LeetCode每日一题(Score of Parentheses)

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该博客讨论了一个算法问题，涉及计算由括号组成的平衡字符串的分数。算法使用深度优先搜索（DFS）和广度优先搜索（BFS）策略，根据给定的规则计算字符串的得分：'()' 得分为 1，AB 结构得分为 A + B，(A) 结构得分为 2 * A。示例展示了不同输入字符串的得分计算过程，所有输入字符串都是平衡的括号串。

Given a balanced parentheses string s, return the score of the string.

The score of a balanced parentheses string is based on the following rule:

“()” has score 1.
AB has score A + B, where A and B are balanced parentheses strings.
(A) has score 2 * A, where A is a balanced parentheses string.

Example 1:

Input: s = “()”
Output: 1

Example 2:

Input: s = “(())”
Output: 2

Example 3:

Input: s = “()()”
Output: 2

Constraints:

2 <= s.length <= 50
s consists of only ‘(’ and ‘)’.
s is a balanced parentheses string.

DFS 逐层乘以 2
BFS 进行加和
()可以看做是’(’, ‘’, ‘)’, ''返回的是 0, 因为 0 * 2 == 0，所以外层的括号要对内层的返回值进行判断，如果返回的是 0，则返回()本身的值，即为 1

impl Solution {
    fn rc(chars: &mut Vec<char>) -> i32 {
        let mut sum = 0;
        while !chars.is_empty() {
            let c = chars.remove(0);
            if c == '(' {
                let next = Solution::rc(chars);
                sum += if next == 0 { 1 } else { next * 2 };
            } else {
                return sum;
            }
        }
        return sum;
    }
    pub fn score_of_parentheses(s: String) -> i32 {
        let mut chars = s.chars().collect();
        Solution::rc(&mut chars)
    }
}