 # Array - 239. 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. Return the max sliding window.

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

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

Could you solve it in linear time?

java：

``````class Solution {

public int[] maxSlidingWindow(int[] nums, int k) {
if (nums == null || k <= 0 ) return new int[]{};

int len = nums.length;
int[] res = new int[len - k + 1];
int index = 0;

Deque<Integer> deque = new ArrayDeque<>();
for(int i = 0; i < len; 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[index++] = nums[deque.peek()];
}
}
return res;
}
}
``````

Nothing just happens, it's all part of a plan.