Also, note that you may be given a proper English sentence "Able was I ere I saw Elba." with punctuation. Your palindrome checker may have to quietly skip punctuation. Also, you may have to quietly match without considering case. This is slightly more complex.

  1. Some leading punctuation. Some letter, x.

  2. Some palindromic substring.

  3. Some letter, x, repeated without regard to case. Some trailing punctuation.

And, by definition, a zero-length string is a palindrome. Also a single-letter string (after removing punctuation) is a palindrome.




回答5:


Since we're posting code anyway, and no one-liner has been posted yet, here goes:

def palindrome(s):
    return len(s) < 2 or s[0] == s[-1] and palindrome(s[1:-1])



回答6:


The function should expect a string. If there is more then one letter in the string compare the first and the last letter. If 1 or 0 letters, return true. If the two letters are equal call the function then again with the string, without the first and the last letter. If they are not equal return false.

 palindrom( word):
   IF length of word 1 or 0 THEN
      return 0;
   IF last and first letter equal THEN
     word := remove first and last letter of word;
     palindrom( word);
   ELSE
     return false;



回答7:


Here's a way you can think of simple recursive functions... flip around the problem and think about it that way. How do you make a palindrome recursively? Here's how I would do it...

def make_palindrome():
    maybe:
        return ""
    elsemaybe:
        return some_char()
    else:
        c = some_char()
        return c + make_palindrome() + c

Then you can flip it around to build the test.




回答8:


My solution

#To solve this I'm using the stride notation within a slice [::]
def amazonPalindrome(input):
    inputB = input
    input = input[::-1]
    #print input
    noPalindrome = inputB + " is not a palindrome"
    isPalindrome = inputB + " is a palindrome"
    #compare the value of the reversed string to input string
    if input[0]!= input[-1]: 
        print noPalindrome
    else:
        print isPalindrome


#invoking the def requires at least 1 value or else it fails
#tests include splitting the string,mixing integers, odd amount palindromes.
#call the def  
amazonPalindrome('yayay')



回答9:


a=raw_input("enter the string:")
b=len(a)
c=0
for i in range(b):
    if a[i]==a[-(i+1)]:
        c=c+1
if c==b:
    print a,"is polindrome"
else:
    print a,"is not polindrome"



回答10:


n=raw_input("Enter a number===>")
n=str(n)
l=len(n)
s=""
for i in range(1,l+1):
    s=s+n[l-i]
if s==n:
    print "Given number is polindrom"
else:
    print "Given number is not polindrom"



回答11:


Here is C version, if anyone happens to land here searching for C code!

int IsPalindrome_Recursive(char *s, int start, int end)
{
    if ((end - start) < 2)
    {
        return 1;
    }
    if (s[start] != s[end])
    {
        return 0;
    }
    return IsPalindrome_Recursive(s, ++start, --end);
}

Call as:

IsPalindrome_Recursive(s, 0, strlen(s) - 1)


来源:https://stackoverflow.com/questions/952110/recursive-function-palindrome-in-python

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!