239. Sliding Window Maximum

Given an array nums, there is a sliding window of size K which are moving from the very left of the array To the very right. You can only see the K numbers in the window. Each of the sliding window moves right by one position.

For example,
Given nums = [1,3,-1,-3,5,3,6,7] , and k = 3.

Window position                Max---------------               -----[1  3  -1]-3  5  3  6  7       3 1 [3  - 1  -3] 5  3  6  7       3 1 3  [-1  -3  5] 3  6  7       5 1  3  -1 [-3  5  3] 6  7       5 1  3  -1  -3 [5  3  6] 7       6 1  3  -1  -3  5 [3  6  7]      7

Therefore, return the max sliding window as [3,3,5,5,6,7] .

Note:
You may assume K are always valid, Ie:1≤k≤input array's size for non-empty array.

Follow up:
Could solve it in linear time?

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265. Paint House II

1Solution 1. Priority Queue2  Public classSolution {3     classPairImplementsComparable<pair>{4          Public intkey;5          Public intidx;6         7          PublicPair (intKintID) {8              This. Key =K;9              This. idx =ID;Ten         } One          A @Override -          Public intcompareTo (Pair p) { -             return  This. Key < P.key? 1: ( This. Key = = P.key? 0:-1);//reverse order, from Large to Small the         } -     } -      -      Public int[] Maxslidingwindow (int[] Nums,intk) { +         if(Nums.length = = 0 | | k <= 0)return New int[0]; -          +Priorityqueue<pair> que =NewPriorityqueue<pair>(); A         int[] ret =New int[Nums.length-k + 1]; at         inti = 0; -          -          for(i = 0; i < nums.length; i++) { -Que.add (NewPair (Nums[i], i)); -              while(Que.peek (). idx < I-k + 1) { - Que.poll (); in             } -             if(I >= k-1) { toRET[I-K+1] =Que.peek (). Key; +             } -         } the          *         returnret; $     }Panax Notoginseng}

239. Sliding Window Maximum