Sliding Window Maximum leetcode

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].

用deque来解决, deque里入他们的下角标, 每当入新数的时候把新数与deque的num[末尾]比较 如果num[末尾]小就把末尾扔掉, 直到num[末尾]大于等于num[新数], 这样一来deque里就是按照num的大小顺序排列的 前边的最大 后边的最小. 每次不用出列不在窗口里德数字, 我们只需要确定deque开头的数是在窗口就可以了. 因为每个数只可能被操作最多两次，一次是加入队列的时候，一次是因为有别的更大数在后面，所以被扔掉，或者因为出了窗口而被扔掉。   所以时间复杂度为O(N).

public 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 - k + 1];
        Deque<Integer> deque = new LinkedList<Integer>();
        for (int i = 0; i < n; i++) {
            //如果最左侧的数不在窗口内 出列
            while (!deque.isEmpty() && deque.peek() < i - k + 1) {
                deque.pollFirst();
            }
            //如果结尾的数小于新数则出列
            while (!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) {
                deque.pollLast();
            }
            //添加入新数
            deque.offerLast(i);
            if (i >= k - 1) {
                res[i - k + 1] = nums[deque.peekFirst()];
            }
        }
        return res;
    }
}