LeetCode OJ：Letter Combinations of a Phone Number（数字字母组合）

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"].

给出相应的按键，然后要求给出所有可能的字母组合，典型的dfs问题，但是这里的重点在于字典的店里，以及怎么来进行组合，代码如下，原理比较简单，这里就不再赘述了。PS:这题一开始其实没有想明白怎么建立字典，但是实际上通过一个map简单的就可以建立数字与字母之间的关联，代码如下所示：

 class Solution {
 public:
     vector<string> letterCombinations(string digits) {
         if(!digits.size())
             return ret;
         createDic(dic);
         dfs(, digits.size(), digits, "");
         return ret;     }     void createDic(map<char, vector<char>> & dic)
     {
         dic[''].push_back('a');
         dic[''].push_back('b');
         dic[''].push_back('c');
         dic[''].push_back('d');
         dic[''].push_back('e');
         dic[''].push_back('f');
         dic[''].push_back('g');
         dic[''].push_back('h');
         dic[''].push_back('i');
         dic[''].push_back('j');
         dic[''].push_back('k');
         dic[''].push_back('l');
         dic[''].push_back('m');
         dic[''].push_back('n');
         dic[''].push_back('o');
         dic[''].push_back('p');
         dic[''].push_back('q');
         dic[''].push_back('r');
         dic[''].push_back('s');
         dic[''].push_back('t');
         dic[''].push_back('u');
         dic[''].push_back('v');
         dic[''].push_back('w');
         dic[''].push_back('x');
         dic[''].push_back('y');
         dic[''].push_back('z');
     }     void dfs(int dep, int maxDep, string & s, string ans)
     {
         if(dep == maxDep){
             ret.push_back(ans);
             return;
         }         for(int i = ; i < dic[s[dep]].size(); ++i) //这里的dep应该得到注意
             dfs(dep+, maxDep, s, ans + dic[s[dep]][i]) ;
     } private:
     map<char, vector<char>> dic;
     vector<string> ret;
 };