[LeetCode] Generate Parentheses 解题报告

最新推荐文章于 2021-02-25 17:37:50 发布

原创最新推荐文章于 2021-02-25 17:37:50 发布 · 239 阅读

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本文提供了一种通过递归生成指定数量的合法括号组合的方法，并详细解释了解题思路和实现过程。

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:

"((()))", "(()())", "(())()", "()(())", "()()()"

» Solve this problem

[解题思路]
典型的递归。一步步构造字符串。当左括号出现次数<n时，就可以放置新的左括号。当右括号出现次数小于左括号出现次数时，就可以放置新的右括号。

[Code]

1:  void CombinationPar(vector<string>& result, string& sample, int deep,   
2:                           int n, int leftNum, int rightNum)  
3:  {  
4:       if(deep == 2*n)  
5:       {  
6:            result.push_back(sample);  
7:            return;  
8:       }  
9:       if(leftNum<n)  
10:       {  
11:            sample.push_back('(');  
12:            CombinationPar(result, sample, deep+1, n, leftNum+1, rightNum);  
13:            sample.resize(sample.size()-1);  
14:       }  
15:       if(rightNum<leftNum)  
16:       {   
17:            sample.push_back(')');  
18:            CombinationPar(result, sample, deep+1, n, leftNum, rightNum+1);  
19:            sample.resize(sample.size()-1);  
20:       }  
21:  }  
22:  vector<string> generateParenthesis(int n) {  
23:       // Start typing your C/C++ solution below  
24:       // DO NOT write int main() function  
25:       vector<string> result;  
26:       string sample;  
27:       if(n!= 0)  
28:            CombinationPar(result, sample, 0, n, 0, 0);  
29:       return result;  
30:  }