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
1 class Solution { 2 public: 3 vector<int> maxSlidingWindow(vector<int>& nums, int k) { 4 vector<int> res; 5 priority_queue<pair<int, int>> q; 6 for(int i = 0; i < k;++i) { 7 q.push({nums[i],i}); 8 } 9 res.push_back(q.top().first); 10 11 for(int i = k;i < nums.size();++i) { 12 q.push({nums[i],i}); 13 while(q.top().second + k <= i) { 14 q.pop(); 15 } 16 res.push_back(q.top().first); 17 } 18 return res; 19 } 20 };