Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Return the max sliding…

class Solution: def maxSlidingWindow(self, nums: 'List[int]', k: int) -> 'List[int]': n = len(nums) if n==0: return [] if k==0: return [] dic = {} for i in range(k): dic.update({i:nums[i]}) maxindex = max(dic,key=dic.get) result = list() result.appen…

Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Return the max sliding…

面试题67 机器人的运动范围 题意: 地上有一个m行和n列的方格.一个机器人从坐标0,0的格子开始移动,每一次只能向左,右,上,下四个方向移动一格,但是不能进入行坐标和列坐标的数位之和大于k的格子. 例如,当k为18时,机器人能够进入方格(,),因为3+++ = .但是,它不能进入方格(,),因为3+++ = .请问该机器人能够达到多少个格子? 解法:回溯法.注意申请的内存要释放掉.delete[] visited; class Solution { public: int movingCoun…

简单题 1. 数据流中的移动平均值 $(leetcode-346) 暂无 2. 最近的请求次数(leetcode-933) 写一个 RecentCounter 类来计算最近的请求. 它只有一个方法:ping(int t),其中 t 代表以毫秒为单位的某个时间. 返回从 3000 毫秒前到现在的 ping 数. 任何处于 [t - 3000, t] 时间范围之内的 ping 都将会被计算在内,包括当前(指 t 时刻)的 ping. 保证每次对 ping 的调用都使用比之前更大的 t 值. 示例:…