题目描述：
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

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

[
“((()))”,
“(()())”,
“(())()”,
“()(())”,
“()()()”
]
代码：

#include<iostream>
#include<vector>
using namespace std ;
void show_vector_char(vector<char> & aim)
{
	for(int i =0 ;i<aim.size() ;i++)
		cout<<aim[i] ;
	cout<<endl ;
	return ;
}
enum {right = 1};
bool judge_fitness(vector<char>& aim)
{
	int sum = 0 ;
	for(int i =0 ;i<aim.size() ;i++)
	{
		if(aim[i]=='(')
			sum++ ;
		else
			sum-- ;
		if(sum <0 )
			break ;
	}
	if(sum == 0)
		return true ;
	else
		return false ;
}

void generate(vector<vector<char> >& result, int left_or_right , int n ,
					int aim_len ,vector<char>&last)
{
	if(aim_len == n)
		return ;
	if(left_or_right == 1)
		last.push_back(')') ;
	else
		last.push_back('(') ;
	if(n == (aim_len-1))
	{
		
		if(judge_fitness(last) == true)
		{
			result.push_back(last) ;
			//show_vector_char(last) ;
		}
		
	}
	vector<char>last1 ;
	vector<char>last2 ;
	last1 = last ; last2 = last ;
	generate(result,0 ,n+1,aim_len,last1) ;
	generate(result,1,n+1,aim_len,last2) ;
	return ;
	
	
}
class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> result_string ;
		vector<vector<char> >result ;
        vector<char>last1 ;
        vector<char>last2 ;
        generate(result,0,0,n*2,last1) ;
        generate(result,1,0,n*2,last2) ;
        string tem ;
        for(int i =0 ;i<result.size() ;i++)
        {
        	tem.clear() ;
        	for(int j = 0 ;j<result[i].size() ;j++)
        		tem.push_back(result[i][j]) ;
        	result_string.push_back(tem) ;
        	
		}
        return result_string ;
        
    }
};
int main(void)
{
	Solution solution ;
	solution.generateParenthesis(3) ;
}