LeetCode 1352. Product of the Last K Numbers
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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 bothadd
andgetProduct
. 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:
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 */
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