调用的函数：（无论类如何）

def partition( pivot, lst ):
    less, same, more = list(), list(), list()
    for val in lst:
        if val < pivot:
            less.append(val)
        elif val > pivot:
            more.append(val)
        else:
            same.append(val)
    return less, same, more


def medianOf3(lst):
"""
From a lst of unordered data, find and return the the median value from
the first, middle and last values.
"""
    finder=[]
    start=lst[0]
    mid=lst[len(lst)//2]
    end=lst[len(lst)-1]
    finder.append(start)
    finder.append(mid)
    finder.append(end)
    finder.sort()
    pivot_val=finder[1]
    return pivot_val

主代码

def quicheSortRec(lst,limit):
"""
A non in-place, depth limited quickSort, using median-of-3 pivot.
Once the limit drops to 0, it uses heapSort instead.
"""
    if limit==0:
        return heapSort.heapSort(lst)
    else:
        pivot=qsPivotMedian3.medianOf3(lst)        # here we select the first element as the pivot
        less, same, more = qsPivotMedian3.partition(pivot, lst)
        return quicheSortRec(less,limit-1) + same + quicheSortRec(more,limit-1)


def quicheSort(lst):
"""
The main routine called to do the sort.  It should call the
recursive routine with the correct values in order to perform
the sort
"""
    N=len(lst)
    end= int(math.log(N,2))
    if len(lst)==0:
        return list()
    else:
        return quicheSortRec(lst,end)

因此，这段代码应该采用一个列表，并使用中位数为3的快速排序实现对其进行排序，直到达到深度递归限制，此时函数将改为使用堆排序。

但是，当我运行代码时，告诉我长度1和10工作正常，但在那之后，它会给出一个列表索引越界错误

start=lst[0]在3个函数的中位数

我不明白为什么。