feat: update solutions to lcci problems: No.16.01,16.25 (doocs#2708)

Author: yanglbme

lcci/16.01.Swap Numbers/README.md

@@ -91,17 +91,4 @@ function swapNumbers(numbers: number[]): number[] {
 
 <!-- tabs:end -->
 
-### 方法二
-
-<!-- tabs:start -->
-
-```ts
-function swapNumbers(numbers: number[]): number[] {
-    [numbers[0], numbers[1]] = [numbers[1], numbers[0]];
-    return numbers;
-}
-```
-
-<!-- tabs:end -->
-
 <!-- end -->

lcci/16.01.Swap Numbers/README_EN.md

@@ -98,17 +98,4 @@ function swapNumbers(numbers: number[]): number[] {
 
 <!-- tabs:end -->
 
-### Solution 2
-
-<!-- tabs:start -->
-
-```ts
-function swapNumbers(numbers: number[]): number[] {
-    [numbers[0], numbers[1]] = [numbers[1], numbers[0]];
-    return numbers;
-}
-```
-
-<!-- tabs:end -->
-
 <!-- end -->

lcci/16.01.Swap Numbers/Solution2.ts

@@ -23,7 +23,7 @@
 
 ### 方法一：哈希表
 
-同时遍历两个字符串，算出对应位置字符相同的个数，累加到 $x$ 中，然后将两个字符串出现的字符以及出现的次数分别记录在哈希表 $cnt1$ 和 $cnt2$ 中。
+我们同时遍历两个字符串，算出对应位置字符相同的个数，累加到 $x$ 中，然后将两个字符串出现的字符以及出现的次数分别记录在哈希表 $cnt1$ 和 $cnt2$ 中。
 
 接着遍历两个哈希表，算出有多少共同出现的字符，累加到 $y$ 中。那么答案就是 $[x, y - x]$。
 

lcci/16.15.Master Mind/README.md

@@ -45,6 +45,13 @@ class Solution:
         return [w for w in words if check(w)]
 ```
 
+```python
+class Solution:
+    def getValidT9Words(self, num: str, words: List[str]) -> List[str]:
+        trans = str.maketrans(ascii_lowercase, "22233344455566677778889999")
+        return [w for w in words if w.translate(trans) == num]
+```
+
 ```java
 class Solution {
     public List<String> getValidT9Words(String num, String[] words) {
@@ -149,17 +156,4 @@ function getValidT9Words(num: string, words: string[]): string[] {
 
 <!-- tabs:end -->
 
-### 方法二
-
-<!-- tabs:start -->
-
-```python
-class Solution:
-    def getValidT9Words(self, num: str, words: List[str]) -> List[str]:
-        trans = str.maketrans(ascii_lowercase, "22233344455566677778889999")
-        return [w for w in words if w.translate(trans) == num]
-```
-
-<!-- tabs:end -->
-
 <!-- end -->

lcci/16.20.T9/README.md

@@ -51,6 +51,13 @@ class Solution:
         return [w for w in words if check(w)]
 ```
 
+```python
+class Solution:
+    def getValidT9Words(self, num: str, words: List[str]) -> List[str]:
+        trans = str.maketrans(ascii_lowercase, "22233344455566677778889999")
+        return [w for w in words if w.translate(trans) == num]
+```
+
 ```java
 class Solution {
     public List<String> getValidT9Words(String num, String[] words) {
@@ -155,17 +162,4 @@ function getValidT9Words(num: string, words: string[]): string[] {
 
 <!-- tabs:end -->
 
-### Solution 2
-
-<!-- tabs:start -->
-
-```python
-class Solution:
-    def getValidT9Words(self, num: str, words: List[str]) -> List[str]:
-        trans = str.maketrans(ascii_lowercase, "22233344455566677778889999")
-        return [w for w in words if w.translate(trans) == num]
-```
-
-<!-- tabs:end -->
-
 <!-- end -->

lcci/16.20.T9/README_EN.md

@@ -41,11 +41,11 @@
 
 ### 方法一：哈希表 + 模拟
 
-我们使用哈希表 $black$ 来记录所有黑色方格的位置，哈希表 $dirs$ 来记录蚂蚁的四个方向。我们使用变量 $x, y$ 来记录蚂蚁的位置，使用变量 $p$ 来记录蚂蚁的方向。我们使用变量 $x1, y1, x2, y2$ 来记录所有黑色方格的最小横坐标、最小纵坐标、最大横坐标、最大纵坐标。
+我们使用哈希表 $\textit{black}$ 来记录所有黑色方格的位置，哈希表 $\textit{dirs}$ 来记录蚂蚁的四个方向。我们使用变量 $x$, $y$ 来记录蚂蚁的位置，使用变量 $p$ 来记录蚂蚁的方向。我们使用变量 $x_1$, $y_1$, $x_2$, $y_2$ 来记录所有黑色方格的最小横坐标、最小纵坐标、最大横坐标、最大纵坐标。
 
