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//! Generate Parentheses [leetcode: generate_parentheses](https://leetcode.com/problems/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: //! //! ***Example:*** //! //! ``` //! [ //! "((()))", //! "(()())", //! "(())()", //! "()(())", //! "()()()" //! ] //! ``` /// # Solutions /// /// # Approach 1: Backtracking /// /// * Time complexity: /// /// * Space complexity: /// /// ```rust /// impl Solution { /// pub fn generate_parenthesis(n: i32) -> Vec<String> { /// let mut result: Vec<String> = vec![]; /// Self::_gen(&mut result, n, n, "".to_string()); /// result /// } /// pub fn _gen(result: &mut Vec<String>, left: i32, right: i32, sublist: String) { /// if left == 0 && right == 0 { /// result.push(sublist); /// return; /// } /// if left > 0 { /// Self::_gen(result, left - 1, right, sublist.clone() + "("); /// } /// if right > left { /// Self::_gen(result, left, right - 1, sublist.clone() + ")"); /// } /// } /// } /// ``` /// pub fn generate_parenthesis(n: i32) -> Vec<String> { let mut result: Vec<String> = vec![]; _gen(&mut result, n, n, "".to_string()); result } fn _gen(result: &mut Vec<String>, left: i32, right: i32, sublist: String) { if left == 0 && right == 0 { result.push(sublist); return; } if left > 0 { _gen(result, left - 1, right, sublist.clone() + "("); } if right > left { _gen(result, left, right - 1, sublist.clone() + ")"); } }