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.
Follow up:
Could you solve it in linear time?
class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
if (nums.empty() || nums.size() < k)
return {};
deque<int> q;
vector<int> res;
for (int i = 0; i < nums.size(); i ++) {
if (!q.empty() && q.front() == i-k)
q.pop_front();
while (!q.empty() && nums[q.back()] < nums[i])
q.pop_back();
q.push_back(i);
if (i >= k-1)
res.push_back(nums[q.front()]);
}
return res;
}
};