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

## 17. Letter Combinations of a Phone Number

## 题目地址

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

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

## 题目描述

```
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.
```

## 代码

### Approach 1: Backtracking

**Complexity Analysis**

* Time complexity : O(3^N×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^N×4^M) since one has to keep 3^*N*×4^*M* solutions.

```java
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) {
      output.add(combination);
    } else {
      String digit = next_digits.substring(0, 1);
      String letters = phone.get(digit);
      for (int i = 0; i < letters.length(); i++) {
        String letter = letters.substring(i, i + 1);
        backtrack(combination + letter, next_digits.substring(1));
      }
    }
  }

}
```


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