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:

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

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Score of Parentheses.

(A) = 2 * A 是表示深度的一個概念。

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;
  }
};