diff --git a/solution/0000-0099/0032.Longest Valid Parentheses/README.md b/solution/0000-0099/0032.Longest Valid Parentheses/README.md
index 26374b354ad89..ce178465d30da 100644
--- a/solution/0000-0099/0032.Longest Valid Parentheses/README.md	
+++ b/solution/0000-0099/0032.Longest Valid Parentheses/README.md	
@@ -73,6 +73,20 @@ $$
 
 时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串的长度。
 
+**方法二：使用栈**
+
+-   使用栈来存储左括号的索引，栈底元素初始化为 `-1`，用于辅助计算有效括号的长度。
+-   遍历字符串，对于每个字符：
+    -   如果是左括号，将当前位置压入栈。
+    -   如果是右括号，弹出栈顶元素表示匹配了一个左括号。
+        -   如果栈为空，说明当前右括号无法匹配，将当前位置压入栈作为新的起点。
+        -   如果栈不为空，计算当前有效括号子串的长度，更新最大长度。
+-   最终返回最大长度。
+
+总结：这个算法的关键在于维护一个线，栈内存放的是左括号的索引，通过弹出和压入的操作来更新有效括号子串的长度。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串的长度。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -95,6 +109,23 @@ class Solution:
         return max(f)
 ```
 
+```python
+class Solution:
+    def longestValidParentheses(self, s: str) -> int:
+        stack = [-1]
+        ans = 0
+        for i in range(len(s)):
+            if s[i] == '(':
+                stack.append(i)
+            else:
+                stack.pop()
+                if not stack:
+                    stack.append(i)
+                else:
+                    ans = max(ans, i - stack[-1])
+        return ans
+```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -168,6 +199,28 @@ func longestValidParentheses(s string) int {
 }
 ```
 
+```go
+func longestValidParentheses(s string) int {
+	ans := 0
+	stack := []int{-1}
+	for i, v := range s {
+		if v == '(' {
+			stack = append(stack, i)
+		} else {
+			stack = stack[:len(stack)-1]
+			if len(stack) == 0 {
+				stack = append(stack, i)
+			} else {
+				if ans < i-stack[len(stack)-1] {
+					ans = i - stack[len(stack)-1]
+				}
+			}
+		}
+	}
+	return ans
+}
+```
+
 ### **C#**
 
 ```cs
@@ -216,6 +269,28 @@ function longestValidParentheses(s: string): number {
 }
 ```
 
+```ts
+function longestValidParentheses(s: string): number {
+    let max_length: number = 0;
+    const stack: number[] = [-1];
+    for (let i = 0; i < s.length; i++) {
+        if (s.charAt(i) == '(') {
+            stack.push(i);
+        } else {
+            stack.pop();
+
+            if (stack.length === 0) {
+                stack.push(i);
+            } else {
+                max_length = Math.max(max_length, i - stack[stack.length - 1]);
+            }
+        }
+    }
+
+    return max_length;
+}
+```
+
 ### **JavaScript**
 
 ```js
@@ -242,6 +317,90 @@ var longestValidParentheses = function (s) {
 };
 ```
 
