方法：回溯—-for循环+递归，不用考虑递归结束条件，for循环结束递归结束且在递归结束时输出结果
    string的用法要注意

    1. class Solution {
    2. public:
    3.      unordered_map<char, string> phoneMap{
    4.             {'2', "abc"},
    5.             {'3', "def"},
    6.             {'4', "ghi"},
    7.             {'5', "jkl"},
    8.             {'6', "mno"},
    9.             {'7', "pqrs"},
    10.             {'8', "tuv"},
    11.             {'9', "wxyz"}
    12.         };
    13.     vector<string> results;
    14.     vector<string> letterCombinations(string digits) {
    15.         if(digits.empty()){
    16.             return results;
    17.         }
    18.         string result;
    19.         combination(digits,0,result,results);
    20.         return results;
    22.     }
    23.     void combination(string& digits,int index,string& result,vector<string>&results){
    24.         if(index==digits.size()){
    25.              results.emplace_back(result);
    26.         }
    27.         char temp=digits[index];
    28.         string pos=phoneMap[temp];
    29.         for(auto p:pos){
    30.             result+=p;
    31.             combination(digits,index+1,result,results);
    32.             result.pop_back();
    33.         }
    34.     }
    35. };