<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 {
public:
    int longestValidParentheses(string s) {
        stack<int> st;
        for(int i=0; i<s.size(); i++){
            if(s[i] == '('){
                st.push(i);
            }
            else{
                if(!st.empty() && s[st.top()] == '('){
                    st.pop();
                }
                else{
                    st.push(i);
                }
            }
        }

        int ans = 0, j = s.size();

        while(!st.empty()){
            int i = st.top();
            st.pop();
            ans = max(ans, j-i-1);
            j = i;
        }

        return max(ans, j);
    }
};
9200