遞迴然後剪枝 其實相當於dfs的思想 （我感覺的，因為也是一條路走到黑然後去重）
思路：一開始把’(‘和’)'分列兩旁 然後開始交換 '(‘只能和’)'交換，然後check檢查兩個情況
1.重複 2.括號不正確

class Solution {
public:
	vector<string> generateParenthesis(int n) {
		vector<string>ans;
		string str;
		//ini
		for (int i = 0; i < n; ++i)str += '(';
		for (int i = 0; i <
 
 n; ++i)str += ')';
        ans.push_back(str);
		generateParenthesis_Recursive(ans, str);
		return ans;
	}
	bool check(vector<string>& ans, string str) {
		//ditto
		for (int i = 0; i < ans.size(); ++i)if (ans[i] == str)return false;

		//correctness
		int Cnt = 0;
		for (int i = 0; i < str.
 
size(); ++i) {
			if (str[i] == '(')++Cnt;
			else if (str[i] == ')') {
				if (Cnt == 0)return false;
				else --Cnt;
			}
		}
		return true;
	}
	void swap_str(string& str, int i, int j) {
		char t = str[i];
		str[i] = str[j];
		str[j] = t;
	}
	void generateParenthesis_Recursive(vector<string>
 
& ans,string str) {
		for (int i = 0; i < str.size(); ++i) {
			for (int j = i + 1; j < str.size(); ++j) {
				if (str[i] == str[j])continue;
				swap_str(str, i, j);
				if (check(ans, str)){
					ans.push_back(str);
					generateParenthesis_Recursive(ans, str);
				}
				swap_str(str, j, i);
			}
		}
	}
};