239. Sliding Window Maximum

You are given an array of integers 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.

Return the max sliding window.

Example 1:

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

Example 2:

Input: nums = [1], k = 1
Output: [1]

TC: O(n), SC: O(n)

public class Solution {
class MaxQueue {

    Deque<Integer> maxs, nums;

    MaxQueue() {
        maxs = new LinkedList<>();
        nums = new LinkedList<>();
    }

    void offer(int num) {
        nums.offer(num); // add method throws an exception and offer doesn't.
        while (!maxs.isEmpty() && maxs.peekLast() < num) {
            maxs.pollLast();
        }
        maxs.offerLast(num);
    }

    Integer poll() {
        Integer res = null;
        if (!nums.isEmpty()) {
            res = nums.poll();
            if (res.equals(maxs.peekFirst())) {
                maxs.pollFirst();
            }
        }
        
        return res;
    }

    Integer getMax() {
        Integer res = null;
        if (!maxs.isEmpty()) {
            res = maxs.peekFirst();
        }
        
        return res;
        }
}
public int[] maxSlidingWindow(int[] nums, int k) {
    int n = nums.length;
    int[] res = new int[n-k+1];
    
    if (n == 0 || k == 0) {
        return new int[0];
    }
    
    MaxQueue mq = new MaxQueue();
    
    for (int i = 0; i < k; i++) {
        mq.offer(nums[i]);
    }

    res[0] = mq.getMax();
    
    for (int i = k; i < n; i++) {
        mq.poll();
        mq.offer(nums[i]);
        res[i-k+1] = mq.getMax();
    }
    
    return res;
}
}

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