使用合并排序的反转次数

诺林格

我编写了这段代码,以获取使用的反转次数merge_sort,但未获得正确的输出。有人可以告诉我为什么输出错误吗?

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

当我跑步时:

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

输出为:

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

但是输出应该是

[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)

输出为:

[1,2,3,4 ] 3 

哪个是对的。哪里出问题了?

chqrlie

您的代码中存在一些问题:

  • 要删除列表的第一个元素,应使用pop(),而不是remove()
  • 逆转的次数,当你从要素C就是len(B),不是1
  • 您应该将每个半数的反转数量和合并阶段的数量相加
  • intial测试中sort_and_num还应该测试一个空列表,并返回一个计数为0的列表。

这是修改后的版本:

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)

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