[LC] 239. 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. Return the max sliding window.

Follow up:
Could you solve it in linear time?

Example:

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
 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

Solution 1: brute force O(NK)

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        int len = nums.length;
        int[] res = new int[len - k + 1];
        for (int i = 0; i < res.length; i++) {
            int tmpMax = nums[i];
            for (int j = i; j < i + k; j++) {
                tmpMax = Math.max(tmpMax, nums[j]);
            }
            res[i] = tmpMax;
        }
        return res;
    }
}

 

Solution 2: Deque O(N)

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        int[] res = new int[nums.length - k + 1];
        Deque<Integer> deque = new LinkedList<>();
        for (int i = 0; i < nums.length; i++) {
            while (!deque.isEmpty() && deque.peekFirst() == i - k) {
                deque.pollFirst();
            }
            while (!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) {
                deque.pollLast();
            }
            deque.offerLast(i);
            if (i - k + 1 >= 0) {
                // from big to small, left to right
                res[i - k + 1] = nums[deque.peekFirst()];
            }
        }
        return res;
    }
}