Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
Output: [3,3,5,5,6,7]
Explanation:

Window position                Max
---------------               -----
[1  3  -1] -3  5  3  6  7       3    012 2
 1 [3  -1  -3] 5  3  6  7       3    123 3
 1  3 [-1  -3  5] 3  6  7       5    234 4
 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

import java.util.ArrayDeque;
import java.util.Deque;

/*
 * @lc app=leetcode id=239 lang=java
 *
 * [239] Sliding Window Maximum
 */
class Solution {
    public static int[] maxSlidingWindow(int[] nums, int k) {
        if (nums.length <= 0) {
            return nums;
        }

        Deque<Integer> window = new ArrayDeque<>();// 用于储存窗口的下标
        int[] res = new int[nums.length - k + 1];
        // 从左边开始前进
        for (int i = 0; i < nums.length; i++) {
            //1.窗口界限 2.窗口左边最大的剔除

            // 1
            if (!window.isEmpty() && window.peek() <= i - k) {
                window.pollFirst();
            }


            // 2
            // 比窗口左边的小，继续，比窗口的大，删除左边的
            while (!window.isEmpty() && nums[window.peekLast()] < nums[i]) {
                window.pollLast();
            }

            window.offer(i);


            System.out.println(window);
            if (i >= k - 1) {
                res[i - k + 1] = nums[window.peek()];
            }

        }

        return res;

    }
}