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

本文详细解析了LeetCode经典题目“Generate Parentheses”的解决方案，采用递归法生成所有合法的括号组合，适用于n对括号的情况。通过深度优先搜索策略，确保左括号与右括号正确配对，提供了C++实现代码。

【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
   
    }
   
    };