一个简单的递归方法会给你你需要的答案。注意，无论传递给这个函数的值是什么，结果总是是'four'

MAP = [
    'zero',
    'one',
    'two',
    'three',
    'four',
    'five',
    'six',
    'seven',
    'eight',
    'nine'
]

def as_string(n):
    if n == 0:
        return MAP[0]
    s = []
    while n > 0:
        s.append(MAP[n%10])
        n //= 10
    return ''.join(s)

def number_of_letters(n):
    s = as_string(n)
    if (ls := len(s)) == n:
        return s
    return number_of_letters(ls)

print(number_of_letters(60))

• • 输出：**

four

• • 注：**

as_string（）函数将有效地反向返回字符串，例如，对于值60，它将返回'zerosix'。但是，这并不重要，因为我们只关心返回值的长度。
as_string（）函数中的整数运算性能明显优于任何涉及与str int之间转换的解决方案<-> int

• • 备选案文：**

使用相同的MAP列表：

def as_string(n):
    def _next(n):
        if n == 0:
            yield 0
        while n > 0:
            yield n % 10
            n //= 10
    return ''.join(reversed([MAP[i] for i in _next(n)]))

def number_of_letters(n, rv=None):
    if rv is None:
        rv = list()
    rv.append(s := as_string(n))
    if (ls := len(s)) == n:
        return rv
    return number_of_letters(ls, rv)

print(number_of_letters(60))

• • 输出：**

['sixzero', 'seven', 'five', 'four']

方法略有不同，每一步使用两个帮助函数：

class Solution(object):
    def numbers_of_letters(self,n):
        while True:
            new_n = words_to_number(number_to_words(n))
            if new_n == n:
                break
            else:
                n = new_n
        return number_to_words(new_n)

def words_to_number(words):
    print("".join(words))
    return sum(map(len, words))

def number_to_words(n):
    print(n)
    arr = {0:'zero',1:'one',2:'two',3:'three',4:'four',5:'five',
              6:'six',7:'seven',8:'eight',9:'nine'}
    to_char_array = list(map(int, str(n)))
    words = list(map(lambda x: arr[x], to_char_array))
    return words

if __name__ == "__main__":
    n=60
    print(Solution().numbers_of_letters(n))

因此，当结束条件不为真时，OP形式的答案只是循环：

class Solution:
    
    def numbers_of_letters(self, n):
        arr = ['zero','one','two','three','four','five','six','seven','eight','nine']

        while True:
           number_as_str = str(n)
           text = ''
           for digit in number_as_str:
               text += arr[int(digit)]
           if len(text) == n:
               return text
           n = len(text)

if __name__ == "__main__":
    n=463829
    print(Solution().numbers_of_letters(n))

输出：
four
正如@Fred所指出的，简单的版本是：

class Solution:
    def numbers_of_letters(self, n):
        return 'four'