題目：在陣列中的兩個數字如果前面一個數字大於後面一個數字 則這兩個數字組成一個逆序對 輸入一個數組 求這個陣列中逆序對的個數

先把陣列分割成子陣列 先統計出子陣列內部的逆序對的數目 然後再統計出兩個相鄰子陣列之間的逆序對的數目 統計逆序對的過程中 還需要對陣列進行排序 這類似與歸併排序 演算法的時間複雜度為O（nlogn) 空間複雜度為O（n)

歸併排序

#include<iostream>
using namespace std;
void merge(int *A, int *temp, int start, int mid, int end)
{
	int a = start;
	int b = mid+1;
	int c = end;
	int d = start;

	while (a <= mid && b <=end )
	{
		if (A[a] < A[b])
			temp[d++] = A[a++];
		else
			temp[d++] = A[b++];
	}
	while (a != mid + 1)
		temp[d++] = A[a++];
	while (b != end + 1)
		temp[d++] = A[b++];
	for (int i = start;i <= end;i++)
		A[i] = temp[i];
}

void sortg(int *A,int *temp ,int a,int b)
{
	if (a < b)
	{
		int mid = a + (b - a) / 2;
		sortg(A, temp, a, mid);
		sortg(A, temp, mid + 1, b);
		merge(A, temp, a, mid, b);
	}
}
int main()
{
	int A[] = { 7,4,10,6,9,9,2,3 };
	int n = 8;
	int *t = new int[n];
	memset(t, 0, sizeof(int)*n);
	sortg(A, t, 0, n - 1);
	for (int i = 0;i < n;i++)
		cout << A[i] << ' ';
	delete[]t;
		return 0;
}
 

#include<iostream>
using namespace std;

int merge(int *A, int *temp, int start, int mid, int end)
{
	int a = end;
	int b = mid;
	int c = end;
	int t = 0;
	while (b >= start && a >= mid + 1)
	{
		if (A[b] > A[a])
		{
			temp[c--] = A[b--];
			t += a - mid;
		}
		else
			temp[c--] = A[a--];
	}
	while (a >= mid + 1)
		temp[c--] = A[a--];
	while (b >= start)
		temp[c--] = A[b--];
	for (int i = start;i <= end;i++)
		A[i] = temp[i];
	return t;
}
int  Nimum(int *A, int *temp, int a, int b)
{

	if (a==b)
		return 0;
	else
	{
		int mid = a + (b - a) / 2;
		int left=Nimum(A, temp, a, mid);
		int right=Nimum(A, temp, mid + 1, b);
		int num=merge(A, temp, a, mid, b);
		return left + right + num;
	}
}
int main()
{
	int A[] = { 7,4,5,3,8,2};
	int n =6;
	int *t = new int[n];
	memset(t, 0, sizeof(int)*n);
	int num=Nimum(A, t, 0, n - 1);
	for (int i = 0;i < n;i++)
		cout << A[i] << ' ';

	cout << endl<< num << endl;
	delete[]t;
	return 0;
}