# Sliding Window Maximum

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.

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].

Note:
You may assume k is always valid, ie: 1 ≤ k ≤ input array's size for non-empty array.

Could you solve it in linear time?

``````public class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return new int;
}

int[] result = new int[nums.length - k + 1];

for (int i = 0; i < nums.length; i++) {
if (!deque.isEmpty() && deque.peek() == i - k) {
deque.poll();
}

while (!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) {
deque.pollLast();
}

deque.offer(i);

if (i >= k - 1) {
result[i - k + 1] = nums[deque.peek()];
}
}

return result;
}
}
``````

Hope this helps,
Michael 