leetcode---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]

class Solution {public:    map<string, int> m;    vector<int> dfs(string input)    {        vector<int> v;        for(int i=0; i<input.size(); i++)        {            if(input[i]=='+' || input[i]=='-' || input[i]=='*')            {                string s1 = input.substr(0, i);                string s2 = input.substr(i+1);                vector<int> r1, r2;                int rr1, rr2;                if(m.find(s1) != m.end())                    rr1 = m[s1];                else                    r1 = dfs(s1);                if(m.find(s2) != m.end())                    rr2 = m[s2];                else                    r2 = dfs(s2);                for(int j=0; j<r1.size(); j++)                {                    for(int k=0; k<r2.size(); k++)                    {                        if(input[i] == '+')                            v.push_back(r1[j] + r2[k]);                        else if(input[i] == '-')                            v.push_back(r1[j] - r2[k]);                        else                            v.push_back(r1[j] * r2[k]);                    }                }            }        }        if(v.size() == 0)            v.push_back(atoi(input.c_str()));        return v;    }    vector<int> diffWaysToCompute(string input)     {        return dfs(input);    }};