</pre><pre name="code" class="java">public class Solution {
    public String longestPalindrome(String s) {
        int down = 0,up = 0,count = 0;   //分别记录扫字符串时的上下区间及字符串长度
        int sl = s.length();
        int maxl = 0;
        int subDown = 0, subUp = 0;      //目标子字符串的上下区间
        String subS = new String("");
        
        if(sl==0)
        	return "";
        //1. 找出目标子字符串的中间位置及上下限
        for(double i=0; i<=sl-1; i=i+0.5)
        {
            down = (int)Math.floor(i);
            up = (int)Math.ceil(i);
            if((i%1==0.5)&&sl!=0)        //判断i所属情况（A或B）
            {
              count = 0;
            }else                     
            { 
              count = 1;
              down--;
              up++;
            }
        
        //2. 以i为中心向两边扫描
            while(!(down<0||up>sl-1))
            {
               if(s.charAt(down)!=(s.charAt(up)))
               {
                   break;                //2.1. 当以i为中心对称的俩字符不相等则跳出扫描
               }else{                    //2.2. 当以i为中心对称的俩字符相等则继续分别向两边移动
                  count+=2;
                  down--;
                  up++;
                }
            }
            
        //3. 更新最长字符串 
            if(count>maxl)             
            {
                maxl = count;
                subDown = down;
                subUp = up;
            }
        }
        //3.1 由于在扫描时，先移动后再判断，停止时其实前后都多移动了一个单位，这里进行一个单位的恢复
        subDown++;
        subUp--;
        subS = s.substring(subDown,subUp+1);
       
    return subS;//返回对称子字符串
    }
}

扩展学习：该文章介绍了几种方法Java Longest Palindromic Substring(最长回文字符串)

           Leetcode – Longest Palindromic Substring (Java)