[leetcode]856. Score of Parentheses

[leetcode]856. Score of Parentheses

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Analysis

心疼奶茶妹妹~—— [马云爸爸恐成最大赢家 哈哈哈]

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.
类似于括号匹配，但是加了一点运算，用栈实现~

Implement

class Solution {
public:
    int scoreOfParentheses(string S) {
        stack<string> num;
        for(auto s:S){
            if(s=='(')
                num.push("(");
            else{
                if(num.top()=="("){
                    num.pop();
                    num.push("1");
                }
                else{
                    int t = 0;
                    while(num.top() != "("){
                        t += stoi(num.top());
                        num.pop();
                    }
                    num.pop();
                    num.push(to_string(2*t));
                }
            }
        }
        int res = 0;
        while(!num.empty()){
            res += stoi(num.top());
            num.pop();
        }
        return res;
    }
};