LeetCode 241. Different Ways to Add Parentheses

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]

#include <string>
#include <vector>
#include <iostream>
using namespace std;
// this coding practise teaches me that I am not good at divide and conquer method....
vector<int> diffWaysToCompute(string str) {
  if(str.size() == 0) return {};
  vector<int> result;
  for(int i = 0; i < str.size(); ++i) {
    if(str[i] != '+' && str[i] != '*' && str[i]!= '-') continue;
    auto front = diffWaysToCompute(str.substr(0, i));
    auto end = diffWaysToCompute(str.substr(i+1));
    for(auto val1 : front) {
      for(auto val2 : end) {
        if(str[i] == '+') result.push_back(val1 + val2);
        else if(str[i] == '*') result.push_back(val1 * val2);
        else if(str[i] == '-') result.push_back(val1 - val2);
      }
    }
  }
  return result.empty() ? vector<int>{stoi(str)} : result;
}

int main(void) {
  vector<int> res = diffWaysToCompute("1+6*3");
  for(int i = 0; i < res.size(); ++i) cout << res[i] << endl;
}