LeetCode括号生成（/dfs）

20230205 通过
20230206 通过
20230209 通过
20230220 通过
20230304 通过
原题解

题目

数字 n 代表生成括号的对数，请你设计一个函数，用于能够生成所有可能的并且有效的括号组合。

约束

解法

解法一

class Solution {
    bool valid(const string& str) {
        int balance = 0;
        for (char c : str) {
            if (c == '(') {
                ++balance;
            } else {
                --balance;
            }
            if (balance < 0) {
                return false;
            }
        }
        return balance == 0;
    }

    void generate_all(string& current, int n, vector<string>& result) {
        if (n == current.size()) {
            if (valid(current)) {
                result.push_back(current);
            }
            return;
        }
        current += '(';
        generate_all(current, n, result);
        current.pop_back();
        current += ')';
        generate_all(current, n, result);
        current.pop_back();
    }
public:
    vector<string> generateParenthesis(int n) {
        vector<string> result;
        string current;
        generate_all(current, n * 2, result);
        return result;
    }
};

解法二

class Solution {
    void backtrack(vector<string>& ans, string& cur, int open, int close, int n) {
        if (cur.size() == n * 2) {
            ans.push_back(cur);
            return;
        }
        if (open < n) {
            cur.push_back('(');
            backtrack(ans, cur, open + 1, close, n);
            cur.pop_back();
        }
        if (close < open) {
            cur.push_back(')');
            backtrack(ans, cur, open, close + 1, n);
            cur.pop_back();
        }
    }
public:
    vector<string> generateParenthesis(int n) {
        vector<string> result;
        string current;
        backtrack(result, current, 0, 0, n);
        return result;
    }
};

不需要另外弄一个变量存即时答案数组长度
右边比左边小

解法三

class Solution {
    shared_ptr<vector<string>> cache[100] = {nullptr};
public:
    shared_ptr<vector<string>> generate(int n) {
        if (cache[n] != nullptr)
            return cache[n];
        if (n == 0) {
            cache[0] = shared_ptr<vector<string>>(new vector<string>{""});
        } else {
            auto result = shared_ptr<vector<string>>(new vector<string>);
            for (int i = 0; i != n; ++i) {
                auto lefts = generate(i);
                auto rights = generate(n - i - 1);
                for (const string& left : *lefts)
                    for (const string& right : *rights)
                        result -> push_back("(" + left + ")" + right);
            }
            cache[n] = result;
        }
        return cache[n];
    }
    vector<string> generateParenthesis(int n) {
        return *generate(n);
    }
};