发表于
2017-03-26 16:04:43
[原].Net 代码性能 - (1)
提前准备测试环境。确保所有的bottleneck处理妥当。a)使用多线程来处理CPU bound问题b)使用异步解决I/O bound问题如果不确定，使用stopwatch测试代码段的运行时间。使用ants performance profiler工具对整体性能进行评价。http://www.red-gate.com/products/dotnet-development/ants-performance-profiler/使用as替代is。使用ajax和提前拉取数据提高用户体验。web服务返回文件时记得flush。对前端文件进行合并减少HTTP连接数。使用LINQ时确保知道生成的SQL是什么。并发简单的处理方式：异步，请求队列，无锁缓存。使用using及时释放资源。确保理解传递引用和传递值的区别，尤其对JIT动态编译时的执行过程的影响。单例与静态的合理使用。全局唯一VS线程无关。确保理解多线程环境下的锁，volatile的使用。最小化WEB请求，最小化数据库连接。IQuerable Vs IEnumerable的合理使用。继续build query vs 立刻拉数据到内存。EF查询时使用AsNoTracking提高性能。context.Entites.Where(x=>x.Name = "xxx").AsNoTracking().ToList();LINQ（SQL查询）时只查询关
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(11)
发表于
2017-03-17 10:10:24
[原]SQL SERVER 2014 进行Trouble Shooting的一些查询
将Isolate级别设为Read Uncommitted的两种方式 （无锁读，但允许读取脏数据）--giving it the NOLOCK hint.
SELECT *
FROM Production.Product WITH (NOLOCK)
WHERE ProductID = 1
...
--SET TRANSACTION ISOLATION LEVEL READ UNCOMMITTED
SELECT *
FROM Production.Product
WHERE ProductID = 1
...查询系统表SELECT * FROM sys.tables
SELECT * FROM INFORMATION_SCHEMA.TABLES查询计划SELECT cp.plan_handle, qp.query_plan, st.[text]
FROM sys.dm_exec_cached_plans cp
CROSS APPLY sys.dm_exec_query_plan(cp.plan_handle) AS qp
CROSS APPLY sys.dm_exec_sql_text(cp.plan_handle) AS st
WHERE qp.[dbid] = DB_ID('AdventureWorks2014')当前阻塞的会话SELECT *
FROM sys.dm_tran
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(25)
发表于
2017-03-02 00:16:29
[原]使用面向对象重构之-依赖注入
本文承接使用面向对象重构之-把抽象控制在一处：提高内聚性。上次遗留的问题是，使用了工厂模式重构之后，创建工厂对象的位置依然没有确定。...
public interface IStaffFactorywhere T: StaffBase{
T CreateStaff();
}
public class StaffFactory: IStaffFactory{
public LocalStaff CreateStaff();
}
public class TravelStaff: IStaffFactory{
public TravelStaff CreateStaff();
}
...本文要强调的就是依赖注入，和大多数项目一样，可以选择一种DI框架来完成依赖注入。一般DI的过程分两部分，一个是定义，一个是注入。几乎每个Prj都会选择程序启动的位置来做依赖注入。下面以.net中的Unit框架为例。1. 定义要做DI的接口public interface IStaffFactorywhere T: StaffBase{
T CreateStaff();
}
public class StaffFactory: IStaffFactory{
public LocalStaff CreateStaff();
}
public class TravelStaffFactory:
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(11)
发表于
2017-03-02 00:14:16
[原]使用面向对象重构之-把抽象控制在一处：提高内聚性
承接上文使用面向对象重构之-使用接口抽象完成不同维度的扩展。本文所介绍的内容可能不是面向对象编程所特有的，在其他编程（函数式或过程式）中也会常用的技巧——使用统一的抽象。上文遗留的问题就是，太多类型的Staff要创建：public class StaffBase{
...
}
public class LocalStaff :StaffBase{
public Staff(ITransportClaim transportClaimProvider){
...
}
}
public class OTStaff: LocalStaff{
public class OTStaff(ITransportClaim transportClaimProvider, IOverTimeFoodClaim otClaimProvider): base(transportClaimProvider){
...
