<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/hellobinfeng</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; hellobinfeng]]></copyright><item><title><![CDATA[挑战最强大脑——来自全球的14个编码社区]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/42731035</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/42731035</guid><author>hellobinfeng</author><pubDate>Thu, 15 Jan 2015 02:23:41 +0800</pubDate><description><![CDATA[摘要：史蒂夫·乔布斯说过，每个人都应该学习给电脑编写程序的技术，因为这一过程能够教你如何去思考！文中搜集了14个不错的学习资源，帮助你挑战自我，领略并探索计算机领域无穷奥秘。史蒂夫·乔布斯说过，每个人都应该学习给电脑编写程序的技术，因为这一过程能够教你如何去思考！众所周知，编程已成为开发者生命中至关重要的一部分。很多事实表明，越来越多的人不管男女老少都将参与编程这个行业。 学习编程的渠道有很多种，]]></description><category></category></item><item><title><![CDATA[Find Minimum in Rotated Sorted Array II 旋转数组中找最小值（有重复元素） @LeetCode]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/41009717</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/41009717</guid><author>hellobinfeng</author><pubDate>Tue, 11 Nov 2014 12:19:53 +0800</pubDate><description><![CDATA[递归public class Solution {
    public int findMin(int[] num) {
        return helper(num, 0, num.length-1);
    }
    
    //with duplicate
    public static int helper(int[] a, int left, int right){
 ]]></description><category></category></item><item><title><![CDATA[Find Minimum in Rotated Sorted Array 旋转数组中找最小值 @LeetCode]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/41009239</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/41009239</guid><author>hellobinfeng</author><pubDate>Tue, 11 Nov 2014 11:41:30 +0800</pubDate><description><![CDATA[O(n)的算法就不说了，这题主要考查的是 O(logn)的算法。有序数组容易想到使用二分查找解决，这题就是在二分基础上做一些调整。数组只有一次翻转，可以知道原有序递增数组被分成两部分，这俩部分都是有序递增的（这题只需要考虑有序数组的递增情况）。假如翻转后的数组以第 x 个结点分为两部分 A[0..x] 和 A[x+1..n]。则 A[0..x] 这一段是有序递增的， A[x+1..m] 这一段也是]]></description><category></category></item><item><title><![CDATA[如何使破解网页的禁止复制黏贴]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/40476095</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/40476095</guid><author>hellobinfeng</author><pubDate>Sun, 26 Oct 2014 13:06:08 +0800</pubDate><description><![CDATA[有些网页无法复制黏贴，最好的解决办法是用Firefox，然后：输入 about:config查找 dom.event.clipboardevents.enabled双击把值改成 "false"]]></description><category></category></item><item><title><![CDATA[LeetCode 题目总结/分类]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/40292735</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/40292735</guid><author>hellobinfeng</author><pubDate>Mon, 20 Oct 2014 04:49:05 +0800</pubDate><description><![CDATA[注：此分类仅供大概参考，没有精雕细琢。有不同意见欢迎评论~利用堆栈：http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/http://oj.leetcode.com/problems/longest-valid-parentheses/ （也可以用一维数组，贪心）http://oj.leetcode.com/proble]]></description><category></category></item><item><title><![CDATA[Mergesort]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/40190799</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/40190799</guid><author>hellobinfeng</author><pubDate>Fri, 17 Oct 2014 20:37:14 +0800</pubDate><description><![CDATA[Mergesort:import java.util.*;

public class MergerSort
{
	public static void main(String[] args)
	{
		Integer[] a = {2, 6, 3, 5, 1};
		mergeSort(a);
		System.out.println(Arrays.toString(a));
	}

