https://leetcode.com/problems/sliding-window-maximum/discuss/65881/O(n)-solution-in-Java-with-two-simple-pass-in-the-array publicint[] maxSlidingWindow2(int[] nums,intk) {if(nums ==null|| k <= 0)returnnewint[0];intn =nums.length;int[] r =newint[n - k + 1];intri = 0; ...
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. For example, Given nums =[1,3,-1,-3,5,3,6,7...
**SolutionJava(deque) ** ** 8ms, beats 87.35% ** **49.5, beats 6.25% **publicclassSolution{publicint[]maxSlidingWindow(int[] nums,intk){if(nums ==null|| nums.length < k || k <1)returnnewint[0]; Deque<Integer> deque =newArrayDeque<>();intindex =0;int[] ans =newint[nums....
维护一个双端队列window其大小为k,从尾入队列从头出队列 Python3实现 双端队列 # @author:leacoder# @des: 双端队列 滑动窗口最大值classSolution:defmaxSlidingWindow(self,nums:List[int],k:int)->List[int]:ifnotnums:return[]length=len(nums)window,result=[],[]#window 存在窗口中的数 result用于存...
更新于 6/9/2020, 7:03:46 PM java 单调的双端队列 // 九章算法强化班版本 // deque中存储位置 public class Solution { int[] a; /** * @param nums: A list of integers. * @return: The maximum number inside the window at each moving. */ void inQueue(Deque<Integer> deque, int pos)...
所以只需要保持一个fixed sliding window的长度为短字符串的长度然后扫长字符串来寻找起始位置 class Solution{ public int strStr(String long, String short) { //sanity check if(long == null || short == null) return -1; int i = 0;
sliding-window Updated Sep 29, 2022 Java java-leetcode-classroom / java_sliding_window_maximum Star 0 Code Issues Pull requests https://useful-journey-3db.notion.site/239-Sliding-Window-Maximum-6378a7fc56c941cbbc9b2325fc0ad08e leetcode sliding-window Updated Aug 26, 2022 Java ahmed...
这个题是239. Sliding Window Maximum的变形。 题目的意思很简单,要求滑动窗口中的元素的中位数。如果是求平均数,就简单很多了:每次滑动窗口移动,都是增加一个元素、删除一个元素,因此窗口内元素的和是非常好维护的,再除以窗口的大小就能得到平均数。
Could you solve it in linear time? Again 先上暴力解法。双层嵌套的循环,时间复杂度是O(N*k)=O(N) classSolution{/** * Brute force: every time traverse the whole window to find the maximum. Time complexity: O(N*k) **/publicint[]maxSlidingWindow(int[]nums,intk){if(nums==null||nums...
Java 代码实现 class Solution { public int[] maxSlidingWindow(int[] nums, int k) { if(nums == null || nums.length < 2) return nums; // 双端队列 保存当前窗口最大值的数组位置 保证队列中数组位置的数值按从大到小排序 LinkedList<Integer> queue = new LinkedList(); // 结果数组 int[] re...