leetcode 22. 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:

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

思路一：对于给定的n，有效的括号串长度为2n，我们可以遍历所有的串，如果是有效的，就保存下来。

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> res;
        if (n == 0)
            return res;
        generateParenthesis(0, n, "", res);
        return res;
    }
private:
    void generateParenthesis(int current, const int n, string str, vector<string> &res) {
        if (current == 2 * n) {
            if (isValid(str))
                res.push_back(str);
            return;
        }
        char c[2] = {'(', ')'};
        for (int i = 0; i < 2; i++) { //当前位置有两种选择
            generateParenthesis(current + 1, n, str + c[i], res);
        }
    }
    bool isValid(string s) {
        int len = s.length();
        int distance = 0;
        for (int i = 0; i < len; i++) {
            if (s[i] == '(') {
                distance++;
            } else {
                if (distance <= 0)
                    return false;
                distance--;
            }
        }
        return (distance == 0);
    }
};

思路二：对于某个位置，可以放 '(' 的条件是当前'('放的个数小于n，可以放')'的条件是'('的个数大于')

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> res;
        if (n == 0)
            return res;
        generateParenthesis(0, 0, n, "", res);
        return res;
    }
private:
    void generateParenthesis(int open, int close, const int n, string str, vector<string> &res) {
        if (str.length() == 2 * n) {
            res.push_back(str);
            return;
        }
        if (open < n)
            generateParenthesis(open + 1, close, n, str + '(', res);
        if (open > close)
            generateParenthesis(open, close + 1, n, str + ')', res);
    }
};

 符合条件的个数是卡特兰数：