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] <= 1041 <= k <= nums.lengthclass Solution {
public:
struct comparator{
bool operator()(pair<int, int> a, pair<int, int> b){
return a.first<b.first;
}
};
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
if(k==1)
return nums;
vector<int> res;
priority_queue<pair<int, int>, vector<pair<int, int>>, comparator> pq;
for(int i=0; i<k; i++)
pq.push({nums[i], i});
res.push_back(pq.top().first);
for(int i=k; i<nums.size(); i++){
while(pq.top().second<=i-k)
pq.pop();
pq.push({nums[i], i});
res.push_back(pq.top().first);
}
return res;
}
};