1352. Product of the Last K Numbers

Implement the class ProductOfNumbers that supports two methods:

1. add(int num)

• Adds the number num to the back of the current list of numbers.

2. getProduct(int k)

• Returns the product of the last k numbers in the current list.
• You can assume that always the current list has at least k numbers.

At any time, the product of any contiguous sequence of numbers will fit into a single 32-bit integer without overflowing.

 

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

class ProductOfNumbers {
    private List<Integer> list;
    public ProductOfNumbers() {
        list = new ArrayList();
    }
    
    public void add(int num) {
        list.add(num);
    }
    
    public int getProduct(int k) {
        int res = 1;
        for(int i = 0;i < k; i++) res *= list.get(list.size() - 1 - i);
        return res;
    }
}

brute force, TLE了，我怎么不意外呢fuck

class ProductOfNumbers {

   ArrayList<Integer> A;
    public ProductOfNumbers() {
        add(0);
    }

    public void add(int a) {
        if (a > 0)
            A.add(A.get(A.size() - 1) * a);
        else {
            A = new ArrayList();
            A.add(1);
        }
    }

    public int getProduct(int k) {
        int n = A.size();
        return k < n ? A.get(n - 1) / A.get(n - k - 1)  : 0;
    }
}

prefix product，如果是0就重置arraylist，把当前设为1，否则后k位就如上所示