Given a balanced parentheses string S, compute the score of the string based on the following rule:
()has score 1-
ABhas scoreA + B, where A and B are balanced parentheses strings. -
(A)has score2 * A, where A is a balanced parentheses string.
Example 1:
Input: 1
Example 2:
Input: 2
Example 3:
Input: 2
Example 4:
Input: 6
Note:
-
Sis 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;
}
};