import java.util.ArrayList;
public class Solution {
    public ArrayList<Integer> maxInWindows(int [] num, int size)
    {    
        //结果集
        ArrayList<Integer> res = new ArrayList<>();
        
        if(num.length < size || size <= 0) return res;
        
        int left = 0;
        int right = size - 1;
        while(right < num.length){
            //先将窗口最左边的元素 给到max
            int max = num[left];
            //遍历窗口的每一个元素，找到最大值
            for(int i = left;i <= right; i++){
               if(max < num[i]){
                   max = num[i];
               }
            }
            //将当前窗口的最大值加入结果集
            res.add(max);
            //整体往前滑
            left++;
            right++;
        }
        return res;
    }
}

这样的话每次都要在窗口中找最大值，时间复杂度较高，剑指大神 Krahets 给出了O（1）时间的效率，截图如下

https://leetcode-cn.com/problems/hua-dong-chuang-kou-de-zui-da-zhi-lcof/solution/mian-shi-ti-59-i-hua-dong-chuang-kou-de-zui-da-1-6/