【LeetCode】【Rust】【Java】【DFS】括号生成

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。

示例 1：

输入：n = 3
输出：["((()))","(()())","(())()","()(())","()()()"]
示例 2：

输入：n = 1
输出：["()"]

提示：

1 <= n <= 8

难得一次排名有点高。。

解答成功: 执行耗时:1 ms,击败了95.79% 的Java用户 内存消耗:38.7 MB,击败了31.94% 的Java用户

Java

import java.util.LinkedList;
import java.util.List;

public class GenerateParentheses{
    public static void main(String[] args) {
        Solution solution = new GenerateParentheses().new Solution();
        System.err.println(solution.generateParenthesis(5));
        
    }

//leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        int max;
        List<String> result = new LinkedList<>();
        public List<String> generateParenthesis(int n) {
            max = n;
            dfs("",0,0);
            return result;
        }
    
        void dfs(String s, int r,int l){
            if (s.length() == max*2){
                result.add(s);
                return;
            }
            if (r<max){
                dfs(s+"(",r+1,l);
            }
    
            if (l<r){
                dfs(s+")",r,l+1);
            }
            return;
        }
    }
//leetcode submit region end(Prohibit modification and deletion)

}

Rust

impl Solution {
    pub fn generate_parenthesis(n: i32) -> Vec<String> {
        let mut result = vec![];
        Self::dfs(String::new(), 0, 0, n as usize, &mut result);
        result
    }

    fn dfs(s: String, r: usize, l: usize, n: usize, result: &mut Vec<String>){
        if (s.len() == 2*n ) {
			result.push(s);
			return;
		}

        if (r<n) {
            let mut s = s.to_owned();
            s.push('(');
            Self::dfs(s, r+1, l, n, result);
        }

        if (l<r) {
            let mut s = s.to_owned();
            s.push(')');
            Self::dfs(s, r, l+1, n, result);
        }
    }
}