The median of three has you look at the first, middle and last elements of the array, and choose the median of those three elements as the pivot. To get the “full effect” of the median of three, it’s also important to sort those three items, not just use the median as the pivot — … Read more
You mean something like this? quicksort [] = [] quicksort (x:xs) = quicksort less ++ (x : equal) ++ quicksort more where (less, equal, more) = partition3 x xs partition3 _ [] = ([], [], []) partition3 x (y:ys) = case compare y x of LT -> (y:less, equal, more) EQ -> (less, y:equal, more) … Read more
In terms of (asymptotic) time complexity – they are both the same. “Recursive is slower then iterative” – the rational behind this statement is because of the overhead of the recursive stack (saving and restoring the environment between calls). However -these are constant number of ops, while not changing the number of “iterations”. Both recursive … Read more
I think people are confusing Quicksort the partition-based sorting algorithm, and “qsort” the various library implementations. I prefer to see Quicksort the algorithm as having a pluggable pivot selection algorithm, which is quite essential in analyzing its behavior. If the first element is always chosen as the pivot, then an already sorted list is the … Read more
The most likely reason: quicksort is not stable, i.e. equal entries can change their relative position during the sort; among other things, this means that if you sort an already sorted array, it may not stay unchanged. Since primitive types have no identity (there is no way to distinguish two ints with the same value), … Read more
Quicksort’s performance is dependent on your pivot selection algorithm. The most naive pivot selection algorithm is to just choose the first element as your pivot. It’s easy to see that this results in worst case behavior if your data is already sorted (the first element will always be the min). There are two common algorithms … Read more
The true quicksort has two beautiful aspects: Divide and conquer: break the problem into two smaller problems. Partition the elements in-place. The short Haskell example demonstrates (1), but not (2). How (2) is done may not be obvious if you don’t already know the technique!
give a try to fork/join framework by Doug Lea: public class MergeSort extends RecursiveAction { final int[] numbers; final int startPos, endPos; final int[] result; private void merge(MergeSort left, MergeSort right) { int i=0, leftPos=0, rightPos=0, leftSize = left.size(), rightSize = right.size(); while (leftPos < leftSize && rightPos < rightSize) result[i++] = (left.result[leftPos] <= right.result[rightPos]) … Read more
Choosing a random pivot minimizes the chance that you will encounter worst-case O(n2) performance (always choosing first or last would cause worst-case performance for nearly-sorted or nearly-reverse-sorted data). Choosing the middle element would also be acceptable in the majority of cases. Also, if you are implementing this yourself, there are versions of the algorithm that … Read more
def sort(array=[12,4,5,6,7,3,1,15]): “””Sort the array by using quicksort.””” less = [] equal = [] greater = [] if len(array) > 1: pivot = array[0] for x in array: if x < pivot: less.append(x) elif x == pivot: equal.append(x) elif x > pivot: greater.append(x) # Don’t forget to return something! return sort(less)+equal+sort(greater) # Just use the … Read more