Am Dienstag 23 Februar 2010 14:54:36 schrieb Jonas Almström Duregård: > You are correct of course. Still, it will probably be a bit less > inefficient if the length of the lists are compared (as opposed to the > elements): > > noneRepeated xs = length xs == length (nub xs)
Only if no repeated elements appear early. For xs = 1 : [1 .. 10^7], xs == nub xs will return False without noticeable delay, length xs == length (nub xs) will take VERY long. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
