Update 713._Subarray_Product_Less_Than_K.md

Author: 0xkookoo

docs/Leetcode_Solutions/713._Subarray_Product_Less_Than_K.md

@@ -64,11 +64,13 @@ class Solution(object):
 > 思路 2
 ******- 时间复杂度: O(N)******- 空间复杂度: O(1)******
 
-``
+参考[gabbu](https://leetcode.com/problems/subarray-product-less-than-k/discuss/108846/Python-solution-with-detailed-explanation)
+```
 Initialize start and end to index 0. Initialize prod to 1. Iterate end from 0 to len(nums)-1. 
 Now if prod * nums[end] is less than k, then all subarray between start and end contribute to the solution. 
 Since we are moving from left to right, we would have already counted all valid subarrays from start to end-1. 
 How many new subarrays with nums[end]? Answer: end-start. What will be the updated prod? Answer: prod * nums[end].
+
 ```
 
 ```python