[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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