[抄题]:
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
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
以为要用kmp:其实有时候简单拼一下就行了
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[一句话思路]:
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
[复杂度]:Time complexity: O(1) Space complexity: O(1)
[算法思想:迭代/递归/分治/贪心]:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
[是否头一次写此类driver funcion的代码] :
[潜台词] :
public boolean rotateString(String A, String B) { return A.length() == B.length() && (A + A).contains(B); }
原文地址:https://www.cnblogs.com/immiao0319/p/9457745.html
时间: 2024-11-05 22:38:40