feat: add rust solution to lc problem: No.3330

solution/3300-3399/3330.Find the Original Typed String I/README.md

@@ -75,7 +75,15 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：直接遍历
+
+根据题目描述，如果所有相邻字符都不相同，那么只有 1 种可能的输入字符串；如果有 1 对相邻字符相同，例如 "abbc"，那么可能的输入字符串有 2 种："abc" 和 "abbc"。
+
+依此类推，如果有 $k$ 对相邻字符相同，那么可能的输入字符串有 $k + 1$ 种。
+
+因此，我们只需要遍历字符串，统计相邻字符相同的对数再加 1 即可。
+
+时间复杂度 $O(n)$，其中 $n$ 为字符串的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -144,6 +152,16 @@ function possibleStringCount(word: string): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn possible_string_count(word: String) -> i32 {
+        1 + word.as_bytes().windows(2).filter(|w| w[0] == w[1]).count() as i32
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/3300-3399/3330.Find the Original Typed String I/README_EN.md

@@ -73,7 +73,15 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Direct Traversal
+
+According to the problem description, if all adjacent characters are different, there is only 1 possible original input string. If there is 1 pair of adjacent identical characters, such as "abbc", then there are 2 possible original strings: "abc" and "abbc".
+
+By analogy, if there are $k$ pairs of adjacent identical characters, then there are $k + 1$ possible original input strings.
+
+Therefore, we just need to traverse the string, count the number of pairs of adjacent identical characters, and add 1.
+
+The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -142,6 +150,16 @@ function possibleStringCount(word: string): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn possible_string_count(word: String) -> i32 {
+        1 + word.as_bytes().windows(2).filter(|w| w[0] == w[1]).count() as i32
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/3300-3399/3330.Find the Original Typed String I/Solution.rs

@@ -0,0 +1,5 @@
+impl Solution {
+    pub fn possible_string_count(word: String) -> i32 {
+        1 + word.as_bytes().windows(2).filter(|w| w[0] == w[1]).count() as i32
+    }
+}