59 - I. maximum value of sliding window

The finger of the sword Offer 59 - I. Maximum sliding window

Ideas ： A monotonous queue

The data structure corresponding to the window is a double ended queue , When traversing an array , Each round is guaranteed to be monotonous deque:

• deque Contains only the elements in the window $\Rightarrow$ Every window slide removes elements nums[i-k]

• deque Inner non strict decreasing $\Rightarrow$ Each round of window sliding requires that the number of windows in the queue is less than nums[i] Element pop-up

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> ans;
        deque<int> q;
        for(int i=0;i<nums.size();++i){
            if(q.size()&&q.front()<=i-k) q.pop_front();
            while(q.size()&&nums[q.back()]<nums[i]) q.pop_back();
            q.push_back(i);
            if(i>=k-1){
                ans.push_back(nums[q.front()]);
            }
        }
        return ans;
    }
};

Time complexity O(n)

Spatial complexity O(k)