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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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LeetCode: LetterCombinations 题解,布布扣,bubuko.com
LeetCode: LetterCombinations 题解
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原文地址：http://www.cnblogs.com/double-win/p/3807124.html
