leetcode713 Subarray Product Less Than K

"""
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.
"""
"""
正确做法，双指针加滑动窗口
很是巧妙
pro为乘积，pro不断乘以nums[j]得到子数组乘积，满足条件小于k时
!!!
res +=（j-i+1），res是计数器。
滑动窗口内元素间的相互组合的数量
"""
class Solution1:
    def numSubarrayProductLessThanK(self, nums, k):
        res, i = 0, 0
        pro = 1 #乘积
        if k <= 1:  # 对于案例nums=[1, 2, 3] k=0  和 nums=[1, 1, 1] k=1
            return 0
        for j in range(len(nums)):
            pro = pro * nums[j]
            while pro >= k:
                pro = pro // nums[i]
                i += 1
            res += j - i + 1 #!!!这个很关键
        return res

"""
我的解法超时，
对于输入nums = [1, 1, 1, 1......,1] k=5
"""
class Solution2:
    def numSubarrayProductLessThanK(self, nums, k):
        res = 0
        for i in range(len(nums)):
            pro = 1
            j = i
            while j < len(nums):
                if nums[j] * pro < k:
                    res += 1
                    pro = nums[j] * pro
                    j += 1
                else:
                    break
        return res