}
}
public class TravelStaff :StaffBase{
public class TravelStaff(ITravelClaim travelClaimProvider){
}
}问题来了，哪个类要做这件事情？如果没有类来做这件事情的话，将会所有调用者的责任，即创建对象的代码(new XXStaff())将会到处都是。这看起来并不是我
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(19)
发表于
2017-03-02 00:11:35
[原]使用面向对象重构之-使用接口抽象完成不同维度的扩展
本文承接上文使用面向对象重构之-使用接口完成行为抽象。这次要面对的需求变化是报销流程不仅与员工类别相关，也与报销种类相关，每个报销种类的流程要去不同部门办理，流程差别较大，需要分别扩展。车费分为打车费和公共交通;打车费需要小票而公共交通则不需要；加班可以报销晚餐；出差的人可以有出差补助，不同城市也有不同的流程。为了让代码体系适应这个需求，就需要引入新的抽象，让代码依赖这个新的抽象体系。以下代码为之前的实现：public interface IClaim
{
void Claim();
}
public class StaffClaimProvider : IClaim
{
public void Claim()
{
throw new NotImplementedException();
}
}
public class SalesClaimProvider : IClaim
{
public void Claim()
{
throw new NotImplementedException();
}
}我们只是根据员工种类提供不同报销逻辑的provide
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(16)
发表于
2017-02-23 23:00:23
[原]使用面向对象重构之-使用接口完成行为抽象
本文承接上文使用面向对象重构之-继承中的抽象—模板方法。新需求来了，员工需要有报销功能，并且基础员工和经理报销的逻辑不太一样。直接的做法是延续现在的继承结构，把不一样的部分使用虚函数完成抽象。 public class Employee
{
...
private void CommonSteps()
{
...
}
public void Claim()
{
CommonSteps();
ExtraSteps();
}
public virtual void ExtraSteps()
{
...
}
}
public class Manager : Employee
{
public override void ExtraSteps()
{
...
}
}现在公司改制度了，报销是分类别的：基础报销，电话费。普通员工和经理的报销体制相同，而销售只可以报销电话费。如果不想改现在的结构，就至少
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(14)
发表于
2017-02-23 22:57:53
[原]使用面向对象重构之-继承中的抽象—模板方法
本文承接上文，使用面向对象重构之-从过程式设计到面向对象。新需求来了，Manager申请Leave的过程和普通员工不一样，这个时候就需要修改几处地方：1. 把ApplyLeave的这个函数virtual一下，abstract awaypublic class Employee
{
...
public virtual bool ApplyLeave(Leave leave)
{
...
}
...
}2. 在manager类中override public class Manager : Employee
{
...
public override bool ApplyLeave(Leave leave)
{
//... manager apply leave logic
}
...
}发现一个问题，申请leave的大部分流程的code是可以重用的，可以把不一样的提取出来，virtual, abstract away。这也是设计模式之一模板方法，在项目中对于简单直接的继承模型关系非常好用。public class Employee
{
...
private void ApplyLeaveStep1()
{
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(12)
发表于
2017-02-23 22:52:36
[原]使用面向对象重构之-从过程式设计到面向对象
写在前面最近两年接手的几个C#和java的project，发现目前很多程序员的编程思维中，几乎没有用到对象和抽象的力量，扩展和修改起来非常困难，甚至阅读都有困难。决定写一个面向对象基础专栏，笔者希望能把这几次重构中所发现的问题，尤其是涉及到面向对象几个重要的基础知识说明清楚，让初学者能够明白并应用到项目中去。本系列文章的所有示例代码为C#。第一步：从过程设计到对象以下是一种典型的面向过程设计的编程风格。类名以management，Processing为结尾的通常都是过程式设计产生的。 public class LeaveManagement
{
public IEnumerableGetLeavesByUserId(string userId)
{
...
}
public Leave GetLeave(Guid id)
{
...
}
public bool ApproveLeave(Leave leave)
{
...
