Haskell 字符串数字到字符串字表示法

我正在做一个小练习，帮我做一个更大的练习。现在，我正在尝试编写一个名为“digitStr”的函数，它的签名是（我认为）

digitStr:：String->String

我在这里试图做的是，当我键入以下内容时，我应该得到以下输出：

> digitStr "23"
"twenty three"

现在，假设数字只上升到25

我自己试过的：

我的第一个想法是向数据库中添加数字单词

wrdtxt :: txt
wrdtxt = " one"," two"," three"," four"," five","six"," seven"," eight"," nine"," ten"," eleven"," twelve","thirteen"," fourteen","fifteen"," sixteen"," seventeen"," eighteen"," nineteen", " twenty"

下一步是创建一个函数，该函数接受字符串并将其输出到字符串列表中：

numWords :: String -> [String]
numWords "" = []
numWords x = words x

下一步是计算函数的长度。如果它的长度是2，我们看第一个数字来确定它是10秒还是20秒。如果是“25”，不知何故，做“20”+“5”

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如你所知，我对该做什么有点困惑。在过去的两天里，我一直在考虑这个问题，这就是我能想到的

以下是模式匹配字符串的方法：

numWords :: String -> [String]
numWords []  =  ...    -- matches only the empty string
numWords [a] =  ...    -- matches when argument is just  a single char;
                          a = the digit
numWords [a,b] = ...   -- matches when there are exactly two chars;
                          a = first chars, b = second char
numWords ['1',b] = ... -- matches when there are exactly two characters and
                          the first char is '1'; b is set to the second char
numWords (a:as) = ...  -- matches when there is at least one character;
                          a = first character, as = second through last characters

但您可能还需要考虑更改NUMITS的签名以获取int：

numWords :: Int -> [String]
numWords x
  | x < 10  = ...     -- x is a single digit
  | x < 20  = ...     -- x is between 11 and 19
  | x < 30  = ...     -- x is between 20 and 29
  | ...

numWords:：Int->[String]
纽姆沃兹x
|x<10=…--x是一个位数
|x<20=…--x在11到19之间
|x<30=…--x在20到29之间
| ...

您已经有算法了吗？每个任务解决方案都首先考虑算法（这是必须的）。然后，您只需编写代码（这通常很简单）。