713. Subarray Product Less Than K**

https://leetcode.com/problems/subarray-product-less-than-k/

题目描述

Your are given an array of positive integers nums.

Count and print the number of (contiguous) subarrays where the product of all the elements in the subarray is less than k.

Example 1:

Input: nums = [10, 5, 2, 6], k = 100
Output: 8
Explanation: The 8 subarrays that have product less than 100 are: [10], [5], [2], [6], [10, 5], [5, 2], [2, 6], [5, 2, 6].
Note that [10, 5, 2] is not included as the product of 100 is not strictly less than k.

Note:

• 0 < nums.length <= 50000.
• 0 < nums[i] < 1000.
• 0 <= k < 10^6.

C++ 实现 1

使用滑动窗口. 注意两点:

1. k <= 1 时需要考虑;
2. 对于 [j, ... i] 范围内的元素, 满足条件的数组个数为 i + 1 - j. 比如 [A, B, C], 如果 j = 0, i = 2, 那么从 C 开始, 只能组成 [C], [B, C], [A, B, C] 三个符合题意的数组, [A, C] 不符合题意, 不连续.

class Solution {
public:
    int numSubarrayProductLessThanK(vector<int>& nums, int k) {
        if (k <= 1) return 0;
        int count = 0, prod = 1, j = 0;
        for (int i = 0; i < nums.size(); ++ i) {
            prod *= nums[i];
            while (prod >= k) prod /= nums[j++];
            count += i - j + 1;
        }
        return count;
    }
};