减少这个Haskell功能

Question

我想加倍列表中的每一个元素.这是代码 -

doubleSec n [] = []
doubleSec n (x:xs)
    | n==1 = x*2 : doubleSec 0 xs
    | otherwise = x : doubleSec 1 xs 

doubleSecond xs =
    doubleSec 0 xs

如何在单个函数中压缩此逻辑？

Answer 1

您可以像这样匹配列表中的模式

doubleSec :: [Int] -> [Int]
doubleSec [] = []
doubleSec [x] = [x]
doubleSec (x : y : xs) = x : 2* y : doubleSec xs

让你对第二个元素做特定的事情