class Solution {
public:
    vector<int> advantageCount(vector<int>& A, vector<int>& B) {
        int max = 0;
        vector<int> maxVector;
        maxVector.insert(maxVector.begin(), A.begin(), A.end());
        sort(A.begin(), A.end());
        do
        {
            int count = 0;
            for(int i = 0; i < A.size(); i ++)
            {
                if(A[i] > B[i] ) count++;
            }
            if(count > max)
            {
                max = count;
                maxVector.clear();
                maxVector.insert(maxVector.begin(), A.begin(), A.end());            
            }
        } while(next_permutation(A.begin(), A.end()));
        return maxVector;
    }
};

class Solution {
public:
    vector<int> advantageCount(vector<int>& A, vector<int>& B) {
        vector<pair<int, int> > b_matrix;
        vector<int> result;  //存储结果索引值
        result.resize(B.size());
        int flag = -1;
        for(int i = 0; i < B.size(); i++)
        {
            b_matrix.push_back(make_pair(B[i],i));
        }
        sort(A.begin(), A.end());
        sort(b_matrix.begin(), b_matrix.end());
        int j = A.size()-1;
        int k = A.size()-1;
        int i = 0;
        while(j>=0)
        {
            if(A[k] > b_matrix[j].first) {
                result[b_matrix[j].second] = A[k];
                k--;
                j--;
            } else {
                result[b_matrix[j].second] = A[i];
                i++;
                j--;
            }
            
        }
         
        return result;
    }
};
/*********example ******

A[k]:          1 2  2 4 6
b_matrix[j]:0 2 2 4 5
6取代5原先的索引，  
4取代2原先的索引,
2取代0原来的索引值。
其他位置则随意
****************/