<p>Given a string <code>s</code>, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abab&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> It is the substring &quot;ab&quot; twice.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aba&quot;
<strong>Output:</strong> false
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcabcabcabc&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> It is the substring &quot;abc&quot; four times or the substring &quot;abcabc&quot; twice.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    bool repeatedSubstringPattern(string s) {
        if(s.length()==1) return false;
        char first=s.front();
        char last=s.back();
        int pos=s.find(last,0);
        if(pos==s.length()-1){// the pos found is the last one
            if(last==first){
                for(int i=0;i<=pos;i++){
                    if(s[i]!=first) return false;
                }
                return true;
            }
            else return false;
        }
        while(pos!=string::npos){
            int p=pos;
            string sub=s.substr(0,pos+1);
            // cout<<sub<<endl;
            int lenSub=pos+1;
            while(sub==s.substr(pos+1,lenSub)){
                pos=pos+lenSub;
                if(pos==s.length()-1) return true;
                if(pos>=s.length()) break;
            }
            pos=s.find(last,p+1);
        }
        return false;
    }
};
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