leetcode 241. Different Ways to Add Parentheses

/* 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] 解题思路：分治，遇到一个运算符,就计算两边的结果，然后把所有可能的结果进行运算，压入vector中 */

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cstdlib>

using namespace std;

class Solution {
public:
    vector<int> diffWaysToCompute(string input) 
    {
        vector<int> res;
        int size = input.size();
        for (size_t i = 0; i < size; ++i)
        {
            if (input[i] == '+' || input[i] == '-' || input[i] == '*')
            {
                vector<int> tmp1 = diffWaysToCompute(input.substr(0, i));
                vector<int> tmp2 = diffWaysToCompute(input.substr(i + 1));
                for (auto n1 : tmp1)
                {
                    for (auto n2 : tmp2)
                    {
                        if (input[i] == '+')
                            res.push_back(n1 + n2);
                        else if (input[i] == '-')
                            res.push_back(n1 - n2);
                        else
                            res.push_back(n1*n2);
                    }
                }
            }
        }

        //如果为空，说明没有运算符号，是一个数字
        if (res.empty())
            res.push_back(atoi(input.c_str()));

        return res;
    }
};

void test()
{
    Solution sol;
    auto res = sol.diffWaysToCompute("2*3-4*5");
    for (auto i : res)
        cout << i << " ";
    cout << endl;
}

int main()
{
    test();

    return 0;
}