Haskell中如何用Integer处理运行最小/最大问题（无需非空输入）

方法一：用Maybe类型包装结果 #

直接用Maybe区分空输入与非空输入的结果：空输入返回Nothing，非空输入返回Just 最小值/最大值。初始化状态设为Nothing，遍历元素时动态确定初始值，完全避开边界值或无穷值的依赖。

示例代码：

import Data.List (foldl')

-- 计算运行最小值
runMin :: [Integer] -> Maybe Integer
runMin = foldl' updateMin Nothing
  where
    updateMin Nothing x = Just x
    updateMin (Just currentMin) x = Just $ min currentMin x

-- 计算运行最大值
runMax :: [Integer] -> Maybe Integer
runMax = foldl' updateMax Nothing
  where
    updateMax Nothing x = Just x
    updateMax (Just currentMax) x = Just $ max currentMax x

该方案贴合Haskell惯用模式，无需额外定义类型，空输入的处理自然融入逻辑。

方法二：自定义带无穷值的扩展整数类型 #

若需明确表示“无穷”状态（比如固定滑动窗口未填满时的初始状态），可自定义包含无穷值的代数数据类型，实现Ord实例后即可像普通整数一样比较：

data ExtendedInteger = Finite Integer | PositiveInfinity | NegativeInfinity
  deriving (Show, Eq)

instance Ord ExtendedInteger where
  compare PositiveInfinity PositiveInfinity = EQ
  compare PositiveInfinity _ = GT
  compare _ PositiveInfinity = LT
  compare NegativeInfinity NegativeInfinity = EQ
  compare NegativeInfinity _ = LT
  compare _ NegativeInfinity = GT
  compare (Finite a) (Finite b) = compare a b

-- 空输入返回PositiveInfinity
runMinExt :: [Integer] -> ExtendedInteger
runMinExt = foldl' updateMin PositiveInfinity
  where
    updateMin current x = min current (Finite x)

-- 空输入返回NegativeInfinity
runMaxExt :: [Integer] -> ExtendedInteger
runMaxExt = foldl' updateMax NegativeInfinity
  where
    updateMax current x = max current (Finite x)

此方案适合需要保留“空窗口对应无穷值”语义的场景，滑动窗口问题中未填满的初始状态可直接用无穷值参与后续比较。

方法三：结合foldr的惰性处理（针对全局最值场景） #

若仅需计算全局最小/最大值而非逐步骤运行结果，用foldr可实现惰性处理，同时支持空输入：

runMinLazy :: [Integer] -> Maybe Integer
runMinLazy [] = Nothing
runMinLazy (x:xs) = Just $ foldr min x xs

runMaxLazy :: [Integer] -> Maybe Integer
runMaxLazy [] = Nothing
runMaxLazy (x:xs) = Just $ foldr max x xs

写法简洁，适合处理大型或惰性列表，可提前终止不必要的计算。

内容的提问来源于stack exchange，提问作者Brendan Langfield