Re: [Haskell-cafe] recursion problem part 2

On Feb 6, 2015 3:41 PM, "Jeffrey Brown" wrote:

Here's a solution: lastDigit x = mod x 10 remainingDigits x = (x - lastDigit x) `div` 10 listDigits x | x < 10 = [x] | otherwise = (listDigits $ remainingDigits x) ++ [lastDigit x]

Here's a faster one: listDigits2 x | x < 10 = [x] | otherwise = (listDigits $ remainingDigits) ++ [lastDigit] where lastDigit = mod x 10 remainingDigits = (x - lastDigit) `div` 10

These give somewhat peculiar results when the argument is negative. That can be rectified, of course. Unfortunately, all of these proposed solutions have a serious performance problem: successively appending single elements to build a list of length n is O(n^2). There's an easy fix, which I'll let you come up with. David