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
Solution 1: brute force O(NK)
class Solution { public int[] maxSlidingWindow(int[] nums, int k) { int len = nums.length; int[] res = new int[len - k + 1]; for (int i = 0; i < res.length; i++) { int tmpMax = nums[i]; for (int j = i; j < i + k; j++) { tmpMax = Math.max(tmpMax, nums[j]); } res[i] = tmpMax; } return res; } }
Solution 2: Deque O(N)
class Solution { public int[] maxSlidingWindow(int[] nums, int k) { int[] res = new int[nums.length - k + 1]; Deque<Integer> deque = new LinkedList<>(); for (int i = 0; i < nums.length; i++) { while (!deque.isEmpty() && deque.peekFirst() == i - k) { deque.pollFirst(); } while (!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) { deque.pollLast(); } deque.offerLast(i); if (i - k + 1 >= 0) { // from big to small, left to right res[i - k + 1] = nums[deque.peekFirst()]; } } return res; } }