Given a string containing digits from 2-9 inclusive, 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. Note that 1 does not map to any letters.
Example:
Input: "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：
给定一个数字串，看看有多少种对应的字母串。
思路：
Assumption：
字符串的合法性，是否包含1这个数字？如果包含，怎么处理？同样，输入是否考虑 * 或 #？
空字符串怎么处理？
多个解按什么顺序返回?
High Level带着面试官walk through：
take "23" for example:
when index = 0, pointing to '2' in "23"
String letters = "abc"
for each char in letters,
add to path
move index forward, pointing to '3' in "23", recursively call the helper function to add every char of "def" to the path
code
class Solution {
private static String[] keyboard =
new String[]{ " ", "", "abc", "def", // '0','1','2',...'9'
"ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
public List<String> letterCombinations(String digits) { //"23"
List<String> result = new ArrayList<>();
if(digits.length() == 0 ) return result;
helper(digits, 0, new StringBuilder(), result);
return result;
}
private void helper(String digits, int index, StringBuilder sb, List<String> result){
if(index == digits.length()){
result.add(sb.toString());
return;
}
String letters = keyboard[digits.charAt(index) - '0'];
for(int i = 0; i < letters.length(); i++ ){
char c = letters.charAt(i);
sb.append(c);
helper(digits, index+1, sb, result);
sb.deleteCharAt(sb.length() - 1);
}
}
}
注意： 这个题也可以用String path来存每条路径。 可是我自己觉得用StringBuilder这么动态伸缩，更显得有“Backtracking”的味道。