找出反序的个数

给定一整型数组，若数组中某个下标值大的元素值小于某个下标值比它小的元素值，称这是一个反序。
即：数组a[]; 对于i < j 且 a[i] > a[j],则称这是一个反序。

给定一个数组，要求写一个函数，计算出这个数组里所有反序的个数。

#include <iostream>
using namespace std;

#define N 5

int merge(int* arr, int low, int mid, int high) {
	int count = 0;
	int firstLen = mid - low + 1;
	int secondLen = high - mid;

	int* tmp1 = new int[firstLen];
	int* tmp2 = new int[secondLen];

	for (int i = 0; i < firstLen; ++i) {
		tmp1[i] = arr[low + i];
	}

	for (int i = 0; i < secondLen; ++i) {
		tmp2[i] = arr[mid + 1 + i];
	}

	int k = low;
	int i = 0, j = 0;
	for (; i < firstLen & j < secondLen;) {
		if (tmp1[i] <= tmp2[j]) {
			arr[k++] = tmp1[i++];
		} else { //逆序对情况
			/*由于前半部分升序,若tmp1[i]与tmp2[j]为逆序对,
			 * 则前半部分从i开始(包含tmp1[i])的后续元素
			 * 与tmp2[j]也为逆序对: firstLen - i*/
			arr[k++] = tmp2[j++];
			count += firstLen - i;
		}
	}

	//将剩余元素放入
	while (i < firstLen) {
		arr[k++] = tmp1[i++];
	}

	while (j < secondLen) {
		arr[k++] = tmp2[j++];
	}

	return count;
}

int mergeSort(int* arr, int low, int high) {
	int count = 0;
	if (low < high) {
		int mid = (low + high) / 2;
		count += mergeSort(arr, low, mid);
		count += mergeSort(arr, mid + 1, high);
		count += merge(arr, low, mid, high);
	}
	return count;
}