ubuntu 一键更换软件源

使用 Ubuntu 可以直接在设置中测试最快的软件源并更换,但是有时候没有图形化界面,比如在 Server 版,或者 Docker File 中想要批量替换软件源,这时候可以使用 sed 命令,因为 sed 命令在大多数环境中不用安装都可运行,而在最简 Ubuntu Docker 镜像中是没有 vim 的。

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Lombok 笔记


Lombok 可以通过注解的形式,来帮助我们在开发的时候省略掉一些重复的工作。比如设置 get、set 方法,设置无参构造函数或者全参构造函数。

如果使用的是 IntelliJ IDEA,可以先安装 Lombok Plugin 插件。 最新版已经自带插件可以支持。

然后在项目中引入 jar 文件,如果使用 Maven 构建项目,可以通过在pom.xml中添加以下代码来引用。

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What's Next?


What’s Next?

Johnny is playing with a large binary number, B. The number is so large that it needs to be compressed into an array of integers, A, where the values in even indices (0, 2, 4,...) represent some number of consecutive 1 bits and the values in odd indices (1, 3, 5,...) represent some number of consecutive 0 bits in alternating substrings of B.

For example, suppose we have array A = {4, 1, 3, 2, 4}. A0 represents "1111", A1 represents "0", A2 represents "111", A3 represents "00", and A4 represents "1111". The number of consecutive binary characters in ith the substring of B corresponds to integer Ai, as shown in this diagram:

When we assemble the sequential alternating sequences of 1‘s and 0‘s, we get B = 11110111001111.

We define setCount(B) to be the number of 1‘s in a binary number, B. Johnny wants to find a binary number, D, that is the smallest binary number > B where setCount(B) = setCount(D). He then wants to compress D into an array of integers, C (in the same way that integer array A contains the compressed form of binary string B).

Johnny isn’t sure how to solve the problem. Given array A, find integer array C and print its length on a new line. Then print the elements of array C as a single line of space-separated integers.

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