所以我试图在Python 2.7中实现QuickSort，并且我已经成功地使用了递归函数，但是当我调用quicksort（range（1000）[:: - 1]）时，可以理解这会让我感到沮丧。也就是说它最大限度地提高了Python中允许的递归深度。
所以现在我想＆＃34;压扁＆＃34;这个函数，也就是说，使用尽可能少的（或没有）递归调用。我一直试图通过一个没有返回列表的分区函数来实现这个功能，但只是在将所有内容排序到左右之后才返回数据库的位置。我已经写过并测试过它，它似乎完美无缺。但是，如果有人对此感兴趣，则代码如下。
现在的诀窍似乎是让这个东西反复分区，直到整个列表被排序，并且我很难看到如何让这个部分发生。我有一个想法是创建一个有序的索引列表，这些索引是用于数据透视的所有索引。然后我可以运行一个while循环，当索引列表与range（n）相同时会中断。我不确定这个想法有多聪明或优雅，因为我正在制作与输入列表长度相同的第二个列表 - 看起来有点像一个可以避免的记忆力。另一方面，你似乎需要一种方法来跟踪你已经排序的列表中的哪些段，为此我只能想到递归或这个列表的想法。
所以问题是，列表的想法是好还是坏，如果不好，什么是更好的主意？
def partition(x, start, end, pivotIndex):
'''
x is a list, start to end is the portion of the list to sort, and
pivotIndex is the index at which pull a value and sort everything above
and below this.
'''
low = start # Low will track the lower bound of the position in the list
# where "small" elements get switched into. It starts at
# start.
high = end-1
p = x[pivotIndex] # Pivot value
x[pivotIndex] = x[end] # We basically move the pivot entry out of the way
# and we'll put it back in at the end.
while True:
while low <= high and x[low] < p: # We cycle up the bottom of the list
# until we've looked at everything or until we hit
# a pice that's out of place.
low += 1
while low <= high and x[high] >= p: # And do likewise from the top of
# the list.
high -= 1
if low > high: # If we've covered the whole list, the partition is
# done. Otherwise, there must be pieces out of place
# so swap them.
break
x[low], x[high] = x[high], x[low]
x[end] = x[low] # The value at end stores a duplicate of some value,
# since we copied it in, so now we want to over-ride that.
# We choose to put the element now at low there, and put
# since the pivot value was taken out of the list, we now
# put it in the low position.
x[low] = p
return low