【ARTS】01_19_左耳听风-20190318~20190324

ARTS：

Algrothm: leetcode算法题目
Review: 阅读并且点评一篇英文技术文章
Tip/Techni: 学习一个技术技巧
Share: 分享一篇有观点和思考的技术文章

Algorithm

【leetcode】13. Roman to Integer

https://leetcode.com/problems/roman-to-integer/

1）problem

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.

Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

2）answer

预想结果；

 result = M + ( M - C )  + (C - X) + ( V - I) = 1000 + 900 + 90 + 4

由于罗马数字通常从左到右从最大到最小。

"MCMXCIV"的实际值应该是：

M = 1000，CM = 900，XC = 90，IV = 4。

具体计算是；

 result = M + ( M - C )  + (C - X) + ( V - I) = 1000 + 900 + 90 + 4

C 100 被放在M 1000前面的时候，应该用减数M-C，然后：

# result = M + C
# temp = M - C
# result =  M + C + temp = M + C + M - C = M + M

因为+C和-C抵消了，所以M+M=2000,和计划中的result = M + （M-C）就少了一个-C的步骤。这是错误的。

为了使当小数放在大数的结果正确，所以就要将-C的值补充回来。也就是 *2 的由来，才是正确的解法。

result += numral_map[s[i]] - 2 * numral_map[s[i-1]]

3）solution

class Solution(object):
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        numral_map ={ "I":1,"V":5, "X":10,  "L":50, "C":100, "D":500, "M":1000 }
        result = 0

        for i in range(len(s)):
            if i > 0 and numral_map[s[i]] > numral_map[s[i-1]]:
                result += numral_map[s[i]] - 2 * numral_map[s[i-1]]
            else:
                result += numral_map[s[i]]

        return result

Review

【漏洞挖掘】10个绕过反病毒的恶意用户技巧

1）场景

概述10个需要注意的问题。绕过杀毒软件

2）问题难点

10个绕过反病毒的恶意用户技巧

3）解决问题的方法

介绍
添加防病毒软件信任名单策略
通过GUI禁用反病毒
终止反病毒软件进程
停止并禁用反病毒服务
通过调试设置禁用反病毒软件
卸载反病毒软件
从UNC路径或可移动媒体执行（U盘）
从备用数据流执行
从DLL执行
从文件系统外部执行
总结

4）方法细节

10个绕过反病毒的恶意用户技巧

https://www.cnblogs.com/17bdw/p/10575815.html

Tip

【安全开发】爬虫基础

1）场景

爬虫知识基础知识

2）问题难点

基础框架整理

3）解决思路

0x1、基础框架原理
1.1、爬虫基础
1.1、基础原理
1.2、发起HTTP请求-Request
1.3、获取响应内容-Response
1.4、练手库-Urllib

4）方法细节

爬虫基础

https://www.cnblogs.com/17bdw/p/10735127.html

【业务】极客时间-左耳听风-程序员攻略-软件设计

1）场景

软件设计学习

2）问题难点

软件设计学习的资源

3）解决思路

程序员练级攻略：软件设计

编程范式
一些软件设计的相关原则
一些软件设计的读物

4）方法细节

极客时间-左耳听风-程序员攻略-软件设计

https://www.cnblogs.com/17bdw/p/10591364.html