【LeetCode & 剑指offer刷题】回溯法与暴力枚举法题4：Generate Parentheses

【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）

Generate Parentheses

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:

[

"((()))",

"(()())",

"(())()",

"()(())",

"()()()"

]

---

C++

 

//问题：产生括号对

//方法一：回溯法（并没有回溯，应该就叫普通递归法）

/*

递归，并用两个变量（open, close）记录当前左括号和右括号的个数

 open < n (n为括号对数量)时添加左括号，close<open时添加右括号

Once we add a '(' we will then discard it and try a ')' which can only close a valid '('. Each of these steps are recursively called

举例：（n=3时的递归树）

                                --> (((,3,0,3--> (((),3,1,3--> ((()),3,2,3--> ((())),3,3,3 return

                    --> ((,2,0,3

                                              --> (()(, 3,1,3--> (()(), 3,2,3--> (()()), 3,3,3 return

"",0,0,3 --> (,1,0,3            --> ((), 2,1,3

                                              --> (()), 2,2,3--> (())(, 3,2,3--> (())(), 3,3,3 return

                   

                                              --> ()((,3,1,3--> ()((),3,2,3--> ()(()),3,3,3 return

                    --> (),1,1,3--> ()(, 2,1,3

                                              --> ()(),2,2,3--> ()()(,3,2,3--> ()()(),3,3,3 return

                               

*/

class Solution

{

public:

    vector<string> generateParenthesis(int n)

    {

        vector<string> ans;

        recursion(ans, "", 0, 0, n);

        return ans;

    }

   

private:

    void recursion(vector<string>& ans, string s, int open, int close, int n)

    {

        if(s.length() == 2*n)

        {

            ans.push_back(s);

            return;

        }

       

        //深度优先搜索，分支为加左括号和右括号，深度方向左括号和右括号保证匹配

        if(open < n) recursion(ans, s+'(', open+1, close, n); //递归树有很多分支，遍历到所有可能的情况 

        if(close < open) recursion(ans, s+')', open, close+1, n); //加右括号，直到左右括号相匹配

    }

};