【leetcode】Generate Parentheses

For example, given n = 3, a solution set is:

"((()))", "(()())", "(())()", "()(())", "()()()"

分析：

  void dfs(vector<string> &re, string& cur_re, int unmatched_lefts, int added_lefts, int n)
{
	
	if(unmatched_lefts == 0 && added_lefts == n)
	{
		re.push_back(cur_re);
	}

	if(unmatched_lefts > 0){//insert ')' is ok
		cur_re.append(1, ')');
		dfs(re, cur_re, unmatched_lefts - 1, added_lefts, n);
	}

	if(cur_re.size() > 0 && added_lefts < n)//can add another '(' 
	{
		cur_re.append(1,'(');
		dfs(re, cur_re, unmatched_lefts + 1, added_lefts + 1, n);	
	}
	cur_re.pop_back();
}
vector<string> generateParenthesis(int n) {

	vector<string> re;
	if(n <= 0)
		return re;
	string one_solve;
	one_solve.append(1, '(');
	dfs(re, one_solve, 1, 1, n);
	return re;
}