856. Score of Parentheses

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 strings

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

Approach #1. DFS. [Java]

class Solution {
    public int scoreOfParentheses(String S) {
        return helper(S, 0, S.length() - 1);
    }
    
    public int helper(String S, int l, int r) {
        if (l + 1 == r) return 1;
        int b = 0;
        for (int i = l; i < r; ++i) {
            if (S.charAt(i) == '(') b++;
            else b--;
            if (b == 0) return helper(S, l, i) + helper(S, i+1, r);
        }
        return 2 * helper(S, l + 1, r - 1);
    }

}

　　

Approach #2: Stack. [Java]

class Solution {
    public int scoreOfParentheses2(String S) {
        boolean mode = true;
        int ret = 0;
        Stack<Character> stack = new Stack<>();
        for (int i = 0; i < S.length(); ++i) {
            if (S.charAt(i) == ')' && mode) {
                ret += Math.pow(2, stack.size() - 1);
                mode = false;
                stack.pop();
            } else if (S.charAt(i) == '(') {
                stack.push('(');
                mode = true;
            } else {
                stack.pop();
            }
        }
        return ret;
    }

}