LeetCode 856. Score of Parentheses 栈顶

最新推荐文章于 2022-10-09 21:16:49 发布

原创最新推荐文章于 2022-10-09 21:16:49 发布 · 273 阅读

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本文介绍了一种计算平衡括号字符串分数的方法，通过解析输入字符串并利用栈数据结构，实现高效计算。文章提供了两种实现方式，详细解释了如何通过栈记录括号层次，最终计算出总分数。

Given a balanced parentheses string S, compute the score of the string 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: "()"
Output: 1

Example 2:

Input: "(())"
Output: 2

Example 3:

Input: "()()"
Output: 2

Example 4:

Input: "(()(()))"
Output: 6

Note:

S is a balanced parentheses string, containing only ( and ).
2 <= S.length <= 50

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反应有点慢，刚开始没有想清楚栈里放哪些东西，扔了4项，后面减成2项。

class Solution:
    def scoreOfParentheses(self, S: str) -> int:
        sta,l = [],len(S)
        for i in range(l):
            if (S[i] == '('):
                sta.append((0,0)) #type,val
            elif (S[i] == ')'):
                cur = 0
                while (True):
                    type,val = sta.pop()
                    if (type == 0 and cur == 0):
                        sta.append((1,1)) #type,val
                        break
                    elif (type == 0 and cur > 0):
                        sta.append((1,cur*2)) #type,val
                        break
                    else:
                        cur += val
        return sum(item[1] for item in sta)

这种问题还可以用栈顶元素来控制层次，以下是一种更简洁的解法：

class Solution:
    def scoreOfParentheses(self, S: str) -> int:
        sta,l = [],len(S)
        for i in range(l):
            if (S[i] == '('):
                sta.append(0)
            elif (S[i] == ')'):
                cur = 0
                while (sta[-1] > 0):
                    cur += sta.pop()
                sta[-1] = 1 if cur == 0 else cur*2
        return sum(sta)