LeetCode 1352. Product of the Last K Numbers

原题链接在这里：https://leetcode.com/problems/product-of-the-last-k-numbers/

题目：

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

题解：

Have ArrayList to record previous product.

If num == 0, all the product including this 0 would be 0. Thus clear the ArrayList.

For last k, if k < n, then we could do list.get(n - 1) / list.get(n - k - 1). Otherwise, it would include previous 0 and return 0.

Time Complexity: add, O(1). getProduct, O(1).

Space: O(n). 

AC Java: 

class ProductOfNumbers {
    ArrayList<Integer> que = new ArrayList<>();
    
    public ProductOfNumbers() {
        que.add(1);    
    }
    
    public void add(int num) {
        if(num != 0){
            que.add(que.get(que.size() - 1) * num);
        }else{
            que = new ArrayList<>();
            que.add(1);
        }
    }
    
    public int getProduct(int k) {
        int n = que.size();
        return k < n ? que.get(n - 1) / que.get(n - 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);
 */