-我们模拟蚂蚁的行走过程。如果蚂蚁所在的方格是白色的，那么蚂蚁向右转 $90$ 度，将方格涂黑，向前移动一个单位。如果蚂蚁所在的方格是黑色的，那么蚂蚁向左转 $90$ 度，将方格涂白，向前移动一个单位。在模拟的过程中，我们不断更新 $x1, y1, x2, y2$ 的值，使得它们能够包含蚂蚁走过的所有方格。
+我们模拟蚂蚁的行走过程。如果蚂蚁所在的方格是白色的，那么蚂蚁向右转 $90$ 度，将方格涂黑，向前移动一个单位。如果蚂蚁所在的方格是黑色的，那么蚂蚁向左转 $90$ 度，将方格涂白，向前移动一个单位。在模拟的过程中，我们不断更新 $x_1$, $y_1$, $x_2$, $y_2$ 的值，使得它们能够包含蚂蚁走过的所有方格。
 
-模拟结束后，我们根据 $x1, y1, x2, y2$ 的值，构造出答案矩阵 $g$。然后，我们将蚂蚁所在的位置涂上蚂蚁的方向，同时将所有黑色方格涂上 $X$，最后返回答案矩阵。
+模拟结束后，我们根据 $x_1$, $y_1$, $x_2$, $y_2$ 的值，构造出答案矩阵 $g$。然后，我们将蚂蚁所在的位置涂上蚂蚁的方向，同时将所有黑色方格涂上 $X$，最后返回答案矩阵。
 
 时间复杂度 $O(K)$，空间复杂度 $O(K)$。其中 $K$ 是蚂蚁行走的步数。
 

lcci/16.22.Langtons Ant/README.md

@@ -61,13 +61,13 @@
 
 ### Solution 1: Hash Table + Simulation
 
-We use a hash table $black$ to record the positions of all black squares, and a hash table $dirs$ to record the four directions of the ant. We use variables $x, y$ to record the position of the ant, and variable $p$ to record the direction of the ant. We use variables $x1, y1, x2, y2$ to record the minimum horizontal coordinate, minimum vertical coordinate, maximum horizontal coordinate, and maximum vertical coordinate of all black squares.
+We use a hash table `black` to record the positions of all black squares, and a hash table `dirs` to record the four directions of the ant. We use variables $x$, $y$ to record the position of the ant, and variable $p$ to record the direction of the ant. We use variables $x_1$, $y_1$, $x_2$, $y_2$ to record the minimum x-coordinate, minimum y-coordinate, maximum x-coordinate, and maximum y-coordinate of all black squares, respectively.
 
-We simulate the ant's walking process. If the square where the ant is located is white, the ant turns right by $90$ degrees, paints the square black, and moves forward one unit. If the square where the ant is located is black, the ant turns left by $90$ degrees, paints the square white, and moves forward one unit. During the simulation, we continuously update the values of $x1, y1, x2, y2$ so that they can contain all the squares the ant has walked through.
+We simulate the walking process of the ant. If the square where the ant is located is white, then the ant turns right by $90$ degrees, paints the square black, and moves forward one unit. If the square where the ant is located is black, then the ant turns left by $90$ degrees, paints the square white, and moves forward one unit. During the simulation process, we continuously update the values of $x_1$, $y_1$, $x_2$, $y_2$ so that they can include all the squares that the ant has walked through.
 
-After the simulation is over, we construct the answer matrix $g$ based on the values of $x1, y1, x2, y2$. Then, we paint the direction of the ant on the square where the ant is located, paint all black squares with $X$, and finally return the answer matrix.
+After the simulation, we construct the answer matrix $g$ based on the values of $x_1$, $y_1$, $x_2$, $y_2$. Then, we paint the direction of the ant at the ant's position, and paint all black squares with $X$, and finally return the answer matrix.
 
-The time complexity is $O(K)$, and the space complexity is $O(K)$. Here, $K$ is the number of steps the ant walks.
+The time complexity is $O(K)$, and the space complexity is $O(K)$. Where $K$ is the number of steps the ant walks.
 
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