https://leetcode.com/problems/sliding-window-maximum/ 题目内容: Given an arraynums, there is a sliding window of sizekwhich is moving from the very left of the array to the very right. You can only see theknumbers in the window. Each time the sliding window moves right by one position...
因为nums中-3比-1小,所以按照step3的逻辑直接将-3对应的索引3压入window即可,如下。同理,跟step4一样,需要把当前滑窗中的最大值存到res中,而window中最左边的元素在nums中的值(nums[window[0]])依然是最大值 Step 5 执行完的状态 step 6:从nums取出第5个元素,此时Index=4,重复step3的过程。在重复step3...
代码 importjdk.jshell.spi.ExecutionControl;importjava.util.ArrayList;importjava.util.Deque;importjava.util.LinkedList;classSolution{publicint[] maxSlidingWindow(int[] nums,intk) {if(nums.length<=0)thrownewRuntimeException("nums是空的!");//创建双端队列Deque<Integer> deque =newLinkedList<>();//...
2、此时window中为[ -1 -3 ]的下标,循环比较 新数 5 大于 -1 -3 顾pop掉。此时window为空 跳出循环 3、将新数 5下标为四 append入window (存放下标) 此时window为[四] 1 3 [-1 -3 5] 3 6 7 4、这时窗口中最大值 为window[0]为下标的数 5、新进数3下标五 进入,-1 已不在window内了不...
publicclassSolution{/* * @param nums: A list of integers * @param k: An integer * @return: The maximum number inside the window at each moving */publicArrayList<Integer>maxSlidingWindow(int[]nums,intk){// write your code hereArrayList<Integer>ans=newArrayList<>();Deque<Integer>deque=new...
""" 同向双指针-滑动窗口 left和right控制窗口,窗口中的元素就是需要的元素 """ class Solution(object): def maxSlidingWindow(self, nums, k): """ :type nums: List[int] :type k: int :rtype: List[int] """ left = 0 right = left + k res = [] while right < len(nums) + 1: ...
1 老师,求滑动窗口最大值这个问题。我的代码如下,是调用了另外一个函数。我觉得是O(n2)级别的算法,为什么还accepted了呢? class Solution { public: vector<int> maxSlidingWindow(vector<int>& nums, int k) {vector<int> res; int n=nums.size();...
self.queue.pop(0) def front(self): return self.queue[0] class Solution: def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]: queue = MonotonicQueue() result = [] for i in range(k): queue.push(nums[i])
Solution：def getMaxSum(arr, k): maxSum = 0 windowSum = 0 start = 0 ...
Sliding Window Maximum 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. ...