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65 lines
1.6 KiB
Markdown
65 lines
1.6 KiB
Markdown
# [713. Subarray Product Less Than K](https://leetcode.com/problems/subarray-product-less-than-k/)
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## 题目
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Your are given an array of positive integers nums.
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Count and print the number of (contiguous) subarrays where the product of all the elements in the subarray is less than k.
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**Example 1**:
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```
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Input: nums = [10, 5, 2, 6], k = 100
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Output: 8
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Explanation: The 8 subarrays that have product less than 100 are: [10], [5], [2], [6], [10, 5], [5, 2], [2, 6], [5, 2, 6].
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Note that [10, 5, 2] is not included as the product of 100 is not strictly less than k.
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```
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**Note**:
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- 0 < nums.length <= 50000.
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- 0 < nums[i] < 1000.
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- 0 <= k < 10^6.
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## 题目大意
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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 。
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## 解题思路
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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。
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## 代码
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```go
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package leetcode
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func numSubarrayProductLessThanK(nums []int, k int) int {
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if len(nums) == 0 {
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return 0
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}
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res, left, right, prod := 0, 0, 0, 1
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for left < len(nums) {
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if right < len(nums) && prod*nums[right] < k {
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prod = prod * nums[right]
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right++
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} else if left == right {
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left++
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right++
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} else {
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res += right - left
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prod = prod / nums[left]
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left++
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}
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}
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return res
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}
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``` |