Longest Palindromic Substring

Description:

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.

算法1：

动态规划。复杂度是 O(n^2)

f[i][j] 表示 字串[i---j]是否是一个回文。

    f[i][j] = true,         i = j

             s[i] == s[j],  j = i+1

            (s[i] == s[j]) && f[i+1][j-1], j > i+1;

    string longestPalindrome(string s) {        const int n=s.size();         int start(0), max_len(1);        bool f[n][n];        fill_n(&f[0][0], n*n, false);                for(int i=n-1; i>=0; --i)        {          f[i][i] = true;          for(int j=i+1; j<n; ++j)          {              f[i][j] = (s[i]==s[j] ? true:false) && (j-i < 2 || f[i+1][j-1]);                               if(f[i][j] && j-i+1 > max_len){                  start = i;                  max_len = j-i+1;              }          }        }        return s.substr(start, max_len);    }};

算法2： 

枚举回文的中心，向两边扩展。。复杂度 O(N^2)

class Solution {public:// start from centerstring longestPalindrome(string s){  if(s.size() < 2) return s;  string longest;  string p;  for(int i=0; i<s.size()-1; ++i){    //palindrome is odd    p = expandFromCenter(s, i, i);if(p.size() > longest.size())  longest = p;//palindrome is even    p = expandFromCenter(s, i, i+1);if(p.size() > longest.size())  longest = p;  }  return longest;}private:string expandFromCenter(string s, int c1, int c2){  int l(c1), r(c2);  while(l >= 0 && r < s.size() && s[l] == s[r]){    -- l;    ++ r;  }  return s.substr(l+1, r-l-1);}};

3. Manacher 算法， 复杂度 O(n)

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

// Manacher algorithm// Transform S into T// For example, S = "abba", T = "^#a#b#b#a#$"// ^ and $ signs are sentinels appended to each string// to avoid bound checkingstring 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 longestPalindromeManacher(string s){  string T = preProcess(s);  int n = T.length();  vector<int> p(n, 0);  int C = 0, R = 0;  for (int i = 1; i < n-1; ++i){    int i_mirror = 2*C-i;p[i] = (i < R) ? min(R-i, p[i_mirror]) : 0; // attempt to expand palindrome centered at iwhile(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(p[i] + 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;}  }  return s.substr((centerIndex - 1 - maxLen)/2 , maxLen);}