1. Description

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.

Note: You may assume k is always valid, ie: 1 ≤ k ≤ input array's size for non-empty array.

2. 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].

3. Code

import java.util.Arrays;
import java.util.LinkedList;

public class LeetCode0239 {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if(k == 0 || nums == null || nums.length == 0){
            return new int[0];
        }
        int[] result = new int[nums.length - k + 1];

        LinkedList<Integer> deque = new LinkedList<Integer>();

        for(int i = 0; i< nums.length; i++){

            if(!deque.isEmpty()&&deque.peekFirst()==i-k){
                deque.pollFirst();
            }

            while(!deque.isEmpty() && nums[deque.peekLast()] < nums[i]){
                deque.removeLast();
            }

            deque.offer(i);

            if(i >= k-1){
                result[i - k + 1] = nums[deque.peekFirst()];
            }
        }
        return result;
    }

    public static void main(String[] args) {
        LeetCode0239 leetcode = new LeetCode0239();
        int[] nums = new int[] {1,3,1,2,0,5};
        System.out.println(Arrays.toString(leetcode.maxSlidingWindow(nums, 3)));
    }
}