# LeetCode – Letter Combinations of a Phone Number (Java)

Given a digit string, return all possible letter combinations that the number could represent. (Check out your cellphone to see the mappings) Input:Digit string "23", Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Analysis

This problem can be solves by a typical DFS algorithm. DFS problems are very similar and can be solved by using a simple recursion. Check out the index page to see other DFS problems.

Java Solution

```public List<String> letterCombinations(String digits) { HashMap<Integer, String> map = new HashMap<Integer, String>(); map.put(2, "abc"); map.put(3, "def"); map.put(4, "ghi"); map.put(5, "jkl"); map.put(6, "mno"); map.put(7, "pqrs"); map.put(8, "tuv"); map.put(9, "wxyz"); map.put(0, "");   ArrayList<String> result = new ArrayList<String>();   if(digits == null || digits.length() == 0) return result;   ArrayList<Character> temp = new ArrayList<Character>(); getString(digits, temp, result, map);   return result; }   public void getString(String digits, ArrayList<Character> temp, ArrayList<String> result, HashMap<Integer, String> map){ if(digits.length() == 0){ char[] arr = new char[temp.size()]; for(int i=0; i<temp.size(); i++){ arr[i] = temp.get(i); } result.add(String.valueOf(arr)); return; }   Integer curr = Integer.valueOf(digits.substring(0,1)); String letters = map.get(curr); for(int i=0; i<letters.length(); i++){ temp.add(letters.charAt(i)); getString(digits.substring(1), temp, result, map); temp.remove(temp.size()-1); } }```

I often found that I write a solution differently each time I solve a problem. Here is another way of writing the solution.

```public List<String> letterCombinations(String digits) { HashMap<Character, char[]> map = new HashMap<Character, char[]>(); map.put('2', new char[]{'a','b','c'}); map.put('3', new char[]{'d','e','f'}); map.put('4', new char[]{'g','h','i'}); map.put('5', new char[]{'j','k','l'}); map.put('6', new char[]{'m','n','o'}); map.put('7', new char[]{'p','q','r','s'}); map.put('8', new char[]{'t','u','v'}); map.put('9', new char[]{'w','x','y','z'});   List<String> result = new ArrayList<String>(); if(digits.equals("")) return result;   helper(result, new StringBuilder(), digits, 0, map);   return result;   }   public void helper(List<String> result, StringBuilder sb, String digits, int index, HashMap<Character, char[]> map){ if(index>=digits.length()){ result.add(sb.toString()); return; }   char c = digits.charAt(index); char[] arr = map.get(c);   for(int i=0; i<arr.length; i++){ sb.append(arr[i]); helper(result, sb, digits, index+1, map); sb.deleteCharAt(sb.length()-1); } }```
Category >> Algorithms >> Interview >> Java
If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:
```<pre><code>
String foo = "bar";
</code></pre>
```
• sunandan

What is the time complexity of this ? Is it n^2 ?

• gao can

Thanks!

• sunandan

your solution is much better . @author : consider changing the code to use this snippet. Much clearer.

• gao can

Consider using a StringBuilder?