力扣——为运算表达式设计优先级

给定一个含有数字和运算符的字符串，为表达式添加括号，改变其运算优先级以求出不同的结果。你需要给出所有可能的组合的结果。有效的运算符号包含 +, - 以及 * 。

示例 1:

输入: "2-1-1"
输出: [0, 2]
解释: 
((2-1)-1) = 0 
(2-(1-1)) = 2

示例 2:

输入: "2*3-4*5"
输出: [-34, -14, -10, -10, 10]
解释: 
(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

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> result;
        // 递归
        // 第一位和最后一位不为运算符，故跳过
        for (size_t i = 1; i < input.size() - 1; 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 (size_t r1 = 0; r1 < result1.size(); r1++){
                    for (size_t r2 = 0; r2 < result2.size(); r2++){
                        if (input[i] == '+'){
                            result.push_back(result1[r1] + result2[r2]);
                        }
                        else if (input[i] == '-'){
                            result.push_back(result1[r1] - result2[r2]);
                        }
                        else{
                            result.push_back(result1[r1] * result2[r2]);
                        }
                    }
                }
            }
        }
        
        if (result.empty() && !input.empty()){
            result.push_back(stoi(input));
        }
        
        return result;
    }
};