LeetCode-Rotate String

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Description：
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.

题意：定义对字符串的一次shift为——将字符串最左端的字符与最右端的字符交换位置；现在，给定两个字符串A与B，计算是否可以通过对A进行多次的shift得到字符串B；

解法：要使得A可以通过多次的shift得到字符串B，那么在A中一定可以找到一个位置，从当前位置循环遍历字符串的长度得到的结果就是字符串B；因此，我们只需要遍历字符串A，每次碰到与字符B.charAt(0)相同的字符时，我们进行上述的处理，判断是否可以得到字符串B；

Java

class Solution {
    public boolean rotateString(String A, String B) {
        if (A.length() != B.length()) return false;
        if (A.length() == 0 && B.length() == 0) return true;
        for (int i = 0; i < A.length(); i++) {
            if (A.charAt(i) != B.charAt(0)) continue;
            int index = i;
            boolean right = true;
            for (int j = 0; j < B.length(); j++) {
                if (A.charAt(index % A.length()) != B.charAt(j)) {
                    right = false;
                    break;
                }
                index++;
            }
            if (index - i == B.length() && right) {
                return true;
            }
        }
        return false;
    }
}