Created
March 16, 2013 02:23
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Given a digit string, return all possible letter combinations that the number could represent.
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| /* | |
| 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. | |
| */ | |
| public class Solution { | |
| public ArrayList<String> letterCombinations(String digits) { | |
| // Start typing your Java solution below | |
| // DO NOT write main() function | |
| 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'}}; | |
| ArrayList<String> ret = new ArrayList<String>(); | |
| if(digits.length() == 0) { | |
| ret.add(""); | |
| return ret; | |
| } | |
| StringBuffer str = new StringBuffer(); | |
| combines(0, digits, str, map, ret); | |
| return ret; | |
| } | |
| public void combines(int start, String digits, StringBuffer str, char[][] map, ArrayList<String> ret) { | |
| int curNum = digits.charAt(start) - '0'; | |
| for(int i = 0; i < map[curNum].length; i++) { | |
| str.append(map[curNum][i]); | |
| if(start == (digits.length() - 1)) { | |
| ret.add(str.toString()); | |
| } | |
| if(start < digits.length() - 1) | |
| combines(start+1, digits, str, map, ret); | |
| // delete the last one | |
| str.deleteCharAt(str.length() - 1); | |
| } | |
| } | |
| } |
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