Sliding Window Maximum

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.

For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

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

Therefore, return the max sliding window as [3,3,5,5,6,7].

Note:
You may assume k is always valid, ie: 1 ≤ k ≤ input array’s size for non-empty array.

Follow up:
Could you solve it in linear time?

Use a heap. heap stores a index, nums[index] pair.

Heap is sorted by nums[index], it’s a maximum heap.

When queue.top().first is within the window, answer is the queue.top().second,

otherwise pop the queue until queue.top().first is located in the window.

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
    	auto cmp = [](pair<int, int>& a, pair<int, int>& b){
    		return a.second < b.second;	
     	};
    	vector<int> ans; 
        priority_queue<pair<int, int>, vector<pair<int, int>>, decltype(cmp)> q(cmp);
    	if(nums.size() == 0 || k == 0) return ans;
        for(int i = 0; i < k - 1; i++){
    	    q.push(make_pair(i, nums[i]));
        }
        for(int i = 0; i < nums.size() - k + 1; i++){
    	    while(!q.empty() && q.top().first < i){
    	        q.pop();
            }
            q.push(make_pair(i + k - 1, nums[i + k - 1]));
            ans.push_back(q.top().second);
        }
        return ans;
    }
};