<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[欲买桂花同载酒，终不似，少年游。]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/htx931005</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; htx931005]]></copyright><item><title><![CDATA[[LeetCode]050-Pow(x,n)]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/51339521</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/51339521</guid><author>htx931005</author><pubDate>Sat, 07 May 2016 19:23:11 +0800</pubDate><description><![CDATA[题目： 
Implement pow(x, n).  
实现x的n次幂Solution： 
类似于二分的方法来优化“x*x*…x“。 
要注意的是当n = INT_MIN时，反过来-n会越界，这时候需要pow(x,INT_MIN) * x 即可。代码如下：class Solution {
public:
    double myPow(double x, int n) 
    {]]></description><category></category></item><item><title><![CDATA[[LeetCode]049-Group Anagrams]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/51338598</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/51338598</guid><author>htx931005</author><pubDate>Sat, 07 May 2016 16:40:26 +0800</pubDate><description><![CDATA[题目: 
Given an array of strings, group anagrams together.For example, given: [“eat”, “tea”, “tan”, “ate”, “nat”, “bat”], 
Return:[ 
  [“ate”, “eat”,”tea”], 
  [“nat”,”tan”], 
  [“bat”] 
]Note:For the re]]></description><category></category></item><item><title><![CDATA[[LeetCode]048-Rotate Image]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/51338390</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/51338390</guid><author>htx931005</author><pubDate>Sat, 07 May 2016 16:06:40 +0800</pubDate><description><![CDATA[题目： 
You are given an n x n 2D matrix representing an image.Rotate the image by 90 degrees (clockwise).Follow up: 
Could you do this in-place?Solution: 
其实就是矩阵的旋转，找到规律即可，比较简单，代码如下：class Solution {
publ]]></description><category></category></item><item><title><![CDATA[[LeetCode]047-Permutations II]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/51337956</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/51337956</guid><author>htx931005</author><pubDate>Sat, 07 May 2016 14:54:13 +0800</pubDate><description><![CDATA[题目： 
 Given a collection of numbers that might contain duplicates, return all possible unique permutations.For example, 
[1,1,2] have the following unique permutations: 
[1,1,2], [1,2,1], and [2,1,1].]]></description><category></category></item><item><title><![CDATA[[LeetCode]046-Permutations]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/51335198</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/51335198</guid><author>htx931005</author><pubDate>Fri, 06 May 2016 22:35:08 +0800</pubDate><description><![CDATA[题目：Given a collection of distinct numbers, return all possible permutations.For example, 
[1,2,3] have the following permutations: 
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1]. Solutions:]]></description><category></category></item><item><title><![CDATA[[数据结构]散列表初识]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50931492</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50931492</guid><author>htx931005</author><pubDate>Sat, 19 Mar 2016 16:07:29 +0800</pubDate><description><![CDATA[散列表也叫哈希表，是根据关键码值(Key value)而直接访问的数据结构，也就是说，它通过把关键码值映射到表中的一个位置来访问记录，以加快查找速度，这个映射函数叫散列函数，存放记录的数组叫散列表。给定表M，存在函数f(key)，对任意给定的关键字值key，代入函数若能得到包含该关键字的记录在表中的地址，则称表M为哈希表，函数f(key)为哈希函数。一、基本概念1、若关键字为k，则其值存放在f(k)]]></description><category></category></item><item><title><![CDATA[[iOS]iOS Category]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50789829</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50789829</guid><author>htx931005</author><pubDate>Thu, 03 Mar 2016 16:54:27 +0800</pubDate><description><![CDATA[转载：http://blog.sina.com.cn/s/blog_923fdd9b0101b9da.html


无论一个类设计的如何完美，都不可避免的会遇到没有预测到的需求，那怎么扩展现有的类呢？当然，继承是个不错的选择。但是Objective-C提供了一种特别的方式来扩展类，叫Catagory，可以动态的为已经存在的类添加新的行为。这样可以保证类的原原来的基础上，较小的改动就可以增加需]]></description><category></category></item><item><title><![CDATA[[iOS]编写高质量的Objective-C代码]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50782746</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50782746</guid><author>htx931005</author><pubDate>Wed, 02 Mar 2016 20:11:24 +0800</pubDate><description><![CDATA[本文由“海水的味道”编译点标记语法 
属性和幂等方法（多次调用和一次调用返回的结果相同）使用点标记语法访问，其他的情况使用方括号标记语法。良好的风格： 
view.backgroundColor = [UIColor orangeColor]; 
[UIApplication sharedApplication].delegate;不良的风格： 
[view setBackgroundColor:[]]></description><category></category></item><item><title><![CDATA[[LeetCode]045-Jump Game II]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50437826</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50437826</guid><author>htx931005</author><pubDate>Wed, 30 Dec 2015 22:47:11 +0800</pubDate><description><![CDATA[题目： 
