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康波周期学习

发表于 2019-07-03 | 分类于 经济学
| 字数统计 937 | 阅读时长 3

康波期是世界经济运动中最长的周期,全称康德拉季耶夫周期,一个循环60年,分为回升、繁荣、衰退、萧条。

康德拉季耶夫认为市场、政策、资源等匀速不是决定长期经济周期的主要因素,因科学技术提高的生产力才是长周期经济的原动力。

科学原理提出到大规模应用需要十年以上的时间,随之而来的是经济十数年的繁荣,科技发展速度跟不上,经济进入衰退和萧条期,而下一代科技往往出现在衰退和萧条期。

重要经济轴周期理论开创者是两个,第一个康波周期,实际上它是全球经济运动的决定力量,也是在座各位人生财富规划的根本理论。

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InterviewBit-Day9

发表于 2019-06-27 | 分类于 InterviewBit
| 字数统计 700 | 阅读时长 4

Practice

Length of Last Word

Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string.

If the last word does not exist, return 0.

Note: A word is defined as a character sequence consists of non-space characters only.

Example:

Given s = "Hello World",

return 5 as length("World") = 5.

Please make sure you try to solve this problem without using library functions. Make sure you only traverse the string once.

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InterviewBit-Day9

发表于 2019-06-26 | 分类于 InterviewBit
| 字数统计 505 | 阅读时长 3

Practice

Palindrome String

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

Example:

"A man, a plan, a canal: Panama" is a palindrome.

"race a car" is not a palindrome.

Return 0 / 1 ( 0 for false, 1 for true ) for this problem

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面试-翼果科技

发表于 2019-06-24 | 分类于 面试
| 字数统计 832 | 阅读时长 2

翼果(深圳)科技有限公司

介绍

翼果(深圳)科技有限公司成立于2015年,公司核心成员来自华为等知名企业,主要合伙人具有500强企业10年以上工作经验和海外工作经历,团队技术实力强,具备国际化视野。

公司研发由多名顶级技术专家领衔,产品和技术氛围浓郁,作风务实,注重研发基础能力建设和技术积累,员工技能培训完善,技术能力成长快。公司已与众多企业客户建立业务合作关系,并获得知名机构投资,业务和团队处于蓬勃发展期,员工职业晋升快,发展空间大。

公司人际关系简单,倡导多劳多得和责任结果导向,实行骨干员工全员持股,提供具备行业竞争力的综合福利待遇。

如果您有梦想、有意愿、有能力,希望通过自己的拼搏获得快速的成长和丰厚的回报,我们热切欢迎您加入我们的队伍,共同奋斗成长!

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InterviewBit-Day8

发表于 2019-06-18 | 分类于 InterviewBit
| 字数统计 377 | 阅读时长 2

Practice

Greatest Common Divisor

Given 2 non negative integers m and n, find gcd(m, n)

GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n.
Both m and n fit in a 32 bit signed integer.

Example

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m : 6
n : 9

GCD(m, n) : 3

NOTE : DO NOT USE LIBRARY FUNCTIONS

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Oauth2 初步

发表于 2019-06-18 | 分类于 Oauth2
| 字数统计 905 | 阅读时长 4

搭建一个最简单的 Oauth2 认证服务

基于 Springboot2,在 pom.xml 中引入 Oauth2:

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<dependency>
<groupId>org.springframework.security.oauth</groupId>
<artifactId>spring-security-oauth2</artifactId>
<version>2.3.3.RELEASE</version>
</dependency>

<!-- 完成认证服务器的自动配置 -->
<dependency>
<groupId>org.springframework.security.oauth.boot</groupId>
<artifactId>spring-security-oauth2-autoconfigure</artifactId>
<version>2.1.5.RELEASE</version>
</dependency>
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Vue前端开发流程总结

发表于 2019-06-17 | 分类于 Vue教程
| 字数统计 278 | 阅读时长 1

Vue 开发流程总结

由于主要做后端开发,对于前端开发不怎么熟悉,在最近学习 vue 后,下载了一个 vue-admin 快速开发框架,框架中把基本需要的组件都集成好了,在实现功能的时候,对相关的页面进行一定的修改就能达到目的。

由此,各个不同业务对应的修改流程是差不多的,这里做一个小的总结。

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InterviewBit-Day7

发表于 2019-06-17 | 分类于 InterviewBit
| 字数统计 580 | 阅读时长 3

Practice

MAXSPPROD

You are given an array A containing N integers. The special product of each i^th^ integer in this array is defined as the product of the following:

  1. LeftSpecialValue: For an index i, it is defined as the index j such that A[j]>A[i] (i>j). If multiple A[j]’s are present in multiple positions, the LeftSpecialValue is the maximum value of j.

  2. RightSpecialValue: For an index i, it is defined as the index j such that A[j]>A[i] (j>i). If multiple A[j]s are present in multiple positions, the RightSpecialValue is the minimum value of j.

Write a program to find the maximum special product of any integer in the array.

Input: You will receive array of integers as argument to function.

Return: Maximum special product of any integer in the array modulo 1000000007.

Note: If j does not exist, the LeftSpecialValue and RightSpecialValue are considered to be 0.

Constraints 1 <= N <= 10^5^ 1 <= A[i] <= 10^9^

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InterviewBit-Day6

发表于 2019-06-14 | 分类于 InterviewBit
| 字数统计 528 | 阅读时长 3

Practice

Set Matrix Zeros

Given an m x n matrix of 0s and 1s, if an element is 0, set its entire row and column to 0.

Do it in place.

Example

Given array A as

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1 0 1
1 1 1
1 1 1

On returning, the array A should be :

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0 0 0
1 0 1
1 0 1
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InterviewBit-Day5

发表于 2019-06-13 | 分类于 InterviewBit
| 字数统计 525 | 阅读时长 3

Practice

Merge Intervals

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:

Given intervals [1,3],[6,9] insert and merge [2,5] would result in [1,5],[6,9].

Example 2:

Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] would result in [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

Make sure the returned intervals are also sorted.

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