Link: https://leetcode.cn/problems/product-of-the-last-k-numbers/

Question

difficulty: mid
adj diff: 4

Design an algorithm that accepts a stream of integers and retrieves the product of the last k integers of the stream.

Implement the ProductOfNumbers class:

    ProductOfNumbers() Initializes the object with an empty stream.
    void add(int num) Appends the integer num to the stream.
    int 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.

The test cases are generated so that, 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:

    0 <= num <= 100
    1 <= k <= 4 * 104
    At most 4 * 104 calls will be made to add and getProduct.
    The product of the stream at any point in time will fit in a 32-bit integer.

Code

以下代码“超出时间限制”，但是应该是 word 的。

class MyInteger {
    int val;
    public MyInteger(int a) {
        val = a;
    }
}

List<MyInteger> products;

public ProductOfNumbers() {
    products = new ArrayList<MyInteger>();
}

public void add(int num) {
    if (num == 0) {
        products.clear();
    }
    products.add(new MyInteger(1));
    for (MyInteger mi: products) {
        mi.val = mi.val * num;
    }
}

public int getProduct(int k) {
    if (k > products.size()) {
        // because existances of '0', all products are 0
        return 0;
    } else {
        return products.get(products.size() - k).val;
    }
}

1352. Product of the Last K Numbers

Question

Code