<p>Given a string containing just the characters <code>&#39;(&#39;</code> and <code>&#39;)&#39;</code>, return <em>the length of the longest valid (well-formed) parentheses </em><span data-keyword="substring-nonempty"><em>substring</em></span>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;(()&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> The longest valid parentheses substring is &quot;()&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;)()())&quot;
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest valid parentheses substring is &quot;()()&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;&quot;
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= s.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>s[i]</code> is <code>&#39;(&#39;</code>, or <code>&#39;)&#39;</code>.</li>
</ul>

class Solution:
    def longestValidParentheses(self, s: str) -> int:
        errors = []
        for i in range(len(s)):
            if s[i] == "(" or len(errors) == 0 or s[errors[-1]] == ")":
                errors.append(i)
            else:
                errors.pop()
        ret = 0
        last = -1
        for i in errors:
            print(ret)
            if i-last - 1 > ret:
                ret = i-last - 1
            last = i
        if len(errors) == 0:
            return len(s)
        if len(s) - 1 - last > ret:
            return len(s)-1-last
        return ret
17700