【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

    }

};