LeetCode--Letter Combinations of a Phone Number

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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;    }};

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