发表于
2017-01-16 10:58:13
[原]cognitive services系列 -- 使用微软机器学习cognitive-services中的文本分析模块
发表于
2017-01-16 10:53:15
[原]使用azure machine learning studio+神经网络回归完成晚点航班预测
发表于
2017-01-16 10:49:43
[原]使用azure machine learning studio+多元神经网络学习完成血型预测
发表于
2017-01-07 11:47:03
[原]Azure Bot中的Attribute+Enum+流畅接口实现
在玩bot framework时，发现他的formflow这部分的设计不错，尤其是attribute和enum的结合使用非常完美。以下为其代码设计： public enum SandwichOptions
{
BLT, BlackForestHam, BuffaloChicken, ChickenAndBaconRanchMelt, ColdCutCombo, MeatballMarinara,
OvenRoastedChicken, RoastBeef, RotisserieStyleChicken, SpicyItalian, SteakAndCheese, SweetOnionTeriyaki, Tuna,
TurkeyBreast, Veggie
};
public enum LengthOptions { SixInch, FootLong };
public enum BreadOptions { NineGrainWheat, NineGrainHoneyOat, Italian, ItalianHerbsAndCheese, Flatbread };
public enum CheeseOptions { American, MontereyCheddar, Peppe
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(13)
发表于
2017-01-05 22:34:37
[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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(9)
发表于
2017-01-05 22:33:48
[原]LeetCode -- Rotate Function
题目描述：Given an array of integers A and let n to be its length.Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow:F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1].Calculate the maximum value of F(0), F(1), ..., F(n-1).Note:n is guaranteed to be less than 105.Example:A = [4, 3, 2, 6]F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23F(3) = (0 * 3) + (1 * 2) + (2
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(9)
发表于
2017-01-05 22:32:52
[原]LeetCode -- Wiggle Sort II
题目描述：Given an unsorted array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3]....Example:(1) Given nums = [1, 5, 1, 1, 6, 4], one possible answer is [1, 4, 1, 5, 1, 6]. (2) Given nums = [1, 3, 2, 2, 3, 1], one possible answer is [2, 3, 1, 3, 1, 2].Wiggle排序II，给定数组，进行排序后，要求生成的序列An满足：A1<A2>A3<A4...>An-1<An，可假设有解。思路：对数组排序，使用两个索引，一个从中间开始:i，一个从最后:j，向前依次遍历即可。实现代码：public class Solution {
public void WiggleSort(int[] nums) {
if(nums == null){
return;
}
var len = nums.Length;
if(len == 1){
return ;
}
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(7)
发表于
2017-01-05 22:31:57
[原]LeetCode -- Remove Duplicate Letters
题目描述：Given a string which contains only lowercase letters, remove duplicate letters so that every letter appear once and only once. You must make sure your result is the smallest in lexicographical order among all possible results.Example:Given "bcabc"Return "abc"Given "cbacdcbc"Return "acdb"移除字符串中所有重复字符，字符串只包含小写字母。返回字符只出现一次组成的字符串。但题目有个前提，需要保持字符原本的相对位置。 （本题不至于简单到只使用一个大小为26的标记数组一次遍历解决）思路：使用字典存每个字符的出现次数。遍历字符串，使用栈来维护字符顺序。栈底元素最小，栈顶元素最大。如果栈顶元素小于当前字符，字典字符次数-1，继续遍历；如果栈顶元素大于当前字符，循环弹出栈元素，直到栈顶元素小于当前字符或字符在字典中没有“库存”为止。反序遍历栈就是结果。实现代码：public class Solution {
publi
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(6)
发表于
2017-01-05 22:31:05
[原]LeetCode -- Divide Two Integers
题目描述Divide two integers without using multiplication, division and mod operator.If it is overflow, return MAX_INT.整除两个整数，不能使用乘除，取模。思路：题目考的是位运算，求出除数的最大位移数max，使用被除数-1<<max，使用剩余的数继续计算。注意特殊值的情况。public class Solution {
public int Divide(int dividend, int divisor) {
if(dividend == 0){
return 0;
}
if(divisor == 0){
return dividend > 0 ? int.MaxValue : int.MinValue;
}
long x1 = dividend;
long x2 = divisor;
var negtive = false;
if(x1 < 0){
