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:
[ "((()))", "(()())", "(())()", "()(())", "()()()" ]class Solution { public: vector<string> generateParenthesis(int n) { vector<string> res; if(n==0) return res; string s; dfs(res,s,n,n); return res; } void dfs(vector<string>& res, string& s, int left, int right) { if(left>right) return; if(left==0&&right==0) { res.push_back(s); return; } if(left>0) { s.push_back('('); dfs(res,s,left-1,right); s.pop_back(); } if(right>0) { s.push_back(')'); dfs(res,s,left,right-1); s.pop_back(); } } };