}
public bool RejectLeave(Leave leave)
{
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(8)
发表于
2017-02-19 22:45:19
[原]LeetCode -- Wildcard Matching
题目描述：Implement wildcard pattern matching with support for '?' and '*'.'?' Matches any single character.'*' Matches any sequence of characters (including the empty sequence).The matching should cover the entire input string (not partial).The function prototype should be:bool isMatch(const char *s, const char *p)Some examples:isMatch("aa","a") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "*") → trueisMatch("aa", "a*") → trueisMatch("ab", "?*") → trueisMatch("aab", "c*a*b") → false实现字符串匹配。输入字符串s,和字符串匹配p，?可被替换为任何非空字符，*可替换空字符或任何字符。设，arr[i,j]表示s的第i个字符是否和p的第j个字符匹配。对于s[i]和p
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(22)
发表于
2017-02-19 22:44:10
[原]LeetCode -- First Unique Character in a String
题目描述：Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.Examples:s = "leetcode"return 0.s = "loveleetcode",return 2.求第一个不重复的字符。比较直接的做法是用哈希，由于题目提示了s中完全是小写字母，可以使用int[26]的数组来代替哈希。实现代码：public class Solution {
public int FirstUniqChar(string s) {
var arr = new int[26];
var len = s.Length;
for (var i = 0;i < len; i++){
arr[s[i]-97]++;
}
for (var i = 0;i < len; i++){
if(arr[s[i]-97] == 1){
return i;
}
}
return -1;
}
}
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(13)
发表于
2017-02-19 22:42:58
[原]LeetCode -- Arranging Coins
题目描述：You have a total of n coins that you want to form in a staircase shape, where every k-th row must have exactly k coins.Given n, find the total number of full staircase rows that can be formed.n is a non-negative integer and fits within the range of a 32-bit signed integer.Example 1:n = 5The coins can form the following rows:¤¤ ¤¤ ¤Because the 3rd row is incomplete, we return 2.其实就是求小于N的最大M，M为1+...+K。思路：遍历1到n/2即可。实现代码：public int ArrangeCoins(int n) {
if(n == 0){
return 0;
}
if(n == 1){
return 1;
}
var half =
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(12)
发表于
2017-02-19 22:41:32
[原]LeetCode -- Minimum Number of Arrows to Burst Balloons
题目描述：There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of start and end of the diameter suffice. Start is always smaller than end. There will be at most 104 balloons.An arrow can be shot up exactly vertically from different points along the x-axis. A balloon with xstart and xend bursts by an arrow shot at x if xstart ≤ x ≤ xend. There is no limit to the number of arrows that can be shot. An arrow once
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(17)
发表于
2017-02-19 22:35:19
[原]LeetCode -- Path Sum III
题目描述：You are given a binary tree in which each node contains an integer value.Find the number of paths that sum to a given value.The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.给定一个二叉树，遍历过程中收集所有可能路径的和，找出和等于X的路径树。思路：设当前节点为root,分别收集左右节点路径和的集合，merge到当前集合中；将当前节点添加到数组中，构成新的可能路径。实现代码：/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public Tre
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(14)
发表于
2017-02-19 22:30:00
[原]LeetCode -- Contains Duplicate II
题目描述：Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the difference between i and j is at most k.给定一个数组，遍历每个数，找出是否存在nums[i]和nums[j]相等并且i和j的距离小于等于k。思路：本题考查的还是哈希表的使用，直接用哈希表存每个数字的所有位置。实现代码：public class Solution {
public bool ContainsNearbyDuplicate(int[] nums, int k) {
if(nums == null || nums.Length == 0){
return false;
}
var dict = new Dictionary>();
for (var i = 0;i < nums.Length; i++){
if(!dict.ContainsKey(nums[i])){
dict.Add(nums[i], new List
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(17)
发表于
2017-02-19 22:29:10
[原]LeetCode -- Dungeon Game
题目描述：The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or
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(15)
发表于