x1 = -x1;
negtive = !negtive;
}
if(x2 < 0){
x2 = -x2;
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(7)
发表于
2017-01-05 22:29:46
[原]使用bot service创建azure bot
发表于
2017-01-02 21:58:42
[原]SQL -- 使用PARTITION和RANK函数完成统计
SQL -- 使用PARTITION和RANK函数完成统计有时需要对查询结果集进行内部分组，或统计排名，这时PARTITION和RANK非常有用。下例以学生成绩表为例，演示一下用法：1 创建测试表：CREATE TABLE [dbo].[StudentScore](
[Id] [int] NOT NULL,
[Name] [nvarchar](20) NULL,
[ClassId] [int] NOT NULL,
[Subject] [nvarchar](20) NULL,
[Score] [decimal](18, 0) NULL
) ON [PRIMARY]
GO表说明：name： 学生名字classId： 班级idsubject： 科目score ： 分数查询实例1： 统计学生总分，科目总分，班级总分。-- use SUM + Partition to stastic total score
select s.Name,classId,
SUM(s.Score) OVER (PARTITION BY name) as StudentTotalScore,
SUM(s.Score) OVER (PARTITION BY Subject) as SubjectTotalScore,
SUM(s.Score) OVER (PARTITION
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(3)
发表于
2017-01-02 21:20:51
[原]Sql Server递归查询实例
有时需要完成递归查询，即不断对数据库进行查询，后一个查询条件需要使用前一个查询结构来构建。这时就需要使用CTE来完成。1. 创建测试表：CREATE TABLE NameList(
[Id] [int] NOT NULL,
[ParentId] [int] NOT NULL,
[Name] [nvarchar](20) NULL
) ON [PRIMARY]说明：为演示递归查询，需要创建一个父子表。id： 主键parentId ： 父idname： 名称查询：从根节点（没有父节点）出发，显示出每行的层数。由于这个查询使用前一个查询结果来构造查询语句，因此需要使用CTE来完成递归查询。declare @id int
set @id=1;
with CTE as (
select id,name, parentid, 0 as Level
from namelist
where ParentId=@id
UNION ALL
-- Recursive
select n.id,n.name, n.parentid, Level+1 as Level
from namelist n
join CTE on n.ParentId=cte.id
)
select * from CTE
option(MAXRECURSION 500)上例的递归层数设为500
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(5)
发表于
2016-12-29 12:22:46
[原]JS Scheduler - DayPilot
JS Scheduler - DayPilot 库链接：http://javascript.daypilot.org/demo/scheduler/demo代码(保存html即可运行)：<!DOCTYPE html>
<html>
<head>
<link type="text/css" rel="stylesheet" href="http://javascript.daypilot.org/demo/media/layout.css" />
<link type="text/css" rel="stylesheet" href="http://javascript.daypilot.org/demo/themes/calendar_g.css" />
<link type="text/css" rel="stylesheet" href="http://javascript.daypilot.org/demo/themes/calendar_green.css" />
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(4)
发表于
2016-12-29 10:20:19
[原]使用pako.js实现gzip的压缩和解压
<script type="text/javascript" src="pako.js"></script>
<script >
var b64Data = 'H4sIAAAAAAAAAJ3UMQ7CMAwF0KugP2ewEzdpcxXUAbWAOiHUMqCqdyeVQAobfGXIYL8hP5ZXnEdkeNEk6vUgXTbLonC4zMjHFY/5Wm511ekdTsOCLKVp2rlIKOA2jTuBot/cr7BhobEwsbAloY8kDGyqoQ5H/oHsdwQ21cCmaspCz0L2jcYOgLHhNGw4TT1yVmBpuS9PZHWY35siqnxvimEvpE9FY4peQhfbhO0FDnuFqWAEAAA=';
var ticketData = '{"ed":"20170121 09:44:01","fs":[{"usg":[1,1,1,1,1,1,1],"act":0,"fid":"003","oids":["1"]},{"usg":[1,1,1,1,1,1,
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(5)
发表于
2016-12-26 16:20:53
[原]LeetCode -- Integer Break
题目描述：Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get.For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4).Note: You may assume that n is not less than 2 and not larger than 58.给定一个数字，将其拆分为a1+a2+...an的和，使得a1*a2...an最大。一道数学题目。本题主要是找规律。实现代码以及分析过程：public class Solution {
public int IntegerBreak(int n) {
if(n < 2){
throw new ArgumentException("n should be more than 2");
}
if(n == 2){return 1
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(5)
发表于
2016-12-26 16:19:37
[原]LeetCode --Shortest Palindrome
