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.

滑动窗口的题目，我们可以借助双向队列deque来解决。一边维护窗口的大小，一边将每个窗口的中的最大元素放在队列的头部，这样每次取队列的第一个元素就可以了。代码如下：

public class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if(nums == null || nums.length == 0) return new int[0];
        int[] result = new int[nums.length - k + 1];
        Deque<Integer> deque = new LinkedList<Integer>();
        int index = 0;
        for(int i = 0; i < nums.length; i++) {
            //查看之前的最大元素是否在窗口中，如果不在就删除
            if(!deque.isEmpty() && (deque.getFirst() == (i - k)))
                deque.removeFirst();
            
            //添加新的元素到队尾
             while(!deque.isEmpty() && nums[deque.getLast()] <= nums[i]) 
                deque.removeLast();
            deque.addLast(i);  
            
            //将最大元素加入到结果中
            if(i >= k - 1) result[index ++] = nums[deque.getFirst()];
        }
        return result;
    }
}

如果不用队列也可以，思路是一样的，维护一个窗口，当窗口移动之后，检查之前窗口的最大元素是否在移动后的窗口中，如果在只需要比较最大元素与新近的元素就可以；如果不存在就在新窗口中找最大元素。代码如下：

public class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if(nums == null || nums.length == 0) return new int[0];
        int[] result = new int[nums.length - k + 1];
        int maxIndex = 0;
        int max = Integer.MIN_VALUE;
        int index = 0;
        for(int i = 0; i < result.length; i++) {
            if(i == 0 || maxIndex == i - 1) {
                max = Integer.MIN_VALUE;
                for(int j = i; j < i + k; j++) 
                    if(nums[j] > max) {
                        max = nums[j];
                        maxIndex = j;
                    }
            }else {
                    if(nums[i + k - 1] > max) {
                        max = nums[i + k - 1];
                        maxIndex = i + k - 1;
                    }
                }
            result[index ++] = max;
        }
        return result;
    }
}