python - 使用归并排序的反转次数

Question

我编写了这段代码来使用 merge_sort 来获取反转次数。 ，但我没有得到正确的输出。有人能告诉我为什么它给出错误的输出吗？

def merge(B,C):  # B,C are sorted
  D = list()
  count = 0
  # empty list to get the merged list
  while (B != []) and (C != []):  # breaks if any one of B or C becomes empty

    b = B[0]  # to get first element of B
    print("b",b)
    c = C[0]  # to get first element of C
    print("c",c)
    
    if b<=c:
      B.remove(b)  # if b<c remove b from B and add to D
      D.append(b)
    else:
      C.remove(c)  # if b>c remove c from C and add to D
      D.append(c)
      count += 1
      print("count_in_merge",count)

  # now if either one of B or C has become empty so while loop is exited
  # so we can add remaining elements of B and C as it is to D as they are
  # already sorted
  if B != []:                                       
    for i in B:
      D.append(i)
  
  if C != []:
     for i in C:
      D.append(i)

  return D, count 

def sort_and_num(A):
  if len(A) == 1:
    return A,0

  m = len(A) // 2
    
  B,count_b = sort_and_num(A[0:m])
  C,count_c = sort_and_num(A[m:])
    
  A_,count = merge(B,C)
  count += (count_b + count_c)
  
  return A_, count

当我运行时:

A = [ 9, 8 ,7, 3, 2, 1] 
A_,c = sort_and_num(A)
print(A_,c) 

输出是:

[1, 2, 3, 7, 8, 9] 9

但输出应该是

[1, 2, 3, 7, 8, 9] 15

另一方面，如果我输入:

A = [3,1,2,4] 
A_, count = sort_and_num(A)
print(A_, count)

输出是:

[1,2,3,4 ] 3 

哪个是正确的。哪里出错了？

Answer 1

您的代码中存在一些问题:

要删除列表的第一个元素，您应该使用 pop() ，不是 remove()

从 C 取元素时的反转次数是 len(B) ，不是 1 .

您应该将每一半的反转次数和合并阶段的反转次数相加

sort_and_num 中的初始测试还应该测试一个空列表并返回它的计数为 0。

这是一个修改后的版本:

def merge(B, count_b, C, count_c):  # B,C are sorted
   D = []
   count = count_b + count_c
   # empty list to get the merged list
   while (B != []) and (C != []):  # breaks if any one of B or C becomes empty
      if B[0] <= C[0]:
         D.append(B.pop())  # if b<=c remove b from B and add it to D
      else:
         D.append(C.pop())  # if b>c remove c from C and add it to D
         count += len(B)    # moving c before all remaining elements of B

  # now if either one of B or C has become empty so while loop is exited
  # so we can add remaining elements of B and C as it is to D as they are
  # already sorted
  for i in B:
     D.append(i)
  
  for i in C:
     D.append(i)

  return D, count 

def sort_and_num(A):
   if len(A) <= 1:
      return A,0

   m = len(A) // 2
    
   B,count_b = sort_and_num(A[:m])
   C,count_c = sort_and_num(A[m:])
    
   return merge(B, count_b, C, count_c)