字符串轮转

给定两个字符串s1和s2，请编写代码检查s2是否为s1旋转而成（比如，waterbottle是erbottlewat旋转后的字符串）。

示例1:

•  输入：s1 = "waterbottle", s2 = "erbottlewat"
•  输出：True

示例2:

•  输入：s1 = "aa", s2 = "aba"
•  输出：False

示例代码1：

class Solution(object):
    def isFlipedString(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: bool
        """
        if len(s1) != len(s2):
            return False
        s3 = s1 + s1
        return s2 in s3


# s1 = "waterbottle"
# s2 = "erbottlewat"
s1 = "aa"
s2 = "aba"
a = Solution()
b = a.isFlipedString(s1, s2)
print(b)

示例代码2：

class Solution(object):
    def isFlipedString(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: bool
        """
        return len(s1) == len(s2) and s2 in s1 * 2


# s1 = "waterbottle"
# s2 = "erbottlewat"
s1 = "aa"
s2 = "aba"
a = Solution()
b = a.isFlipedString(s1, s2)
print(b)

运行效果：

思路解析：

方法一：
倍增s2，若长度相同的情况下，s1一定为s3的一个子串
复杂度分析

• 时间复杂度：O(N)
• 空间复杂度：O(N)