Created
May 16, 2013 17:00
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Letter Combinations of a Phone Number
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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. | |
| Algorithm | |
| 1. It needs to be done using recursively with the length of the numbers and the chars for each number. | |
| */ | |
| public class Solution { | |
| public ArrayList<String> letterCombinations(String digits) { | |
| // Start typing your Java solution below | |
| // DO NOT write main() function | |
| ArrayList<String> ret = new ArrayList<String>(); | |
| if(digits.length() == 0) { | |
| ret.add(""); | |
| return ret; | |
| } | |
| StringBuffer sb = new StringBuffer(); | |
| String[] numStrMap = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; | |
| letterCombinationsHelper(digits, ret, 0, sb, numStrMap); | |
| return ret; | |
| } | |
| public void letterCombinationsHelper(String digits, ArrayList<String> ret, int start, StringBuffer sb, String[] numStrMap) { | |
| if(sb.length() == digits.length()) { | |
| ret.add(sb.toString()); | |
| return; | |
| } | |
| for(int i = start; i < digits.length(); i++) { | |
| int index = digits.charAt(i)-'0'; | |
| for(int j = 0; j < numStrMap[index].length(); j++) { | |
| sb.append(numStrMap[index].charAt(j)); | |
| letterCombinationsHelper(digits, ret, i+1, sb, numStrMap); | |
| sb.deleteCharAt(sb.length() - 1); | |
| } | |
| } | |
| return; | |
| } | |
| } |
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