Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent. Return the answer in any order.
A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
Example 1:
Input: digits = "23"
Output: ["ad","ae","af","bd","be","bf","cd","ce","cf"]
Example 2:
Input: digits = ""
Output: []
Example 3:
Input: digits = "2"
Output: ["a","b","c"]
Constraints:
0 <= digits.length <= 4
digits[i] is a digit in the range ['2', '9'].
Solution:
Approach 1: Iterative
class Solution {
public List<String> letterCombinations(String digits) {
LinkedList<String> ans = new LinkedList<String>();
if(digits.isEmpty()) return ans;
String[] combinations = new String[] {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
ans.add("");
for(int i =0; i<digits.length();i++){
int x = Character.getNumericValue(digits.charAt(i));
while(ans.get(0).length()==i){
String t = ans.remove(0);
for(char s : combinations[x].toCharArray())
ans.add(t+s);
}
}
return ans;
}
}
Approach 2: DFS
class Solution {
String[] combinations = new String[]{"0","1","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
public List<String> letterCombinations(String digits) {
LinkedList<String> ans = new LinkedList<String>();
if(digits.isEmpty()) return ans;
dfs(0,new StringBuilder(),digits,ans);
return ans;
}
private void dfs(int i, StringBuilder current, String digits,List<String> ans)
{
if(i==digits.length())
{
ans.add(current.toString());
return;
}
char curchar = digits.charAt(i);
for(char neigh:combinations[curchar-'0'].toCharArray())
{
current.append(neigh);
dfs(i+1,current,digits,ans);
current.deleteCharAt(current.length()-1);
}
}
}