Leetcode - 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].
[分析] 首先要正确理解题意，结果是返回每个长度为k的窗口中的最大值。当窗口滑动一格时如何确定窗口中的最大值呢？直接的方法是遍历，则复杂度是O（k * n），选用最大堆存储窗口，则找到当前堆中最大值时O（1），窗口滑动时，需要添加一个新元素并删除窗口左边界元素，耗费为O（logk），因此总的复杂度是O（n*logk）,当k很大时效率提升比较明显

更优的思路是使用Deque, 时间复杂度为O（N），之前都没用过这种数据结构，JDK文档上说ArrayDeque用作stack时比Stack数据结构快，用作queue时比LinkedList快，好东西呀~
参考

https://leetcode.com/discuss/46578/java-o-n-solution-using-deque-with-explanation

// Method 2: O(N), using Deque
public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null || nums.length == 0)
            return new int[0];
        int n = nums.length;
        int[] max = new int[n - k + 1];
        int idx = 0;
        Deque<Integer> deque = new ArrayDeque<Integer>();
        for (int i = 0; i < n; i++) {
            if (!deque.isEmpty() && deque.peek() == i - k)
                deque.pollFirst();
            while (!deque.isEmpty() && nums[deque.peekLast()] <= nums[i])
                deque.pollLast();
            deque.offer(i);
            if (i >= k - 1) max[idx++] = nums[deque.peek()];
        }
        return max;
    }

// Method 1: O(N * logk), using PriorityQueue
public class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null)
            return null;
        if (nums.length == 0)
            return new int[0];
        int n = nums.length;
        int[] result = new int[n - k + 1];
        PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(k, new LargerComparator());
        for (int i = 0; i < k; i++) {
            maxHeap.offer(nums[i]);
        }
        result[0] = maxHeap.peek();
        int start = 0;
        for (int i = k; i < n; i++) {
            maxHeap.offer(nums[i]);
            maxHeap.remove(nums[start++]);
            result[i - k + 1] = maxHeap.peek();
        }
        return result;
    }
}
class LargerComparator implements Comparator<Integer> {
    public int compare(Integer o1, Integer o2) {
        return o2 - o1;
    }
}