Given a balanced parentheses string S
, compute the score of the string based on the following rule:
()
has score 1AB
has scoreA + B
, where A and B are balanced parentheses strings.(A)
has score2 * 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
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Score of Parentheses.
(A) = 2 * A 是表示深度的一个概念。
class Solution { public: int scoreOfParentheses(string S) { int ret = 0, cnt = 0; char last = ‘ ‘; for(auto ch : S){ if(ch == ‘(‘){ cnt++; }else { cnt--; if(last == ‘(‘){ ret += (1<<cnt); } } last = ch; } return ret; } };
原文地址:https://www.cnblogs.com/ethanhong/p/10192920.html
时间: 2024-10-08 19:24:55