Leetcode 315. 计算右侧小于当前元素的个数

class Solution {
public:
    vector<int> count;
    
    void merge_sort_two_vec(vector<pair<int, int>>& sbu_vec1, vector<pair<int, int>> &sbu_vec2, vector<pair<int, int>>& vec)
    {
        int i = 0, j = 0;
        while(i<sbu_vec1.size() && j<sbu_vec2.size())
        {
            if(sbu_vec1[i].first <= sbu_vec2[j].first)
            {
                count[sbu_vec1[i].second] += j;
                vec.push_back(sbu_vec1[i]);
                i++;
            }
            else
            {
                vec.push_back(sbu_vec2[j]);
                j++;
            }
        }
        for(;i<sbu_vec1.size(); ++i)
        {
            count[sbu_vec1[i].second] += j;
            vec.push_back(sbu_vec1[i]);
        }
         for(;j<sbu_vec2.size(); ++j)
        {
            vec.push_back(sbu_vec2[j]);
        }   
    }
    
    
    void merge_sort(vector<pair<int, int>>& vec)
    {
        if(vec.size() < 2)
            return ;
        int mid = vec.size()/2;
        vector<pair<int, int>> sbu_vec1;
        vector<pair<int, int>> sbu_vec2;
        for(int i=0;  i<mid; ++i)
        {
            sbu_vec1.push_back(vec[i]);
        }
        for(int i=mid; i<vec.size(); ++i)
        {
            sbu_vec2.push_back(vec[i]);
        }
        
        merge_sort(sbu_vec1);
        merge_sort(sbu_vec2);
        vec.clear();
        
        merge_sort_two_vec(sbu_vec1, sbu_vec2, vec);
        
    }
    vector<int> countSmaller(vector<int>& nums) {
        vector<pair<int, int>> vec;
        
        for(int i=0; i<nums.size();  ++i)
        {
            vec.push_back(make_pair(nums[i], i));
        }
        count.resize(nums.size(), 0);
        
        merge_sort(vec);
        
        return count;
    }
};