leetcode#796. Rotate String

We are given two strings, A and B.

A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = ‘abcde’, then it will be ‘bcdea’ after one shift on A. Return True if and only if A can become B after some number of shifts on A.

Example 1:
Input: A = ‘abcde’, B = ‘cdeab’
Output: true

Example 2:
Input: A = ‘abcde’, B = ‘abced’
Output: false

Note:

A and B will have length at most 100.

题目大意：给定两个字符串A和B，让你判断B是否能由A循环左移得到

思路：

暴力枚举起点i （就是把i移动到初始位置）然后看看和B是否相同。复杂度O(n^2)

其实这个过程和在A+A中寻找B是一样的。

class Solution {
public:
    bool rotateString(string A, string B) {
        return A.size() == B.size() && (A + A).find(B) != string::npos;
    }
};