Python 如何修改合并排序代码以从最大到最小排序

我知道这很琐碎，但我找不到解决方法。我有一个从最小到最大排序的合并排序代码，现在我被要求修改代码，从最大到最小排序，但似乎找不到我的方法。请帮忙，谢谢

以下是注释代码：

def mergesort(lst):

    #Then, what does it do? mergesort should recursively
    #run mergesort on the left and right sides of lst until
    #it's given a list only one item. So, if lst has only
    #one item, we should just return that one-item list.

    if len(lst) <= 1:
        return lst

    #Otherwise, we should call mergesort separately on the
    #left and right sides. Since each successive call to
    #mergesort sends half as many items, we're guaranteed
    #to eventually send it a list with only one item, at
    #which point we'll stop calling mergesort again.
    else:

        #Floor division on the length of the list will
        #find us the index of the middle value.
        midpoint = len(lst) // 2

        #lst[:midpoint] will get the left side of the
        #list based on list slicing syntax. So, we want
        #to sort the left side of the list alone and
        #assign the result to the new smaller list left.
        left = mergesort(lst[:midpoint])

        #And same for the right side.
        right = mergesort(lst[midpoint:])


        #So, left and right now hold sorted lists of
        #each half of the original list. They might
        #each have only one item, or they could each
        #have several items.

        #Now we want to compare the first items in each
        #list one-by-one, adding the smaller to our new
        #result list until one list is completely empty.

        newlist = []
        while len(left) and len(right) > 0:

            #If the first number in left is lower, add
            #it to the new list and remove it from left
            if left[0] < right[0]:
                newlist.append(left[0])
                del left[0]

            #Otherwise, add the first number from right
            #to the new list and remove it from right
            else:
                newlist.append(right[0])
                del right[0]

        #When the while loop above is done, it means
        #one of the two lists is empty. Because both
        #lists were sorted, we can now add the remainder
        #of each list to the new list. The empty list
        #will have no items to add, and the non-empty
        #list will add its items in order.

        newlist.extend(left)
        newlist.extend(right)

        #newlist is now the sorted version of lst! So,
        #we can return it. If this was a recursive call
        #to mergesort, then this sends a sorted half-
        #list up the ladder. If this was the original
        #call, then this is the final sorted list.

        return newlist

#Let's try it out!
print(mergesort([2, 5, 3, 8, 6, 9, 1, 4, 7]))

但我预计会出版：

[9, 8, 7, 6, 5, 4, 3, 2, 1]

---

提前感谢

您只在代码中的一个位置进行比较：在合并期间，当您比较

左[0]

和

右[0]

时。颠倒比较的意义将颠倒整体排序。因此，为了按降序排序，将

left[0]

和

right[0]

中的较大者添加到输出中，即如果

left[0]>right[0]：最新的.append（left[0]）

只需在

left[0]

和

right[0]之间交换
如果这不明显，我应该这么做……我是否应该相信你编写了代码和解释，但却无法将代码更改为代码？@StefanPochmann他们没有声称自己编写了代码。这显然是个家庭作业。@user8964866-为了保持稳定性，比较应该是
left[0]>=right[0]
。如果元素相等，则要移动左侧元素以保持相等元素的原始顺序。
[9, 8, 7, 6, 5, 4, 3, 2, 1]