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‎Index/模拟.md‎

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| [762. 二进制表示中质数个计算置位](https://leetcode-cn.com/problems/prime-number-of-set-bits-in-binary-representation/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/prime-number-of-set-bits-in-binary-representation/solution/by-ac_oier-w50x/) | 简单 | 🤩🤩🤩🤩 |
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| [766. 托普利茨矩阵](https://leetcode-cn.com/problems/toeplitz-matrix/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/toeplitz-matrix/solution/cong-ci-pan-du-qu-cheng-ben-fen-xi-liang-f20w/) | 简单 | 🤩🤩🤩 |
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| [794. 有效的井字游戏](https://leetcode-cn.com/problems/valid-tic-tac-toe-state/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/valid-tic-tac-toe-state/solution/gong-shui-san-xie-fen-qing-kuang-tao-lun-pikn/) | 中等 | 🤩🤩🤩🤩 |
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| [796. 旋转字符串](https://leetcode-cn.com/problems/rotate-string/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/rotate-string/solution/by-ac_oier-bnkx/) | 简单 | 🤩🤩🤩 |
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| [846. 一手顺子](https://leetcode-cn.com/problems/hand-of-straights/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/hand-of-straights/solution/gong-shui-san-xie-shu-ju-jie-gou-mo-ni-t-4hxw/) | 中等 | 🤩🤩🤩 |
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| [859. 亲密字符串](https://leetcode-cn.com/problems/buddy-strings/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/buddy-strings/solution/gong-shui-san-xie-jian-dan-zi-fu-chuan-m-q056/) | 简单 | 🤩🤩🤩🤩🤩 |
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| [867. 转置矩阵](https://leetcode-cn.com/problems/transpose-matrix/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/transpose-matrix/solution/yi-you-wei-jin-huo-xu-ni-neng-kan-kan-zh-m53m/) | 简单 | 🤩🤩🤩🤩 |
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### 题目描述
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这是 LeetCode 上的 **[796. 旋转字符串](https://leetcode-cn.com/problems/rotate-string/solution/by-ac_oier-bnkx/)** ,难度为 **简单**。
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Tag : 「模拟」
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给定两个字符串,`s` 和 `goal`。如果在若干次旋转操作之后,`s` 能变成 `goal`,那么返回 `true`。
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`s` 的旋转操作就是将 `s` 最左边的字符移动到最右边。
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* 例如, 若 `s = 'abcde'`,在旋转一次之后结果就是`'bcdea'` 。
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示例 1:
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```
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输入: s = "abcde", goal = "cdeab"
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输出: true
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```
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示例 2:
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```
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输入: s = "abcde", goal = "abced"
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输出: false
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```
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提示:
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* 1ドル <= s.length, goal.length <= 100$
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* `s` 和 `goal` 由小写英文字母组成
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---
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### 模拟
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由于每次旋转操作都是将最左侧字符移动到最右侧,因此如果 `goal` 可由 `s` 经过多步旋转而来,那么 `goal` 必然会出现在 `s + s` 中,即满足 `(s + s).contains(goal)`,同时为了 `s` 本身过长导致的结果成立,我们需要先确保两字符串长度相等。
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代码:
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```Java
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class Solution {
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public boolean rotateString(String s, String goal) {
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return s.length() == goal.length() && (s + s).contains(goal);
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}
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}
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```
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* 时间复杂度:$O(n)$
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* 空间复杂度:$O(n)$
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---
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### 最后
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这是我们「刷穿 LeetCode」系列文章的第 `No.796` 篇,系列开始于 2021年01月01日,截止于起始日 LeetCode 上共有 1916 道题目,部分是有锁题,我们将先把所有不带锁的题目刷完。
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在这个系列文章里面,除了讲解解题思路以外,还会尽可能给出最为简洁的代码。如果涉及通解还会相应的代码模板。
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为了方便各位同学能够电脑上进行调试和提交代码,我建立了相关的仓库:https://github.com/SharingSource/LogicStack-LeetCode 。
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在仓库地址里,你可以看到系列文章的题解链接、系列文章的相应代码、LeetCode 原题链接和其他优选题解。
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