LeetCode 22. Generate Parentheses (Catalan数字 + 递归)

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本文介绍了一种利用Catalan数解决括号组合问题的方法。通过递归定义和计算Catalan数，实现了生成所有有效的括号组合。文章提供了具体的C++实现代码，并列举了Catalan数的前几项。

这题看的第一眼，这不就是Catalan Number么 - - 这不就是递归么。这是我之前学Catalan Number的教材的部分摘录，看完就知道怎么写啦。

Catalan numbers:

Set P of balanced parentheses strings are recursively defined as

• λ ∈ P (λ is empty string)

• If α, β ∈ P, then (α) β ∈ P

Every nonempty balanced paren string can be obtained via Rule 2 from a unique α, β pair.

Cn : number of balanced parentheses strings with exactly n pairs of parenthesesC0 = 1 empty string

Cn+1? Every string with n + 1 pairs of parentheses can be obtained in a unique way via rule 2.

One paren pair comes explicitly from the rule.k pairs from α, n−k pairs from β

Cn+1 = sigma Ck · Cn−k n≥0 // 优快云的编辑器用不惯，直接写sigma，就是二项式求和嘛 T^T 想用$$写数学公式都显示不了？

C0 =1

C1 = C02 =1

C2 = C0C1 + C1C0 = 2

C3 = ··· = 5

1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, 18367353072152, 69533550916004, 263747951750360, 1002242216651368

然后哼哧哼哧马上写好了，一次通过，可是妈蛋啊，只打败了 28% 的人- -

等我什么时候有空了再去看看别人很快的代码吧～

题目：

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:

[
  "((()))",
  "(()())",
  "(())()",
  "()(())",
  "()()()"
]

先贴上我的只打败了28%的代码- - ，只有8个测试用例，废话要是有40个，岂不算到下一次宇宙大爆炸。

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> res;
        if(n < 0)
            return res;
        if (n == 0){          // #0 = 1
            res.push_back("");
            return res;
        }
        if( n == 1){          // #1 = 1
            res.push_back("()");
            return res;
        }
        res = iter(n);
        return res;
     }
    
    vector<string> iter(int N){
        vector<string> res;
        if(N == 0){
            res.push_back("");
            return res;                       //递归出口
        }
        if(N == 1){
            res.push_back("()");
            return res;                       //递归出口
        }
        for(int k = 0; k <= (N-1); k++){
            string tempStr;
            vector<string> temp1 = iter(k);       //递归
            vector<string> temp2 = iter(N-1-k);   //递归
            for(int i = 0; i < temp1.size(); i++){
                for(int j = 0; j < temp2.size(); j++){
                    tempStr = "(" + temp1[i] + ")" + temp2[j];
                    res.push_back(tempStr);
                }
             } 
        }
        return res;
    }
 
};