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一、直接设置系统环境变量

1. 临时代理

在终端中输入

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export http_proxy="http://127.0.0.1:1087"   
export https_proxy="http://127.0.0.1:1087"
或者直接
export all_proxy="http://127.0.0.1:1087"// 这是直接设置了http/https省得麻烦，当然你要是有分开设置的需求就分开设置

这是代理本身走的https/http

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Docker是目前最具代表性的容器技术之一，对云计算及虚拟化技术产生了颠覆性的影响。本文对Docker容器在应用中可能面临的安全问题和风险进行了研究，并将Docker容器应用环境中的安全机制与相关解决方案分为容器虚拟化安全、容器安全管理、容器网络安全三部分进行分析。

一、从虚拟化安全到容器安全

1、传统虚拟化技术

虚拟化技术是实现硬件基础设施资源的充分利用、合理分配和有效调度的重要技术手段。例如，在基于OpenStack的典型IaaS服务中，云服务提供商可通过搭建设备集群建立资源池，并将服务器、存储和网络等底层资源进行弹性虚拟化提供给租户。

传统虚拟化技术以虚拟机为管理单元，各虚拟机拥有独立的操作系统内核，不共用宿主机的软件系统资源，因此具有良好的隔离性，适用于云计算环境中的多租户场景。

2、容器技术

容器技术可以看作一种轻量级的虚拟化方式，将应用与必要的执行环境打包成容器镜像，使得应用程序可以直接在宿主机（物理机或虚拟机）中相对独立地运行。容器技术在操作系统层进行虚拟化，可在宿主机内核上运行多个虚拟化环境。相比于传统的应用测试与部署，容器的部署无需预先考虑应用的运行环境兼容性问题；相比于传统虚拟机，容器无需独立的操作系统内核就可在宿主机中运行，实现了更高的运行效率与资源利用率。

Docker是目前最具代表性的容器平台之一，它模糊了传统的IaaS和PaaS的边界，具有持续部署与测试、跨云平台支持等优点。在基于Kubernetes等容器编排工具实现的容器云环境中，通过对跨主机集群资源的调度，容器云可提供资源共享与隔离、容器编排与部署、应用支撑等功能。

Docker容器技术以宿主机中的容器为管理单元，但各容器共用宿主机内核资源，分别通过Linux系统的Namespaces和CGroups机制实现资源的隔离与限制。

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Description

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Note: Given n will be a positive integer.

Example 1:

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Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

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Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

Difficulty: Easy

Code:

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class Solution {
    public int climbStairs(int n) {
        
    }
}

题意

有楼梯台阶为n级，每次可以爬1或者2步台阶，问有多少种不同的方式到达台阶顶端。

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Description

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

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Input: 4
Output: 2

Example 2:

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Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.

Difficulty: Easy

Code:

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class Solution {
    public int mySqrt(int x) {
        
    }
}

题意

实现求x的平方根，x为非负整数，返回结果也为整数，小数位截取掉。

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Description

Given an array of words and a width maxWidth, format the text such that each line has exactly maxWidth characters and is fully (left and right) justified.

You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ‘ ‘ when necessary so that each line has exactly maxWidth characters.

Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line do not divide evenly between words, the empty slots on the left will be assigned more spaces than the slots on the right.

For the last line of text, it should be left justified and no extra space is inserted between words.

Note:

  • A word is defined as a character sequence consisting of non-space characters only.
  • Each word’s length is guaranteed to be greater than 0 and not exceed maxWidth.
  • The input array words contains at least one word.

Example 1:

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Input:
words = ["This", "is", "an", "example", "of", "text", "justification."]
maxWidth = 16
Output:
[
   "This    is    an",
   "example  of text",
   "justification.  "
]

Example 2:

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Input:
words = ["What","must","be","acknowledgment","shall","be"]
maxWidth = 16
Output:
[
  "What   must   be",
  "acknowledgment  ",
  "shall be        "
]
Explanation: Note that the last line is "shall be    " instead of "shall     be",
             because the last line must be left-justified instead of fully-justified.
             Note that the second line is also left-justified becase it contains only one word.

Example 3:

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Input:
words = ["Science","is","what","we","understand","well","enough","to","explain",
         "to","a","computer.","Art","is","everything","else","we","do"]
maxWidth = 20
Output:
[
  "Science  is  what we",
  "understand      well",
  "enough to explain to",
  "a  computer.  Art is",
  "everything  else  we",
  "do                  "
]

Difficulty: Hard

Code:

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class Solution {
    public List<String> fullJustify(String[] words, int maxWidth) {
        
    }
}

题意

在给定每行宽度为maxWidth实现文本左右对齐，空格要尽量均匀的分布。

若一行只有一个单词或者最后一行则左对齐，单词间隔为1个空格，没达到宽度则补全空格。

若空格在单词间不能均匀分布，则左侧的空格要比右侧多谢。

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