题目:
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.
采用dfs深度搜索.
1 #include <iostream> 2 #include <map> 3 #include <vector> 4 #include <string> 5 6 class Solution{ 7 private: 8 map<char,vector<char> > dict; 9 vector<string> res; 10 public: 11 void createDict() 12 { 13 dict.clear(); 14 dict['2'].push_back('a'); 15 dict['2'].push_back('b'); 16 dict['2'].push_back('c'); 17 dict['3'].push_back('d'); 18 dict['3'].push_back('e'); 19 dict['3'].push_back('f'); 20 dict['4'].push_back('g'); 21 dict['4'].push_back('h'); 22 dict['4'].push_back('i'); 23 dict['5'].push_back('j'); 24 dict['5'].push_back('k'); 25 dict['5'].push_back('l'); 26 dict['6'].push_back('m'); 27 dict['6'].push_back('n'); 28 dict['6'].push_back('o'); 29 dict['7'].push_back('p'); 30 dict['7'].push_back('q'); 31 dict['7'].push_back('r'); 32 dict['7'].push_back('s'); 33 dict['8'].push_back('t'); 34 dict['8'].push_back('u'); 35 dict['8'].push_back('v'); 36 dict['9'].push_back('w'); 37 dict['9'].push_back('x'); 38 dict['9'].push_back('y'); 39 dict['9'].push_back('z'); 40 } 41 42 void dfs(int dep,int maxDep,string &s,string ans){ 43 if(dep==maxDep){ 44 res.push_back(ans); 45 return; 46 } 47 48 for(int i=0;i<dict[s[dep]].size();i++){ 49 dfs(dep+1,maxDep,s,ans+dict[s[dep]][i]); 50 } 51 } 52 53 vector<string> letterCombinations(string digits){ 54 res.clear(); 55 createDict(); 56 dfs(0,digits.size(),digits,""); 57 } 58 59 };