算法导论2-4 O(nlgn)时间复杂度求逆序对

def mergesort(nums,le,ri):
    if le>ri-2:
        return 0
    mi=le+(ri-le)//2
    a=mergesort(nums,le,mi)
    b=mergesort(nums,mi,ri)
    c=merge(nums,le,mi,ri)
    return a+b+c
def merge(nums,le,mi,ri):
    i,j=le,mi
    data=[]
    count=0
    while i<mi and j<ri:
        if nums[i]<nums[j]:
            data.append(nums[i])
            i+=1
        else:
            print(nums[i],nums[j])
            data.append(nums[j])
            j+=1
            count+=mi-i
    while i<mi:
        data.append(nums[i])
        i+=1
    while j<ri:
        data.append(nums[j])
        j+=1
    nums[le:ri]=data
    return count
x=mergesort(a,0,len(a))
print(a)
print(x)

解释：就是在merge里加一个计数器，若A[I]>A[J]则A[J]和A[I]到A[MID-1]的所有元素都构成逆序对，即count+=（mid-1）-i+1=mid-i