LeetCode 1071. Greatest Common Divisor of Strings

1071. Greatest Common Divisor of Strings (GCD string)

link

https://leetcode-cn.com/problems/greatest-common-divisor-of-strings

topic

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: ""
 
Tip:

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

Thinking

Be it math, if str1 + str2 and str2 + str1 same, the divisor would like to request the presence of it, and the length already know, is the common divisor of string length, only the output of a long string interception that section can be.

Code

  public int gcd(int a, int b) {
    if (b == 0) {
      return a;
    } else {
      return gcd(b, a % b);
    }
  }

  public String gcdOfStrings(String str1, String str2) {
    if ((str1 + str2).equals(str2 + str1)) {
      return str1.substring(0, gcd(str1.length(), str2.length()));
    } else {
      return "";
    }
  }