## 题目描述：

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.)
```

## 解题思路：

```记字符串长度为size

next数组具有如下性质：

str[ 0 : next[i] ] == str[ i + 1 - next[i] : i + 1 ]

a, b, c, d, a, b, c, a, b, c, d, a, b, c, d, c

0, 0, 0, 0, 1, 2, 3, 1, 2, 3, 4, 5, 6, 7, 4, 0```

## Python代码：

``````class Solution(object):
def repeatedSubstringPattern(self, str):
"""
:type str: str
:rtype: bool
"""
size = len(str)
next = [0] * size
for i in range(1, size):
k = next[i - 1]
while str[i] != str[k] and k:
k = next[k - 1]
if str[i] == str[k]:
next[i] = k + 1
p = next[-1]
return p > 0 and size % (size - p) == 0
``````

```若字符串可以由其子串重复若干次构成，则子串的起点一定从原串的下标0开始

## Python代码：

``````class Solution(object):
def repeatedSubstringPattern(self, str):
"""
:type str: str
:rtype: bool
"""
size = len(str)
for x in range(1, size / 2 + 1):
if size % x:
continue
if str[:x] * (size / x) == str:
return True
return False
``````

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