vector<int> maxSlidingWindow(vector<int>& nums,intk) { deque<int>window;for(inti =0; i < k; ++i) {if(window.empty()) window.push_back(i);else{while(!window.empty() && nums[window.back()] <=nums[i]) window.pop_back(); window.push_back(i); } } vector<int>res;for(inti ...
最坏case时间复杂度为2N,所以O(N)复杂度。 vector<int> maxSlidingWindow(vector<int> &nums,intk) {//write your code heredeque<int>my_deq; vector<int>res;for(inti =0; i < nums.size(); i++) {if(i - k >=0) { // 出队列踢头节点判断if(!my_deq.empty() && nums[i - k] ==...
""" 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: ...
classSolution:defmaxSlidingWindow(self,nums,k):ifnotnums:return[]res=[]foriinrange(len(nums)-k+1):temp=max(nums[i:i+k])res.append(temp)returnres 这里是直接用一个max函数来求滑窗内元素的最大值。而在一个无序数组(滑窗)中求最大值的时间复杂度是O(k),k是滑窗的长度,所以上面解法的整体...
https://leetcode-cn.com/problems/sliding-window-maximum/ 题目内容 给定一个数组 nums,有一个大小为 k 的滑动窗口从数组的最左侧移动到数组的最右侧。你只可以看到在滑动窗口内的 k 个数字。滑动窗口每次只向右移动一位。 返回滑动窗口中的最大值。
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. ...
首先，初始化一个名为window的双端队列用于保存滑窗内元素的下标。然后，从数组中取出元素并压入队列，同时更新滑窗内的最大值。对于每个新元素，若其大于队列最左侧元素，则弹出队列最左侧元素，直至队列为空或当前元素小于等于队列最左侧元素。最后，队列最左侧的元素即为当前滑窗内的最大值，将最大值...
classSolution{public:vector<int>maxSlidingWindow(vector<int>&nums,intk){vector<int>result;deque<int>Q;//队列记录的是编号if(nums.size()<k||k<=0)returnresult;for(inti=0;i<k;i++){while(!Q.empty()&&nums[i]>nums[Q.back()])Q.pop_back();Q.push_back(i);}for(inti=k;i<nums.siz...
【Leetcode】Sliding Window Maximum https://leetcode.com/problems/sliding-window-maximum/ 题目: 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 ...
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