[Leetcode 241, Medium] Different Ways to Add Parentheses

Problem:

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, -and *.

Example 1

Input: "2-1-1".

((2-1)-1) = 0
(2-(1-1)) = 2

Output: [0, 2]

Example 2

Input: "2*3-4*5"

(2*(3-(4*5))) = -34
((2*3)-(4*5)) = -14
((2*(3-4))*5) = -10
(2*((3-4)*5)) = -10
(((2*3)-4)*5) = 10

Output: [-34, -14, -10, -10, 10]

Analysis:

Solutions:

C++:

    int SToI(const string& str)
    {
        bool is_neg = false;
        if(str[0] == '-')
            is_neg = true;
            
        int start = is_neg;
        int value = 0;
        for(; start < str.size(); ++start) {
            value *= 10;
            value += str[start] - '0';
        }
        
        return (is_neg ? -1 : 1) * value;
    }

    vector<int> diffWaysToCompute(string input) {
        vector<int> results;
        if(input.size() == 0) 
            return results;
            
        for(int i = 0; i < input.size(); ++i) {
            if(input[i] == '+' || input[i] == '-' || input[i] == '*') {
                vector<int> left_results;
                left_results = diffWaysToCompute(input.substr(0, i));
                vector<int> right_results;
                right_results = diffWaysToCompute(input.substr(i + 1));
                
                for(int left = 0; left < left_results.size(); ++left) {
                    for(int right = 0; right < right_results.size(); ++right) {
                        if(input[i] == '+')
                            results.push_back(left_results[left] + right_results[right]);
                        else if(input[i] == '-')
                            results.push_back(left_results[left] - right_results[right]);
                        else
                            results.push_back(left_results[left] * right_results[right]);
                    }
                }
            }
        }
        
        if(results.empty())
            results.push_back(SToI(input));
        
        return results;
    }

Java:

Python: