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...
You are given an array of integers 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 window. ...
classSolution {public: 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); } } vec...
then the window move one step forward. [1, |2, 7 ,7|, 8], return the maximum7; then the window move one step forward again. [1, 2, |7, 7, 8|], return the maximum8; Challenge o(n) time and O(k) memory Solution Key to the solution is to maintain a deque (size <= k)...
二、Possible Solution 1、借助大顶堆 这里注意随着窗口移动,维护大顶堆(加入新元素,删除离开),最后结果为堆顶元素。一个元素维护大顶堆的时间复杂度为O(logk),最后结果时间复杂度为O(1),最后的n个元素的时间复杂度为O(n*logk) 2、双端队列 维护一个双端队列window其大小为k,从尾入队列从头出队列 ...
from collections import deque class Solution: """ @param nums: A list of integers. @param k: An integer @return: The maximum number inside the window at each moving. """ def maxSlidingWindow(self, nums, k): # write your code here if not nums or not k: return [] queue, ans = ...
classSolution{public:vector<int>maxSlidingWindow(vector<int>&nums,intk){vector<int>result;priority_queue<pair<int,int>>Q;//pair.second记录原始数据的序号,从而便于识别最大值是否在当前窗口内,其实也可以只存储编号if(nums.size()<k||k<1)returnresult;for(inti=0;i<k-1;i++)Q.push(pair<int,...
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 ...[LeetCode] 239. Sliding Window Maximum @ python 一.题目: 给定一个数组.还有...
239. Sliding Window Maximum class Solution { public: vector<int> maxSlidingWindow(vector<int>& nums, int k) { vector<int> result; //use ... 查看原文 c++中set和multiset容器的区别和用法 ,1,1,10,10,7,6,9,11,2}; multiset<;int, greater<int>;>save2; for (vector<;int>;:...
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();...