<p>You are given a string <code>s</code> that consists of lower case English letters and brackets.</p>

<p>Reverse the strings in each pair of matching parentheses, starting from the innermost one.</p>

<p>Your result should <strong>not</strong> contain any brackets.</p>

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

<pre>
<strong>Input:</strong> s = &quot;(abcd)&quot;
<strong>Output:</strong> &quot;dcba&quot;
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;(u(love)i)&quot;
<strong>Output:</strong> &quot;iloveu&quot;
<strong>Explanation:</strong> The substring &quot;love&quot; is reversed first, then the whole string is reversed.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;(ed(et(oc))el)&quot;
<strong>Output:</strong> &quot;leetcode&quot;
<strong>Explanation:</strong> First, we reverse the substring &quot;oc&quot;, then &quot;etco&quot;, and finally, the whole string.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code> only contains lower case English characters and parentheses.</li>
	<li>It is guaranteed that all parentheses are balanced.</li>
</ul>

class Solution {
public:
    string reverseParentheses(string s) 
    {
        // Stack
        // Step 1: prepare stack
        stack<char> data;

        // Step 2:
        int n = s.size();
        for(int i = 0; i<n; i++)
        {
            if(s[i] != ')')
            {
                data.push(s[i]);
            }
            else
            {
                // store all chars above '(' into a queue
                queue<char> temp;
                while(data.top() != '(')
                {
                    temp.push(data.top());

                    data.pop();
                }

                // pop '('
                data.pop();

                // repush temp into data
                while(!temp.empty())
                {
                    data.push(temp.front());
                    temp.pop();
                }
            }
        }    

        // Step 3: output stack
        string output;
        while(!data.empty())
        {
            output += data.top();
            data.pop();
        }

        // reverse output
        reverse(output.begin(), output.end());
        return output;
    }
};
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