Generate Parentheses



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:

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

Tag：使用回溯法。

queue<char> q;int left_num;int right_num;void help(vector<string> &ans, string str, int n){if(str.size() == 2 * n){ans.push_back(str);return;}if(left_num < n){left_num++;q.push('(');str.push_back('(');help(ans, str, n);left_num--;q.pop();str = str.substr(0, str.size() - 1);}if(right_num < n && (!q.empty() && q.front() == '(')){right_num++;q.pop();str.push_back(')');help(ans, str, n);right_num--;q.push('(');str = str.substr(0, str.size() - 1);}}vector<string> generateParenthesis(int n){//q.clear();left_num = 0;right_num = 0;vector<string> ans;string str;help(ans, str, n);return ans;}int main(){/* code */vector<string> ans = generateParenthesis(3);for(int i = 0; i < ans.size(); i++){cout<<ans[i]<<endl;}return 0;}