题目描述：Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation.For example:Given "aacecaaa", return "aaacecaaa".Given "abcd", return "dcbabcd".给定字符串s，如果不是回文，添加最短前缀，使得s成为回文。思路：两个索引两边扫i,j：扫描长度为len=s.Length-1. i=0,j=len如果不相等，i++,j--如果相等，len-1。i与j分别重置为i和len最后将[len+1,s.Length]之间的长度截取，旋转作为前缀。实现代码：public class Solution {
public string ShortestPalindrome(string s)
{
if(string.IsNullOrEmpty(s)){
return s;
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(2)
发表于
2016-11-24 22:23:49
[原]LeetCode -- Divide Two Integers
Divide Two Integers题目描述Divide two integers without using multiplication, division and mod operator.If it is overflow, return MAX_INT.整除两个整数，不能使用乘除，取模。思路：题目考的是位运算，求出除数的最大位移数max，使用被除数-1<<max，使用剩余的数继续计算。注意特殊值的情况。public class Solution {
public int Divide(int dividend, int divisor) {
if(dividend == 0){
return 0;
}
if(divisor == 0){
return dividend > 0 ? int.MaxValue : int.MinValue;
}
long x1 = dividend;
long x2 = divisor;
var negtive = false;
if(x1 < 0){
x1 = -x1;
negtive = !negtive;
}
if(x2 < 0){
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(10)
发表于
2016-11-17 22:28:40
[原]Java Gzip类 - base64压缩和解压
public final class ZipUtil
{
public static String CompressToBase64(String string){
try {
ByteArrayOutputStream os = new ByteArrayOutputStream(string.length());
GZIPOutputStream gos = new GZIPOutputStream(os);
gos.write(string.getBytes());
gos.close();
byte[] compressed = os.toByteArray();
os.close();
String result = Base64.encodeToString(compressed, Base64.DEFAULT);
return result;
} catch (IOException e) {
e.printStackTrace();
}
catch (Exception
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(5)
发表于
2016-11-17 22:25:21
[原]C# Windows form application 播放小视频
1. 下载direcly-show lib DLL点击打开链接2. DxPlay.cs （可以在下载的例子中找到）: public class DxPlay : IDisposable
{
enum GraphState
{
Stopped,
Paused,
Running,
Exiting
}
#region Member variables
// File name we are playing
private string m_sFileName;
// graph builder interfaces
private IFilterGraph2 m_FilterGraph;
private IMediaControl m_mediaCtrl;
private IMediaEvent m_mediaEvent;
// Used to grab current snapshots
ISampleGrabber m_sampGrabber = null
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(7)
发表于
2016-11-17 22:23:29
[原]Android -- 读取NFC卡号
1. menifest文件中需要添加：<uses-permission android:name="android.permission.NFC" />
<uses-feature
android:name="android.hardware.nfc"
android:required="true" />在activity中添加：<intent-filter>
<action android:name="android.nfc.action.TAG_DISCOVERED" />
<category android:name="android.intent.category.DEFAULT" />
</intent-filter>2. 初始化时代码： Intent nfcIntent = new Intent(this, getClass());
nfcIntent.addFlags(Intent.FLAG_ACTIVITY_SINGLE_TOP);
mPendingIntent =
PendingIntent.getActivity(this, 0,nfc
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(9)
发表于
2016-11-17 22:20:37
[原]Android -- 打开时隐藏软键盘
当activity中有edittext时，软键盘会自动弹出，有时希望隐藏软键盘（并保留edittext的focus，比如扫码）可以在activity onCreate时，使用以下代码来实现：this.getWindow().setSoftInputMode(WindowManager.LayoutParams.SOFT_INPUT_STATE_ALWAYS_HIDDEN);
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(4)
发表于
2016-11-17 22:19:29
[原]android - asynTask中打开多选alertDialog
private class xxxAsyncTask extends AsyncTask<String, Void, Boolean> {
...
AlertDialog.Builder alertBuilder ;
...
@Override
protected void onPreExecute() {
...
alertBuilder = new AlertDialog.Builder(ResultActivity.this);
...
}
@Override
protected Boolean doInBackground(String... params) {
...
