[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。

 1 class Solution {
 2 public:
 3     char vi[10][4];
 4     vector<string > out;
 5     int length;
 6     void getVi()
 7     {
 8         int i,j;
 9         for(i=0;i<=9;i++)
10             for(j=0;j<4;j++)
11                 vi[i][j]=0;
12         for(i=2;i<=7;i++)
13         {
14             for(j=0;j<3;j++)
15             {
16                 vi[i][j]= 'a'+ 3*(i-2)+j;
17             }
18         }
19         vi[0][0]=' ';
20         vi[7][3]='s';
21         for(i=8;i<=9;i++)
22         {
23             for(j=0;j<3;j++)
24             {
25                 vi[i][j]= 'b' + 3*(i-2)+j;
26             }
27         }
28         vi[9][3]='z';
29     }
30     
31     void DFS(int len,string digits, string c)
32     {
33         if(len==length)
34         {
35             out.push_back(c);
36             return ;
37         }
38         int count = digits[len]-'0';
39         for(int i=0;i<4;i++)
40         {
41            if(vi[count][i]!=0)
42                DFS(len+1,digits,c+vi[count][i]);
43         }
44     }
45     vector<string> letterCombinations(string digits) {
46         out.clear();
47         length = digits.size();
48         if(length==0) 
49         {
50             out.push_back("");
51             return out;
52         }
53         if(digits.find('1')!=string::npos) return out;
54         getVi();
55         DFS(0,digits,"");
56         return out;
57     }
58 };

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