【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: ...
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 +=...
【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...
Github 同步地址: https://github.com/grandyang/leetcode/issues/856 类似题目: Valid Parentheses 参考资料: https://leetcode.com/problems/score-of-parentheses/ https://leetcode.com/problems/score-of-parentheses/discuss/141777/C%2B%2BJavaPython-O(1)-Space [LeetCode All in One 题目讲解汇总(持续更...
https://github.com/grandyang/leetcode/issues/856 类似题目: Valid Parentheses 参考资料: https://leetcode.com/problems/score-of-parentheses/ https://leetcode.com/problems/score-of-parentheses/discuss/141777/C%2B%2BJavaPython-O(1)-Space [LeetCode All in One 题目讲解汇总(持续更新中...)](...
考虑到效率问题,两层for循环两两组合数组中的元素为最基本方法,时间复杂度为O(n²),在example3中明显会导致运行时间太长 由于容器大小和两个因素有关,width和height,假设初始最大盛水量的状态为width = len(arr) - 1,height = min(arr[0], arr[len[arr] - 1]) ...
856 + ```java 857 + // Welcome to create a PR to complete the code of this language, thanks! 858 + ``` 859 + 589 860 Dear LeetCoders! For a better LeetCode problem-solving experience, please visit website [LeetCodePython.com](https://leetcodepython.com): Dare to ...
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For example, in the given tree above, the leaf value sequence is(6, 7, 4, 9, 8). Two binary trees are considered leaf-similar if their leaf value sequence is the same. Returntrueif and only if the two given trees with head nodesroot1androot2are leaf-similar. ...
python代码如下: class Solution(object): def minAreaFreeRect(self, points): """ :type points: List[List[int]] :rtype: float """ N = len(points) # (l^2, x#, y#) : [(0,1), (1,2)] d = collections.defaultdict(list)