22. Generate Parentheses

最新推荐文章于 2024-01-18 13:52:44 发布

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最新推荐文章于 2024-01-18 13:52:44 发布

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分类专栏： Leetcode

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本文介绍了一个递归算法，用于生成所有合法的n对括号组合。通过使用两个计数器来跟踪剩余的左括号和右括号数量，确保生成的括号序列总是合法的。

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:

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

求出所以可能的括号对
递归
The idea is intuitive. Use two integers to count the remaining left parenthesis (n) and the right parenthesis (m) to be added. At each function call add a left parenthesis if n >0 and add a right parenthesis if m>0. Append the result and terminate recursive calls when both m and n are zero.

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> res;
        addingpar(res, "", n, 0);
        return res;
    }
    void addingpar(vector<string> &v, string str, int n, int m){
        if(n==0 && m==0) {
            v.push_back(str);
            return;
        }
        if(m > 0){ addingpar(v, str+")", n, m-1); }
        if(n > 0){ addingpar(v, str+"(", n-1, m+1); }
    }
};