}
@Override
protected void onPostExecute(Boolean result) {
super.onPostExecute(result);
alertBuilder.setTitle("Choose Attraction");
... //
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(6)
发表于
2016-10-31 22:53:19
[原]leetcode -- Count Numbers with Unique Digits
题目描述：Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.Example:Given n = 2, return 91. (The answer should be the total numbers in the range of 0 ≤ x < 100, excluding [11,22,33,44,55,66,77,88,99])给定数位n，返回n位数字集合中，数位各自不重复的数字。分析：使用a[n]数组保存n位数字时多少数字不重复。a[1] = 10a[2] = 9*9a[3] = 9*9*8a[4] = 9*9*8*7 ....a[k] = 9*...*[11-k]此题为全排列问题，即对于给定数位X，a[x]为9*B[N].其中B[N] = [M*(M-1)*(M-2)....]，而M的值总为9。对于X，数列B[N]的个数为X-1实现：public int CountNumbersWithUniqueDigits(int n)
{
if(n == 0){
return 1;
}
if(n == 1){
return 10;
}
if(n == 2){
return 91;
}
v
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(9)
发表于
2016-10-31 22:32:34
[原]SQL server replication的三种方式
事务在初始化snapshot之后，一旦有数据修改，subscriber会从publisher收到实时的数据同步，事务一致性也能保证。事务类型的replication主要用于：单向同步。服务器-服务器这样的环境。订阅端数据增量同步。实时或低延时。快照定时对数据库进行快照并同步。适用场景：单向同步。对实时性没有要求。数据同步量较大，因此带宽适用很大。mid-night做数据同步。归并订阅者和发布者定时同步并合并数据。适用场景：（1个或多个）客户端-服务器环境的数据同步。离线模式的（若干）客户端数据与数据库进行同步（一旦连上网络环境）。数据Merge操作产生的冲突由Snapshot Agent和Merge Agent解决，需要配置rule。总结： 项目中事务类的数据同步使用较多；快照类在一些业务场景非常有用，做选择时还是要根据具体业务而定；需要注意的是在使用Merge同步时，数据冲突时如何解决，最好分析一下哪些列可能会出现同步，这样在定义rule时才能考虑全面。
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(16)
发表于
2016-10-31 22:30:50
[原]leet code - Third Maximum Number
题目描述Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).返回数组中第三大的数字，可能会包含重复。思路：直接遍历过程中存最大值求出结果。public class Solution {
public int ThirdMax(int[] nums)
{
if(nums == null || nums.Length == 0){
throw new ArgumentException();
}
if(nums.Length == 1){
return nums[0];
}
if(nums.Length == 2){
return Math.Max(nums[0],nums[1]);
}
int max = nums[0];
for(var i = 1;i < nums.Length; i++){
if(nums[i] > max){
max = nums[i];
}
}
int? sec
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(27)
发表于
2016-10-31 22:29:57
[原]LeetCode -- Range Sum Query - Immutable
LeetCode -- Range Sum Query - Immutable题目描述Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.Example:Given nums = [-2, 0, 3, -5, 2, -1]sumRange(0, 2) -> 1sumRange(2, 5) -> -1sumRange(0, 5) -> -3Note:You may assume that the array does not change.There are many calls to sumRange function.传入起始索引，返回数组求和的值。思路：使用缓存前n项的和。public class NumArray {
private int[] _cacheSum ;
private int[] _nums;
public NumArray(int[] nums) {
_nums = nums;
_cacheSum = new int[nums.Length];
var s = 0;
for (var i = 0;i < nums.Lengt
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(7)
发表于
2016-10-31 22:28:47
[原]leet code -- Swap Nodes in Pairs
题目描述：Given a linked list, swap every two adjacent nodes and return its head.For example,Given 1->2->3->4, you should return the list as 2->1->4->3.Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.Subscribe to see which companies asked this question就是成对的交换链表的相邻元素。比较直接的思路是两次遍历：第一次遍历使用数据结构存链表节点对（本例使用的是sector类）第二次遍历节点生成新链表并返回其头指针即可。/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int x) { val = x; }
* }
*/
publi
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(10)
发表于
2016-10-31 22:27:05
[原]LeetCode -- Bulb Switcher
题目描述：There are n bulbs that are initially off. You first turn on all the bulbs. Then, you turn off every second bulb. On the third round, you toggle every third bulb (turning on if it's off or turning off if it's on). For the ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb. Find how many bulbs are on after n rounds.Example:Given n = 3. At first, the three bulbs are [off, off, off].After first round, the three bulbs are [on, on, on].After second round, the three bulbs are [on, off, on].After third round, the three bulbs are [on, off, off]. So you s
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(9)
发表于
2016-10-29 19:41:29
[原]Windows 10 File Explorer打不开的问题解决方案
可能是VM的原因，在parallell里面的windows10打不开文件浏览器。1. 管理员身份运行 command2. netsh3. winsock reset4. 重启
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(17)
发表于
2016-10-26 22:30:21
[原]Android - popup text input
Android popup 输入文本框Button btnManual = (Button) findViewById(R.id.btnManual);
btnManual.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
AlertDialog.Builder alertDialog = new AlertDialog.Builder(ResultActivity.this);
alertDialog.setTitle("Ticket number");
alertDialog.setMessage("Enter ticket number");
final EditText txtManualTicketNo = new EditText(ResultActivity.this);
LinearLayout.LayoutParams lp = new LinearLayout.LayoutParams(
LinearLayout.LayoutParams.MATCH_PARENT
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