leetcode 22 -- Generate Parentheses

Posted 2022-12-02 夏天的技术博客

Generate Parentheses

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:
“((()))”, “(()())”, “(())()”, “()(())”, “()()()”

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题意：
给出括号对数，任意组合，找出所有满足括号匹配原则的字符串。

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思路：
考虑出规律即可，规律就是我们要在字符串的任意时刻保证左括号“（”的数目大于右括号的数目“）”即可。^_^

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代码：

class Solution 
public:
    void addRetStr(int leftNum, int rightNum, string s, vector<string>&ivecs)
        if(leftNum == 0 && rightNum == 0)
            ivecs.push_back(s);
        
        //左括号数目等于右括号数目立即添加左括号来保证规则。
        if(leftNum == rightNum)
            s += "(";
            addRetStr(leftNum-1, rightNum, s, ivecs);
        else
            if(leftNum > 0)
                //注意这里参数必须是new string而不能s += "("然后传递参数s，因为还有别的string要用s。
                string new_str = s + "(";
                addRetStr(leftNum-1, rightNum, new_str, ivecs);
            
            if(rightNum > 0)
                string new_str = s + ")";
                addRetStr(leftNum, rightNum-1, new_str, ivecs);
            
        
    

    vector<string> generateParenthesis(int n) 
        string s;
        vector<string>ret;
        addRetStr(n, n, s, ret);
        return ret;
    
;

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虽然很晚了强迫症还是要写完今天的博客，哈哈
最后 祝自己20岁生日快乐^_^，感谢父母带我来这个世界，感谢我的每一位家人，朋友，加油
life's battle doesn't always go to the stronger or faster man
but sooner or later the man who wins
is the man who thinks he can...

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