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:

["((()))","(()())","(())()","()(())","()()()" ]
解題思路：本題就是個(gè)遞歸。
當(dāng)左括號(hào)大于右括號(hào)的時(shí)候，可以選擇左或者右
當(dāng)左括號(hào)等于右括號(hào)的時(shí)候選擇右=左 class Solution { public:void helper(vector<string>&res, int sum, int left, int right,string str){if(left+right==sum){if(left==right){res.push_back(str);}return ;}if(left<right)return;helper(res,sum,left+1,right,str+"(");if(left>right){helper(res,sum,left,right+1,str+")");}}vector<string> generateParenthesis(int n) {vector<string>res;helper(res,2*n,0,0,"");return res;} };

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轉(zhuǎn)載于:https://www.cnblogs.com/tsunami-lj/p/7600583.html

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