459.Repeated Substring Pattern

459.Repeated Substring Pattern

难度:Easy

Given a non-empty string check if it can be constructed by taking a substring of it and appending multiple copies of the substring together. You may assume the given string consists of lowercase English letters only and its length will not exceed 10000.

Example 1:

Input: "abab"
Output: True
Explanation: It's the substring "ab" twice.
Example 2:

Input: "aba"
Output: False
Example 3:

Input: "abcabcabcabc"
Output: True
Explanation: It's the substring "abc" four times. (And the substring "abcabc" twice.)

找出是否是循环字串，可以先统计26个字母出现的频次，然后找到所有出现次数不为0的最小公约数(至少为2)，即最少的循环次数。然后比较。

lass Solution {
int max_factor(int a, int b)
{
    if(a<b) swap(a,b);
    if(b==0) return a;
    for(int i=2;i<=b;i++)
        if(a%i==0 && b%i ==0 )return i;
    return 1;
}
public:
    bool repeatedSubstringPattern(string s) {
        vector<int>alpha(26,0);
        for(auto c:s)
            ++alpha[c-'a'];
        int len=0;
        for(auto i:alpha)
//        {
            len=max_factor(len,i);
     //       cout<< i<<" ";
   //     }
   //     cout<<endl;
  //     cout<< len << " "<<endl;

        if(len == 1  )return false;
        
        int one_size=s.length()/len;
//  cout<<one_size<<endl;
     for(int j=0;j<one_size;j++)
            for(int i=1;i<len;i++)
                if(s[i*one_size+j] !=s[j]) return false;
        return true;
        
    }
};

执行用时 :28 ms, 在所有 C++ 提交中击败了97.08%的用户 内存消耗 :11.5 MB, 在所有 C++ 提交中击败了94.35%的用户