796. Rotate String* (旋转字符串)

796. Rotate String* (旋转字符串)

https://leetcode.com/problems/rotate-string/

题目描述

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.

C++ 实现 1

最简洁的实现方法是, 利用 A + A 组成的字符串可以表示出 A 的所有经 shift 生成的字符串, 比如:

A = 'abc'
A + A = 'abcabc'

在该字符串中, 'abc', 'bca', 'cab' 均存在

此时只需要判断 B 是否在 A + A 中:

class Solution {
    
    
public:
    bool rotateString(string A, string B) {
    
    
        return (A.size() == B.size()) && (A + A).find(B) != -1;
    }
};

C++ 实现 2

如果实在没有想到上面的解法, 使用很直接的思路, 将 shift 实现, 然后判断每个 shift 生成的字符串是否和 B 相等.

class Solution {
    
    
private:
    string shift(const string &s) {
    
    
        if (s.empty()) return s;
        auto size = s.size();
        return s.substr(size - 1, 1) + s.substr(0, size - 1);
    }
public:
    bool rotateString(string A, string B) {
    
    
        if (A == B) return true;
        auto num = A.size();
        for (int i = 1; i < num; ++ i) {
    
    
            A = shift(A);
            if (A == B) return true;
        }
        return false;
    }
};