在Python中反转这个编码算法

Question

我需要帮助将这个编码算法转换为解码算法。我理解swap函数，但我对其余代码有问题。

from string import ascii_letters, digits

def shift(text, shift: int = 0):
    SPACE = ' '
    letters = ascii_letters + digits + SPACE
    letters_length = len(letters)
    shifted_chars = []
    for char in text:
        if char in letters: 
            shifted_chars.append(letters[(letters.index(char) + shift) % letters_length])
        else:
            shifted_chars.append(char)
    return ''.join(shifted_chars)

def swap(text):
    middle = len(text) // 2
    return text[middle:] + text[:middle]
def encode(text):
    if len(text) < 2:
        return shift(text, 7)
    for i in range(31):
        text = swap(shift(text, i))
    return text

Answer 1

固定由于paria . h。R的有用注释

当len(text)为奇数时，deswap(text)函数均衡swap(text)的不对称结果。

from string import ascii_letters, digits

def shift(text, shift: int = 0):
    SPACE = ' '
    letters = ascii_letters + digits + SPACE
    letters_length = len(letters)
    shifted_chars = []
    for char in text:
        if char in letters: 
            shifted_chars.append(letters[(letters.index(char) + shift) % letters_length])
        else:
            shifted_chars.append(char)
    return ''.join(shifted_chars)

def swap(text):
    middle = len(text) // 2
    return text[middle:] + text[:middle]

def deswap(text):
    middle =  (len(text) + (len(text) % 2)) // 2
    return text[middle:] + text[:middle]

def encode(text):
    if len(text) < 2:
        return shift(text, 7)
    for i in range(31):
        text = swap(shift(text, i))
    return text

def decode(text):
    if len(text) < 2:
        return shift(text, -7)
    for i in range(30, -1, -1):
        text = deswap(shift(text, -i))
        # text = shift(deswap(text), -i)
    return text

:

decode(encode('opening'))   
# 'opening'
decode(encode('openin'))
# 'openin'
decode(encode('Quora'))
# 'Quora'
decode(encode('Quoran'))
# 'Quoran'
decode(encode(ascii_letters + digits)) == ascii_letters + digits
# True
decode(encode(ascii_letters + digits + ' ')) == ascii_letters + digits + ' '
# True