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.
Note:
0 < nums.length <= 50000.
0 < nums[i] < 1000.
0 <= k < 10^6.
利用sliding window的頭尾兩個pointers去找到一個符合條件的最大乘積的區間，每次找到一個新的j pointer就代表可以多形成(j - i + 1)個以nums[j]結尾的subarrays。
// Sliding Window
int numSubarrayProductLessThanK(vector<int>& nums, int k) { // time: O(n); space: O(1)
int res = 0, n = nums.size(), pro = 1;
for (int i = 0, j = 0; j < n; ++j) {
pro *= nums[j];
while (i <= j && pro >= k) {
pro /= nums[i++];
}
res += (j - i + 1); // number of subarrays ending with nums[j]
}
return res;
}