题目：
For strings S and T, we say "T divides S" if and only if S = T + ... + T (T concatenated with itself 1 or more times)
Return the largest string X such that X divides str1 and X divides 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: ""
Note:
1 <= str1.length <= 10001 <= str2.length <= 1000str1[i] and str2[i] are English uppercase letters.
解题思路：
输入为str1和str2，定义：对于字符串T来说，若可以分解为字符串X的级联形式，如X X X ...，则称X为T的divisor分解串。输出时找到str1和str2的最大divisor并输出，若str1和str2不包含divisor则输出空字符串“”。
假设str2为较短的字串，先对它寻找divisor，然后将divisor与str1进行判断，如果不符合divisor条件则继续寻找，符合则返回输出。
代码：
class Solution { public: bool checkDivisor(string& T, string& X) { if(T.length() % X.length()) return false; int times = T.length() / X.length(); for(int i = 0; i<times; i++) { string subT = T.substr(i*X.length(), X.length()); if(subT != X) return false; } return true; } string gcdOfStrings(string str1, string str2) { if(str1.length() < 1 || str2.length() < 1) return ""; if(str1.length() < str2.length()) { swap(str1, str2); } string X = ""; for(int i = 1; i <= str2.length(); i++) { if(str2.length() % i) continue; string x = str2.substr(0, str2.length() / i); if(!checkDivisor(str2, x)) continue; if(checkDivisor(str1, x)) { X = x; break; } } return X; } };
忘记string可以直接比较= =，也不太熟悉string的取子串substr接口，所以debug了一段时间。Anyway，最终还是AC啦~
