239. 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.

For example,
Givennums=[1,3,-1,-3,5,3,6,7], andk= 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 is always valid, ie: 1 ≤ k ≤ input array's size for non-empty array.

Solution:

Using a deque to maintain a descending order value of k elements, deque holds the position i of the value in array.

• establish a deque using linkedlist
• iterate through the array, when the head element (the most left element) is out of range (current index - head _element _index >= k) remove the head element
• while the current element value > the tail element of deque, remove the tail element
• for each index+1-k >= 0, update the result array

 public int[] maxSlidingWindow(int[] nums, int k) {
        if(nums == null || nums.length == 0){
            return new int[0];
        }
        LinkedList<Integer> deque =  new LinkedList<>();
        int[] res = new int[nums.length - k + 1];

        for(int i = 0; i < nums.length; i++){
            if(!deque.isEmpty() && i-k == deque.peekFirst()){
                deque.poll();
            }
            while(!deque.isEmpty() && nums[i] > nums[deque.peekLast()]){
                deque.pollLast();
                //deque.add(i);
            }
            deque.offer(i);
            if(i+1-k >=0){
                res[i+1-k] = nums[deque.peekFirst()];
            }
        }
        return res;
    }