双指针,滑动窗口
数组长度n
n = len(nums)
p记录乘积
p = 1
i为begin,窗口左端. j为end,窗口右端
i = 0 j = 0
total记录符合条件subarray个数
total = 0
while j < n:
p *= nums[j]
while i <= j and p >= k:
p /= nums[i]
i += 1
total += j - i + 1
j += 1
尽量扩展窗口
最后返回total
return total
双指针,滑动窗口数组长度nn = len(nums)p记录乘积p = 1i为begin,窗口左端. j为end,...
文章作者：Tyan博客：noahsnail.com | CSDN | 简书 1. Description 2. S...
1 The value of describing your product in less than 10 wo...
713. 乘积小于K的子数组[https://leetcode-cn.com/problems/subarray-...
152. Maximum Product Subarray
LeetCode 152. Maximum Product Subarray Description Given ...
152. Maximum Product Subarray Given an integer array nums...
Question from lintcodeFind the contiguous subarray within...
Medium, Dynamic Programming Question 找到积最大的子序列 For exampl...
less than一点都不 毫不 less than grateful一点都不感激 white collar 白领...
本文标题：713. Subarray Product Less Than
本文链接：https://www.haomeiwen.com/subject/ufuhmxtx.html
网友评论