 Given an array of non-negative integers, you are initially positioned at the first index of the array.Each element in the array represents your maximum jump length at that position.Your goal is t]]></description><category></category></item><item><title><![CDATA[[LeetCode]044-Wildcard Matching]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50437439</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50437439</guid><author>htx931005</author><pubDate>Wed, 30 Dec 2015 21:24:25 +0800</pubDate><description><![CDATA[题目：‘?’ 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]]></description><category></category></item><item><title><![CDATA[[LeetCode]043-Multiply Strings]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50432967</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50432967</guid><author>htx931005</author><pubDate>Wed, 30 Dec 2015 10:38:43 +0800</pubDate><description><![CDATA[题目： 
Given two numbers represented as strings, return multiplication of the numbers as a string.Note: The numbers can be arbitrarily large and are non-negative.Solution: 
思路： 
乘数为num1和num2，则模拟手算，每次更新nu]]></description><category></category></item><item><title><![CDATA[[LeetCode]042-Trapping Rain Water]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50428747</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50428747</guid><author>htx931005</author><pubDate>Tue, 29 Dec 2015 19:06:01 +0800</pubDate><description><![CDATA[题目： 
 Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.For example, 
Given [0,1,0,2,1,0,1,3,2,1,2,1]]]></description><category></category></item><item><title><![CDATA[[LeetCode]041-First Missing Positive Integer]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50427697</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50427697</guid><author>htx931005</author><pubDate>Tue, 29 Dec 2015 16:44:51 +0800</pubDate><description><![CDATA[题目： 
Given an unsorted integer array, find the first missing positive integer.For example, 
Given [1,2,0] return 3, 
and [3,4,-1,1] return 2.Your algorithm should run in O(n) time and uses constant spa]]></description><category></category></item><item><title><![CDATA[[LeetCode]040-Combination Sum II]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50426609</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50426609</guid><author>htx931005</author><pubDate>Tue, 29 Dec 2015 14:45:05 +0800</pubDate><description><![CDATA[题目： 
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.Each number in C may only be used once in the combinati]]></description><category></category></item><item><title><![CDATA[[LeetCode]039-Combination Sum]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50426369</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50426369</guid><author>htx931005</author><pubDate>Tue, 29 Dec 2015 14:18:12 +0800</pubDate><description><![CDATA[题目： 
 Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.The same repeated number may be chosen from C unlimited numbe]]></description><category></category></item><item><title><![CDATA[[LeetCode]038-Count And Say]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50402766</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50402766</guid><author>htx931005</author><pubDate>Fri, 25 Dec 2015 13:56:02 +0800</pubDate><description><![CDATA[题目：The count-and-say sequence is the sequence of integers beginning as follows: 
1, 11, 21, 1211, 111221, …1 is read off as “one 1” or 11. 
11 is read off as “two 1s” or 21. 
21 is read off as “one 2,]]></description><category></category></item><item><title><![CDATA[[回溯法]八皇后问题的递归与非递归算法]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50364014</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50364014</guid><author>htx931005</author><pubDate>Sun, 20 Dec 2015 14:02:45 +0800</pubDate><description><![CDATA[八皇后的问题非常有名，初次理解可能稍有难度，不过多看书，看博客和代码，几遍下来，也基本清晰。首先不用想初始的情况，先假设前面已经排列好了几个皇后，即将排列下一个皇后。依次遍历八个位置，然后与之前的进行判断这个位置是否可行，如可行则进行下一个皇后，否则则移动位置继续判断。很简单。但是有两个个问题：1、不全，某个位置有八种方法排列，你只用了一种，当然你可以每个位置遍历，但太耗时，回溯法可以解决很大的一部]]></description><category></category></item><item><title><![CDATA[[LeetCode]037-Sudoku Solver]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50359933</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50359933</guid><author>htx931005</author><pubDate>Sat, 19 Dec 2015 22:03:41 +0800</pubDate><description><![CDATA[题目： 
Write a program to solve a Sudoku puzzle by filling the empty cells.Empty cells are indicated by the character ‘.’.You may assume that there will be only one unique solution. Solution: 
思路，类似于回溯法，]]></description><category></category></item><item><title><![CDATA[[LeetCode]036-Valid Sudoku]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50324039</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50324039</guid><author>htx931005</author><pubDate>Wed, 16 Dec 2015 10:06:48 +0800</pubDate><description><![CDATA[题目： 
Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.The Sudoku board could be partially filled, where empty cells are filled with the character ‘.’. 
Solution: 
（1）同一行没有相同的数字。]]></description><category></category></item><item><title><![CDATA[[LeetCode]035-Search Insert Position]]></title><link>https://blog.youkuaiyun.com/htx931005/article/details/50319947</link><guid>https://blog.youkuaiyun.com/htx931005/article/details/50319947</guid><author>htx931005</author><pubDate>Tue, 15 Dec 2015 20:02:24 +0800</pubDate><description><![CDATA[题目： 
Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.You may assume no duplicates in the array.]]></description><category></category></item></channel></rss>