leetcode22.括号生成

一道典型的回溯，可以使用以下2种方法：

1. 生成所有的情况，判断每种情况是否符合题意

 vector<string> generateParenthesis(int n) {
        vector<string> res;
        string s;
        generateAllParenthesis(n, s, res);
        return res;
    }

    void generateAllParenthesis(int n, string s, vector<string> &res) {
        if (s.size() == 2 * n) {
            if (isValid(s)) {
                res.push_back(s);
            }
        } else {
            generateAllParenthesis(n, s + "(", res);
            generateAllParenthesis(n, s + ")", res);
        }
    }

    bool isValid(string s) {
        int cnt = 0;
        for (int i = 0; i < s.size(); i++) {
            if (s[i] == '(') cnt++;
            else if (s[i] == ')') cnt--;
            if (cnt < 0) return false;
        }
        return cnt == 0;
    }

2. 限定生成括号的类型，必须保证左括号的要和右括号一一匹配

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