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# 32. Longest Valid Parentheses

<https://leetcode.com/problems/longest-valid-parentheses/>

## solution

* 动态规划
  * 以s\[i]结尾的字符串能够构成的最长的匹配串的长度

```python
# https://zhuanlan.zhihu.com/p/110240060


```

时间复杂度：O()\
空间复杂度：O()

* 经典做法
  * 一个栈维护左括号的下标，然后遇到匹配的右括号时弹出左括号并通过下标计算长度

```python
class Solution:
    def longestValidParentheses(self, s: str) -> int:
        max_len = 0
        stack = [-1]
        for i in range(len(s)):

            if s[i] == '(':
                stack.append(i)
            else:
                stack.pop()
                if not stack:
                    stack.append(i)
                else:
                    max_len = max(max_len, i - stack[-1])
        return max_len
```

## follow op

[括号类小结](/mle-interview/01_leetcode/07_dfs/22.-generate-parentheses.md)


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