剑指offer-数组中的逆序对-数组-python

class Solution:
    def InversePairs(self, data):
        self.count =0
        self.MergeSort(data)
        # print(self.MergeSort(data))#归并排序
        return self.count % 1000000007
    def MergeSort(self,lists):
        # 递归结束条件
        if len(lists) <= 1:
            return lists
        # 分治进行递归
        num = int(len(lists)/2)
        left = self.MergeSort(lists[:num])
        right = self.MergeSort(lists[num:])
        # 将两个有序数组进行合并
        r, l=0, 0
        result=[]
        # 将较小值放入到result中
        while l<len(left) and r<len(right):
            if left[l] < right[r]:
                result.append(left[l])
                l += 1
            else:
                result.append(right[r])
                r += 1
                #剩余的左边的数都大于右边的那个数
                self.count += len(left)-l
        # 将未被扫描到的直接追加到result后面
        result += right[r:]
        result += left[l:]
        return result

print(Solution().InversePairs([1,2,3,4,5,6,7,0]))