# 239 Sliding Window Maximum

### Problem:

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.

Follow up: Could you solve it in linear time?

### Solutions:

``````public class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return new int[0];
}
int[] res = new int[nums.length - k + 1];
PriorityQueue<Integer> q = new PriorityQueue<Integer>(Collections.reverseOrder());
for (int i = 0; i < k; i ++) {
}
res[0] = q.peek();
for (int j = k; j < nums.length; j ++) {
int remove = nums[j - k];
q.remove((Integer)remove);
res[j - k + 1] = q.peek();
}
return res;
}
}
``````
``````public class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return new int[0];
}
int[] res = new int[nums.length - k + 1];
for (int i = 0; i < nums.length; i ++) {
if (!q.isEmpty() && q.peek() == i - k) {
q.removeFirst();
}
while (!q.isEmpty() && nums[q.peekLast()] < nums[i]) {
q.removeLast();
}