Python 选择中间元素作为轴心的快速排序 我有这个代码，但是雇主让我在中间旋转。 如果有人能提供帮助，请编辑此代码 def quicksort(sequence, low, high): if low < high: pivot = partition(sequence, low, high) quicksort(sequence, low, pivot - 1) quicksort(sequence, pivot + 1, high) def partition(sequence, low, high): pivot = sequence[low] i = low + 1 for j in range(low + 1, high + 1): if sequence[j] < pivot: sequence[j], sequence[i] = sequence[i], sequence[j] i += 1 sequence[i-1], sequence[low] = sequence[low], sequence[i-1] return i - 1 def快速排序（顺序、低、高）： 如果低

帕特尔的答案从问题的罗密托方案切换到霍尔方案。对问题代码最简单的修复方法是将中间元素与低元素交换为分区中的第一行

def partition(sequence, low, high):
    sequence[low],sequence[(low+high)//2] = sequence[(low+high)//2],sequence[low] #change
    pivot = sequence[low]
      ...

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帕特尔的回答从问题的罗密托方案转向霍尔方案。对问题代码最简单的修复方法是将中间元素与低元素交换为分区中的第一行

def partition(sequence, low, high):
    sequence[low],sequence[(low+high)//2] = sequence[(low+high)//2],sequence[low] #change
    pivot = sequence[low]
      ...

---

按以下方式更改分区方法

def partition(sequence, low, high):
    sequence[low],sequence[(low+high)//2] = sequence[(low+high)//2],sequence[low]
    pivot = sequence[low]

---

D:

按以下方式更改分区方法

def partition(sequence, low, high):
    sequence[low],sequence[(low+high)//2] = sequence[(low+high)//2],sequence[low]
    pivot = sequence[low]

D: