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.

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));
      }
    }
  }

}