题目：
Given an array of integers?nums?and an integer?k, return?the number of contiguous subarrays where the product of all the elements in the subarray is strictly 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.
Example 2:
Input: nums = [1,2,3], k = 0 Output: 0
Constraints:
1 <= nums.length <= 3 * 1041 <= nums[i] <= 10000 <= k <= 106
思路：
左右指针滑动窗口，初始化左右指针都是0，之后右指针一路右移，那么每一次右移，只要积小于k，那么路径上的都符合结果，即一共有right - left + 1种：假设窗口为[left, right]，且nums[left] * nums[left + 1] * ...... * nums[right - 1] * nums[right]的积小于k，那么nums[left], nums[left] * nums[left + 1]，........， nums[left] * nums[left + 1] *...... nums[right - 1] * nums[right]都是合规的答案，总共就是right - left + 1种。?当积大于等于k的时候，则需要左指针往右移，同时当前积要除以nums[l]。
代码：
class Solution {
public:
? ? int numSubarrayProductLessThanK(vector<int>& nums, int k) {
? ? ? ? if (k <= 1) return 0;
? ? ? ? int cur = 1;
? ? ? ? int l = 0, r = 0;
? ? ? ? int ans = 0;
? ? ? ? while (r < nums.size()) {
? ? ? ? ? ? cur *= nums[r];
? ? ? ? ? ? while (cur >= k) {
? ? ? ? ? ? ? ? cur /= nums[l];
? ? ? ? ? ? ? ? l++;
? ? ? ? ? ? }
? ? ? ? ? ??
? ? ? ? ? ? ans += r - l + 1;
? ? ? ? ? ? r++;
? ? ? ? }
? ? ? ? return ans;
? ? }
};
