Why don't you just swap the pattern match order? remdups :: (Eq a) => [a] -> [a] remdups (x : xx : xs) = if x == xx then remdups (x : xs) else x : remdups (xx : xs) remdups xs = xs
This should cover all cases no? Also I prefer guards, but I guess that is personal remdups (x1:x2:xs) | x1 == x2 = remdups (x2 : xs) | otherwise = x1 : remdups (x2 : xs) remdups xs = xs 2009/3/15 R J <rj248...@hotmail.com> > I need to write an implementation using foldl, and a separate > implementation using foldr, of a function, "remdups xs", that removes > adjacent duplicate items from the list xs. For example, remdups > [1,2,2,3,3,3,1,1]= [1,2,3,1]. > > My approach is first to write a direct recursion, as follows: > > remdups :: (Eq a) => [a] -> [a] > remdups [] = [] > remdups (x : []) = [x] > remdups (x : xx : xs) = if x == xx then remdups (x : xs) else x : > remdups (xx : xs) > > This code works, but it has three cases, not usual two, namely [] and (x : > xs). > > What, if any, is the implementation using only two cases? > > Also, if three cases are required, then how can it be implemented using > foldr, and how using foldl? > > Thanks. > > ------------------------------ > Express your personality in color! Preview and select themes for Hotmail®. See > how.<http://www.windowslive-hotmail.com/LearnMore/personalize.aspx?ocid=TXT_MSGTX_WL_HM_express_032009#colortheme> > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > >
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