LeetCode22——Generate Parentheses（给定n对括号，然后看有多少正确的括号组合）

题目：

解法：

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); }}
};

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