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:

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

public class Solution {
    public List
  
    generateParenthesis(int n) {
        int left=0;
        int right=0;
        String str="";
        List
   
     list =new ArrayList
    
     (); gen(left, right, n, str, list); return list; } public void gen(int left, int right, int n, String str, List
     
       list){ if( left==n && right==n){ list.add(str); return; } if( left
      
       right ){ gen(left+1, right, n,str+"(", list); gen(left, right+1, n,str+")", list); } else if(left==n && right
       
        主要用到了递归的思想，递归时保证字符串中左括号数大于等于右括号数即可。