原题：

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.

思路：涉及到排列组合问题，使用递归专门处理组合问题。一开始老是runtime error，主要问题出在string char以及const char的转换上，比较烦人。思路比较简单的。

代码：

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> res, tmp;
        if(digits.size()==0) {
            string str = "";
            res.push_back(str);
            return res;
        }
        for (int i = 0; i<digits.size(); i++){
            string accord = digitString(digits[i]);
            tmp.push_back(accord);
        }
        return makecombine(tmp.begin(),tmp.end());
    }
    
    vector<string> makecombine(vector<string>::iterator first, vector<string>::iterator last){
        vector<string> res,tmp;
        if(first==last) return res;
        tmp = makecombine(first+1,last);
        for(string::iterator it = (*first).begin(); it!=(*first).end(); it++){
            if(tmp.empty())  {
                string t = "";
                t = t + *it;
                res.push_back(t);
            }
            else {
                for(int j = 0; j<tmp.size(); j++){
                    string t = "";
                    t = *it;
                    t = t+tmp[j];
                    res.push_back(t);
                }
            }
        }
        return res;
    }
    
    string digitString(char a){
        string str= "";
        if (a-'0'==0) str =' ';
        else if(a -'0'== 1) ;
        else if(a -'0'== 2) str = "abc";
        else if(a -'0'== 3) str = "def";
        else if(a -'0'== 4) str = "ghi";
        else if(a -'0'== 5) str = "jkl";
        else if(a -'0'== 6) str = "mno";
        else if(a -'0'== 7) str = "pqrs";
        else if(a -'0'== 8) str = "tuv";
        else if(a -'0'== 9) str = "wxyz";
        return str;
    }
};