<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Palindrome on 禹过留声</title><link>//blog.haojunyu.com/tags/palindrome/</link><description>Recent content in Palindrome on 禹过留声</description><generator>Hugo -- gohugo.io</generator><managingEditor>haojunyu2012@gmail.com (hjy)</managingEditor><webMaster>haojunyu2012@gmail.com (hjy)</webMaster><lastBuildDate>Tue, 25 Aug 2015 17:54:04 +0800</lastBuildDate><atom:link href="//blog.haojunyu.com/tags/palindrome/index.xml" rel="self" type="application/rss+xml"/><item><title>解题报告之最长回文子串</title><link>//blog.haojunyu.com/coder/report_longpalindrome/</link><pubDate>Tue, 25 Aug 2015 17:54:04 +0800</pubDate><author>haojunyu2012@gmail.com (hjy)</author><guid>//blog.haojunyu.com/coder/report_longpalindrome/</guid><description>&lt;p>题目很简单，就是求解一个字符串中最长的回文子串，回文字符串是指颠倒之后的字符串和原来的字符串完全一样。网上很多人的博文，像
&lt;a href="http://blog.csdn.net/kangroger/article/details/37742639" target="_blank">kangroger&lt;/a>,还有把最牛b的 Manacher 算法说得很清晰的
&lt;a href="http://blog.csdn.net/pi9nc/article/details/9251455" target="_blank">pi9nc&lt;/a>和
&lt;a href="http://blog.csdn.net/yzl_rex/article/details/7908259" target="_blank">yzl_rex&lt;/a>，介绍了四种解法：分别是暴力法（$O(n^3)$），动态规划（$O(n^2)$），中心扩展（$O(n^2)$）和Manache法（$O(n)$）。本博文只是对这四种方法的自我实现。&lt;/p></description></item></channel></rss>