[LeetCode]Longest Palindromic Substring

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.

思考：遍历求每个字符的左右延伸的最长字符串。

class Solution {
public:
    string compare(string s,int left,int right)
    {
        while(left>=0&&right<s.size())
        {
            if(s[left]==s[right])
            {
                left--;
                right++;
            }
            else break;
        }
        return s.substr(left+1,right-left-1);
    }
    string longestPalindrome(string s) {
        int len=0;
        int maxlen=0;
        string ans,res;
        for(int i=0;i<s.size();i++)
        {
            ans=compare(s,i,i);
            if(ans.size()>res.size()) res=ans;
            ans=compare(s,i,i+1);
            if(ans.size()>res.size()) res=ans;
        }
        return res;
    }
};

 网上有个更好的Manacher算法，只要O（n）：

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

http://www.felix021.com/blog/read.php?2040

class Solution {
public:
	string preProcess(string s) 
	{  
		int n = s.length();  
		if (n == 0) 
			return "^$";  
		string ret = "^";  
		for (int i = 0; i < n; i++)    
			ret += "#" + s.substr(i, 1);   
		ret += "#$";  return ret;
	} 
	string longestPalindrome(string s) {
		// IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
		string T = preProcess(s);  
		int n = T.length();  
		int *P = new int[n];  
		int C = 0, R = 0;  
		for (int i = 1; i < n-1; i++) 
		{    
			int i_mirror = 2*C-i; // equals to i' = C - (i-C)        
			P[i] = (R > i) ? min(R-i, P[i_mirror]) : 0;        
			// Attempt to expand palindrome centered at i    
			while (T[i + 1 + P[i]] == T[i - 1 - P[i]])      
				P[i]++;     
			// If palindrome centered at i expand past R,    
			// adjust center based on expanded palindrome.    
			if (i + P[i] > R) 
			{      
				C = i;      
				R = i + P[i];    
			}  
		}   
		// Find the maximum element in P.  
		int maxLen = 0;  
		int centerIndex = 0;  
		for (int i = 1; i < n-1; i++) 
		{    
			if (P[i] > maxLen) 
			{      
				maxLen = P[i];      
				centerIndex = i;    
			}  
		}  
		delete[] P;    
		return s.substr((centerIndex - 1 - maxLen)/2, maxLen);
	}
};