Brutal Force: O(n^2)

public class Solution {    public String longestPalindrome(String s) {        String maxPal="";        for(int i=0;i
     
      maxPal.length()) maxPal=str1; String str2=lenofPal(i, i+1, s); if(str2.length()>maxPal.length()) maxPal=str2; } return maxPal; } private String lenofPal(int l, int r, String s) { while(l>=0&&r
     

 

 

Manacher: O(n)

public class Solution {    public String longestPalindrome(String s) {        char[] arr=new char[s.length()*2+1];        for(int i=0;i
     
      i)                len[i]=Math.min(mx-i, len[2*po-i]);            else                len[i]=1;            while(i-len[i]>=0&&i+len[i]
      
       mx)            {                mx=i+len[i];                po=i;            }            if(len[i]-1>ret.length())                ret=s.substring((i-len[i]+1)/2,(i+len[i]-1)/2);        }        return ret;    }}