241. Different Ways to Add Parentheses

The description of the problem

Given a string experssion of numbers and operators, return 
all possible results from computing all the different 
possible ways to group number and operators.
You may return the answer in any order. 
The test cases are generated such that the output values fit in a 32 bit integer and the number of different results does not exceed $10^4$

an example

Input: expression = "2-1-1"
Output: [0,2]
Explanation:
((2-1)-1) = 0 
(2-(1-1)) = 2

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/different-ways-to-add-parentheses

The codes for this

#include <vector>
#include <iostream>
#include <string>
using namespace std;
class Solution {
    
public:
    vector<int> diffWaysToCompute(string expression) {
    
        vector<int> res;
        int n = expression.size();
        for (int i = 0; i < n; i++) {
    
            if (expression[i] == '+' || expression[i] == '-' || expression[i] == '*') {
    
                vector<int> left = diffWaysToCompute(expression.substr(0,i));
                vector<int> right = diffWaysToCompute(expression.substr(i + 1));
                for (int l : left) {
    
                    for (int r : right) {
    
                        switch (expression[i]) {
    
                            case '+' : res.emplace_back(l + r); break;
                            case '-' : res.emplace_back(l - r); break;
                            case '*' : res.emplace_back(l * r); break;
                        }
                    }
                }
            }
        }
        if (res.empty()) {
    
            res.emplace_back(stoi(expression));
        }
        return res;
    }
};
int main() 
{
    
    Solution s;
    string s1 = "2-1-1";
    vector<int> res = s.diffWaysToCompute(s1);
    for (int i : res) {
    
        cout << i << " ";
    }
    cout << endl;
    return 0;
}

The corresponding results

$ ./test
2 0