leetcode 22 -- Generate Parentheses

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

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。

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。^_^

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;
    }
};

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
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but sooner or later the man who wins
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