/
NikolayIvkin
/
TheAlgorithms
Обзор
Документация
Войти
/
NikolayIvkin
/
TheAlgorithms
Код
Запросы
0
Задачи
Вики
Пакеты
0
Релизы
0
Аналитика
Безопасность
master
src/main/java/com/thealgorithms/strings/LetterCombinationsOfPhoneNumber.java
65 строк
2 KB
Samuel Facchinello
refactor: redesign `LetterCombinationsOfPhoneNumber` (#5221)
13 июн 2024, 20:40
Не верифицирован
13 июн 2024, 20:40
51fcc66
Код
Авторство
О чём код?
package com.thealgorithms.strings; import java.util.ArrayList; import java.util.Collections; import java.util.List; public final class LetterCombinationsOfPhoneNumber { private static final char EMPTY = '\0'; // Mapping of numbers to corresponding letters on a phone keypad private static final String[] KEYPAD = new String[] {" ", String.valueOf(EMPTY), "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; private LetterCombinationsOfPhoneNumber() { } /** * Generates a list of all possible letter combinations that the provided * array of numbers could represent on a phone keypad. * * @param numbers an array of integers representing the phone numbers * @return a list of possible letter combinations */ public static List<String> getCombinations(int[] numbers) { if (numbers == null) { return List.of(""); } return generateCombinations(numbers, 0, new StringBuilder()); } /** * Recursive method to generate combinations of letters from the phone keypad. * * @param numbers the input array of phone numbers * @param index the current index in the numbers array being processed * @param current a StringBuilder holding the current combination of letters * @return a list of letter combinations formed from the given numbers */ private static List<String> generateCombinations(int[] numbers, int index, StringBuilder current) { // Base case: if we've processed all numbers, return the current combination if (index == numbers.length) { return new ArrayList<>(Collections.singletonList(current.toString())); } final var number = numbers[index]; if (number < 0 || number > 9) { throw new IllegalArgumentException("Input numbers must in the range [0, 9]"); } List<String> combinations = new ArrayList<>(); // Iterate over each letter and recurse to generate further combinations for (char letter : KEYPAD[number].toCharArray()) { if (letter != EMPTY) { current.append(letter); } combinations.addAll(generateCombinations(numbers, index + 1, current)); if (letter != EMPTY) { current.deleteCharAt(current.length() - 1); // Backtrack by removing the last appended letter } } return combinations; } }