# 17.Letter-Combinations-of-a-Phone-Number

## ้ข็ฎๅฐๅ

https://leetcode.com/problems/letter-combinations-of-a-phone-number/

https://www.jiuzhang.com/solutions/palindrome-partitioning-ii/

## ้ข็ฎๆ่ฟฐ

``````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.``````

## ไปฃ็ 

### Approach 1: Backtracking

Complexity Analysis

• Time complexity : O(3^Nร4^M) where `N` is the number of digits in the input that maps to 3 letters (e.g. `2, 3, 4, 5, 6, 8`) and `M` is the number of digits in the input that maps to 4 letters (e.g. `7, 9`), and `N+M` is the total number digits in the input.

• Space complexity : O(3^Nร4^M) since one has to keep 3^Nร4^M solutions.

``````class Solution {
Map<String, String> phone = new HashMap<String, String>(){{
put("2", "abc");
put("3", "def");
put("4", "ghi");
put("5", "jkl");
put("6", "mno");
put("7", "pqrs");
put("8", "tuv");
put("9", "wxyz");
}};

List<String> output = new ArrayList<String>();

public List<String> letterCombinations(String digits) {
if (digits.length() != 0) {
backtrack("", digits);
}
return output;
}

private void backtrack(String combination, String next_digits) {
if (next_digits.length() == 0) {