[LeetCode] Generate Parentheses

 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:

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

括号匹配。使用递归的方法。

设定left、right表示左右括号的剩余数，

如果left > right表示匹配失衡，直接返回。

如果left < right表示已经存在一个左括号，需要匹配下一次左括号或者右括号，进行递归。

如果left == right == 0，表示括号匹配完毕且合法。把当前括号序列放入结果数组中。

参考代码如下：

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

另一种方法，提前剪枝，思路更清晰,left、right表示当前左右括号的数量

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> res;
        string out;
        dfs(n, 0, 0, out, res);
        return res;
    }
    void dfs(int n, int left, int right, string out, vector<string>& res)
    {
        if (left < n)
        {
            out.push_back('(');
            dfs(n, left+1, right, out, res);
            out.pop_back();
        }
        if (left > right)
        {
            out.push_back(')');
            dfs(n, left, right+1, out, res);
            out.pop_back();
        }
        if (out.size() == n*2)
            res.push_back(out);
    }
};