Description:

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:

Example:

[
"((()))",
"(()())",
"(())()",
"()(())",
"()()()"
]

Link:

https://leetcode.com/problems/generate-parentheses/#/description

解题方法：

使两个int：left和right各表示(和)还剩下多少个，用递归生成字符串，过程中需遵守的规则是left 不能大于 right。并且left和right都要大于0。

Tips:

字符串可以直接用运算符'+'

helper(builder + '(', left-1, right);
helper(builder + ')', left, right-1);

Time Complexity：

O(N^2)

完整代码：

vector<string> result;
    vector<string> generateParenthesis(int n) 
    {
        int left = n, right = n;
        helper("", left, right);
        return result;
    }
    void helper(string builder, int left, int right)
    {
        if(left > right || left < 0 || right < 0)
            return;
        if(left == 0 && right == 0)
        {
            result.push_back(builder);
            return;
        }
        helper(builder + '(', left-1, right);
        helper(builder + ')', left, right-1);
    }