leetcode Sliding Window Maximum

A long array A[] is given to you. There is a sliding window of size w which is moving from the very left of the array to the very right. You can only see the wnumbers in the window. Each time the sliding window moves rightwards by one position. Following is an example:
The array is [1 3 -1 -3 5 3 6 7], and w is 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

Input: A long array A[], and a window width w
Output: An array B[], B[i] is the maximum value of from A[i] to A[i+w-1]
Requirement: Find a good optimal way to get B[i]

// Type your Java code and click the "Run Code" button!
// Your code output will be shown on the left.
// Click on the "Show input" button to enter input data to be read (from stdin).

public class Interpreter {
    public static void main(String[] args) {
        // Start typing your code here...
        int a[]={1,3,-1,-3,5,3,6,7};
        int b[]=new int[6];
        maxSlidingWindow(a,8,3,b);
        for(int i=0;i<6;i++)    System.out.println(b[i]);
    }
    public static void maxSlidingWindow(int a[],int n,int w,int b[]){
        if(n<=0||n<w) return;
        LinkedList<Integer> l=new LinkedList<Integer>();
        for(int i=0;i<w;i++){
            while(l.size()>0&&a[l.peekLast()]<=a[i]) l.pollLast();
            l.add(i);
        }
        b[0]=a[l.peekFirst()];
        for(int i=w;i<n;i++){
            while(l.size()>0&&l.peekFirst()<=i-w) l.removeFirst();
            while(l.size()>0&&a[l.peekLast()]<=a[i]) l.pollLast();
            l.add(i);
            b[i-w+1]=a[l.peekFirst()];
        }
        
    }
}