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:
"((()))", "(()())", "(())()", "()(())", "()()()"
给定变量n，表示有n对括号，输出n对括号所有正确匹配的情况。
用递归来实现，left和right表示左右括号各剩下多少个，left >= right，即左括号剩下的多，则不能插入右括号。
public class Solution { private List<string> result = new List<string>(); public IList<string> GenerateParenthesis(int n) { string current = ""; generate(n, n, current); return result; } public void generate(int left, int right, string current)//left 和 right 表示还剩下的左右括号的个数 { if(left == 0 && right == 0) { result.Add(current); return; } if(left > 0) { generate(left - 1, right, current + "("); } if(right > 0 && left < right) { generate(left, right - 1, current + ")"); } } }
原文：http://www.cnblogs.com/Anthony-Wang/p/5413366.html