如何使用合并排序来计算倒置？

Question

我正在尝试使用def count_inversion计算反转的总次数

def count_inversion(alist):
count = 0
if len(alist)>1:
    mid = len(alist)//2
    lefthalf = alist[:mid]
    righthalf = alist[mid:]

    a=count_inversion(lefthalf)
    b=count_inversion(righthalf)

    i=0
    j=0
    k=0
    track = 0
    while i < len(lefthalf) and j < len(righthalf):
        if lefthalf[i] < righthalf[j]:
            alist[k]=lefthalf[i]
            i=i+1

        else:
            alist[k]=righthalf[j]
            j=j+1
            count+=len(righthalf[i:])
        k=k+1

    while i < len(lefthalf):
        alist[k]=lefthalf[i]
        i=i+1
        k=k+1


    while j < len(righthalf):
        alist[k]=righthalf[j]
        j=j+1
        k=k+1

return count

def main():
    alist = [10,9,8,7,6,5,4,3,2,1]
    inversion = count_inversion(alist)
    print(alist)
    print(inversion)    

main()

我确实得到了一个排序列表1，2，3，4，5，6，7，8，9，10，但是对于倒置计数，它显示它是25而不是45。我想我可能在我的代码中犯了一些错误，但我不知道如何修复it...It，如果有人能帮助我的话……

Answer 1

# store inversion count 
count+=count_inversion(lefthalf)
count+=count_inversion(righthalf)

# updated line > Instead of righthalf count length of lefthalf
count+=len(lefthalf[i:])

在你的代码中更新这两行

Answer 2

下面是我在一个类函数中的实现。这也计入了倒置。您可以取消对语句的注释，以查看跟踪反转的另一种方法。

class MergeSort(object):
'''
Instantiates an array into the object from which the method 'merge_sort' can be called.
Returns number of inversions and sorted array.
>>>x = MergeSort([1,6,5,2,10,8,7,4,3,9])
>>>x.merge_sort()
(20, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
'''
def __init__(self, array):
    self.array = array

def merge_sort(self):
    count = 0
    #count_ = []
    if len(self.array) > 1:
        m = len(self.array)//2
        left = self.array[:m]
        right = self.array[m:]

        leftsorter = MergeSort(left)
        leftsorter = leftsorter.merge_sort()
        rightsorter = MergeSort(right)
        rightsorter = rightsorter.merge_sort()

        # Two different ways to track inversions

        # numeric counter, better way
        count += leftsorter[0]
        count += rightsorter[0]

        # list counter
        #count_.append(leftsorter[0])
        #count_.append(rightsorter[0])

        i = 0
        j = 0
        k = 0

        while i < len(left) and j < len(right):
            if left[i] < right[j]:
                self.array[k] = left[i]
                i += 1
            else:
                self.array[k] = right[j]
                j += 1
                count += len(left[i:])
                #count_.append(len(left[i:]))
            k += 1

        while i < len(left):
            self.array[k] = left[i]
            i += 1
            k += 1

        while j < len(right):
            self.array[k] = right[j]
            j += 1
            k += 1
    return count, self.array, #sum(count_)

它是这样运行的。

array = [1,6,5,2,10,8,7,4,3,9]
x = MergeSort(array)
x.merge_sort()

Out[ ]:
(20, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10])

Answer 3

你能不能不使用这个示例代码？https://www.geeksforgeeks.org/counting-inversions/