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:
"((()))", "(()())", "(())()", "()(())", "()()()"
思路：这道题给你n对括号，生成所有合法的括号规则。有如下规则：左括号和右括号的数量都为n时，将生成的括号push到result里；左括号的数量小于n，插入str，然后递归；当右括号的数量小于左括号的数量，将‘)‘插入str中，然后递归。
class Solution { public: void MatchNums(int n,vector<string> &result,string &str,int nLeftBrackts,int nRightBrackts) { if(nLeftBrackts==n&&nRightBrackts==n) { result.push_back(str); return; } if(nLeftBrackts<n) { str.push_back(‘(‘); MatchNums(n,result,str,nLeftBrackts+1,nRightBrackts); str.pop_back(); } if(nRightBrackts<nLeftBrackts) { str.push_back(‘)‘); MatchNums(n,result,str,nLeftBrackts,nRightBrackts+1); str.pop_back(); } } vector<string> generateParenthesis(int n) { vector<string> result; string str; MatchNums(n,result,str,0,0); return result; } };
Generate Parentheses,布布扣,bubuko.com
原文：http://www.cnblogs.com/awy-blog/p/3618674.html