Question

以下代码是Python中的快速排序。如何计算算法中比较运行的次数？

虽然我首先指定count = 0，但由于递归，它会重置为0。

def QuickSort(lst):
    if len(lst) > 1:
        pivot_idx = len(lst) // 2
        smaller_nums, larger_nums = [], []

        for idx, num in enumerate(lst):
            if idx != pivot_idx:
                if num < lst[pivot_idx]:
                    smaller_nums.append(num)

                else:
                    larger_nums.append(num)

        QuickSort(smaller_nums)
        QuickSort(larger_nums)
        lst[:] = smaller_nums + [lst[pivot_idx]] + larger_nums

    return lst

Answer 1

编辑，我删除了这个答案，因为我认为这是不正确的，但我认为这毕竟是正确的

如建议的那样，如果函数是无状态的，我认为也会更好：您可以返回lst和呼叫数量：

def QuickSort(lst, ncalls=0):
  ncalls += 1

  if len(lst) > 1:
    pivot_idx = len(lst) // 2
    smaller_nums, larger_nums = [], []

    for idx, num in enumerate(lst):
        if idx != pivot_idx:
            if num < lst[pivot_idx]:
                smaller_nums.append(num)

            else:
                larger_nums.append(num)
    lst1, ncalls = QuickSort(smaller_nums, ncalls)
    lst1, ncalls = QuickSort(larger_nums, ncalls)
    lst[:] = smaller_nums + [lst[pivot_idx]] + larger_nums



  return lst,ncalls


QuickSort([1,3,52,4,6,5])
=> [1, 3, 4, 5, 6, 52],7

Answer 2

将其递归传递为参数：

def QuickSort(lst, count=0):
    if len(lst) > 1:
        pivot_idx = len(lst) // 2
        smaller_nums, larger_nums = [], []

        for idx, num in enumerate(lst):
            if idx != pivot_idx:
                if num < lst[pivot_idx]:
                    smaller_nums.append(num)

                else:
                    larger_nums.append(num)

        count = QuickSort(smaller_nums, count+1)[1]
        count = QuickSort(larger_nums, count+1)[1]
        lst[:] = smaller_nums + [lst[pivot_idx]] + larger_nums


    return (lst,count)

Answer 3

将count指定为全局变量并将其设置为零。也就是说，将count=0设置在函数定义之外，并在每次比较时递增它。

如何计算递归函数？

3 个答案: