[LeetCode 题解]: LetterCombinations

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.

题解： 递归。首先将输入字符串解析成数字集合，记录字符串大小为Len。

生成的字符串长度依旧为Len。从首位开始判断该位可以填充的字符。

注意：本题首先需要对手机键盘的字符进行映射，作为辅助。

 class Solution {
 public:
     char vi[][];
     vector<string > out;
     int length;
     void getVi()
     {
         int i,j;
         for(i=;i<=;i++)
             for(j=;j<;j++)
                 vi[i][j]=;
         for(i=;i<=;i++)
         {
             for(j=;j<;j++)
             {
                 vi[i][j]= 'a'+ *(i-)+j;
             }
         }
         vi[][]=' ';
         vi[][]='s';
         for(i=;i<=;i++)
         {
             for(j=;j<;j++)
             {
                 vi[i][j]= 'b' + *(i-)+j;
             }
         }
         vi[][]='z';
     }

     void DFS(int len,string digits, string c)
     {
         if(len==length)
         {
             out.push_back(c);
             return ;
         }
         int count = digits[len]-'';
         for(int i=;i<;i++)
         {
            if(vi[count][i]!=)
                DFS(len+,digits,c+vi[count][i]);
         }
     }
     vector<string> letterCombinations(string digits) {
         out.clear();
         length = digits.size();
         if(length==)
         {
             out.push_back("");
             return out;
         }
         if(digits.find('')!=string::npos) return out;
         getVi();
         DFS(,digits,"");
         return out;
     }
 };

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