面试题36—数组中的逆序对

**题目：在数组中的两个数如果前面一个数字大于后面的数字，则这两个数字组成一个逆序对。求一个数组的逆序对的总数。
代码示例：**

#include<iostream>using namespace std;int Merge2(int a[], int low, int mid, int high)//二路归并算法{    int InverseNum = 0;    int i = low;    int j = mid + 1;    int k = 0;    int *temp = new int[high - low + 1];    while (i <= mid&&j <= high)    {        if (a[i] <= a[j])        {            temp[k] = a[i];            i++;            k++;        }        else        {            for (int h = i; h <= mid;h++)                cout << "逆序对：" << a[h] << " " << a[j] << endl;            InverseNum += mid - i + 1;            temp[k] = a[j];            k++;            j++;        }    }    while (i <= mid)    {        temp[k] = a[i];        i++;        k++;    }    while (j <= high)    {        temp[k] = a[j];        k++;        j++;    }    //    for (int k = 0; k < high-low + 1; k++)    {        a[low + k] = temp[k];    }    delete temp;    return InverseNum;}int MergePass2(int a[], int n, int len){    int InverseNum = 0;    int i;    for (i = 0; i + 2*len - 1 < n; i = i + 2 * len)    {        InverseNum += Merge2(a, i, i + len - 1, i + 2 * len - 1);    }    if (i+len-1<n)        InverseNum += Merge2(a, i, i + len - 1, n - 1);    return InverseNum;}int CountInverseNum(int a[], int n){    if (n <= 1 || a == NULL)        return -1;    int InverseNum = 0;    int len = 1;    for (; len < n; len = len * 2)    {        InverseNum += MergePass2(a, n, len);    }    return InverseNum;}int main(){    const int n = 8;    int a[n] = { 5,4,3,3,3,3,2,1};    cout << "原数组：";    for (int i = 0; i < n; i++)        cout << a[i] << " ";    cout << endl << endl;    int InverseNum = 0;    InverseNum = CountInverseNum(a, n);    cout << endl<<"逆序对数：" << InverseNum << endl;}