【LeetCode】18.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"].

递归的题老是做不好。拿到此题一直纠结于字母与数字的关系，可以再怎么纠结，也是得一个个遍历才可以。用DFS，注意if(ans.size()==0) ans.add("");

public class Solution {
     public ArrayList<String> letterCombinations(String digits) {
         ArrayList<String> ans = new ArrayList<String>();
         ArrayList<Character> chars = new ArrayList<Character>();
         HashMap<Character,String> map = new HashMap<Character,String>();
         init(map);
         dfs(map,digits,0,chars,ans);
         if(ans.size()==0)
             ans.add("");
         return ans;
     }
     private void dfs(HashMap<Character, String> map, String digits, int idx, ArrayList<Character> chars, ArrayList<String> arr){
         if(idx >= digits.length()) return;
         char c = digits.charAt(idx);
         String s = map.get(c);
         for(int i = 0; i<s.length();i++){
             if(idx == digits.length()-1){
                 chars.add(s.charAt(i));
                 StringBuilder sb = new StringBuilder();
                 for(char cc: chars){
                     sb.append(cc);
                 }
                 arr.add(sb.toString());
                 chars.remove(chars.size()-1);
             }else{
                 chars.add(s.charAt(i));
                 dfs(map,digits,idx+1,chars,arr);
                 chars.remove(chars.size()-1); //char.remove(index)
             }
         }
     }
      
     private void init(HashMap<Character,String> map){
         map.put('1', "");
         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");
     }
}