1071. GCD string
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
Thinking
str1 + str2 == str2 + str1 certainly present common factors. And the string must be the greatest common factor of the length of the largest common factor.
My answer
class Solution {
public:
string gcdOfStrings(string str1, string str2) {
int lenStr1 = str1.length();
int lenStr2 = str2.length();
if (str1 + str2 == str2 + str1) {
return str1.substr(0,gcd(lenStr1, lenStr2));
}
else return "";
}
};
inline int gcd(int a, int b) {
if (b) while ((a %= b) && (b %= a));
return a + b;
}