2017-02-19 22:28:08
[原]LeetCode -- Range Sum Query 2D - Immutable
题目描述：Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).Range Sum Query 2DThe above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.Example:Given matrix = [ [3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4, 1, 0, 1, 7], [1, 0, 3, 0, 5]]sumRegion(2, 1, 4, 3) -> 8sumRegion(1, 1, 2, 2) -> 11sumRegion(1, 2, 2, 4) -> 12Note:You may assume that the matrix does not change.There are many calls to sumRegion function.Yo
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(21)
发表于
2017-02-19 22:27:03
[原]LeetCode -- Longest Palindrome
题目描述：Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.This is case sensitive, for example "Aa" is not considered a palindrome here.Note:Assume the length of given string will not exceed 1,010.Example:Input:"abccccdd"Output:7Explanation:One longest palindrome that can be built is "dccaccd", whose length is 7.给定一个字符串，利用字符串中的字符组成回文，返回最大回文长度。一开始以为本题是判断字符串存在的最大回文长度，准备使用分治。后来仔细看发现不是，直接使用哈希统计每个字符出现次数即可：如果为偶数：直接累加。如果是奇数，判断-1后是否为偶数，如果为偶数，累加这个偶数。标记是否出现过奇数（放在回文中间，结果累加1）实现代码：public class Solution {
public i
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(16)
发表于
2017-02-19 22:25:50
[原]LeetCode -- Russian Doll Envelopes
题目描述：You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope.What is the maximum number of envelopes can you Russian doll? (put one inside other)Example:Given envelopes = [[5,4],[6,4],[6,7],[2,3]], the maximum number of envelopes you can Russian doll is 3 ([2,3] => [5,4] => [6,7]).给定一个二维数组，对于[i,j]和[a,b]若满足in)；内层循环从后向前查找(j:i->0)：若a[i,0]>a[j,0] && a[0,i] > a[0,j]则取 Max(a[i]当前的结果,a[j]的结果+1)因此可使用DP，用result[i]来存a[i]的
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(13)
发表于
2017-02-19 22:24:31
[原]LeetCode -- Valid Perfect Square
题目描述：Given a positive integer num, write a function which returns True if num is a perfect square else False.Note: Do not use any built-in library function such as sqrt.Example 1:Input: 16Returns: TrueExample 2:Input: 14Returns: False本题属于数学分类。思路：任意完全平方数可通过1+3+5...+K的和得到。实现代码：public class Solution {
public bool IsPerfectSquare(int num) {
var start = 1;
while(num > 0){
num -= start;
start += 2;
}
return num == 0;
}
}
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(31)
发表于
2017-02-19 22:23:37
[原]LeetCode -- Count of Smaller Numbers After Self
题目描述：You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].Example:Given nums = [5, 2, 6, 1]To the right of 5 there are 2 smaller elements (2 and 1).To the right of 2 there is only 1 smaller element (1).To the right of 6 there is 1 smaller element (1).To the right of 1 there is 0 smaller element.Return the array [2, 1, 1, 0].给定一个数组a[n]，返回数组b[n]，每个元素对应当前元素a[i]右边小于a[i]的个数。思路：本题比较直接的实现是创建二叉搜索树。从右向左遍历数组a[n]，i∈[n-1,0]，创建二叉树。二叉树的每个节点node存放a[i]以及小于当前节点的元素个数。如果a[i]小于node
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(12)
发表于
2017-02-19 22:22:32
[原]关于面向对象中的代码审查（Code Review）
在进行面向对象编程语言的代码审查的工作中，对于一般项目而言，通常考虑以下几点是否满足：能否快速被理解（可读性）最重要的一点，主要考虑以下几个部分。命名以及代码格式是否整洁逻辑是否清晰注释：(仅当代码动机（why，NOT what ,NOT how）不确定的话需要写)单一职责代码职责是否清晰并单一类或方法是否太大需要拆分代码重复重复代码需要被提出来可测试性代码依赖是否可注入，每个功能是否容易做单元测试所有实现是否都是可替换异常处理异常有没有被handle并log性能代码会不会有性能问题（多线程环境或高并发情况）安全是否存在安全漏洞（网络，数据库）对代码质量要求更高的项目，可能还需考虑以下几点SOLID单一职责(Single Responsibility)开闭原则(Open close)里氏代换（Liskov substitution）对象的可替代性接口分离(Interface Segregation）对抽象进行细分。大接口拆分成多个功能更小的接口。依赖反转（Dependency Inverion）依赖抽象而非实现。GRASP原则Creator(对象的创建）对象统一创建（工厂）高内聚（高扇入）提高类（库）的被使用次数。低耦合（低扇出）对外部依赖较少（少于3个接口）Information Expert (信息专家)单一信息来源。每种信息的来源是一致的是否把相关的信息都放在了一个位置可扩展
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(10)
发表于
2017-02-19 22:16:36
[原]Android Studio 编译错误DuplicateFileException
在使用android studio编译遇到错误Execution failed for task ':bcms:transformResourcesWithMergeJavaResForDebug'.
