ARTS-第三周

Algorithm

二叉树深度 给定一个二叉树，找出其最大深度。

二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。

说明: 叶子节点是指没有子节点的节点。

示例： 给定二叉树 [3,9,20,null,null,15,7]，

3
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/
9 20 /
15 7 返回它的最大深度 3 。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int maxDepth(TreeNode root) {
        if(root == null){
            return 0;
        }
        
        int number = 1 ;
        int leftDepth = 0,rightDepth = 0;
        if(root.left != null){
            leftDepth = maxDepth(root.left);
        }
        
        if(root.right != null){
            rightDepth = maxDepth(root.right);
        }

        number += leftDepth > rightDepth ? leftDepth : rightDepth;
        return number;
    }
}
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Review

The gender gap is also a confidence gap link

Tip

以前有个朋友在写sql的时候碰到一个问题写sql查询的时候碰到了问题：

#sql1
select  u.real_name,u.phone,b.order_no,b.amount as borrow_amount,b.create_time as borrow_time,r.id,r.user_id,r.borrow_id,r.state,r.amount  as repay_amount,r.repay_time,r.penalty_amout,r.penalty_day,
         b.fee ,b.real_amount,channel.name as channel_name,cmro.state as allotState,u.living_img,u.front_img,u.back_img, r.remark
        from  cl_borrow_repay r  left join  cl_user_base_info u on  u.user_id=r.user_id  join cl_borrow b on r.borrow_id=b.id
         left join cl_user user2 on user2.id = u.user_id
         left join cl_channel channel on user2.channel_id = channel.id
         left join cl_manual_repay_order cmro on r.id = cmro.borrow_repay_id;

#sql2
select  u.real_name,u.phone,b.order_no,b.amount as borrow_amount,b.create_time as borrow_time,r.id,r.user_id,r.borrow_id,r.state,r.amount  as repay_amount,r.repay_time,r.penalty_amout,r.penalty_day,
         b.fee ,b.real_amount,channel.name as channel_name,cmro.state as allotState,u.living_img,u.front_img,u.back_img, r.remark
        from  cl_borrow_repay r  left join  cl_user_base_info u on  u.user_id=r.user_id  join cl_borrow b on r.borrow_id=b.id
         left join cl_user user2 on user2.id = u.user_id
         left join cl_channel channel on user2.channel_id = channel.id
         left join cl_manual_repay_order cmro on r.id = cmro.borrow_repay_id
ORDER BY r.id desc;
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sql1与sql2上基本（是否是应该用专业点的属于表示）没有改动，仅仅是在末位加上了order by ，可是最后的查询时间有着天壤之别。explain以后的结果以下 sql1:

sql2:

发如今Extra中多了Using temporary；Using filesort。 原来查询慢的缘由是由于使用orderby 后致使sql查询新增长了临时表及进行了文件排序。大量时间花在这上面了。 那么怎么优化？ 一、能够考虑在条件中把主键id加入进去左右一块儿的条件，由于若是order by 的字段也在where里面这样就不会进入文件排序 二、能够参考一次mysql 优化 （Using temporary ; Using filesort） link

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