LeetCode - Letter Combinations of a Phone Number

Letter Combinations of a Phone Number

2013.12.6 19:52

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.

Solution:

　　Given a certain digit string and the corresponding mapping between digits [0-9] and letters [a-z],  return all the possible letter combinatios that map to this digit string. Since all the combinations must be enumerated, using DFS is efficient in coding. A mapping array between  digits and letters is needed.

　　Code segment is quite short and need no more explanation.

　　Time complexity is BIG_PI(i in range(0, len(digit_str))){number of mapping for digit_str[i]}, space complexity is O(len(digit_str)).

Accepted code:

 1 // 7CE, 2RE, 1WA, 1AC, too bad~

 2 #include <string>

 3 using namespace std;

 4 

 5 class Solution {

 6 public:

 7     vector<string> letterCombinations(string digits) {

 8         // IMPORTANT: Please reset any member data you declared, as

 9         // the same Solution instance will be reused for each test case.

10         result.clear();

11         

12         dfs(digits, 0, "");

13         

14         return result;

15     }

16 private:

17     vector<string> result;

18     string dict[10] = {

19         " ", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"

20     };

21     

22     void dfs(string &digits, int idx, string str) {

23         if(idx == digits.length()){

24             result.push_back(str);

25             // 1RE here, didn't return

26             return;

27         }

28         

29         // 7CE here, dict[digits[idx] - '0'], type mismatch

30         for(int i = 0; i < dict[digits[idx] - '0'].length(); ++i){

31             // 1RE here, 1WA here, wrong index

32             dfs(digits, idx + 1, str + dict[digits[idx] - '0'][i]);

33         }

34     }

35 };