+```js
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var longestValidParentheses = function (s) {
+    let ans = 0;
+    const stack = [-1];
+    for (i = 0; i < s.length; i++) {
+        if (s.charAt(i) === '(') {
+            stack.push(i);
+        } else {
+            stack.pop();
+            if (stack.length === 0) {
+                stack.push(i);
+            } else {
+                ans = Math.max(ans, i - stack[stack.length - 1]);
+            }
+        }
+    }
+    return ans;
+};
+```
+
+### Rust
+
+```rust
+impl Solution {
+    pub fn longest_valid_parentheses(s: String) -> i32 {
+        let mut ans = 0;
+        let mut f = vec![0; s.len() + 1];
+        for i in 2..=s.len() {
+            if
+                s
+                    .chars()
+                    .nth(i - 1)
+                    .unwrap() == ')'
+            {
+                if
+                    s
+                        .chars()
+                        .nth(i - 2)
+                        .unwrap() == '('
+                {
+                    f[i] = f[i - 2] + 2;
+                } else if
+                    (i as i32) - f[i - 1] - 1 > 0 &&
+                    s
+                        .chars()
+                        .nth(i - (f[i - 1] as usize) - 2)
+                        .unwrap() == '('
+                {
+                    f[i] = f[i - 1] + 2 + f[i - (f[i - 1] as usize) - 2];
+                }
+                ans = ans.max(f[i]);
+            }
+        }
+        ans
+    }
+}
+```
+
+```rust
+impl Solution {
+    pub fn longest_valid_parentheses(s: String) -> i32 {
+        let mut stack = vec![-1];
+        let mut res = 0;
+        for i in 0..s.len() {
+            if let Some('(') = s.chars().nth(i) {
+                stack.push(i as i32);
+            } else {
+                stack.pop().unwrap();
+                if stack.is_empty() {
+                    stack.push(i as i32);
+                } else {
+                    res = std::cmp::max(res, (i as i32) - stack.last().unwrap());
+                }
+            }
+        }
+        res
+    }
+}
+```
+
 ### **...**
 
 ```
diff --git a/solution/0000-0099/0032.Longest Valid Parentheses/README_EN.md b/solution/0000-0099/0032.Longest Valid Parentheses/README_EN.md
index 78a9d2b669aa0..e4d4b974df5e9 100644
--- a/solution/0000-0099/0032.Longest Valid Parentheses/README_EN.md	
+++ b/solution/0000-0099/0032.Longest Valid Parentheses/README_EN.md	
@@ -67,6 +67,21 @@ Finally, we only need to return $max(f)$.
 
 The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the string.
 
+**Solution 2: Using Stack**
+
+-   Maintain a stack to store the indices of left parentheses. Initialize the bottom element of the stack with the value -1 to facilitate the calculation of the length of valid parentheses.
+-   Iterate through each element of the string:
+    -   If the character is a left parenthesis, push the index of the character onto the stack.
+    -   If the character is a right parenthesis, pop an element from the stack to represent that we have found a valid pair of parentheses.
+        -   If the stack is empty, it means we couldn't find a left parenthesis to match the right parenthesis. In this case, push the index of the character as a new starting point.
+        -   If the stack is not empty, calculate the length of the valid parentheses and update it.
+
+Summary:
+
+The key to this algorithm is to maintain a stack to store the indices of left parentheses and then update the length of the valid substring of parentheses by pushing and popping elements.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the string.
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -87,6 +102,23 @@ class Solution:
         return max(f)
 ```
 
+```python
+class Solution:
+    def longestValidParentheses(self, s: str) -> int:
+        stack = [-1]
+        ans = 0
+        for i in range(len(s)):
+            if s[i] == '(':
+                stack.append(i)
+            else:
+                stack.pop()
+                if not stack:
+                    stack.append(i)
+                else:
+                    ans = max(ans, i - stack[-1])
+        return ans
+```
+
 ### **Java**
 
 ```java
@@ -158,6 +190,28 @@ func longestValidParentheses(s string) int {
 }
 ```
 
+```go
+func longestValidParentheses(s string) int {
+	ans := 0
+	stack := []int{-1}
+	for i, v := range s {
+		if v == '(' {
+			stack = append(stack, i)
+		} else {
+			stack = stack[:len(stack)-1]
+			if len(stack) == 0 {
+				stack = append(stack, i)
+			} else {
+				if ans < i-stack[len(stack)-1] {
+					ans = i - stack[len(stack)-1]
+				}
+			}
+		}
+	}
+	return ans
+}
+```
+
 ### **C#**
 
 ```cs
@@ -206,6 +260,28 @@ function longestValidParentheses(s: string): number {
 }
 ```
 
