Givennpairs of parentheses, write a function to generate all combinations of well-formed parentheses.
For example, givenn= 3, a solution set is:
"((()))", "(()())", "(())()", "()(())", "()()()"
题意：给你左右括号的个数，问你可以组成完全匹配的个数。
思路：递归：也就两种选择，要么左括号，要么右括号，分别记录个数；只要左括号的个数大于0的话，那么就可以填写左括号；右括号的话，那么要确保此时的序列，左括号的个数要大于右括号的个数。
class Solution {public:void print(int left, int right, string s, vector<string> &result) {if (left == 0 && right == 0) {result.push_back(s);}if (left > 0) {print(left-1, right, s+'(', result);}if (right > 0 && left < right) {print(left, right-1, s+')', result);}}vector<string> generateParenthesis(int n) {vector<string> result;string tmp = "";print(n, n, tmp, result);return result;}};
，当明知不可挽回，唯一补偿的方法就是怀念，
