Repeated Substring Pattern

题目

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.)  

思路

start : 代表要比较的子串的开始索引

length : 代表当前的字串长度, length最多为str.length()/2长度，而且还得是偶数字节

循环不定式
如果duplicate的length [1, str.length()/2]

那么我们可以实时得到默认的子串个数就是 str.length() / length = compare_times

这样就是说我们要比较compare_times次，其中每一次比较要比较length长度

class Solution {
public:
    bool repeatedSubstringPattern(string s) {
        int len = 1, start=0, cmp_times=0;
        for(int len = 1; len <= (floor)(s.length() / 2); ++len)
        {
            if(s.length() % len != 0)
                continue;
            cmp_times = s.length() / len;
            bool bCurDup = true;
            for(int i=1; i<cmp_times; ++i)
            {
                if(bCurDup == true)
                    for(int j=0; j<len; ++j)
                    {
                        if(s[j] != s[i*len+j])
                        {
                            bCurDup = false;
                            break;
                        }
                    }
                else
                {
                    break;
                }
            }
            if(bCurDup == true)
            {
                return true;
            }
        }
        
        return false;
    }
};