实现代码:
#include<vector>
#include<iostream>
using namespace std;
int InversePairs(vector<int> data) {
if (data.size() <= 1){
return 0;
}
int count = 0;
for (int i = 0; i < data.size(); i++){
for (int j = i + 1; j < data.size(); j++){
if (data[i]>data[j]){
count++;
}
}
}
return count%1000000007;
}
int main(){
vector<int> v1;
v1.push_back(7);
v1.push_back(8);
v1.push_back(6);
v1.push_back(5);
v1.push_back(4);
v1.push_back(2);
int count = InversePairs(v1);
cout << count << endl;
system("pause");
return 0;
}
此方法虽然可以得到正确结果,但是时间复杂度是O(n^2),所以还有一种归并排序思想后面再更新