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:

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

递归与深搜。

1 class Solution { 2 public: 3     void getRes(vector
     
       &res, int st, string str, int count, int n) { 4         if (str.length() > 2 * n || st > n) return; 5         if (count == n && st == 0) { 6             res.push_back(str); 7             return; 8         } 9         getRes(res, st + 1, str + '(', count, n);10         if (st > 0) getRes(res, st - 1, str + ')', count + 1, n);11     }12     13     vector
      
        generateParenthesis(int n) {14         vector
       
         res;15         getRes(res, 0, "", 0, n);16         return res;17     }18 };