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每次我对函数进行递归时,我都想刷新一个变量值。为了简单起见,我会给你一个例子。

假设我们给一个函数一个数字 (n),它将返回它可以拥有的最大的 mod,其数字小于自身。

{- Examples:
  n=5 `mod` 5
  n=5 `mod` 4
  n=5 `mod` 3
  n=5 `mod` 2
  n=5 `mod` 1
-}

example :: Integer -> Integer
example n
    |  n `mod` ...  >  !The biggest `mod` it found so far!  && ... > 0
       = !Then the biggest `mod` so far will change its value.
    |  ... = 0 !The number we divide goes 0 then end! = 0

其中... =递归(我认为)

我不知道如何更好地描述它。如果你能帮助我,那就太好了。:)

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1 回答 1

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您可以按照您的描述编写它:

example :: Integer -> Integer
example n = biggestRemainder (abs n) 0
    where
        biggestRemainder 0 biggestRemainderSoFar = biggestRemainderSoFar
        biggestRemainder divisor biggestRemainderSoFar = biggestRemainder (divisor - 1) newBiggestRemainder
            where
                thisRemainder = n `mod` divisor
                newBiggestRemainder = case thisRemainder > biggestRemainderSoFar of
                    True -> thisRemainder
                    False -> biggestRemainderSoFar

这个函数也可以更容易地写成

example2 :: Integer -> Integer
example2 0 = 0
example2 n = maximum $ map (n `mod`) [1..(abs n)]
于 2013-11-07T15:53:19.667 回答