求一个数组中逆序对的个数(牛客网)

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

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转载自blog.csdn.net/cx2479750196/article/details/80486257