1352 Product of the Last K Numbers 最后 K 个数的乘积
Description:
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 * 10^4
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.
题目描述：
请你实现一个「数字乘积类」ProductOfNumbers，要求支持下述两种方法：
- add(int num)
将数字 num 添加到当前数字列表的最后面。
- getProduct(int k)
返回当前数字列表中，最后 k 个数字的乘积。
你可以假设当前列表中始终 至少 包含 k 个数字。
题目数据保证：任何时候，任一连续数字序列的乘积都在 32-bit 整数范围内，不会溢出。
示例：
输入：
["ProductOfNumbers","add","add","add","add","add","getProduct","getProduct","getProduct","add","getProduct"]
[[],[3],[0],[2],[5],[4],[2],[3],[4],[8],[2]]
输出：
[null,null,null,null,null,null,20,40,0,null,32]
解释：
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); // 返回 20 。最后 2 个数字的乘积是 5 * 4 = 20
productOfNumbers.getProduct(3); // 返回 40 。最后 3 个数字的乘积是 2 * 5 * 4 = 40
productOfNumbers.getProduct(4); // 返回 0 。最后 4 个数字的乘积是 0 * 2 * 5 * 4 = 0
productOfNumbers.add(8); // [3,0,2,5,4,8]
productOfNumbers.getProduct(2); // 返回 32 。最后 2 个数字的乘积是 4 * 8 = 32
提示：
add 和 getProduct 两种操作加起来总共不会超过 40000 次。
0 <= num <= 100
1 <= k <= 40000
思路：
前缀积
如果 add 的数为 0, 清空数组
如果数组为空, 先加入哨兵 1
然后加入数组最后一个数和当前数的乘积
返回最后一个数和倒数第 k 个数的商
add 函数时间复杂度为 O(n), getProduct 函数时间复杂度为 O(1), 空间复杂度为 O(n)
代码:
C++:
class ProductOfNumbers
{
private:
vector<int> pre;
public:
ProductOfNumbers() {}
void add(int num)
{
if (!num)
{
pre.clear();
return;
}
if (pre.empty()) pre.emplace_back(1);
pre.emplace_back(pre.back() * num);
}
int getProduct(int k)
{
return k < pre.size() ? pre.back() / pre[pre.size() - 1 - k] : 0;
}
};
/**
* Your ProductOfNumbers object will be instantiated and called as such:
* ProductOfNumbers* obj = new ProductOfNumbers();
* obj->add(num);
* int param_2 = obj->getProduct(k);
*/
Java:
class ProductOfNumbers {
private List<Integer> pre;
public ProductOfNumbers() {
pre = new ArrayList<>();
}
public void add(int num) {
if (num == 0) {
pre.clear();
return;
}
if (pre.isEmpty()) pre.add(1);
pre.add(pre.get(pre.size() - 1) * num);
}
public int getProduct(int k) {
return k < pre.size() ? pre.get(pre.size() - 1) / pre.get(pre.size() - 1 - k) : 0;
}
}
/**
* Your ProductOfNumbers object will be instantiated and called as such:
* ProductOfNumbers obj = new ProductOfNumbers();
* obj.add(num);
* int param_2 = obj.getProduct(k);
*/
Python:
class ProductOfNumbers:
def __init__(self):
self.pre = []
def add(self, num: int) -> None:
if not num:
self.pre = []
return
if not self.pre:
self.pre.append(1)
self.pre.append(self.pre[-1] * num)
def getProduct(self, k: int) -> int:
return 0 if k >= len(self.pre) else self.pre[-1] // self.pre[-1 - k]
# Your ProductOfNumbers object will be instantiated and called as such:
# obj = ProductOfNumbers()
# obj.add(num)
# param_2 = obj.getProduct(k)
