# 713.Subarray Product Less Than K

**713.Subarray Product Less Than K**

难度:Medium 给定一个正整数数组 nums。

找出该数组内乘积小于 k 的连续的子数组的个数。

```
示例 1:

输入: nums = [10,5,2,6], k = 100
输出: 8
解释: 8个乘积小于100的子数组分别为: [10], [5], [2], [6], [10,5], [5,2], [2,6], [5,2,6]。
需要注意的是 [10,5,2] 并不是乘积小于100的子数组。
说明:

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

Method: 因为是连续子数组，可以以左右指针来计数。左边起始点为left，每向右增加一个，实际上增加的连续子数组的数目是(right-left+1)，因为增加的连续子数组数目必须包含right，否则就重复了。 然后从左往右遍历，如果乘积超过K了，则left右移，直到小于k为止。

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

    }
};
```

> 执行用时 :264 ms, 在所有 C++ 提交中击败了97.70%的用户\
> 内存消耗 :83.7 MB, 在所有 C++ 提交中击败了8.27%的用户


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