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?

public class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        int len = nums.length;
        int[] result = new int[len - k + 1];
        if(nums.length == 0) return new int[0];
        Queue<Integer> queue = new PriorityQueue<Integer>(new Comparator<Integer>(){
            @Override
            public int compare(Integer i1, Integer i2){
                return Integer.compare(i2, i1);
            }
        });
 
        for(int i = 0; i < k; i ++){
            queue.add(nums[i]);
        }
        result[0] = queue.peek();
        for(int i = k; i < len; i ++){
            queue.remove(nums[i - k]);
            queue.add(nums[i]);
            result[i - k + 1] = queue.peek();
        }
 
        return result;
    }
}

========

public class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null || nums.length == 0 || k == 0) {
            return new int[0];
        }
        Deque<Integer> deque = new ArrayDeque<>();
        int n = nums.length;
        int[] res = new int[n - k + 1];
        int j = 0;
        for (int i = 0; i < n; i++) {
            if (!deque.isEmpty() && deque.peek() <= i - k) {
                deque.poll();
            }
            while (!deque.isEmpty() && nums[i] > nums[deque.peekLast()]) {
                deque.pollLast();
            }
            deque.offer(i);
            if (i >= k - 1) {
                res[j++] = nums[deque.peek()];
            }
        }
        return res;
    }
}