我最近开始学习Haskell,目前正在尝试编写基本的Haskell函数。
我已经写了一个名为intToRoman的函数,它可以将整数转换为罗马数字。它将整数列表( 1400 ->1,4,0,0)中的数字除以,并将每个数字转换为罗马数字,同时考虑到列表的帐户长度,以确定何时是千或百。
但是,它也不会停止并检查零。例如,数字1400将返回:
MCD** Exception: Map.!: given key is not an element in the map
CallStack (from HasCallStack)下面是代码本身:
mapInttoRoman:: M.Map Int String
mapInttoRoman = M.fromList
[(1,"I"),(4,"IV"),(5,"V"),(9,"IX"),(10,"X"),(40,"XL")
,(50,"L"),(100,"C"),(400,"CD"),(900,"CM"),(500,"D"),(1000,"M")]
listOfInt :: Int -> [Int]
listOfInt 0 = [ ]
listOfInt c = listOfInt(c`div`10) ++ [c`mod`10]
duplicate :: String -> Int -> String
duplicate string n = concat $ replicate n string
intToRoman :: Int -> String
intToRoman 0 = ""
intToRoman c = createInt x (len-1)
where x = listOfInt c
len = length x
createInt y xz = findRoman (head y) xz ++ createInt(drop 1 y)(xz-1)
where findRoman b l
| l < 1 = mapInttoRoman M.! b
| b == 0 = " "
| l == 3 = duplicate (mapInttoRoman M.! (10^l)) b
| b == 1 = mapInttoRoman M.! (10^l)
| otherwise = mapInttoRoman M.! (b*10^l)发布于 2020-03-01 04:09:31
我认为你使用的算法是不正确的,而且你让它变得比需要的更复杂。
而且列表也不完整,我添加了90和900个案例。
看一下这段代码..我认为它很简单,你会很容易得到它。
import qualified Data.Map as M
mapInttoRoman:: M.Map Int String
mapInttoRoman = M.fromList
[(1,"I"),(4,"IV"),(5,"V"),(9,"IX"),(10,"X"),(40,"XL")
,(50,"L"),(90,"XC"),(100,"C"),(400,"CD"),(500,"D"),(900,"CM"),(1000,"M")]
intToRoman :: Int -> String
intToRoman 0 = ""
intToRoman n | M.member n mapInttoRoman = mapInttoRoman M.! n
| otherwise = val ++ intToRoman (n-key)
where
keys = reverse (M.keys mapInttoRoman)
key = maxValidKey n keys
val = mapInttoRoman M.! key
maxValidKey n (k:ks)
| n >= k = k
| otherwise = maxValidKey n ks 测试:

https://stackoverflow.com/questions/46693478
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