leetcode-Letter Combinations of a Phone Number-17

输入一组数字，根据手机九宫格上每个数字对应的字母，输出这组数字能代表的所有字母的组合。

dfs，回溯。

 1 class Solution {
 2 private:
 3     map<char,vector<char> > v;
 4 public:
 5     Solution(){
 6         v['2'].push_back('a');
 7         v['2'].push_back('b');
 8         v['2'].push_back('c');
 9         v['3'].push_back('d');
10         v['3'].push_back('e');
11         v['3'].push_back('f');
12         v['4'].push_back('g');
13         v['4'].push_back('h');
14         v['4'].push_back('i');
15         v['5'].push_back('j');
16         v['5'].push_back('k');
17         v['5'].push_back('l');
18         v['6'].push_back('m');
19         v['6'].push_back('n');
20         v['6'].push_back('o');
21         v['7'].push_back('p');
22         v['7'].push_back('q');
23         v['7'].push_back('r');
24         v['7'].push_back('s');
25         v['8'].push_back('t');
26         v['8'].push_back('u');
27         v['8'].push_back('v');
28         v['9'].push_back('w');
29         v['9'].push_back('x');
30         v['9'].push_back('y');
31         v['9'].push_back('z');
32     }
33     void dfs(string s,int i,int j,vector<string> &ans,string ss){
34         if(i==s.size()||j==v[s[i]].size()) return;
35         ss+=v[s[i]][j];
36         if(i==s.size()-1){
37             ans.push_back(ss);
38             return;
39         }
40         for(int k=0;k<v[s[i+1]].size();k++) dfs(s,i+1,k,ans,ss);
41     }
42     vector<string> letterCombinations(string digits) {
43         vector<string> ans;
44         if(digits.size()==0) return ans;
45         string s;
46         for(int i=0;i<v[digits[0]].size();i++) dfs(digits,0,i,ans,s);
47         return ans;
48     }
49 };

　　

posted @ 2016-08-29 22:13  0_summer  阅读(130)  评论(0)    收藏  举报