feat: update solutions to lc problems: No.0189,0238

Author: yanglbme

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

@@ -38,7 +38,7 @@ tags:
 <pre>
 <strong>输入：</strong>nums = [-1,-100,3,99], k = 2
 <strong>输出：</strong>[3,99,-1,-100]
-<strong>解释:</strong> 
+<strong>解释:</strong>
 向右轮转 1 步: [99,-1,-100,3]
 向右轮转 2 步: [3,99,-1,-100]</pre>
 
@@ -248,23 +248,4 @@ public class Solution {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def rotate(self, nums: List[int], k: int) -> None:
-        k %= len(nums)
-        nums[:] = nums[-k:] + nums[:-k]
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

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

@@ -37,7 +37,7 @@ rotate 3 steps to the right: [5,6,7,1,2,3,4]
 <pre>
 <strong>Input:</strong> nums = [-1,-100,3,99], k = 2
 <strong>Output:</strong> [3,99,-1,-100]
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 rotate 1 steps to the right: [99,-1,-100,3]
 rotate 2 steps to the right: [3,99,-1,-100]
 </pre>
@@ -246,23 +246,4 @@ public class Solution {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### Solution 2
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def rotate(self, nums: List[int], k: int) -> None:
-        k %= len(nums)
-        nums[:] = nums[-k:] + nums[:-k]
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/0100-0199/0189.Rotate Array/Solution2.py

@@ -61,15 +61,15 @@ tags:
 
 ### 方法一：两次遍历
 
-我们定义两个变量 $left$ 和 $right$，分别表示当前元素左边所有元素的乘积和右边所有元素的乘积。初始时 $left=1$, $right=1$。定义一个长度为 $n$ 的答案数组 $ans$。
+我们定义两个变量 $\textit{left}$ 和 $\textit{right}$，分别表示当前元素左边所有元素的乘积和右边所有元素的乘积。初始时 $\textit{left}=1$, $\textit{right}=1$。定义一个长度为 $n$ 的答案数组 $\textit{ans}$。
 
-我们先从左到右遍历数组，对于遍历到的第 $i$ 个元素，我们用 $left$ 更新 $ans[i]$，然后 $left$ 乘以 $nums[i]$。
+我们先从左到右遍历数组，对于遍历到的第 $i$ 个元素，我们用 $\textit{left}$ 更新 $\textit{ans}[i]$，然后 $\textit{left}$ 乘以 $\textit{nums}[i]$。
 
-然后，我们从右到左遍历数组，对于遍历到的第 $i$ 个元素，我们更新 $ans[i]$ 为 $ans[i] \times right$，然后 $right$ 乘以 $nums[i]$。
+然后，我们从右到左遍历数组，对于遍历到的第 $i$ 个元素，我们更新 $\textit{ans}[i]$ 为 $\textit{ans}[i] \times \textit{right}$，然后 $\textit{right}$ 乘以 $\textit{nums}[i]$。
 
-遍历结束后，数组 `ans` 即为所求的答案。
+遍历结束后，返回答案数组 $\textit{ans}$。
 
-时间复杂度 $O(n)$，其中 $n$ 是数组 `nums` 的长度。忽略答案数组的空间消耗，空间复杂度 $O(1)$。
+时间复杂度 $O(n)$，其中 $n$ 是数组 $\textit{nums}$ 的长度。忽略答案数组的空间消耗，空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -258,22 +258,4 @@ class Solution {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### TypeScript
-
-```ts
-function productExceptSelf(nums: number[]): number[] {
-    return nums.map((_, i) => nums.reduce((pre, val, j) => pre * (i === j ? 1 : val), 1));
-}
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/0200-0299/0238.Product of Array Except Self/README.md

@@ -51,15 +51,15 @@ tags:
 
 ### Solution 1: Two Passes
 
-We define two variables $left$ and $right$, which represent the product of all elements to the left and right of the current element respectively. Initially, $left=1$, $right=1$. Define an answer array $ans$ of length $n$.
+We define two variables $\textit{left}$ and $\textit{right}$ to represent the product of all elements to the left and right of the current element, respectively. Initially, $\textit{left} = 1$ and $\textit{right} = 1$. We define an answer array $\textit{ans}$ of length $n$.
 
-We first traverse the array from left to right, for the $i$th element we update $ans[i]$ with $left$, then $left$ multiplied by $nums[i]$.
+First, we traverse the array from left to right. For the $i$-th element, we update $\textit{ans}[i]$ with $\textit{left}$, then multiply $\textit{left}$ by $\textit{nums}[i]$.
 
-Then, we traverse the array from right to left, for the $i$th element, we update $ans[i]$ to $ans[i] \times right$, then $right$ multiplied by $nums[i]$.
+Next, we traverse the array from right to left. For the $i$-th element, we update $\textit{ans}[i]$ to $\textit{ans}[i] \times \textit{right}$, then multiply $\textit{right}$ by $\textit{nums}[i]$.
 
-After the traversal, the array `ans` is the answer.
+After the traversal, we return the answer array $\textit{ans}$.
 
-The time complexity is $O(n)$, where $n$ is the length of the array `nums`. Ignore the space consumption of the answer array, the space complexity is $O(1)$.
+The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -248,22 +248,4 @@ class Solution {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### Solution 2
-
-<!-- tabs:start -->
-
-#### TypeScript
-
-```ts
-function productExceptSelf(nums: number[]): number[] {
-    return nums.map((_, i) => nums.reduce((pre, val, j) => pre * (i === j ? 1 : val), 1));
-}
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->