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.

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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].

O(n log k) O(k) space heap solution

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public class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null || nums.length < 1) return new int[0];
        int[] result = new int[nums.length - k + 1];
        PriorityQueue<Integer> pq = new PriorityQueue<Integer>(k, (a, b) -> b - a);
        for (int i = 0; i <= k - 1; i++) {
            pq.offer(nums[i]);
        }
        result[0] = pq.peek();
        int left = 1, right = k;
        for (int j = 1; j < result.length; j++) {
            pq.remove(new Integer(nums[left - 1]));
            pq.offer(nums[right]);
            result[j] = pq.peek();
            right++;
            left++;
        }
        return result;
    }
}

O(n) time O(k) space deque solution

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