请你实现一个「数字乘积类」ProductOfNumbers
,要求支持下述两种方法:
1. add(int num)
- 将数字
num
添加到当前数字列表的最后面。
2. 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
package Solution1352;
import java.util.ArrayList;
public class ProductOfNumbers {
ArrayList<Integer> nums;
public ProductOfNumbers() {
nums = new ArrayList<Integer>();
}
public void add(int num) {
nums.add(num);
}
public int getProduct(int k) {
int product = 1;
for (int i = nums.size() - 1; i > nums.size() - 1 - k; i--) {
product = product * nums.get(i);
}
return product;
}
public static void main(String[] args) {
/**
* Your ProductOfNumbers object will be instantiated and called as such:
* ProductOfNumbers obj = new ProductOfNumbers(); obj.add(num); int param_2 =
* obj.getProduct(k);
*/
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
System.out.println(productOfNumbers.getProduct(2));
}
}