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] 7Therefore, 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?
用一个滑窗在数组上滑动，求滑动到每个位置时，滑窗内数字的最大值，线性时间内完成。使用双端队列完成，队列记录数组中的下标，维护多个单调递减区间。
当当前下标和队列首的元素相差超过k-1时，说明滑窗超过了既定大小，需要把前一个元素扔掉。
因为要维护单调递增，所以每次加入一个新的下标时，需要将新加入的数和队列尾的数字比较，扔掉比新加入的数小的数，因为它们不可能是滑窗内的最大值。
当下标大于等于k-1的时候，说明要开始记录滑窗的最大值了。
class Solution { public: vector<int> maxSlidingWindow(vector<int>& nums, int k) { vector<int> res; deque<int> q; for(int i = 0; i < nums.size(); i++) { if(!q.empty() && i - q.front() >= k) q.pop_front(); while(!q.empty() && nums[q.back()] <= nums[i]) q.pop_back(); q.push_back(i); if(i >= k - 1) res.push_back(nums[q.front()]); } return res; } };