leetcode:Generate Parentheses

最新推荐文章于 2019-08-24 20:10:39 发布

martin_liang

最新推荐文章于 2019-08-24 20:10:39 发布

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本文介绍了一个生成有效括号组合的算法。给定一个正整数 n，该算法能生成所有可能的有效括号字符串，每种组合包含 n 对括号。通过递归方式构造合法的左括号和右括号序列，确保任何时候右括号的数量不会超过左括号的数量。

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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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class Solution {
    
private:
    void generateParenthesisHelper(int leftNum, int rightNum, int totalNum, int curIdx, vector<char> &curString, vector<string> &retVtr) {
        
        if (leftNum == totalNum && rightNum == totalNum)
        {
            string str(curString.begin(), curString.end());
            retVtr.push_back(str);
            return;
        }
        
        if (leftNum < totalNum)
        {
            curString[curIdx] = '(';
            generateParenthesisHelper(leftNum+1, rightNum, totalNum, curIdx+1, curString, retVtr);
        }
        
        if (rightNum < leftNum)
        {
            curString[curIdx] = ')';
            generateParenthesisHelper(leftNum, rightNum+1, totalNum, curIdx+1, curString, retVtr);
        }
    }
    
public:
    vector<string> generateParenthesis(int n) {
        
        vector<string> retVtr;
        vector<char> curString(n*2,0);
        generateParenthesisHelper(0, 0, n, 0, curString, retVtr);
        
        return retVtr;
    }
};