Leetcode刷题总结：638. Shopping Offers

题目：

给定商品价格、特价组合、要买的商品数量，给出买到指定数量商品的最小花费

Input: [2,5], [[3,0,5],[1,2,10]], [3,2]
Output: 14
Explanation: 

[2,5]表示有两种商品A和B，售价分别是$2，$5

In special offer 1, you can pay $5 for 3A and 0B
In special offer 2, you can pay $10 for 1A and 2B.

[3,2] 表示要买3A 和 2B, so you may pay $10 for 1A and 2B (special offer #2), and $4 for 2A.

 

题解：

针对每个special offer，计算使用这个special offer后的花费，如果比minPrice少， 则保存为minPrice;

递归方法实现，如果商品数量是k，special offer的数量是n, 因为原题规定了商品数目最大为6，每个商品最多采购6个，special offer最多是100个；

所以时间复杂度是O(n)

class Solution {
public:
    bool checkvalid(vector<int>& needs, const vector<int>& special) {
        for (int j = 0; j < needs.size(); ++j) {
                if (needs[j]-special[j] < 0) {
                    return false;
                }
        }
        return true;
    }
    int shoppingOffers(vector<int>& price, vector<vector<int>>& special, vector<int>& needs) {
        int minPrice = 0;
        for (int i = 0; i < needs.size(); ++i) {
            minPrice += needs[i]*price[i];
        }
       
        for (int i = 0; i < special.size(); ++i) {
            if (checkvalid(needs, special[i])) {
                vector<int> curNeeds;
                for (int j = 0; j < needs.size(); ++j) {
                    curNeeds.push_back(needs[j]-special[i][j]);
                }
                int tempPrice = shoppingOffers(price, special, curNeeds)+special[i][needs.size()];
                minPrice = min(minPrice, tempPrice);
            }
        }

        return minPrice;
    }
};