diff --git a/src/leetcode/0394.decode-string/Cargo.toml b/src/leetcode/0394.decode-string/Cargo.toml
new file mode 100644
index 00000000..2928b583
--- /dev/null
+++ b/src/leetcode/0394.decode-string/Cargo.toml
@@ -0,0 +1,7 @@
+[package]
+name = "lc-0394-decode-string"
+version = "0.1.0"
+edition = "2021"
+publish = false
+
+[dependencies]
diff --git a/src/leetcode/0394.decode-string/index.md b/src/leetcode/0394.decode-string/index.md
new file mode 100644
index 00000000..77a618ba
--- /dev/null
+++ b/src/leetcode/0394.decode-string/index.md
@@ -0,0 +1,4 @@
+
+# 
+
+[问题描述](../problems/)
diff --git a/src/leetcode/0394.decode-string/src/main.rs b/src/leetcode/0394.decode-string/src/main.rs
new file mode 100644
index 00000000..c17d1567
--- /dev/null
+++ b/src/leetcode/0394.decode-string/src/main.rs
@@ -0,0 +1,98 @@
+// Copyright (c) 2024 Xu Shaohua <shaohua@biofan.org>. All rights reserved.
+// Use of this source is governed by General Public License that can be found
+// in the LICENSE file.
+
+// Stack
+pub fn decode_string1(s: String) -> String {
+    assert!(!s.is_empty());
+
+    // 用于存储 '[' 之前的数字, 可能是多位数.
+    let mut num_stack: Vec<i32> = Vec::new();
+    // 存储 '[' 之前的字符串.
+    let mut str_stack: Vec<Vec<char>> = Vec::new();
+
+    let chars: Vec<char> = s.chars().collect();
+    // 存放当前的字符串.
+    let mut parts: Vec<char> = Vec::new();
+    // 存放当前的数字.
+    let mut num: i32 = 0;
+
+    for chr in chars {
+        match chr {
+            chr if chr.is_ascii_digit() => {
+                // 处理数字
+                let digit = chr.to_digit(10).unwrap() as i32;
+                num = num * 10 + digit;
+            }
+            '[' => {
+                // 将 '[' 之前的数字和字符串入栈.
+                num_stack.push(num);
+                str_stack.push(parts.clone());
+
+                // 并重置它们.
+                parts.clear();
+                num = 0;
+            }
+            ']' => {
+                // 收网, 从两个栈中分别取出整数和字符串, 进行一次拼装,
+                // 然后将拼装结果入字符串栈.
+                //
+                // curr_num 是当前字符串重复次数.
+                let curr_num = num_stack.pop().unwrap();
+                // last_parts 是 '[' 之前的字符串, 相当于当前字符串的前缀.
+                let last_parts = str_stack.pop().unwrap();
+                // 合成的新的字符串.
+                // parts = last_parts + curr_parts * curr_num.
+                let curr_parts = parts;
+                parts = last_parts;
+
+                for _i in 0..curr_num {
+                    parts.extend_from_slice(&curr_parts);
+                }
+            }
+            letter => {
+                // 拾取所有字符
+                parts.push(letter);
+            }
+        }
+    }
+
+    // 组装最后的字符串
+    parts.into_iter().collect()
+}
+
+pub type SolutionFn = fn(String) -> String;
+
+fn check_solution(func: SolutionFn) {
+    let s = "3[a]2[bc]".to_owned();
+    assert_eq!(&func(s), "aaabcbc");
+
+    let s = "3[a2[c]]".to_owned();
+    assert_eq!(&func(s), "accaccacc");
+
+    let s = "2[abc]3[cd]ef".to_owned();
+    assert_eq!(&func(s), "abcabccdcdcdef");
+
+    let s = "100[leetcode]".to_owned();
+    assert_eq!(func(s).len(), 800);
+
+    let s = "3[z]2[2[y]pq4[2[jk]e1[f]]]ef".to_owned();
+    assert_eq!(
+        &func(s),
+        "zzzyypqjkjkefjkjkefjkjkefjkjkefyypqjkjkefjkjkefjkjkefjkjkefef"
+    );
+}
+
+fn main() {
+    check_solution(decode_string1);
+}
+
+#[cfg(test)]
+mod tests {
+    use super::{check_solution, decode_string1};
+
+    #[test]
+    fn test_decode_string1() {
+        check_solution(decode_string1);
+    }
+}