解题思路:
一道典型的滑动窗口题型,每次控制窗口内满足乘积小于k,每往前进一次就要加上子集的个数,注意不能重复,所以每次加上r-l+1,就不会重复加了,代码如下:
class Solution {
public:
int numSubarrayProductLessThanK(vector& nums, int k) {
if(k == 0 || k == 1) return 0;
int n = nums.size();
int l = 0, r = 0, mul = 1;
int res = 0;
while(r < n) {
mul *= nums[r];
while(mul >= k) {
mul /= nums[l ++];
}
res += r - l + 1;
r ++;
}
return res;
}
};


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