[Happy DSA] 求解最长回文子字符串

来源：互联网发布：mysql 查询名次编辑：程序博客网时间：2024/05/20 05:55

题目：输入一个字符串，输出该字符串中对称的子字符串的最大长度。

比如输入字符串google，由于该字符串里最长的对称子字符串是goog，因此输出4

关于这个问题，有一个比较好的线性时间复杂度的算法Manacher's ALGORITHM

这2篇技术博客非常好的阐述了它的算法思想：

1) http://www.felix021.com/blog/read.php?2040 (中文)

2) http://www.leetcode.com/2011/11/longest-palindromic-substring-part-ii.html

更多的博客文章可供参考：

http://www.akalin.cx/longest-palindrome-linear-time

C++代码实现：

// longest palindromic substring// http://www.felix021.com/blog/read.php?2040// Manacher's ALGORITHM: O(n)std::string lps(const std::string& str){    std::cout << "original string is " << str << std::endl;    std::string ss = "$";    std::ostringstream ostr;    for (int i = 0; i < str.length(); i++)    {           ostr << '#' << str[i];    }       ostr << "#";    ss += ostr.str();    std::cout << "new string is " << ss << std::endl;    int* p = new int[ss.length()];    *p = 0;    int id = 0, mix = 0;    for (int i = 1; i < ss.length(); i++)    {           p[i] = (mix > i) ? std::min(p[2*id - i], mix-i) : 1;        while (ss[i + p[i]] == ss[i-p[i]]) p[i]++;        if (p[i] + i > mix)            {               mix = p[i] + i;            id = i;        }       }           int* pmax = std::max_element(p, p+ss.length());    int ipos = (pmax - p)/2;    int len = *pmax - 1;   std::string lpsstr = str.substr(ipos-len/2, len);    delete [] p;    return lpsstr;}