【python-leetcode856-子集】括号的分数 给定一个平衡括号字符串 S,按下述规则计算该字符串的分数: () 得 1 分。 AB 得 A + B 分,其中 A 和 B 是平衡括号字符串。 (A) 得 2 * A 分,其中 A 是平衡括号字符串。 示例1: 输入: "()" 输出: 1 示例 2: 输入: "(())" 输出: 2 示例 3: ...
【python-leetcode856-子集】括号的分数 给定一个平衡括号字符串 S,按下述规则计算该字符串的分数: () 得 1 分。 AB 得 A + B 分,其中 A 和 B 是平衡括号字符串。 (A) 得 2 * A 分,其中 A 是平衡括号字符串。 示例1: 输入: "()" 输出: 1 示例2: 输入: "(())" 输出: 2 示例3: 输入...
代码(Python3) class Solution: def scoreOfParentheses(self, s: str) -> int: # score 统计所有括号的分数和 score: int = 0 # 维护每一个括号当前的深度 depth: int = 0 # 带下标遍历每一个括号 for (i, ch) in enumerate(s): if ch == '(': # 如果当前是左括号,则深度加 1 depth +=...
Screenshots of Leetcode Editorials and Premium Problems Table of Contents Editorials Premium Problems About Editorials Leetcode Editorials are in-depth articles which explain how to approach and solve a problem. About 50% of problems have editorials. They are premium-only. Open the below links in ...
Screenshots of Leetcode Editorials and Premium Problems Table of Contents Editorials Premium Problems About Editorials Leetcode Editorials are in-depth articles which explain how to approach and solve a problem. About 50% of problems have editorials. They are premium-only. Open the below links in ...