Maximum value of sliding window

Topic

Given an array and the size of the sliding window, find the maximum value of the values in all the sliding windows.
For example, if the input array {2,3,4,2,6,2,5,1} and the size of the sliding window 3, there are altogether 6 sliding windows, their maximum value is {4,4,6,6,6,5}, the sliding window for the array {2,3,4,2,6,2,5,1} has the following 6:
{[2,3,4],2,6,2,5,1},
{2,[3,4,2],6,2,5,1},
{2,3,[4,2,6],2,5,1},
{2,3,4,[2,6,2],5,1},
{2,3,4,2,[6,2,5],1},
{2,3,4,2,6,[2,5,1]}.

Solving

Method One: Violence
Complexity of Time: O(nk)

Import Java.util.*;public class Solution {public arraylist<integer> maxinwindows (int[] num,int size{arraylist<integer> result = new arraylist<integer> ();intlen = num.length;if(size<=0||size> Len)returnResult for(intI=0; i<len;i++) {int Max= Integer.min_value; for(intk=0; K+i < Len && k<size; k++) {Max=Max&GT;NUM[I+K]?Max: Num[i+k]; } result.add (Max);if(i+size==len) Break; }returnResult }}

Method Two: Using the queue
The queue holds the subscript of the larger element in the sliding window

ImportJava.util.ArrayList;Importjava.util.*; Public  class solution {     PublicArraylist<integer>maxinwindows(int[] num,intSize) {arraylist<integer> result =NewArraylist<> ();if(num = =NULL) {returnResult }if(Num.length < Size | | Size <1) {returnResult } linkedlist<integer> Indexdeque =NewLinkedlist<> (); for(inti =0; i < size; i++) { while(!indexdeque.isempty () && num[i] >= num[indexdeque.getlast ()])            {indexdeque.removelast ();        } indexdeque.addlast (i); } for(inti = size; i < num.length; i++) {Result.add (Num[indexdeque.getfirst ())); while(!indexdeque.isempty () && num[i] >= num[indexdeque.getlast ()])            {indexdeque.removelast (); }if(!indexdeque.isempty () && Indexdeque.getfirst () <= (i-size))            {Indexdeque.removefirst ();          } indexdeque.addlast (i); } result.add (Num[indexdeque.getfirst ()));returnResult }}

Maximum number of sliding windows