LeetCode 239. Sliding Window Maximum

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?

这道题目非常难，需要用到deque

java

class Solution {    public int[] maxSlidingWindow(int[] nums, int k) {        if (nums == null || nums.length == 0 || nums.length < k || k <= 0) {            return new int[]{};        }        int[] result = new int[nums.length - k + 1];        int index = 0;        Deque<Integer> deque = new LinkedList<>();        for (int i = 0; i < k; i++) {            while (!deque.isEmpty() && nums[i] > nums[deque.peekLast()]) {                deque.pollLast();            }            deque.offerLast(i);        }        for (int i = k; i < nums.length; i++) {            result[index++] = nums[deque.peekFirst()];            while (!deque.isEmpty() && nums[i] > nums[deque.peekLast()]) {                deque.pollLast();             }            while (!deque.isEmpty() && deque.peekFirst() <= i - k) {                deque.pollFirst();            }            deque.offerLast(i);        }        result[index++] = nums[deque.peekFirst()];        return result;    }}