LeetCode [239] Sliding Window Maximum

239. Sliding Window Maximum

Hard

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. Return the max sliding window.

Follow up:
Could you solve it in linear time?

Example:

Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
Output: [3,3,5,5,6,7] 
Explanation: 

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

 

Constraints:

• 1 <= nums.length <= 10^5
• -10^4 <= nums[i] <= 10^4
• 1 <= k <= nums.length

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class Solution { public: vector<int> maxSlidingWindow(vector<int>& nums, int k) { vector<int> ret; deque<int> dq;//index. the corresponding numbers are decreasing from front to back int n = nums.size(); for(int i=0; i<n; ++i) { while(!dq.empty() && dq.front()<i-k+1) dq.pop_front(); while(!dq.empty() && nums[dq.back()]<=nums[i]) dq.pop_back(); dq.push_back(i); if(i>=k-1) ret.push_back(nums[dq.front()]); } return ret; } };

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class Solution { public int[] maxSlidingWindow(int[] nums, int k) { List<Integer> list = new ArrayList<>(); int n = nums.length; Deque<Integer> dq = new ArrayDeque<>(); for(int i=0; i<n; ++i){ if(!dq.isEmpty() && dq.getFirst()<i+1-k) dq.removeFirst(); while(!dq.isEmpty() && nums[dq.peekLast()]<=nums[i]){ dq.pollLast(); } dq.addLast(i); if(i>=k-1) list.add(nums[dq.peekFirst()]); } return list.stream().mapToInt(i->i).toArray(); } }