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.

时间复杂度O(n²)

class Solution {public:    string longestPalindrome(string s) {        int n = s.length();        if(n == 0){            return "";        }        string result = s.substr(0,1);        for(int i = 0; i < n - 1; i++){            string s1 = process(s,i,i);     //奇数的情况下            if(s1.length() > result.length()){                result = s1;            }                        string s2 = process(s,i,i+1);   //偶数的情况下            if(s2.length() > result.length()){                result = s2;            }        }        return result;     }     string process(string s, int begin, int end){         while(begin >= 0 && s[begin] == s[end]){             begin--;             end++;         }         return s.substr(begin + 1, end - begin - 1);     }};

时间复杂度O(n)

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

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) {        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);    }};