17. 电话号码的字母组合 

 1 //DFS问题一直很难
 2 class Solution 
 3 {
 4     void dfs(string digits,vector<vector<char>>& d,vector<string> &res,int cur,string& temp)
 5     {
 6         if(cur == digits.size()) 
 7         {
 8             res.push_back(temp);
 9             return;
10         }
11 
12         for(auto a : d[digits[cur] - '0'])
13         {
14             temp.push_back(a);
15             dfs(digits,d,res,cur + 1,temp);
16             temp.pop_back();
17         }
18     }
19 public:
20     vector<string> letterCombinations(string digits) 
21     {
22         vector<vector<char>> d(10);
23         d[0] = {' '};
24         d[1] = {};
25         d[2] = {'a','b','c'};
26         d[3] = {'d','e','f'};
27         d[4] = {'g','h','i'};
28         d[5] = {'j','k','l'};
29         d[6] = {'m','n','o'};
30         d[7] = {'p','q','r','s'};
31         d[8] = {'t','u','v'};
32         d[9] = {'w','x','y','z'};
33         vector<string> res;
34         if(digits.empty()) return res;
35         string temp;
36         dfs(digits,d,res,0,temp);
37         return res;
38     }
39 };

 

 1 //DFS问题一直很难
 2 class Solution 
 3 {
 4     void dfs(string digits,vector<string>& d,vector<string> &res,int cur,string& temp)
 5     {
 6         if(cur == digits.size()) 
 7         {
 8             res.push_back(temp);
 9             return;
10         }
11 
12         for(auto a : d[digits[cur] - '2'])
13         {
14             temp.push_back(a);
15             dfs(digits,d,res,cur + 1,temp);
16             temp.pop_back();
17         }
18     }
19 public:
20     vector<string> letterCombinations(string digits) 
21     {
22         vector<string> d = {"abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
23         vector<string> res;
24         if(digits.empty()) return res;
25         string temp;
26         dfs(digits,d,res,0,temp);
27         return res;
28     }
29 };

 

posted @ 2020-03-15 18:32  Jinxiaobo0509  阅读(144)  评论(0)    收藏  举报