	publ]]></description><category></category></item><item><title><![CDATA[ls背后发生了什么 Down the 'ls' Rabbit Hole]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/38936381</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/38936381</guid><author>hellobinfeng</author><pubDate>Sat, 30 Aug 2014 01:58:27 +0800</pubDate><description><![CDATA[This article was written by Adam Fletcher (@adamfblahblah)FROM LS(1) TO THE KERNEL AND BACK AGAIN.Too often sysadmins are afraid to dive into the source code of our core utilities to see how they real]]></description><category></category></item><item><title><![CDATA[如何提升网站性能 Website Performance  Optimization]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/38936261</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/38936261</guid><author>hellobinfeng</author><pubDate>Sat, 30 Aug 2014 01:53:20 +0800</pubDate><description><![CDATA[This article was written by Bob Feldbauer (@bobfeldbauer)Optimizing website performance can dramatically increase conversions and loyalty, and decrease costs. Although it may be traditionally viewed a]]></description><category></category></item><item><title><![CDATA[一道小题：从一个数组里产生所有可能的乘积组合]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/38558785</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/38558785</guid><author>hellobinfeng</author><pubDate>Thu, 14 Aug 2014 15:10:09 +0800</pubDate><description><![CDATA[比如给定一个数组[2,3,11] 要求产生[1,2,3,6,11,22,33,66]观察可得：[2,3] 产生了[1,2,3,6] 的乘积可能。当加入11时，11会和现有的每一个元素都相乘得到[1,2,3,6,11,22,33,66]public static void allProducts(int[] arr) {
		List list = new ArrayList();
		list.a]]></description><category></category></item><item><title><![CDATA[Reverse Words in a String 翻转一个字符串里的单词顺序 @LeetCode]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/38531289</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/38531289</guid><author>hellobinfeng</author><pubDate>Wed, 13 Aug 2014 06:31:56 +0800</pubDate><description><![CDATA[LeetCode新题，但是比较简单，直接用栈即可Given an input string, reverse the string word by word.For example,Given s = "the sky is blue",return "blue is sky the".click to show clarification.Clarification:What constitut]]></description><category></category></item><item><title><![CDATA[Word Break II 求把字符串拆分为字典里的单词的所有方案 @LeetCode]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/38530983</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/38530983</guid><author>hellobinfeng</author><pubDate>Wed, 13 Aug 2014 05:16:27 +0800</pubDate><description><![CDATA[这道题类似  Word Break 判断是否能把字符串拆分为字典里的单词 @LeetCode 只不过要求计算的并不仅仅是是否能拆分，而是要求出所有的拆分方案。因此用递归。但是直接递归做会超时，原因是LeetCode里有几个很长但是无法拆分的情况，所以就先跑一遍Word Break，先判断能否拆分，然后再进行拆分。递归思路就是，逐一尝试字典里的每一个单词，看看哪一个单词和S的开头部分匹配，如果匹配则]]></description><category></category></item><item><title><![CDATA[MAC OSX 中解决编译“'cc' failed with exit status 1”错误  和clang: error: unknown argument 错误]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/37591051</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/37591051</guid><author>hellobinfeng</author><pubDate>Wed, 09 Jul 2014 04:44:33 +0800</pubDate><description><![CDATA[错误例子：error: command 'cc' failed with exit status 1clang: error: unknown argument: '-multiply_definedsuppress' [-Wunused-command-line-argument-hard-error-in-future]
clang: note: this will be a hard err]]></description><category></category></item><item><title><![CDATA[用递归翻转一个栈 Reverse a stack using recursion]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/37332009</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/37332009</guid><author>hellobinfeng</author><pubDate>Sun, 06 Jul 2014 14:41:08 +0800</pubDate><description><![CDATA[明确递归语句之前的语句都是顺序执行，而递归语句之后的语句都是逆序执行package recursion;

import java.util.Stack;

public class Reverse_a_stack_using_recursion {

	/*
	 Input stack:
	 3
	 2
	 1
	 Output stack:
	 1
	 2
	 3
	 */
	public s]]></description><category></category></item><item><title><![CDATA[判断一个整数是否为回文数 Check if a number is palindrome]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/37185101</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/37185101</guid><author>hellobinfeng</author><pubDate>Sun, 06 Jul 2014 03:08:10 +0800</pubDate><description><![CDATA[一种方法是先翻转当前数，然后把它和原数比较（略）另一种是递归方法,借用一个复制数，对原数递归，使之翻转，然后配合复制数比较package recursion;

