[leetcode] 796. Rotate String @ python

最新推荐文章于 2021-04-19 18:41:12 发布

原创最新推荐文章于 2021-04-19 18:41:12 发布 · 170 阅读

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本文介绍了一种判断两个字符串是否可以通过旋转操作相互转换的方法。通过遍历并比较字符，尝试重构源字符串来匹配目标字符串，实现高效判断。

原题

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.

解法

遍历A, 如果某个字符与B的首字符相等, 尝试将A重构成B, 如果重构成功,则返回True.
Time: O(n)
Space: O(n)

代码

class Solution(object):
    def rotateString(self, A, B):
        """
        :type A: str
        :type B: str
        :rtype: bool
        """
        # base case
        if A == B: return True
        # A takes some shift
        for i, char in enumerate(A):
            if char == B[0]:
                # try to reconstruct B
                left, right = A[:i], A[i:]
                if right + left == B:
                    return True
        return False