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,
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 is always valid, ie: 1 ≤ k ≤ input array's size for non-empty array.

Follow up:
Could you solve it in linear time?

Hint:

1. How about using a data structure such as deque (double-ended queue)?
2. The queue size need not be the same as the window’s size.
3. Remove redundant elements and the queue should store only elements that need to be considered.

The hint suggests us to use Deque. To achieve O(n) complexity, we can keep the queue with capacity of k, then always keep the largest element at top. If the previous element is smaller than the one we need to add, we pop out all previous elements. In order to keep the capacity of k, we delete the i - k + 1 element every time we add in an element. The deque method is implement as that we only delete the element if it is the top of the element. That's because all elements that is smaller than the top has already been deleted, and all elements behind the top is within the k range.

    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums.length == 0) {
            return new int[]{};
        }
        List rst = new ArrayList<>();
        Deque queue = new ArrayDeque<>();
        
        for (int i = 0; i < k - 1; i++) {
            enque(queue, nums[i]);
        }
        
        for (int i = k - 1; i < nums.length; i++) {
            enque(queue, nums[i]);
            rst.add(queue.peek());
            deque(queue, nums[i - k + 1]);
        }
        
        int[] r = new int[rst.size()];
        for (int i = 0; i < rst.size(); i++) {
            r[i] = rst.get(i);
        }
        return r;
    }
    
    private void enque(Deque queue, int n) {
        while (!queue.isEmpty() && queue.peekLast() < n) {
            queue.pollLast();
        }
        queue.offer(n);
    }
    
    private void deque(Deque queue, int n) {
        if (queue.peekFirst() == n) {
            queue.pollFirst();
        }
    }