Quick sort python

Python with quick sort

array = [29,99,27,41,66,28,44,78,87,19,31,76,58,88,83,97,12,21,44]

quick_sort(array, 0, len(array) - 1)
print(array)

Source: stackabuse.com

Quick sort python

def quick_sort(array):
    sort(array, 0, len(array) - 1)


def sort(array, left, right):
    if left >= right:
        return array
    pivot = array[(left + right) // 2]
    index = partition(array, left, right, pivot)

    sort(array, left, index - 1)
    sort(array, index, right)


def partition(array, left, right, pivot):
    while left <= right:
        while array[left] < pivot:
            left += 1
        while array[right] > pivot:
            right -= 1

        if left <= right:
            array[left], array[right] = array[right], array[left]
            left += 1
            right -= 1

    return left


arr = [7, 2, 4, 0, 1, 5, 8, 3, 2, 0]
quick_sort(arr)
print(arr) # [0, 0, 1, 2, 2, 3, 4, 5, 7, 8]

Quick sort python

def quick_sort(array):
    if len(array) < 2:
        return array
    else:
        pivot = array[0]
        less = [i for i in array[1:] if i <= pivot]
        greater = [i for i in array[1:] if i > pivot]
        return quick_sort(less) + [pivot] + quick_sort(greater)


arr = [7, 2, 4, 0, 1, 5, 8, 3, 2, 0]
print(quick_sort(arr)) # [0, 0, 1, 2, 2, 3, 4, 5, 7, 8]

Python with quick sort

[12, 19, 21, 27, 28, 29, 31, 41, 44, 44, 58, 66, 76, 78, 83, 87, 88, 97, 99]

Source: stackabuse.com

Python quick sort

/* low  --> Starting index,  high  --> Ending index */
quickSort(arr[], low, high)
{
    if (low < high)
    {
        /* pi is partitioning index, arr[p] is now
           at right place */
        pi = partition(arr, low, high);

        quickSort(arr, low, pi - 1);  // Before pi
        quickSort(arr, pi + 1, high); // After pi
    }
}
Python3

Source: geeksforgeeks.org

Python with quick sort

def quick_sort(array, start, end):
    if start >= end:
        return

    p = partition(array, start, end)
    quick_sort(array, start, p-1)
    quick_sort(array, p+1, end)

-2

Source: stackabuse.com

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