leetcode 459. Repeated Substring Pattern

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.)

class Solution(object):
    def repeatedSubstringPattern(self, s):
        """
        :type s: str
        :rtype: bool
        """
        #return True if re.match(r"(\w+)*", s) else False
        def is_repeat(s1, s2):            
            return s1 == s2*(len(s1)/len(s2))            
         
        l = len(s)
        for i in xrange(1, l/2+1):
            if l % i == 0:
                if is_repeat(s, s[:i]):
                    return True
        return False

上面是暴力解法，下面是技巧解法：

class Solution(object):
    def repeatedSubstringPattern(self, s):
        """
        :type s: str
        :rtype: bool
        """
        #return True if re.match(r"(\w+)*", s) else False
        if not s:
            return False           
        ss = (s + s)[1:-1]
        return ss.find(s) != -1

解释：

Let's say T = S + S.
"S is Repeated => From T[1:-1] we can find S" is obvious.

If from T[1:-1] we found S at index p-1, which is index p in T and S.
let s1 = S[:p], S can be represented as s1s2...sn, where si stands for substring rather than character.
then we know T[p:len(S) + p] = s2s3...sn-1sns1 = S = s1s2...sn-2sn-1sn.
So s1 = s2, s2 = s3, ..., sn-1 = sn, sn = s1,Which means S is Repeated.