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Longest Palindromic Substring(最长回文子串)
题目描述：
Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.
这是一道很有名的题目，相信很多人面试的时候会被问到。
本人算法能力一般，题目看完毫无头绪。苦思冥想大半天，才想到用动态规划。代码写的很丑，就不贴了。结果一提交，系统提示Time Limit Exceeded。
这说明算法复杂度太高。我又抓耳挠腮大半天，想到了中心扩展法。提交通过，很是开心。
中心扩展法：
string longestPalindrome(string s) { int len = s.size(); if(len < 2) return s; int maxlen = 1; int start = 0; int low,high; for (int i = 1;i < len;++i) { low = i - 1; high = i; while (low >= 0 && high < len && s[low] == s[high]) { --low; ++high; } if (high - low - 1 > maxlen) { maxlen = high - low - 1; start = low + 1; } low = i - 1; high = i + 1; while (low >= 0 && high < len && s[low] == s[high]) { --low; ++high; } if (high - low - 1 > maxlen) { maxlen = high - low - 1; start = low + 1; } } return string(s,start,maxlen);}
走到这一步已经很不容易，上网一搜，发现还存在一个更牛的解法，如下：
Manacher‘s algorithm
string preProcess(const string &s){ int n = s.size(); string res = "$"; for (int i = 0;i < n;++i) { res += "#"; res += s[i]; } res += "#^"; return res;}string longestPalindrome(string s){ if(s.size() < 2) return s; string str = preProcess(s); int n = str.size(); int *p = new int[n]; int id = 0, mx = 0; for (int i = 1;i < n-1;++i) { p[i] = mx > i ? min(p[2*id-i], mx-i) : 1; while(str[i+p[i]] == str[i-p[i]]) ++p[i]; if (i + p[i] > mx) { mx = i + p[i]; id = i; } } int maxlen = 0, index = 0; for (int i = 1;i < n-1;++i) { if (p[i] > maxlen) { maxlen = p[i]; index = i; } } delete [] p; return string(s,(index - maxlen)/2,maxlen-1);}
这里引用别人的一段话：
This algorithm is definitely non-trivial and you won’t be expected to come up with such algorithm during an interview setting.
But, please go ahead and understand it, I promise it will be a lot of fun.
想了解原理的可以参考这篇文章http://www.cnblogs.com/tenosdoit/p/3675788.html
Longest Palindromic Substring(最长回文子串)
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