Given an arraynums, 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.

Follow up:
Could you solve it in linear time?

tag: 单调栈

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null || nums.length == 0) return new int[0];
        int n = nums.length;
        int[] res = new int[n];
        Deque<Integer> dq = new ArrayDeque<Integer>();

        for (int i = 0; i < k - 1; i++){
            inQueue(dq, nums[i]);
        }

        int index = 0;
        for (int i = k - 1; i < n; i++){
            inQueue(dq, nums[i]);
            res[index++] = dq.peekFirst();
            outQueue(dq, nums[i - k + 1]);
        }

        int[] ans = new int[index];
        for (int i = 0; i < index; i++){
            ans[i] = res[i];
        }

        return ans;
    }

    private void inQueue(Deque<Integer> dq, int num){
        while (!dq.isEmpty() && dq.peekLast() < num){
            dq.pollLast();
        }
        dq.offer(num);

    }

    private void outQueue(Deque<Integer> dq, int num){
        if (dq.peekFirst() == num) dq.pollFirst();
    }
}