public class Check_if_a_number_is_palindrome {

	public static void main(String[] args) {
		int num = 121321;
		System.]]></description><category></category></item><item><title><![CDATA[翻转整数 Reverse digits of a number]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/37183799</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/37183799</guid><author>hellobinfeng</author><pubDate>Sun, 06 Jul 2014 03:06:13 +0800</pubDate><description><![CDATA[两种方法翻转一个整数，顺序翻转和递归翻转这里没考虑overflow的情况递归的作用是使得反向处理，即从递归栈的最低端开始处理，通过画图可得。如果是rec(num/10):123451234123121         package recursion;

public class Reverse_digits_of_a_number {

	public static void main(Str]]></description><category></category></item><item><title><![CDATA[MAC OSX下查看某个端口被哪个程序占用及杀进程方法]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/37049405</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/37049405</guid><author>hellobinfeng</author><pubDate>Sat, 05 Jul 2014 13:51:18 +0800</pubDate><description><![CDATA[sudo lsof -i : 9000COMMAND   PID    USER   FD   TYPE             DEVICE SIZE/OFF NODE NAMEjava    61342 a  313u  IPv6 0x1111111111111     0t0  TCP *:cslistener (LISTEN)然后根据PID杀进程：sudo kill -9 61342]]></description><category></category></item><item><title><![CDATA[打印出大小为n的数组（可能有重复元素）里所有可能的组合]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/37030615</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/37030615</guid><author>hellobinfeng</author><pubDate>Sat, 05 Jul 2014 09:17:32 +0800</pubDate><description><![CDATA[Input:	 {1, 2, 3, 4}, r=2	 Output:	 {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4} and {3, 4}.package recursion;

import java.util.ArrayList;
import java.util.Collections;

public class Print_all_possible_com]]></description><category></category></item><item><title><![CDATA[给出一个set的字符和一个正数k，求所有由这个set能组成长度为k的字符串集合 print-all-combinations-of-given-length]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/37021707</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/37021707</guid><author>hellobinfeng</author><pubDate>Sat, 05 Jul 2014 07:21:04 +0800</pubDate><description><![CDATA[// 给出一个set的字符和一个正数k，求所有由这个set能组成长度为k的字符串集合	/*	 Input: 	set[] = {'a', 'b'}, k = 3		Output:	aaa	aab	aba	abb	baa	bab	bba	bbb			Input: 	set[] = {'a', 'b', 'c', 'd'}, k = 1	Output:	a	b	c	dpackage recursion]]></description><category></category></item><item><title><![CDATA[比screen还好用的命令tmux]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/36279639</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/36279639</guid><author>hellobinfeng</author><pubDate>Wed, 02 Jul 2014 02:16:52 +0800</pubDate><description><![CDATA[tmux是GUN screen的替代品，它比screen有很多方便的地方，下面我就来数显tmux的使用方法，我用的是ubuntu，所以安装tmux直接利用apt-get install tmux就可以安装这个服务了，一个tmux服务可以开启多个session，一个session可以开多个窗口，一个窗口可以分多个子窗口，下面是一些常用方法：     tmux默认的前缀操作都是ctrl + b ，当然]]></description><category></category></item><item><title><![CDATA[项目中发现的一些关于JavaScript中JSON的注意点]]></title><link>https://blog.youkuaiyun.com/hellobinfeng/article/details/28485673</link><guid>https://blog.youkuaiyun.com/hellobinfeng/article/details/28485673</guid><author>hellobinfeng</author><pubDate>Wed, 04 Jun 2014 23:12:27 +0800</pubDate><description><![CDATA[一个是如何创建JSON：var obj = {};
obj['name'] = value;
obj['anotherName'] = anotherValue;如果要创建多级的JSON，则：ips[ipId] = {};
ips[ipId]['name'] = value;
ips[ipId]['anotherName'] = anotherValue;注意要用bracket[]而不能用点. 来]]></description><category></category></item></channel></rss>