[]帮忙测试，这个程序还有什么有关问题吗

/** * 程序目的: * 请编程求解两个字符串中的最大子串，例如abcdefg和nbcdyu中的最大子串为bcd。 * */public class $3_MaxSubString {    public static void main(String[] args) {        String longStr = "sadfasdf";        String shortStr ="gasdsadf";        System.out.println(maxSubString(longStr,shortStr));    }        public static String maxSubString(String longStr,String shortStr){        //保证前面的字符串长度大于后面的字符串长度        if(longStr.length()<shortStr.length()){            String temp = longStr;            longStr = shortStr;            shortStr = temp;        }        String maxsub;        String subLong;        for(int i=shortStr.length();i>0;i--){    //定义小字符串取的长度            for(int j=0;j<=shortStr.length()-i;j++){   //定义小串取的位置                maxsub = shortStr.substring(j,i+j);                for(int k=0;k<=longStr.length()-maxsub.length();k++){                    subLong = longStr.substring(k, maxsub.length()+k);                    if(maxsub.equals(subLong))return maxsub;                }            }        }        return null;    }}

public final class MaxSubString {    public static void main(String[] args) {        MaxSubString maxSubString = new MaxSubString();        System.out.println("[" + maxSubString.maxSubString("abcdefg", "nbcdyu")                + "]");    }    private String maxSubStr;    public String maxSubString(final String strA, final String strB) {        String strAA = strA;        String strBB = strB;        if (strA.length() < strB.length()) {            strAA = strB;            strBB = strA;        }        this.maxSubStr = "";        String result = null;        String subStrB = null;        int loops = strBB.length();        for (int i = 0; i < loops; i++) {            subStrB = strBB.substring(i);            for (int j = subStrB.length(); j > 0; j--) {                result = subStrB.substring(0, j);                if (strAA.contains(result)) {                    if (this.maxSubStr.length() < result.length()) {                        this.maxSubStr = result;                    }                }            }        }        return this.maxSubStr;    }}