String GCD

/ * String GCD 
For strings S and T, only S = T + ... + T ( T is connected with itself one or more times), we identified "T S can be divisible." 
Returns the longest string X, X can meet the requirements, and X can be divisible divisible str1 str2. 
Example 1: 
Input: str1 = "ABCABC", str2 = "ABC" 
Output: "ABC" 
Example 2: 
Input: str1 = "ABABAB", str2 = "ABAB" 
Output: "AB" 
link: https: // leetcode- cn.com/problems/greatest-common-divisor-of-strings*/ 
// direct problem-solving ideas (correct result, efficiency is not very good)

public static String gcdOfStrings(String str1, String str2) {
    int len1 = str1.length();
    int len2 = str2.length();
    StringBuilder sb = new StringBuilder();
    if (len1 >= len2) {
        for (int i = 0; i < len2; i++) {
            if (len1 % (len2 - i) == 0 && len2 % (len2 - i) == 0) {
                String substring = str2.substring(0, len2 - i);
                if ("".equals(str1.replaceAll(substring, "")) && "".equals(str2.replaceAll(substring, ""))) {
                    return substring;
                }
            }
        }
    } else {
        for (int i = 0; i < len1; i++) {
            if (len1 % (len1 - i) == 0 && len2 % (len1 - i) == 0) {
                String substring = str1.substring(0, len1 - i);
                if ("".equals(str1.replaceAll(substring, "")) && "".equals(str2.replaceAll(substring, ""))) {
                    return substring;
                }
            }
        }
    }

    return "";
}

//第二种
//辗转相除法
public static int gcd(int a ,int b){
   if(a%b == 0){
      return b;
   }else{
       return gcd(b,a%b);
   }
}
public static String gcdOfStrings2(String str1, String str2) {
    int len1 = str1.length();
    int len2 = str2.length();
    int gcd = 0;
    if (len1 >= len2) {
        gcd = gcd(len1,len2);
    } else {
        gcd = gcd(len2,len1);
    }
    String substring = str1.substring(0, gcd);
    /*if("".equals(str1.replaceAll(substring, "")) && "".equals(str2.replaceAll(substring, ""))){
        return substring;
    }*/
    if(){

    }
    return "";
}