使用quicksort（Python）查找第k个最小项目

Question

I was trying to implement the algorithm discussed in this video and also in this document .

My quicksort code, which depends on picking the middle element in the array as the pivot (see below), as opposed to the approach used by the author of the document above, which uses the first element in the array as the pivot as shown here in the video .

Obviously, my code doesn't work (runs out of the recursion limit eventually). I wonder if it's because of some silly error in my code or that it simply would NOT work as long as I pick the pivot from the middle.

def partition(a, start, end):
    # I pick the pivot as the middle item in the array
    # but the algorithm shown in the video seems to
    # pick the first element in the array as pivot
    piv = (start + end) // 2
    pivotVal = a[piv]

    left = start
    right = end

    while left <= right:

        while a[left] < pivotVal:
            left += 1

        while a[right] > pivotVal:
            right -= 1

        if left <= right:
            temp = a[left]
            a[left] = a[right]
            a[right] = temp
            left += 1
            right -= 1

    return left


def quicksort(a, start, end, k):

    if start < end:
        piv = partition(a, start, end)

        if piv == (k - 1):
            print("Found kth smallest: " + piv)
            return a[piv]
        elif piv > (k - 1):
            return quicksort(a, start, piv, k)
        else:
            return quicksort(a, piv + 1, end, k)

myList = [54, 26, 93, 17, 77, 31, 44, 55, 20]
quicksort(myList, 0, len(myList) - 1, 3)
print(myList)

Answer 1

If you are using inclusive array bounds, which isn't the most convenient trick, you'll have to change the range in the recursive calls to [start, piv - 1] and [piv + 1, end]

Reason being, you are reconsidering the piv element in the each recursion that goes to the left of array.

The code with said changes ran without any stack overflow error.

EDIT The output isn't right for few values of k. You may need to check your partition logic again.