利用归并排序求一序列中inversion的个数

inversion指的是在一个序列a[]中若有i<j, 且a[i]>a[j]，则称(i,j)为一个inversion，现要求在o(nlogn)时间内求解任一序列的inversion的总数。

如此将meger_sort改造一下，注意在meger排序中每一次数据的位置调换都对应了一个inversion，如此，o(nlogn) 时间内求任意序列的inversion的代码如下

#include <iostream>
#include <cstdlib>
#include <limits>
using namespace std;
/*
 * name : merge inversion
 * author : sangoly
 * O(nlgn)
 * S(n)
 * date : 2014/4/15
 */
int merge_inversion(int a[], int start, int middle, int end)
{     
	int inversions = 0;
	bool counted = false;

    int max = numeric_limits<int>::max();
    int tempArray1[middle-start+2];//include the start to middle
    int tempArray2[end-middle+1];//include the middle+1 to end
     
    for (int i=0; i<middle-start+1; i++)
        tempArray1[i] = a[start+i];
    tempArray1[middle-start+1] = max;//As the soldier
     
    for (int i=0; i<end-middle; i++)
        tempArray2[i] = a[middle+1+i];
    tempArray2[end-middle] = max;//As the soldier
     
    int index1 = 0;
    int index2 = 0;
     
	for (int k = start; k <= end; k++) {
		if (counted == false && tempArray2[index2] < tempArray1[index1]) {
			inversions += (middle - start - index1 + 1);//if the first element in tempArray1 is bigger than the j value in tempArray2
			                                                //then all the value follow index i will bigger than the j in tempArray2 too
			counted = true;
		}
		if (tempArray1[index1] <= tempArray2[index2]) {
			a[k] = tempArray1[index1++];
		}
		else {
			a[k] = tempArray2[index2++];
			counted = false;
		}
	}
	return inversions;
}

int count_inversion(int a[], int start, int end)
{
	int inversions = 0;
	if (start < end) {
		int middle = (start + end) / 2;
		inversions += count_inversion(a, start, middle);
		inversions += count_inversion(a, middle + 1, end);
		inversions += merge_inversion(a, start, middle, end);
	}
	return inversions;
}

int main(int argc, const char **argv) {
    int a[] = {2, 3, 8, 6, 1};
    int total_inversions = count_inversion(a, 0, 4);
    for (int i=0; i<5; i++)
        cout<<a[i]<<" ";
    cout<<endl;
	cout << "total_inversions : " << total_inversions << endl;
    system("pause");
    return 0;
}