241. Different Ways to Add Parentheses

https://leetcode.com/problems/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"
Output: [0, 2]
Explanation: 
((2-1)-1) = 0 
(2-(1-1)) = 2

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

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> nums;
        vector<char> ops;
        preprocess(input,  nums, ops);
        
        return diffWaysToCompute(nums, ops, 0, nums.size() - 1);
    }
    inline void preprocess(string &input, vector<int> &nums, vector<char> &ops) {
        int d = input.front() - '0';
        for (int i = 1; i < input.size(); ++i) {
            if (!isdigit(input[i])) {
                nums.push_back(d);
                ops.push_back(input[i]);
                d = 0;
            } else {
                d = (d * 10) + (input[i] - '0');
            }
        }
        nums.push_back(d);
    }
    inline int calc(char op, int n1, int n2) {
        switch (op) {
            case '+':
                return n1 + n2;
            case '-':
                return n1 - n2;
            default:
                return n1 * n2;
        }
    }
    vector<int> diffWaysToCompute(vector<int> &nums, vector<char> &ops, int begin, int end) {
        if (begin == end) return vector<int>{nums[begin]};
        vector<int> results;
        for (int i = begin; i < end; ++i) {
            const vector<int> &left = diffWaysToCompute(nums, ops, begin, i);
            const vector<int> &right = diffWaysToCompute(nums, ops, i + 1, end);
            
            for (const auto l: left) {
                for (const auto r: right) {
                    results.push_back(calc(ops[i], l, r));
                }
            }
        }
        return results;
    }
};

• #recursive