diff --git a/README.md b/README.md
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@@ -37,6 +37,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 28 |[Implement strStr()](https://leetcode.com/problems/implement-strstr/)|[C++](solutions/28.%20Implement%20strStr().md)|Easy|  |
 | 29 |[Divide Two Integers](https://leetcode.com/problems/divide-two-integers)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/29.%20Divide%20Two%20Integers.md)|Medium|  |
 | 31 |[Next Permutation](https://leetcode.com/problems/next-permutation)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/31.%20Next%20Permutation.md)|Medium|  |
+| 32 |[Longest Valid Parentheses](https://leetcode.com/problems/longest-valid-parentheses/)|[C++](solutions/32.%20Longest%20Valid%20Parentheses.md)|Hard|  |
 | 34 |[Find First and Last Position of Element in Sorted ](https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/34.%20Find%20First%20and%20Last%20Position%20of%20Element%20in%20Sorted%20Array.md)|Medium|  |
 | 33 |[Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/33.%20Search%20in%20Rotated%20Sorted%20Array.md)|Medium|  |
 | 35 |[Search Insert Position](https://leetcode.com/problems/search-insert-position)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/35.%20Search%20Insert%20Position.md)|Easy|  |
diff --git a/solutions/32. Longest Valid Parentheses.md b/solutions/32. Longest Valid Parentheses.md
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+# [32. Longest Valid Parentheses](https://leetcode.com/problems/longest-valid-parentheses/)
+
+# 思路
+
+求有效括号匹配子串的最大长度。
+
+## 思路一、栈
+
+括号匹配类的问题往往都能用栈解决，此题也不例外。
+
+从前往后遍历字符串，设下标为i
+* 如果遇到左括号或者栈空，则将当前下标i压入栈；
+* 如果遇到右括号且栈顶为左括号，则遇到了匹配的括号对，将栈顶弹出即可；
+
+然后就可以计算以字符`s[i]`作为结尾的最大长度：`i - stk.top()`。
+
+时空复杂度均为O(n)
+
+## 思路二、动归
+
+由于此问题存在很多子问题，且子问题之前又有很多重复，所可以考虑用动归。
+
+先来看状态定义:
+```
+dp[i]: 以字符s[i]结尾的最大匹配长度
+```
+状态转移方程就很简单了：
+* 如果`s[i] = '('`，那么`dp[i] = 0`；
+* 否则，我们需要看以字符`s[i-1]`为结尾的最大匹配子串的前面那个字符`s[i - dp[i-1] - 1]`是否是左括号：
+    * 若是，说明匹配上了，所以`dp[i] = dp[i-1] + 2 + dp[i - dp[i-1] - 2]`；
+    * 若不是，则没匹配上，`dp[i] = 0`。
+
+时空复杂度均为O(n)
+
+## 思路三
+
+前面两种思路的空间复杂度均为O(n)，此题还有一种常数空间的思路，和前面栈的思想有点类似。
+
+使用两个变量left和right分别用来记录到当前位置时左括号和右括号的出现次数，当遇到左括号时，left自增1；否则right自增1。
+
+* 当`left == right`时，说明匹配成功，此时匹配长度为`2*left`；
+* 一旦遇到`right > left`，说明当前位置不能匹配，则重新开始，即left和right均重置为0。
+
+但是对于"(()"这种时，在遍历结束时左右子括号数都不相等，此时没法更新结果res，但其实正确答案是2，怎么处理这种情况呢？答案是再反向遍历一遍，采取类似的方法，稍有不同的是此时若`right < left`了，则重置0。
+
+时间复杂度O(n)，空间复杂度O(1)
+
+# C++
+
+## 思路一
+``` C++
+class Solution {
+public:
+    int longestValidParentheses(string s) {
+        stack<int>stk;
+        int res = 0, start = 0;
+        for(int i = 0; i < s.size(); i++){
+            if(!stk.empty() && s[stk.top()] == '(' && s[i] == ')') stk.pop();
+            else stk.push(i);
+            // 栈顶为当前匹配子串的起始位置
+            res = max(res, stk.empty() ? i + 1 : i - stk.top());
+        }
+        return res;
+    }
+};
+```
+
+## 思路二
+``` C++
+class Solution {
+public:
+    int longestValidParentheses(string s) {
+        vector<int>dp(s.size(), 0);
+        int res = 0;
+        for(int i = 1; i < s.size(); i++){
+            if(s[i] == '(') continue;
+            int tmp = i - dp[i-1] - 1;
+            if(tmp >= 0 && s[tmp] == '(') 
+                dp[i] = dp[i-1] + 2 + (tmp >= 1 ? dp[tmp - 1] : 0);
+            res = max(res, dp[i]);
+        }
+        return res;
+    }
+};
+```
+
+## 思路三
+``` C++
+class Solution {
+public:
+    int longestValidParentheses(string s) {
+        int res = 0, left = 0, right = 0, n = s.size();
+        for (int i = 0; i < n; i++) {
+            if(s[i] == '(') left++; 
+            else right++;
+            if (left == right) res = max(res, 2 * right);
+            else if (right > left) left = right = 0;
+        }
+        
+        left = right = 0;
+        for (int i = n - 1; i >= 0; i--) {
+            if(s[i] == '(') left++;
+            else right++;
+            if (left == right) res = max(res, 2 * left);
+            else if (left > right) left = right = 0;
+        }
+        return res;
+    }
+};
+```
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