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.
题意：将数字的字符串，转换成所有可能的字符串，放到一个list中。
分析：此题已经有一点难度了，因为我们不确定数字的字符串有多长，所以我们不可以单纯的利用for循环来做这道题来，我们需要用到递归算法来做这道题。
java代码：
class 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);
}
}
}
