ARTS打卡—第3周

本周主要是做了一道关于链表相关操作的算法题目，数据库查询的题目，阅读了一篇关于 UI 自动化测试相关的英文文章，仔细学习了下 ConstraintLayout 相关的资料，后期要仔细总结一下相关的知识。

• Algorithm
  • 题目描述
  • Java
  • Python
• Database
• Review
• Tips
• Share

Algorithm

题目描述

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

> Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) > Output: 7 -> 0 -> 8 > Explanation: 342 + 465 = 807. >

题目相对比较简单，只是简单的链表操作。

Java

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    if (l1 == null || l2 == null) {
        return l1 == null ? l2 == null ? null : l2 : l1;
    }
    int upNum = 0;
    ListNode current = new ListNode(0);
    ListNode res = current;
    while (l1 != null || l2 != null) {
        int tmp = 0;
        if (l1 != null) {
            tmp += l1.val;
        }
        if (l2 != null) {
            tmp += l2.val;
        }
        tmp += upNum;
        ListNode tmpNode= new ListNode(tmp % 10);
        current.next = tmpNode;
        current = tmpNode;
        upNum = tmp / 10;
        if (l1 != null) {
            l1 = l1.next;
        }
        if (l2 != null) {
            l2 = l2.next;
        }
    }
    if (upNum != 0) {
        current.next = new ListNode(upNum);
    }
    return res.next;
}

Python

def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
    if not l1 or not l2:
        if l1:
            return l1
        elif l2:
            return l2
        else:
            return None
    current = ListNode(0)
    res = current
    upNum = 0
    while l1 or l2:
        tmp = 0
        if l1:
            tmp = tmp + l1.val
        if l2:
            tmp = tmp + l2.val
        tmp = tmp + upNum
        upNum = tmp // 10
        tmpNode = ListNode(tmp % 10)
        current.next = tmpNode
        current = tmpNode
        if l1:
            l1 = l1.next
        if l2:
            l2 = l2.next
    if upNum != 0:
        current.next = ListNode(upNum)
    return res.next

Database

181. Employees Earning More Than Their Managers

The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.

> +----+-------+--------+-----------+ > | Id | Name | Salary | ManagerId | > +----+-------+--------+-----------+ > | 1 | Joe | 70000 | 3 | > | 2 | Henry | 80000 | 4 | > | 3 | Sam | 60000 | NULL | > | 4 | Max | 90000 | NULL | > +----+-------+--------+-----------+ >

Given the Employee table, write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.

> +----------+ > | Employee | > +----------+ > | Joe | > +----------+ >

SELECT 
    a.Name AS Employee
FROM 
    Employee AS a,
    Employee AS b
WHERE 
    a.ManagerId = b.Id AND a.Salary > b.Salary

用到的方法就是是把表查询两遍，通过对比两个表的数据来确定最后的结果。

Review

本周主要学习了和 UI 测试相关的文章，文章中用到了 Espress、Mock 框架，对异步刷新 UI 数据的情况进行了测试方法的介绍。

Espresso UI Testing With Android Architecture Components

Tips

1. python 取余%，这一点和java一样，取整数\\,和java有所差异。
2. 谷歌提供的新的布局ConstraintLayout，虽然可以减少布局层次，但是这个新布局measure过程相对传统布局来说较长，尤其是该布局用在ListItem的自定义 View 时，启动速度会较慢。

Share

一篇约束布局相关的文章：https://blog.csdn.net/guolin_blog/article/details/53122387