c++回溯法解決1到9之間插入加減或空使運算結果為100

#include <iostream>
#include <math.h>
#include <vector>
#include <string>
using namespace std;
class Solution{
private:
    vector<string> res;
    vector<int> nums;
    vector<int> eles;
private:
    void _compute(vector<int> vec, int index, int target, string &s){
        if (index == vec.size()){
            if (0 == target)
                res.push_back(s + "=100");
            return;
        }
        //分“+”和“-”兩種情況討論
        for (int i = 0; i < 2; i++){
            if (i == 0){
                string tempStr = s + "+" + to_string(vec[index]);
                _compute(vec, index + 1, target - vec[index], tempStr);
            }
            else if (i == 1){
                string tempStr = s + "-" + to_string(vec[index]);
                _compute(vec, index + 1, target + vec[index], tempStr);
            }
        }
        return;
    }

    //用來得到1-9的不同整數組合，比如{123, 456, 789}，本質是將“”這個空符號加入到數之間
    void _recursion(int index, int target){
        if (index == 9){
            string s = to_string(eles[0]);
            _compute(eles, 1, target - eles[0], s);
            return;
        }

        //為瞭問題的泛化采用i <= 9，如果針對結果為100的可以改成i <= 5
        for (int i = 1; i <= 9; i++){
            if (index + i > 9)
                break;
            int temp = 0;

            //求得分解出來的每個元素的值
            for (int j = 0; j < i; j++){
                temp += nums[index + j] * pow(10, i - j - 1);
            }
            eles.push_back(temp);
            _recursion(index + i, target);
            eles.pop_back();
        }
        return;
    }

public:
    Solution(){
        nums = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    }

    vector<string> recursion(int index, int target){
        _recursion(index, target);
        return res;
    }
};

int main()
{
    Solution s;
    vector<string> res = s.recursion(0, 100);
    cout << "共有" << res.size() << "種情況" << endl;
    for (string s : res){
        cout << s << endl;
    }
    return 0;
}