Given a balanced parentheses string S, compute the score of the string based on the following rule:

() has score 1AB 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: 1Example 2:Input: "(())"Output: 2Example 3:Input: "()()"Output: 2Example 4:Input: "(()(()))"Output: 6Note:S is a balanced parentheses string, containing only ( and ).2 <= S.length <= 50

对于当前字符，如果是'('记录他的位置，如果是')'

有两种状态，要么是() 要么是(()) 所以用个数组去维护与)对应的(之间的数的和。具体见代码

class Solution {public:    int scoreOfParentheses(string S) {        int s[55] = {0};        vector
    
      v(55, 0);        int top = 0;        int left = 0;        int sum = 0;        for (int i = 1; i < S.size(); ++i) {            if (S[i] == '(') s[++top] = i;            else {                    left = s[top--];                if (left == i-1) {                    v[i] = 1;                }                else {                    sum = 0;                    for (int j = left; j < i; ++j) {                        sum += v[j];                        v[j] = 0;                    }                    v[i] = 2*sum;                }                            }        }        return accumulate(v.begin(), v.end(), 0);    }};