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
;
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