### 题目

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.

### two pointers标准暴力过程

10 * 5 * 2 = 100
lo     hi
|<---->|
[10, 5, 2, 6]

#### 代码

class Solution {
public int numSubarrayProductLessThanK(int[] nums, int k) {
int count = 0;
for (int from = 0; from < nums.length; from++) {
int prod = 1;
for (int to = from; to < nums.length; to++) {
prod *= nums[to];
if (prod >= k) break;
count++;
}
}
return count;
}
}

### 窗口法

MAX = 800

10 * 5 * 2 * 6 = 600 < 800
lo         hi
|<------->|
[10, 5, 2, 6, 3, 4, 9, 7]

MAX = 800

10 * 5 * 2 * 6 = 600 < 800
lo         hi
|<------->|
[10, 5, 2, 6, 3, 4, 9, 7]
|<--->|
lo     hi
5 * 2 * 6 = 600 / 10 = 60 < 800

[10, 5, 2, 6, 3, 4, 9, 7]
|<--------->|
lo           hi
5 * 2 * 6 * 3 * 4 = 720 < 800

#### 代码

class Solution {
public int numSubarrayProductLessThanK(int[] nums, int k) {
if (k < 1) return 0;
int lo = 0, hi = 0, prod = 1, count = 0;
while (lo < nums.length) {
if (lo < hi) {
count += (hi - lo);
} else {
prod = 1;
hi = lo;
}
while (hi < nums.length && prod * nums[hi] < k) {
prod *= nums[hi++];
count++;
}
prod /= nums[lo++];
}
return count;
}
}