<p>Given <code>n</code> pairs of parentheses, write a function to <em>generate all combinations of well-formed parentheses</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> n = 3
<strong>Output:</strong> ["((()))","(()())","(())()","()(())","()()()"]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> n = 1
<strong>Output:</strong> ["()"]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 8</code></li>
</ul>

class Solution {
public:
    void dfs(int i, int j, string now, char p, int total){
        if(i + j == 2 * total){
            ans.push_back(now + p);
            return;
        }
        if(i < total) dfs(i + 1, j, now + p, '(', total);
        if(i > j) dfs(i, j + 1, now + p, ')', total);
    }
    vector<string> generateParenthesis(int n) {
        string now = "";
        dfs(1,0, now, '(', n);
        return ans;
    }
private:
    vector<string>ans;
};
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