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website/content/ChapterFour/0713.Subarray-Product-Less-Than-K.md

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+# [713. Subarray Product Less Than K](https://leetcode.com/problems/subarray-product-less-than-k/)
+
+## 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.
+
+## 题目大意
+
+给出一个数组，要求在输出符合条件的窗口数，条件是，窗口中所有数字乘积小于 K 。
+
+## 解题思路
+
+这道题也是滑动窗口的题目，在窗口滑动的过程中不断累乘，直到乘积大于 k，大于 k 的时候就缩小左窗口。有一种情况还需要单独处理一下，即类似 [100] 这种情况。这种情况窗口内乘积等于 k，不小于 k，左边窗口等于右窗口，这个时候需要左窗口和右窗口同时右移。
+
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func numSubarrayProductLessThanK(nums []int, k int) int {
+	if len(nums) == 0 {
+		return 0
+	}
+	res, left, right, prod := 0, 0, 0, 1
+	for left < len(nums) {
+		if right < len(nums) && prod*nums[right] < k {
+			prod = prod * nums[right]
+			right++
+		} else if left == right {
+			left++
+			right++
+		} else {
+			res += right - left
+			prod = prod / nums[left]
+			left++
+		}
+	}
+	return res
+}
+
+```