1071. GCD string

greatest-common-divisor-of-strings

Title Description

For strings S and T, only S = T + ... + time T (T connected to 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"

Example 3:

Input: str1 = "LEET", str2 = "CODE"
Output: ""

prompt:

. 1 <= str1.length <= 1000
. 1 <= str2.length <= 1000
str1 [I] and str2 [i] uppercase letters

Code

package pid1071;

public class Solution {
	public String gcdOfStrings(String str1,String str2){
		// 枚举法
		int length1 = str1.length();
		int length2 = str2.length();
		int maxLength = Math.max(length1, length2);
		for(int i=maxLength;i>=1;i--){
			if(length1%i==0 && length2%i==0){
				String X = str1.substring(0,i);
				if(check(str1,X) && check(str2,X)){
					return X;
				}
			}
		}
		return "";
	}
	
	public boolean check(String str,String substr){
		int lenx = str.length() / substr.length();
		String ans = "";
		for(int i=1;i<=lenx;i++){
			ans = ans + substr;
		}
		return str.equals(ans);
	}
}