1352. Product of the Last K Numbers

问题：

求两个接口，使得，

add接口，向列表中添加元素，

getProduct（k）接口，可得最后添加的k个元素的乘积。

Example:
Input
["ProductOfNumbers","add","add","add","add","add","getProduct","getProduct","getProduct","add","getProduct"]
[[],[3],[0],[2],[5],[4],[2],[3],[4],[8],[2]]

Output
[null,null,null,null,null,null,20,40,0,null,32]

Explanation
ProductOfNumbers productOfNumbers = new ProductOfNumbers();
productOfNumbers.add(3);        // [3]
productOfNumbers.add(0);        // [3,0]
productOfNumbers.add(2);        // [3,0,2]
productOfNumbers.add(5);        // [3,0,2,5]
productOfNumbers.add(4);        // [3,0,2,5,4]
productOfNumbers.getProduct(2); // return 20. The product of the last 2 numbers is 5 * 4 = 20
productOfNumbers.getProduct(3); // return 40. The product of the last 3 numbers is 2 * 5 * 4 = 40
productOfNumbers.getProduct(4); // return 0. The product of the last 4 numbers is 0 * 2 * 5 * 4 = 0
productOfNumbers.add(8);        // [3,0,2,5,4,8]
productOfNumbers.getProduct(2); // return 32. The product of the last 2 numbers is 4 * 8 = 32 
 

Constraints:

There will be at most 40000 operations considering both add and getProduct.
0 <= num <= 100
1 <= k <= 40000

　　

解法：

前缀和法

各个添加元素位置上为，累计到目前为止所有元素的乘积PreSum(cur)

要求则为，

getProduct(k)=

PreSum(last) / Presum(last-k-1)

♻️ 优化：按照乘法规则X0的话，前面的所有结果都=0

getProduct(k)=

PreSum(last) / Presum(last-k-1) = 0/x = 0

因此，若插入新的0元素，则可清空前面的累计Presum

代码参考：

class ProductOfNumbers {
public:
    vector<int> pro;
    ProductOfNumbers() {
        pro={1};
    }
    
    void add(int num) {
        if(num==0){
            pro={1};
        }else{
            pro.push_back(num*pro.back());
        }
    }
    
    int getProduct(int k) {
        return k<pro.size()?pro.back()/pro[pro.size()-k-1]:0;
    }
};

/**
 * Your ProductOfNumbers object will be instantiated and called as such:
 * ProductOfNumbers* obj = new ProductOfNumbers();
 * obj->add(num);
 * int param_2 = obj->getProduct(k);
 */