> com.android.build.api.transform.TransformException: com.android.builder.packaging.DuplicateFileException: Duplicate files copied in APK META-INF/DEPENDENCIES
File1: /...libs/httpclient-4.5.2.jar
File2: /.../libs/httpcore-4.4.4.jar在project级别的gradle文件中添加：packagingOptions {
exclude 'META-INF/DEPENDENCIES.txt'
exclude 'META-INF/DEPENDENCIES'
exclude 'META-INF/dependencies.txt'
exclude 'META-INF/LICENSE.txt'
exclude 'META-INF/LICENSE'
exclude '
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(19)
发表于
2017-02-17 15:52:28
[译]【译】微软 azure -- 机器学习算法选择的步骤
原文：https://azuremlsimpleds.azurewebsites.net/simpleds/预测值实例场景：估计产品需求预测销售数据决定设备服务优先级算法选择：序数回归：类别是有序的泊松回归：预测事件次数快速森林分位数回归：预测一种分布线性回归：快速训练，针对线性模型贝叶斯线性回归：针对线性模型，小数据集神经网络回归：预测较精确，但训练时间长决策森林回归：较精确，训练快速，手工选择特征提升决策树回归：精确，训练快速，占用较大内存找出异常情况实例场景：预测信用风险欺诈检测捕获异常设备读取算法选择：一元SVM：100个特征以内基于PCA的异常检测：训练快速聚类（结构挖掘）实例场景：客户分类预测客户口味决定哪种产品会以同样的原因失败算法选择：K-Means两种类别中预测某类别实例场景：某观点是否乐观客户是否会续约哪种优惠券吸引更多客户算法选择：二元SVM：少于100个特征，线性模型二元平均感知器：快速训练，线性模型二元贝叶斯点机：快速训练，线性模型二元决策森林：精确，训练快速二元逻辑回归：训练快速，线性模型二元提升决策树：精确，训练快速，但占用较多内存二元决策丛林：精确，使用内存较少（相比决策森林）相对于决策树而言，解决了在训练数据非常大的情况下，节点指数增长的问题。更多二元决策丛林的资料二元本地深度SVM：少于100个特征二元神经网络：精确，但训练时间长多个类别中预测某
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(9)
发表于
2017-02-15 22:33:47
[原]C# 单元测试EntityFramework中的DbFunctions中的函数
C# 单元测试中mock EntityFramework中的DbFunctions中的函数1. 写一个挡板类来提供C#代码的实现public static class DbFunctionUnitTestable
{
////// This is for Unit Test.
/// When Running against DB ,EF will generate the SQL and ignore the implemenation
/// in this method . when calling from C#(Unit test) , will exeucte the logic below
/////////[System.Data.Entity.DbFunction("Edm", "TruncateTime")]
public static DateTime? TruncateTime(DateTime? dt)
{
if (dt.HasValue)
{
return dt.Value.Date;
}
return null;
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(14)
发表于
2017-02-06 09:55:33
[原].NET中两种OCR方式对比
1.使用 Tesseract Install-Package Tesseract2.从这里下载实例项目（为了获得训练数据）tessdata目录下的就是训练数据，等下初始化Tesseract引擎对象会使用。测试代码： private static void TesseractSample()
{
var testImagePath = "phototest.tif";
try
{
using (var engine = new TesseractEngine(@"..\..\tessdata", "eng", EngineMode.Default))
{
using (var img = Pix.LoadFromFile(testImagePath))
{
using (var page = engine.Process(img))
{
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(2)
发表于
2017-02-05 16:13:19
[原]SQL Server 2014 查询大全
发表于
2017-01-30 20:41:34
[原]Cognitive Services API 系列-- 使用计算机视觉api完成图片分类
发表于
2017-01-30 20:36:48
[原]VS one-click一键部署
发表于
2017-01-30 20:16:39
[原]使用PSExec时Access is denied 的解决方案
PsExec是微软提供的简单运维工具，可以方便的执行远程计算机命令。下载链接：安装psExechttps://technet.microsoft.com/en-us/sysinternals/pxexec.aspx在执行：psexec.exe \\192.168.1.101 -u [USER] -p [password] [some.exe] 时，遇到错误：“Access is denied”。可考虑在服务器端增加以下注册表项：HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows\CurrentVersion\Policies\System添加类型为DWORD名称为LocalAccountTokenFilterPolicy项，值为1。重启服务器。
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(4)
发表于
2017-01-21 15:59:32
[原]cognitive-services系列 -- 使用微软cognitive-services中的人脸识别模块
