Sliding Window Maximum You are given an array of integers 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 1:
Input: nums = [1,3,-1,-3,5,3,6,7], 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
Example 2:
Input: nums = [1], k = 1 Output: [1]
Constraints:
1 <= nums.length <= 105
-104 <= nums[i] <= 104
1 <= k <= nums.length
Sliding Window Maximum Solutions
✅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
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