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feat: update solutions to leetcode problem: No.0189
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‎README.md

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- [LeetCode 《剑指 Offer（第 2 版）》](/lcof/README.md)
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- [LeetCode 《程序员面试金典（第 6 版）》](/lcci/README.md)
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## 面试高频题
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## 面试高频考题
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### 数组
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1. [数组中重复的数字](./lcof/面试题03.%20数组中重复的数字/README.md)
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1. [旋转数组](./solution/0100-0199/0189.Rotate%20Array/README.md)
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### 链表
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<a href="https://opencollective.com/doocs-leetcode/contributors.svg?width=890&button=true" target="_blank"><img src="https://opencollective.com/doocs-leetcode/contributors.svg?width=890&button=false" /></a>
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## 赞助人
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## 赞助者
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特别感谢以下个人、组织对本项目的赞助！
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<a href="https://opencollective.com/doocs-leetcode/backers.svg?width=890" target="_blank"><img src="https://opencollective.com/doocs-leetcode/backers.svg?width=890"></a>
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> You help the developer community practice for interviews, and there is nothing better we could ask for. -- [Alan Yessenbayev](https://opencollective.com/alan-yessenbayev)
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<a href="https://opencollective.com/doocs-leetcode/sponsors.svg?width=890" target="_blank"><img src="https://opencollective.com/doocs-leetcode/sponsors.svg?width=890"></a>
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## 赞助商
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<a href="https://opencollective.com/doocs-leetcode/sponsors.svg?width=890" target="_blank"><img src="https://opencollective.com/doocs-leetcode/sponsors.svg?width=890"></a>
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> You help the developer community practice for interviews, and there is nothing better we could ask for. -- [Alan Yessenbayev](https://opencollective.com/alan-yessenbayev)
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## 许可证
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‎README_EN.md

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<a href="https://opencollective.com/doocs-leetcode/contributors.svg?width=890&button=true" target="_blank"><img src="https://opencollective.com/doocs-leetcode/contributors.svg?width=890&button=false" /></a>
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## Backers
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## Backers & Sponsors
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Thank you to all our backers!
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Thank you to all our backers and sponsors!
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<a href="https://opencollective.com/doocs-leetcode#backers" target="_blank"><img src="https://opencollective.com/doocs-leetcode/backers.svg?width=890"></a>
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<a href="https://opencollective.com/doocs-leetcode/backers.svg?width=890" target="_blank"><img src="https://opencollective.com/doocs-leetcode/backers.svg?width=890"></a>
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> You help the developer community practice for interviews, and there is nothing better we could ask for. -- [Alan Yessenbayev](https://opencollective.com/alan-yessenbayev)
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## Sponsors
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<a href="https://opencollective.com/doocs-leetcode/sponsors.svg?width=890" target="_blank"><img src="https://opencollective.com/doocs-leetcode/sponsors.svg?width=890"></a>
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Support this project by becoming a sponsor. Your logo will show up here with a link to your website.
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<a href="https://opencollective.com/doocs-leetcode#sponsors" target="_blank"><img src="https://opencollective.com/doocs-leetcode/sponsors.svg?width=890"></a>
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> You help the developer community practice for interviews, and there is nothing better we could ask for. -- [Alan Yessenbayev](https://opencollective.com/alan-yessenbayev)
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## License
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‎solution/0100-0199/0189.Rotate Array/README.md

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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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若 `k=3`，`nums=[1,2,3,4,5,6,7]`。
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先将 `nums` 整体翻转：
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`[1,2,3,4,5,6,7]` -> `[7,6,5,4,3,2,1]`
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再翻转 `0~k-1` 范围内的元素：
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`[7,6,5,4,3,2,1]` -> `[5,6,7,4,3,2,1]`
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最后翻转 `k~n-1` 范围内的元素，即可得到最终结果：
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`[5,6,7,4,3,2,1]` -> `[5,6,7,1,2,3,4]`
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def rotate(self, nums: List[int], k: int) -> None:
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"""
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Do not return anything, modify nums in-place instead.
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"""
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n = len(nums)
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k %= n
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if n < 2 or k == 0:
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return
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nums[:] = nums[::-1]
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nums[:k] = nums[:k][::-1]
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nums[k:] = nums[k:][::-1]
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public void rotate(int[] nums, int k) {
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if (nums == null) {
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return;
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}
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int n = nums.length;
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k %= n;
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if (n < 2 || k == 0) {
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return;
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}
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rotate(nums, 0, n - 1);
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rotate(nums, 0, k - 1);
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rotate(nums, k, n - 1);
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}
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private void rotate(int[] nums, int i, int j) {
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while (i < j) {
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int t = nums[i];
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nums[i] = nums[j];
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nums[j] = t;
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++i;
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--j;
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}
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}
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}
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```
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### **...**

‎solution/0100-0199/0189.Rotate Array/README_EN.md

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<ul>
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<li>Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.</li>
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<li>Could you do it in-place with O(1) extra space?</li>
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</ul>
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### **Python3**
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```python
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class Solution:
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def rotate(self, nums: List[int], k: int) -> None:
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"""
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Do not return anything, modify nums in-place instead.
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"""
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n = len(nums)
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k %= n
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if n < 2 or k == 0:
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return
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nums[:] = nums[::-1]
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nums[:k] = nums[:k][::-1]
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nums[k:] = nums[k:][::-1]
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```
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### **Java**
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```java
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class Solution {
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public void rotate(int[] nums, int k) {
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if (nums == null) {
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return;
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}
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int n = nums.length;
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k %= n;
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if (n < 2 || k == 0) {
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return;
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}
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rotate(nums, 0, n - 1);
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rotate(nums, 0, k - 1);
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rotate(nums, k, n - 1);
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}
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private void rotate(int[] nums, int i, int j) {
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while (i < j) {
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int t = nums[i];
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nums[i] = nums[j];
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nums[j] = t;
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++i;
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--j;
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}
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}
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}
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```
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### **...**

‎solution/0100-0199/0189.Rotate Array/Solution.java

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class Solution {
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public void rotate(int[] nums, int k) {
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if (nums == null) {
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return;
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}
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int n = nums.length;
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k %= n;
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if (n < 2 || k == 0) {
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return;
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}
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rotate(nums, 0, n - 1);
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rotate(nums, 0, k - 1);
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rotate(nums, k, n - 1);
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}
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private void rotate(int[] nums, int start, int end) {
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while (start < end) {
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int t = nums[start];
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nums[start] = nums[end];
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nums[end] = t;
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++start;
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--end;
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private void rotate(int[] nums, int i, int j) {
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while (i < j) {
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int t = nums[i];
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nums[i] = nums[j];
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nums[j] = t;
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++i;
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--j;
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}
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}
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}

‎solution/0100-0199/0189.Rotate Array/Solution.py

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class Solution:
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def rotate(self, nums: List[int], k: int) -> None:
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"""
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Do not return anything, modify nums in-place instead.
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"""
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n = len(nums)
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k %= n
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if n < 2 or k == 0:
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return
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nums[:] = nums[::-1]
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nums[:k] = nums[:k][::-1]
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nums[k:] = nums[k:][::-1]

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