# [leetcode] Sliding Window Maximum

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.

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;
}
};```

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