LeetCode OJ 之最长回文子字符串

题目：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.
最直观的解法，找出该字符串所有子字符串，依次判断该子字符串是否为回文字符串，之后再找出最长的子回文字符串，程序如下：

public class Solution {
    public String longestPalindrome(String s) {
        String tempstring = new String();
        int len = 0;
        int strlen = s.length();
        for(int i = 0; i < strlen; i++){
            for(int j = i+1; j <= strlen; j++){
                String substring = s.substring(i,j);
                if(isPalindrome(substring)){
                    if(tempstring.length() < substring.length()){
                        tempstring = substring;
                    }
                }
            }
        }
        return tempstring;
    }
    public boolean isPalindrome(String s){
        char[] chararr = s.toCharArray();
        int strlen = s.length();
        boolean result = true;
        for(int i = 0; i < strlen/2; i++){
            if(chararr[i] == chararr[strlen-i-1]){
                result = true;
            }
            else{
                result = false;
                break;
            }
        }
        return result;
    }
}

但该程序时间复杂度过高，不会被OJ接受，一个可被OJ接受的算法是，从某一个字符向两边分别进行比较，以找出以该字符为中心的回文子字符串，实现方法如下：

public class Solution {
    private int resultlen = 0;
    private String result = new String();
    public String longestPalindrome(String s) {
        char[] chararr = s.toCharArray();
        int len = chararr.length;
        for(int i = 0; i < len; i++){
            findPalindrome(chararr,len,i,i+1);
            findPalindrome(chararr,len,i,i);
        }
        return result;
    }
    private void findPalindrome(char[] myarr,int len,int left,int right){
        int templen = 0;
        StringBuilder sb = new StringBuilder();
        while(left >= 0 && right < len && myarr[left] == myarr[right]){
            left--;
            right++;
        }
        if ((right-left-1) > resultlen) {
            resultlen = right-left-1;
            result = new String(myarr, left+1, resultlen);
        }
    }
}