设A[1...n]是一个包含n个不同数的数组。如果在i<j的情况下，有A[i]>A[j]，则(i,j)就成为A中的一个逆序对(inversion)。

要确定一个数组中的逆序对的个数，可以采取分治法。将A分为两部分A1和A2，则A中逆序对的数目等于A1中逆序对的数目、A2中逆序对的数目和A1，A2合并时A1中比A2中元素大的数目。

参考代码：

#include <iostream> using namespace std; #define MAX_VALUE 99999 //合并时计算逆序对数目 int MergeCountReverse(int *a,int low,int mid,int high) { int num1 = mid - low + 1; int num2 = high - mid; int count = 0; int* a1 = (int*)malloc((num1 + 1) * sizeof(int)); int* a2 = (int*)malloc((num2 + 1) * sizeof(int)); for(int i = 0;i < num1;++i) *(a1 + i) = *(a + low + i); *(a1 + num1) = MAX_VALUE; for(int i = 0;i < num2;++i) *(a2 + i) = *(a + mid + 1 + i); *(a2 + num2) = MAX_VALUE; int index1 = 0; int index2 = 0; for(int k = low;k <= high;++k) { if(*(a1 + index1) > *(a2 + index2)) { *(a + k) = *(a2 + index2); ++index2; count += num1 - index1;//逆序对数 } else { *(a + k) = *(a1 + index1); ++index1; } } free(a1); free(a2); return count; } //递归计算逆序对数目：采用分治法 int CountReverse(int *a,int low,int high) { int mid,count = 0; if(high > low) { mid = (low + high) / 2; count += CountReverse(a,low,mid); count += CountReverse(a,mid + 1,high); count += MergeCountReverse(a,low,mid,high); } return count; } int main() { int n,temp; cout<<"please input the number of the values:"<<endl; cin>>n; int *a = (int*)malloc(n * sizeof(int)); cout<<"please input each value:"<<endl; for(int i = 0;i < n;++i) { cin>>temp; *(a + i) = temp; } int count = CountReverse(a,0,n - 1); cout<<"the number of reverse pairs :"<<count<<endl; } 数据测试：