LeetCode--Letter Combinations of a Phone Number

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Note:

Although the above answer is in lexicographical order, your answer could be in any order you want.

class Solution {
public:
    vector<string> letterCombinations(string digits)
    {
        string table = "abcdefghijklmnopqrstuvwxyz";
        vector<int> s_int;
        for(int i=0; i<digits.length(); i++)
            s_int.push_back(digits[i]-48);
        vector<string> res;
        vector<vector<char>> temp_res;
        for(int i=0; i<s_int.size(); i++)
        {
            int loc = s_int[i];
            vector<char> temp;
            if(loc < 7)
            {
                temp.push_back(table[3*(loc-1-1)]);
                temp.push_back(table[3*(loc-1-1)+1]);
                temp.push_back(table[3*(loc-1-1)+2]);
            }
            else if(loc == 7)
            {
                temp.push_back(table[3*(loc-1-1)]);
                temp.push_back(table[3*(loc-1-1)+1]);
                temp.push_back(table[3*(loc-1-1)+2]);
                temp.push_back(table[3*(loc-1-1)+3]);
            }
            else if(loc == 8)
            {
                temp.push_back(table[3*(loc-1-1)+1]);
                temp.push_back(table[3*(loc-1-1)+2]);
                temp.push_back(table[3*(loc-1-1)+3]);
            }
            else
            {
                temp.push_back(table[3*(loc-1-1)+1]);
                temp.push_back(table[3*(loc-1-1)+2]);
                temp.push_back(table[3*(loc-1-1)+3]);
                temp.push_back(table[3*(loc-1-1)+4]);
            }
            temp_res.push_back(temp);
        }
        int n = temp_res.size();
        get_res(temp_res,n,0,"",res);
        return res;

    }
    void get_res(vector<vector<char>>& source, int all, int start, string pre, vector<string>& res)
    {
        if(start == all)
        {
            res.push_back(pre);
            return;
        }
        int n = source[start].size();
        for(int i=0; i<n; i++)
        {
            get_res(source, all, start+1, pre+source[start][i],res);
        }
        return;
    }
};