【leetcode刷题笔记】Letter Combinations of a Phone Number

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.

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题解：用一个二维数组map保存数字到字母的映射，然后深度优先递归搜索如下的一棵树（以"23"为例，树不完整）

代码如下：

 1 public class Solution {

 2     public List<String> letterCombinations(String digits) {

 3         ArrayList<String> answer = new ArrayList<String>();

 4         

 5         char[][] map = {{'a','b','c'},{'d','e','f'},{'g','h','i'},{'j','k','l'},{'m','n','o'},{'p','q','r','s'},{'t','u','v'},{'w','x','y','z'}};

 6         String currentPath = new String();

 7         

 8         DeepSearch(map, answer, digits, currentPath);

 9         return answer;

10     }

11     

12     public void DeepSearch(char[][] map,List<String> answer,String digits,String currentPath){

13         if(digits.length() == 0)

14         {

15             String temp = new String(currentPath);

16             answer.add(temp);

17 

18             return;

19         }

20         

21         int now = digits.charAt(0) - '0';

22         for(int j = 0;j < map[now-2].length;j++){

23             currentPath += map[now-2][j];

24             DeepSearch(map, answer, digits.substring(1), currentPath);

25             currentPath = currentPath.substring(0, currentPath.length()-1);

26         }

27     }

28 }