在Haskell中旋转char Matrix图像

Question

我正在为我的班级工作,其中一个要求是创建一个名为rotate90的函数.此功能基本上采用[[Char]]并顺时针旋转90度.

例如:

type Picture = [[Char]]
pic :: Picture
pic = [ "123",
    "456",
    "789" ]

变成:

[ "741",
  "852",
  "963" ]

到目前为止我的代码看起来像这样:

rotate90 :: Picture -> Picture
rotate90 (x:xs)
    | (x:xs) == []          = []
    | xs == [] && x /= []   = formRow ([[]]) (formCol x)
    | xs /= []              = formRow (rotate90 xs) (formCol x)


formCol :: [Char] -> [[Char]]
formCol y = [[a] | a <- y]

formRow :: [[Char]] -> [[Char]] -> [[Char]]
formRow (x:xs) (y:ys)
    | xs == [] || ys == []  = (x++y):[]
    | otherwise             = (x++y):formRow xs ys

现在它只打印矩阵的第一个"线",从示例中,它是"741".如何打印其余部分？

Answer 1

一个简单的实现Data.List.transpose就是

-- | Rotate clockwise
cw = map reverse . transpose
-- | Rotate counter-clockwise
cw = reverse . transpose

转置原始图片会产生

147
258
369

并且反转每一行导致旋转的图像

741
852
963

通常,您可以使用以下三个功能的组合来表达任意方向的镜像和旋转:

transpose
map reverse -- mirror left <-> right
reverse -- mirror top <-> bottom