是否有 Python 不使用递归的快速排序实现?

Is there a Python implementation of quicksort without recursion?

我试图在不使用递归的情况下在 Python 中实现快速排序,但到目前为止我发现的所有参考实现或伪代码都使用递归。

这样做的原因是我将在使用 Numba 的 GPU 上调整此非递归实现以适应 运行,并且我不能在那里进行递归调用。

是否有不使用递归的一维数组(例如,Numpy 数组或 Python 列表)的快速排序实现?

谢谢,

爱德华多

标准 unix/linux qsort 的实现没有递归以提高效率。你可以查一下,或者把这个答案中的代码翻译成 python:

FWIW,这是一个 Python 版本:

# This function is same in both iterative and recursive
def partition(arr,l,h):
    i = ( l - 1 )
    x = arr[h]
  
    for j in range(l , h):
        if   arr[j] <= x:
  
            # increment index of smaller element
            i = i+1
            arr[i],arr[j] = arr[j],arr[i]
  
    arr[i+1],arr[h] = arr[h],arr[i+1]
    return (i+1)
  
# Function to do Quick sort
# arr[] --> Array to be sorted,
# l  --> Starting index,
# h  --> Ending index
def quickSortIterative(arr,l,h):
  
    # Create an auxiliary stack
    size = h - l + 1
    stack = [0] * (size)
  
    # initialize top of stack
    top = -1
  
    # push initial values of l and h to stack
    top = top + 1
    stack[top] = l
    top = top + 1
    stack[top] = h
  
    # Keep popping from stack while is not empty
    while top >= 0:
  
        # Pop h and l
        h = stack[top]
        top = top - 1
        l = stack[top]
        top = top - 1
  
        # Set pivot element at its correct position in
        # sorted array
        p = partition( arr, l, h )
  
        # If there are elements on left side of pivot,
        # then push left side to stack
        if p-1 > l:
            top = top + 1
            stack[top] = l
            top = top + 1
            stack[top] = p - 1
  
        # If there are elements on right side of pivot,
        # then push right side to stack
        if p+1 < h:
            top = top + 1
            stack[top] = p + 1
            top = top + 1
            stack[top] = h
  
# Driver code to test above
arr = [4, 3, 5, 2, 1, 3, 2, 3]
n = len(arr)
quickSortIterative(arr, 0, n-1)
print ("Sorted array is:")
for i in range(n):
    print ("%d" %arr[i])