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?
单调队列。
1 class Solution { 2 public: 3 vector<int> maxSlidingWindow(vector<int>& nums, int k) { 4 vector<int> res; 5 if (k == 0 || nums.size() < k) return res; 6 deque<int> que; 7 for (int i = 0; i < k; ++i) { 8 while (!que.empty() && que.back() < nums[i]) que.pop_back(); 9 que.push_back(nums[i]); 10 } 11 res.push_back(que.front()); 12 for (int i = k; i < nums.size(); ++i) { 13 if (que.front() == nums[i - k]) que.pop_front(); 14 while (!que.empty() && que.back() < nums[i]) que.pop_back(); 15 que.push_back(nums[i]); 16 res.push_back(que.front()); 17 } 18 return res; 19 } 20 };