Leetcode 5. 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.

这里给出两个效率较高的方法：

方法1：动态规划法 O(n2)

class Solution {public:    string longestPalindrome(string s) {        int maxbeg=0,maxlen=0;        if(s.size()==0||s.size()==1)            return s;        bool flag[1000][1000]={false};        for(int i=0;i<1000;i++)            flag[i][i]=true;        for(int i=0;i<s.size()-1;i++){            if(s[i]==s[i+1]){                flag[i][i+1]=flag;                maxbeg=i;                maxlen=2;            }        }                    for(int len=3;len<=s.size();len++){            for(int i=0;i+len-1<s.size();i++){                int beg=i,end=i+len-1;                if(s[beg]==s[end]&&flag[beg+1][end-1]){                    flag[beg][end]=true;                    maxbeg=beg;                    maxlen=len;                }            }        }        return s.substr(maxbeg,maxlen);    }};

方法2：Manacher算法 O(n)，核心思想利用回文的对称性求解回文串的半径。实现要对原串进行改造

class Solution {public:    string longestPalindrome(string s) {        string str;        str+='$';        for(int i=0;i<s.size();i++)            str=str+'#'+s[i];        str+='#';                int mid=0,mx=0,maxi=1;        int p[str.size()];        for(int i=1;i<str.size();i++){            if(mx>=i)                p[i]=min(p[2*mid-i],mx-i+1);            else                p[i]=1;                        int beg=i-p[i],end=i+p[i];            while(str[beg]==str[end]){                p[i]++;                beg--;                end++;            }                        if(p[i]+i-1>mx){                mx=p[i]+i-1;                mid=i;            }                        maxi=p[i]>p[maxi]?i:maxi;        }                return s.substr((maxi-p[maxi]+1)/2,p[maxi]-1);    }};

Manacher算法 具体讲解请看：http://www.cnblogs.com/bitzhuwei/p/Longest-Palindromic-Substring-Part-II.html