713. Subarray Product Less Than K

Posted 2021-02-22 habibah-chang

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：

 1 class Solution {
 2 public:
 3     int numSubarrayProductLessThanK(vector<int>& nums, int k) {
 4         int res=0;
 5         int low=0, high=0;
 6         if(nums.size()==0||k==0)return res;
 7         int protmp=1;
 8         for(low=0,high=0;high<nums.size();high++){
 9             protmp*=nums[high];
10             while(low<=high && protmp>=k){
11                 protmp/=nums[low];
12                 low++;
13             }
14             res+=(high-low+1);
15         }
16         return res;
17     }
18 };