力扣5 ：最长回文数串

方法一：暴力穷举。显然超时了。90/103。

class Solution {
    public boolean isHuiwen(String s){
        for(int i = 0,j = s.length()-1;i <= j;i++,j--){
            if(s.charAt(i)!=s.charAt(j)) return false;
        }
        return true;
    }
    public String longestPalindrome(String s) {
        StringBuffer list = new StringBuffer();
        if(s == null || s.length() == 0) return "";
        if(s.length() == 1) return s;
        for(int i = 0; i < s.length();i++){
            for(int j = i+1; j < s.length();j++ ){
                if(isHuiwen(s.substring(i,j+1))){
                    if(s.substring(i,j+1).length() > list.length()){
                        list = list.delete(0,list.length());
                        list.append(s.substring(i,j+1));
                    }
                }
            }
        }
        if(list.length()==0){
            return new Character(s.charAt(0)).toString();
        }else{
            return list.toString();
        }
    }
}

 对上述方法进行优化，当必须新的子串长度大于已经找到的回文数，才进行判断，否则有无回文都不必判断。（103/103）用例全部通过，超时

class Solution {
    int len = -1;
    public boolean isHuiwen(String s,int m,int n){
        for(int i = 0,j = s.length()-1;i <= j;i++,j--){
            if(s.charAt(i)!=s.charAt(j)){
               
                return false;
            } 
        }
        len = n - m;
        return true;
    }
    public String longestPalindrome(String s) {
        StringBuffer list = new StringBuffer();
        if(s == null || s.length() == 0) return "";
        if(s.length() == 1) return s;
        for(int i = 0; i < s.length();i++){
            for(int j = i+1; j < s.length();j++ ){
                if( j - i < len){
                    continue;
                }
                if(isHuiwen(s.substring(i,j+1),i,j)){
                    if(s.substring(i,j+1).length() > list.length()){
                        list = list.delete(0,list.length());
                        list.append(s.substring(i,j+1));
                    }
                }
            }
        }
        if(list.length()==0){
            return new Character(s.charAt(0)).toString();
        }else{
            return list.toString();
        }
    }
}