+```ts
+function longestValidParentheses(s: string): number {
+    let max_length: number = 0;
+    const stack: number[] = [-1];
+    for (let i = 0; i < s.length; i++) {
+        if (s.charAt(i) == '(') {
+            stack.push(i);
+        } else {
+            stack.pop();
+
+            if (stack.length === 0) {
+                stack.push(i);
+            } else {
+                max_length = Math.max(max_length, i - stack[stack.length - 1]);
+            }
+        }
+    }
+
+    return max_length;
+}
+```
+
 ### **JavaScript**
 
 ```js
@@ -232,6 +308,90 @@ var longestValidParentheses = function (s) {
 };
 ```
 
+```js
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var longestValidParentheses = function (s) {
+    let ans = 0;
+    const stack = [-1];
+    for (i = 0; i < s.length; i++) {
+        if (s.charAt(i) === '(') {
+            stack.push(i);
+        } else {
+            stack.pop();
+            if (stack.length === 0) {
+                stack.push(i);
+            } else {
+                ans = Math.max(ans, i - stack[stack.length - 1]);
+            }
+        }
+    }
+    return ans;
+};
+```
+
+### Rust
+
+```rust
+impl Solution {
+    pub fn longest_valid_parentheses(s: String) -> i32 {
+        let mut ans = 0;
+        let mut f = vec![0; s.len() + 1];
+        for i in 2..=s.len() {
+            if
+                s
+                    .chars()
+                    .nth(i - 1)
+                    .unwrap() == ')'
+            {
+                if
+                    s
+                        .chars()
+                        .nth(i - 2)
+                        .unwrap() == '('
+                {
+                    f[i] = f[i - 2] + 2;
+                } else if
+                    (i as i32) - f[i - 1] - 1 > 0 &&
+                    s
+                        .chars()
+                        .nth(i - (f[i - 1] as usize) - 2)
+                        .unwrap() == '('
+                {
+                    f[i] = f[i - 1] + 2 + f[i - (f[i - 1] as usize) - 2];
+                }
+                ans = ans.max(f[i]);
+            }
+        }
+        ans
+    }
+}
+```
+
+```rust
+impl Solution {
+    pub fn longest_valid_parentheses(s: String) -> i32 {
+        let mut stack = vec![-1];
+        let mut res = 0;
+        for i in 0..s.len() {
+            if let Some('(') = s.chars().nth(i) {
+                stack.push(i as i32);
+            } else {
+                stack.pop().unwrap();
+                if stack.is_empty() {
+                    stack.push(i as i32);
+                } else {
+                    res = std::cmp::max(res, (i as i32) - stack.last().unwrap());
+                }
+            }
+        }
+        res
+    }
+}
+```
+
 ### **...**
 
 ```
diff --git a/solution/0000-0099/0032.Longest Valid Parentheses/Solution.rs b/solution/0000-0099/0032.Longest Valid Parentheses/Solution.rs
new file mode 100644
index 0000000000000..68aa7b6cd430c
--- /dev/null
+++ b/solution/0000-0099/0032.Longest Valid Parentheses/Solution.rs	
@@ -0,0 +1,33 @@
+impl Solution {
+    pub fn longest_valid_parentheses(s: String) -> i32 {
+        let mut ans = 0;
+        let mut f = vec![0; s.len() + 1];
+        for i in 2..=s.len() {
+            if
+                s
+                    .chars()
+                    .nth(i - 1)
+                    .unwrap() == ')'
+            {
+                if
+                    s
+                        .chars()
+                        .nth(i - 2)
+                        .unwrap() == '('
+                {
+                    f[i] = f[i - 2] + 2;
+                } else if
+                    (i as i32) - f[i - 1] - 1 > 0 &&
+                    s
+                        .chars()
+                        .nth(i - (f[i - 1] as usize) - 2)
+                        .unwrap() == '('
+                {
+                    f[i] = f[i - 1] + 2 + f[i - (f[i - 1] as usize) - 2];
+                }
+                ans = ans.max(f[i]);
+            }
+        }
+        ans
+    }
+}