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haskell heap sort
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import Data.List | |
swap :: Int -> Int -> [a] -> [a] | |
swap i j xs | i == j = xs | |
swap i j xs | otherwise = initial ++ (xs !! b) : middle ++ (xs !! a) : end | |
where [a,b] = sort [i,j] | |
initial = take a xs | |
middle = take (b-a-1) (drop (a+1) xs) | |
end = drop (b+1) xs | |
largest :: Ord a => Int -> Int -> [a] -> Int | |
largest i hs xs = | |
let large = if (l < hs) && ((xs !! l) > (xs !! i)) then l else i | |
in if (r < hs) && ((xs !! r) > (xs !! large)) then r else large | |
where l = 2 * i + 1 | |
r = 2 * i + 2 | |
heapify :: Ord a => Int -> Int -> [a] -> [a] | |
heapify i hs xs = | |
if (large /= i) then heapify large hs (swap large i xs) | |
else xs | |
where large = largest i hs xs | |
buildheap :: Ord a => Int -> [a] -> [a] | |
buildheap 0 xs = heapify 0 (length xs) xs | |
buildheap i xs = buildheap (i - 1) (heapify i (length xs) xs) | |
hpsort i xs = let swapped = swap 0 i xs | |
in if i /= 1 then hpsort (i - 1) (heapify 0 i swapped) | |
else (heapify 0 i swapped) | |
heapsort xs = let heap = buildheap (length xs `div` 2) xs | |
in heapsorting (length xs - 1) heap | |
-- heapsort [16,4,10,14,7,9,3,2,8] |
Yes, the array immutability makes it almost infeasible for a direct algorithm such as this to work. I read some of Chris Okasaki's work and I must confess it got me. Anyway, nice code indeed. There are some redundant guards though.
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Hi. Nice code! I am reading "Introduction to Algorithms" and trying to write algorithms on haskell. I reached to heapsort and got a troubles. It is very imperative. So I wanted to say: your decision is done correctly according to the algorithm, but it is not effective. You know swap, take, drop, ++ it is all works for O(n). heapsort ([1..10000]):
"sample.hs: out of memory
[Finished in 138.6s with exit code 1]"
possible solutions
I'll try 3d
PS Also on 33 line heapsorting -> hpsort