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.

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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: 

 1 class ProductOfNumbers {
 2     ArrayList<Integer> que = new ArrayList<>();
 3     
 4     public ProductOfNumbers() {
 5         que.add(1);    
 6     }
 7     
 8     public void add(int num) {
 9         if(num != 0){
10             que.add(que.get(que.size() - 1) * num);
11         }else{
12             que = new ArrayList<>();
13             que.add(1);
14         }
15     }
16     
17     public int getProduct(int k) {
18         int n = que.size();
19         return k < n ? que.get(n - 1) / que.get(n - k - 1) : 0;
20     }
21 }
22 
23 /**
24  * Your ProductOfNumbers object will be instantiated and called as such:
25  * ProductOfNumbers obj = new ProductOfNumbers();
26  * obj.add(num);
27  * int param_2 = obj.getProduct(k);
28  */