• Time:O(n)
• Space:O(n)

## C++

``````class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
vector<int> ans;
deque<int> q;  // max queue

for (int i = 0; i < nums.size(); ++i) {
while (!q.empty() && q.back() < nums[i])
q.pop_back();
q.push_back(nums[i]);
if (i >= k && nums[i - k] == q.front())  // out of bound
q.pop_front();
if (i >= k - 1)
ans.push_back(q.front());
}

return ans;
}
};
``````

## JAVA

``````class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
int[] ans = new int[nums.length - k + 1];
Deque<Integer> q = new ArrayDeque<>(); // max queue

for (int i = 0; i < nums.length; ++i) {
while (!q.isEmpty() && q.peekLast() < nums[i])
q.pollLast();
q.offerLast(nums[i]);
if (i >= k && nums[i - k] == q.peekFirst()) // out of bound
q.pollFirst();
if (i >= k - 1)
ans[i - k + 1] = q.peekFirst();
}

return ans;
}
}
``````

## Python

``````class Solution:
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
ans = []
q = deque()  # max queue

for i, num in enumerate(nums):
while q and q[-1] < num:
q.pop()
q.append(num)
if i >= k and nums[i - k] == q[0]:  # out of bound
q.popleft()
if i >= k - 1:
ans.append(q[0])

return ans
``````