python 数组中的逆序对

# -*- coding:utf-8 -*-
count = 0 
class Solution:  
    def InversePairs(self, data):  
        global count  
        def MergeSort(lists):  
            global count  
            if len(lists) <= 1:  
                return lists  
            num = int( len(lists)/2 )  
            left = MergeSort(lists[:num])  
            right = MergeSort(lists[num:])  
            r, l=0, 0 
            result=[]  
            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 
                    count += len(left)-l  
            result += right[r:]  
            result += left[l:]  
            return result  
        MergeSort(data)  
        return count%1000000007

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转载自blog.csdn.net/qq_32799915/article/details/83064889