LeetCode: LetterCombinations 題解

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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.

題解： 遞迴。首先將輸入字串解析成數字集合，記錄字串大小為Len。

產生的字串長度依舊為Len。從首位開始判斷該位可以填充的字元。

注意：本題首先需要對手機鍵盤的字元進行映射，作為輔助。

 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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