Solution 1: DFS + Pruning

The range of $n$ in the problem is $[1, 8]$ , so we can directly solve this problem through "brute force search + pruning".

We design a function $df s (l, r, t)$ , where $l$ and $r$ represent the number of left and right brackets respectively, and $t$ represents the current bracket sequence. Then we can get the following recursive structure:

If $l > n$ or $r > n$ or $l < r$ , then the current bracket combination $t$ is invalid, return directly;
If $l = n$ and $r = n$ , then the current bracket combination $t$ is valid, add it to the answer array ans, and return directly;
We can choose to add a left bracket, and recursively execute dfs(l + 1, r, t + "(");
We can also choose to add a right bracket, and recursively execute dfs(l, r + 1, t + ")").

The time complexity is $O (2^{n \times 2} \times n)$ , and the space complexity is $O (n)$ .

Python3

class Solution:
    def generateParenthesis(self, n: int) -> List[str]:
        def dfs(l, r, t):
            if l > n or r > n or l < r:
                return
            if l == n and r == n:
                ans.append(t)
                return
            dfs(l + 1, r, t + '(')
            dfs(l, r + 1, t + ')')

        ans = []
        dfs(0, 0, '')
        return ans

Java

class Solution {
    private List<String> ans = new ArrayList<>();
    private int n;

    public List<String> generateParenthesis(int n) {
        this.n = n;
        dfs(0, 0, "");
        return ans;
    }

    private void dfs(int l, int r, String t) {
        if (l > n || r > n || l < r) {
            return;
        }
        if (l == n && r == n) {
            ans.add(t);
            return;
        }
        dfs(l + 1, r, t + "(");
        dfs(l, r + 1, t + ")");
    }
}

C++

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> ans;
        function<void(int, int, string)> dfs = [&](int l, int r, string t) {
            if (l > n || r > n || l < r) return;
            if (l == n && r == n) {
                ans.push_back(t);
                return;
            }
            dfs(l + 1, r, t + "(");
            dfs(l, r + 1, t + ")");
        };
        dfs(0, 0, "");
        return ans;
    }
};

Go

func generateParenthesis(n int) (ans []string) {
    var dfs func(int, int, string)
    dfs = func(l, r int, t string) {
        if l > n || r > n || l < r {
            return
        }
        if l == n && r == n {
            ans = append(ans, t)
            return
        }
        dfs(l+1, r, t+"(")
        dfs(l, r+1, t+")")
    }
    dfs(0, 0, "")
    return ans
}

TypeScript

impl Solution {
    pub fn generate_parenthesis(n: i32) -> Vec<String> {
        let mut ans = Vec::new();

        fn dfs(ans: &mut Vec<String>, l: i32, r: i32, t: String, n: i32) {
            if l > n || r > n || l < r {
                return;
            }
            if l == n && r == n {
                ans.push(t);
                return;
            }
            dfs(ans, l + 1, r, format!("{}(", t), n);
            dfs(ans, l, r + 1, format!("{})", t), n);
        }

        dfs(&mut ans, 0, 0, String::new(), n);
        ans
    }
}

Rust

impl Solution {
    fn dfs(left: i32, right: i32, s: &mut String, res: &mut Vec<String>) {
        if left == 0 && right == 0 {
            res.push(s.clone());
            return;
        }
        if left > 0 {
            s.push('(');
            Self::dfs(left - 1, right, s, res);
            s.pop();
        }
        if right > left {
            s.push(')');
            Self::dfs(left, right - 1, s, res);
            s.pop();
        }
    }

    pub fn generate_parenthesis(n: i32) -> Vec<String> {
        let mut res = Vec::new();
        Self::dfs(n, n, &mut String::new(), &mut res);
        res
    }
}

JavaScript

/**
 * @param {number} n
 * @return {string[]}
 */
var generateParenthesis = function (n) {
    function dfs(l, r, t) {
        if (l > n || r > n || l < r) {
            return;
        }
        if (l == n && r == n) {
            ans.push(t);
            return;
        }
        dfs(l + 1, r, t + '(');
        dfs(l, r + 1, t + ')');
    }
    let ans = [];
    dfs(0, 0, '');
    return ans;
};

PHP

class Solution {
    /**
     * @param Integer $n
     * @return String[]
     */
    function generateParenthesis($n) {
        $ans = [];

        $dfs = function ($l, $r, $t) use ($n, &$ans, &$dfs) {
            if ($l > $n || $r > $n || $l < $r) {
                return;
            }
            if ($l == $n && $r == $n) {
                $ans[] = $t;
                return;
            }
            $dfs($l + 1, $r, $t . '(');
            $dfs($l, $r + 1, $t . ')');
        };

        $dfs(0, 0, '');
        return $ans;
    }
}

Problem