在 PYTHON 中使用递归检查回文

Question

我正在检查一个词是否是回文，但我似乎无法让我的代码响应。我觉得它的声音很漂亮，但显然我错过了一些东西。有人可以指出那可能是什么吗？

def reverse(usrWrd, index):   
    newWord = ""  

    if index == len(usrWrd):
        return newWord
    else:
      newWord += usrWrd[index]
      return reverse(usrWrd, index + 1)

def main(): 
    usrWrd = input("please enter a word to check for palindrome-ness:")

    result = reverse(usrWrd, 0)
    if result == usrWrd:
        print("That word is a palindrome")
    else:
        print("Sorry,",usrWrd,  "is NOT a palindrome")

main()

Answer 1

def isPalindrome(s):
   length = len(s);
   if length <= 1:
       return True
   if s[0] != s[length-1]:
       return False
   return isPalindrome(s[1:length-1])

或者对于那些喜欢更简洁代码的人：

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

Answer 2

def reverse(usrWrd, index):
    newWord = ""                     # This is a local

    if index == len(usrWrd):
        return newWord               # It will always be empty string here

    else:
        newWord += usrWrd[index]
        return reverse(usrWrd, index + 1)

我完全不确定你为什么认为你需要 newWord。你应该只比较第一个和最后一个字母是否相等，然后用你的递归函数检查剩下的子串也是回文。

Answer 3

如前所述，局部变量是问题所在，也许您可以一起发送局部变量。典型的尾递归实现

def reverse(usrWrd, index, newWord=''):

    if index == len(usrWrd):
        return newWord       
    else:
        newWord += usrWrd[index]
        return reverse(usrWrd, index + 1, newWord)

希望对您有所帮助！

Answer 4

def main():
    usrWrd = input("please enter a word to check for palindrome-ness:")

    result=reversed(usrWrd)

    if list(result) == list(usrWrd):
        print("That word is a palindrome")
    else:
        print("Word is NOT a palindrome")

Answer 5

# Correct approach to your solution
def reverse(usrWrd, index, newWord):
    if index < 0:
        return newWord
    else:
      newWord += usrWrd[index]
      return reverse(usrWrd, index - 1, newWord)

def main():
    newWord = ""
    usrWrd = input("please enter a word to check for palindrome-ness:")
    result = reverse(usrWrd, len(usrWrd) - 1, newWord)

    if result == usrWrd:
        print("That word is a palindrome")
    else:
        print("Sorry,",usrWrd,  "is NOT a palindrome")

################################################## ###########

# Alternate approach
def isPalindrome(usrWord):
    sLen = len(usrWord)
    if sLen <= 1:
        return True
    else:
        if usrWord[0] != usrWord[sLen-1]:
            return False
        else:
            return isPalindrome(usrWord[1:sLen-1])

def main():
    newWord = ""
    usrWrd = input("please enter a word to check for palindrome-ness:")
    result = isPalindrome(usrWrd)

    if result:
        print("That word is a palindrome")
    else:
        print("Sorry,",usrWrd,  "is NOT a palindrome")

################################################## ###########

# Pythonic way as suggested by 'Yaman Jain'
if usrWord == usrWord[::-1]:
    return True # Palindrome
else:
    return False # Not Palindrome

Answer 6

def rec_palindrome(S, n):
    if not S:
        return False
    if n == 0:
        return True
    else:
        return S[n] == S[(len(S)-1) - n] and rec_palindrome(S, n - 1)

如有改进请提出建议

Answer 7

def isPal(s):
    if len(s) <= 1:
        print("yes it is")
        return True
    else:
        return s[0] == s[-1] and isPal(s[1:-1])

这是您可能想要检查的另一个解决方案！

Answer 8

    string=string.lower()
    if(len(string)<2):
       return True 
    elif(string[0]!=string[-1]):
       return False
    else:
       return string[1:-1]
palin=is_palindrome("Abbacus")
if(palin):
  print("The string is palindrome")
else:
  print("The string is not Palindrome")

在 PYTHON 中使用递归检查回文

Checking for palindrome using recursion in PYTHON

python

recursion

palindrome