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# 796.Rotate String

**796.Rotate String**

难度:Easy

> 给定两个字符串, A 和 B。

A 的旋转操作就是将 A 最左边的字符移动到最右边。 例如, 若 A = 'abcde'，在移动一次之后结果就是'bcdea' 。如果在若干次旋转操作之后，A 能变成B，那么返回True。

```
示例 1:
输入: A = 'abcde', B = 'cdeab'
输出: true

示例 2:
输入: A = 'abcde', B = 'abced'
输出: false
注意：

A 和 B 长度不超过 100。
```

最直观的方法，把A旋转后的所有字符串存到集合中，然后判断是否有B。 另一个方法构造2A，如果可以通过旋转得到，则B必然在2A的字符串中。\
就写第一种方法了:

```
class Solution {
public:
    bool rotateString(string A, string B) {

        if(A.length()!=B.length()) return false;
        unordered_set<string> clusterA;
        clusterA.insert(A);
        int len=A.length();
        for(int k=1;k<len;k++)
        {
            string tmp;
            for(int i=k;i<len;i++ )
                tmp+=A[i];
            for(int i=0;i<k;i++)
                tmp+=A[i];
            clusterA.insert(tmp);
            //cout<<tmp<<endl;
        }
        return clusterA.count(B);
        
    }
};
```

> 执行用时 :12ms, 在所有 C++ 提交中击败了7.97%的用户\
> 内存消耗 :10.1MB, 在所有 C++ 提交中击败了5.62%的用户


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