[leetcode] 856. Score of Parentheses

原创

是念 2022-08-11 17:46:29 博主文章分类：C++ ©著作权

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Description

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:

Example 2:

Input:

"(())"

Output:

Example 3:

Input:

"()()"

Output:

Example 4:

Input:

"(()(()))"

Output:

Note:

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

分析

题目的意思是：给你一个合法的括号字符串，按照题目给定的规则来计算这个字符串的得分。

这种括号字符串可以用递归实现，写着还是不难，按照上面给的三条规则写就完事了，递归的终止条件是“()”，然后返回1就行了，其他的根据规则写就行了。

代码

class Solution {
public:
    int scoreOfParentheses(string S) {
        return solve(S);
    }
    int solve(string s){
        if(s=="()"){
            return 1;
        }
        int cnt=0;
        for(int i=0;i<s.length()-1;i++){
            if(i>0&&cnt==0){
                return solve(s.substr(0,i))+solve(s.substr(i)); 
            }else if(s[i]=='('){
                cnt++;
            }else if(s[i]==')'){
                cnt--;
            }
        }
        return 2*solve(s.substr(1,s.size()-2));
    }
};