713. Subarray Product Less Than K

问题：

求给定数组的连续子数组个数，使得子数组之乘积，小于给定值 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.

　　

解法：

窗口法，

左窗口low，右窗口high

主要移动 high，每加一个满足条件的high，子数组数+=（high-low）+1

如果乘积大于 k，那么向右移动左窗口low，同时乘积/nums[low]

参考代码：

class Solution {
public:
    int numSubarrayProductLessThanK(vector<int>& nums, int k) {
        int res=0;
        int low=0, high=0;
        if(nums.size()==0||k==0)return res;
        int protmp=1;
        for(low=0,high=0;high<nums.size();high++){
            protmp*=nums[high];
            while(low<=high && protmp>=k){
                protmp/=nums[low];
                low++;
            }
            res+=(high-low+1);
        }
        return res;
    }
};