Problem description:

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 *.

1
2
3
4
5
6
7
Example 1:

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

1
2
3
4
5
6
7
8
9
10
Example 2:

Input: "2*3-4*5"
Output: [-34, -14, -10, -10, 10]
Explanation: 
(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

#Solution:
The key of this problem is that every operators in the string could be the last operator to be operated.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        //output every results of calculation
        vector<int> res;
        
        for(int i= 0; i< input.length(); i++){
            if(input[i] == '+' || input[i] == '-' || input[i] == '*'){
                vector<int> result1= diffWaysToCompute(input.substr(0, i));
                vector<int> result2= diffWaysToCompute(input.substr(i+1));
                
                for(auto n1: result1){
                    for(auto n2: result2){
                        if(input[i] == '+')
                            res.push_back(n1+n2);
                        else if(input[i] == '-')
                            res.push_back(n1-n2);
                        else
                            res.push_back(n1*n2);
                    }
                }
            }
            else
                continue;
        }
        
        if(res.empty())
            res.push_back(atoi(input.c_str()));
        return res;
    }
};