电话号码的字母组合，修正程序，有点要求

const pair<int, string> valueSymbols[] = {
{0,""},
{1,""},
{2,"abc"},
{3,"def"},
{4,"ghi"},
{5,"jkl"},
{6,"mno"},
{7,"pqrs"},
{8,"tuv"},
{9,"wxyz"},
}
class Solution
{
public:
int n = digits.size();
for (int i = 0; i < n; ++i)
{
const auto& [value, symbol]:valueSymbols;
value = digits(i);
string sum[];
sum.push_back[symbol];
}
for (int j = 0; j < n; j++)
{
for (int k = j + 1; k < n; k++)
{
int s = sum[j].size();
int d = sum[k].size();
string ans[];
for (int first = 0; first < s; first++)
{
for (int second = 0; second < d; second++)
{
string bns = sum[j][first] + sum[k][second];
string ans.push_back(bns)
}
}
}
}
return ans;
};
要求：这是力扣上的第十七题，c++，我知道我写的有问题，希望来个大佬，在我写的思路上更改成能在力扣上成功编译的程序，注意：必须在在力扣上能够成功编译，且在我的思路上进行修正，如果我的思路错误或者语法错误，请大佬文字说明一下，来一个负责任的大佬

class Solution {
    unordered_map<int, vector<char>> dict{ {1,{}},
        {2,{'a','b','c'}},
        {3,{'d','e','f'}},
        {4,{'g','h','i'}},
        {5,{'j','k','l'}},
        {6,{'m','n','o'}},
        {7,{'p','q','r','s'}},
        {8,{'t','u','v'}},
        {9,{'w','x','y','z'}}
    };
public:
    vector<string> dfs(string str) {
        vector<string> res;
        if (str == "")
            return res;
        int len = str.size();
        if (len == 1)
        {
            int t = str[0] - '0';
            for (char c : dict[t]) {
                string s;
                s+= c;
                res.push_back(s);
            }
            return res;
        }
        int t = str[0]-'0';
        str = str.substr(1, len - 1);
        vector<string> vtmp = dfs(str);
        for (char c : dict[t]) {
            for (auto s : vtmp) {
                string str = c + s;
                res.push_back(str);
            }
        }
        return res;
    }
    vector<string> letterCombinations(string digits) {
        vector<string> res;
        res=dfs(digits);
        return res;
    }
};

大概是那个点过不去还是代码本身过不去，如果一开始假设错误，在你的代码上改是做无用功

要考虑时间复杂度和空间复杂度吗