leetcode 856 括号的分数

题目:
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

方法一：
使用栈，这个很简单，首先遍历字符串，对于遍历到的每个字符c：
1.c为’(’,入栈
2.c为’(’,出栈并且累加中间结果，直到遇到’(’
3.将累加结果乘以2倍，放回栈顶
Python
def scoreOfParentheses(self, S):
“”"
:type S: str
:rtype: int
“”"
stack=[0]
for c in S:
if c==’(’:
stack.append(0)
else:
v=stack.pop()
stack[-1]+=max(2*v,1)
return stack.pop()

C++
int scoreOfParentheses(string S) {
stack stack;
int cur = 0;
for (char i : S)
if (i == ‘(’) {
stack.push(cur);
cur = 0;
}
else {
cur += stack.top() + max(cur, 1);
stack.pop();
}
return cur;
}