Barbara Shirtcliff <ba...@gmx.com> писал(а) в своём письме Wed, 04 May
2011 16:41:07 +0300:
Also, note that "lexOrder s@[_] = [s]" is nicer than "lexOrder s |
length s == 1 = [s]".
I agree that that initial version was a little clumsy, but your
suggestion doesn't really seem to work:
lexOrder :: [Char] -> [[Char]]
lexOrder s@[_] = s
lexOrder s =
concat $ map (\n -> h n) [0..((length z) - 1)]
where z = sort $ nub s
h :: Int -> [String]
h n = map (z!!n :) $ lexOrder $ filter (\c -> lexI c z /=
n) z
Euler.hs:8:18:
Couldn't match expected type `[Char]' with actual type `Char'
Expected type: [[Char]]
Actual type: [Char]
In the expression: s
In an equation for `lexOrder': lexOrder s@[_] = s
It actually works, you have forgotten square brackets: "lexOrder s@[_] =
[s] --not s!".
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