diff --git a/solution/3400-3499/3407.Substring Matching Pattern/README.md b/solution/3400-3499/3407.Substring Matching Pattern/README.md
index 39ed56168bf5a..53158bf41521f 100644
--- a/solution/3400-3499/3407.Substring Matching Pattern/README.md	
+++ b/solution/3400-3499/3407.Substring Matching Pattern/README.md	
@@ -80,7 +80,13 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：字符串匹配
+
+根据题目描述，`*` 可以被替换为零个或多个字符组成的任意字符序列，因此我们可以将模式字符串 $p$ 按照 `*` 分割成若干个子串，如果这些子串依次出现在字符串 $s$ 中且顺序不变，则说明 $p$ 可以变成 $s$ 的子字符串。
+
+因此，我们首先初始化一个指针 $i$ 指向字符串 $s$ 的起始位置，然后遍历模式字符串 $p$ 按照 `*` 分割得到的每个子串 $t$，在字符串 $s$ 中从位置 $i$ 开始查找子串 $t$，如果找到了，则将指针 $i$ 移动到子串 $t$ 的末尾位置继续查找下一个子串；如果找不到，则说明模式字符串 $p$ 不能变成字符串 $s$ 的子字符串，返回 $\text{false}$。如果所有子串都找到了，则返回 $\text{true}$。
+
+时间复杂度 $O(n \times m)$，空间复杂度 $O(m)$。其中 $n$ 和 $m$ 分别是字符串 $s$ 和模式字符串 $p$ 的长度。
 
 <!-- tabs:start -->
 
diff --git a/solution/3400-3499/3407.Substring Matching Pattern/README_EN.md b/solution/3400-3499/3407.Substring Matching Pattern/README_EN.md
index eeedc9ccf096c..48c6b6f6d8a86 100644
--- a/solution/3400-3499/3407.Substring Matching Pattern/README_EN.md	
+++ b/solution/3400-3499/3407.Substring Matching Pattern/README_EN.md	
@@ -78,7 +78,13 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: String Matching
+
+According to the problem description, `*` can be replaced by any sequence of zero or more characters, so we can split the pattern string $p$ by `*` into several substrings. If these substrings appear in order in the string $s$, then $p$ can become a substring of $s$.
+
+Therefore, we first initialize a pointer $i$ to the start of string $s$, then iterate over each substring $t$ obtained by splitting the pattern string $p$ by `*`. For each $t$, we search for it in $s$ starting from position $i$. If it is found, we move the pointer $i$ to the end of $t$ and continue searching for the next substring. If it is not found, it means the pattern string $p$ cannot become a substring of $s$, and we return $\text{false}$. If all substrings are found, we return $\text{true}$.
+
+The time complexity is $O(n \times m)$, and the space complexity is $O(m)$, where $n$ and $m$ are the lengths of strings $s$ and $p$, respectively.
 
 <!-- tabs:start -->
 
@@ -174,6 +180,24 @@ function hasMatch(s: string, p: string): boolean {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn has_match(s: String, p: String) -> bool {
+        let mut i = 0usize;
+        for t in p.split('*') {
+            if let Some(j) = s[i..].find(t) {
+                i += j + t.len();
+            } else {
+                return false;
+            }
+        }
+        true
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->
diff --git a/solution/3400-3499/3407.Substring Matching Pattern/Solution.rs b/solution/3400-3499/3407.Substring Matching Pattern/Solution.rs
new file mode 100644
index 0000000000000..4aa4cab8c8d68
--- /dev/null
+++ b/solution/3400-3499/3407.Substring Matching Pattern/Solution.rs	
@@ -0,0 +1,13 @@
+impl Solution {
+    pub fn has_match(s: String, p: String) -> bool {
+        let mut i = 0usize;
+        for t in p.split('*') {
+            if let Some(j) = s[i..].find(t) {
+                i += j + t.len();
+            } else {
+                return false;
+            }
+        }
+        true
+    }
+}