713. Subarray Product Less Than K

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.

brute-force解法O(N*N), 超时无法通过。分析可知，在确定起始点后，如果依次往后scanning找到那个临界点后，可以知道这个起始点对应的子序列个数。起始点后移一位，并不需要重新开始逐个算乘积确定新的临界点。因为上一个起始点的计算可以再次利用。也就是sliding window的思想。

代码：

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

        return cnt;
    }
};