卡在java电话号码字生成器上
我有一项任务，从7位数的电话号码生成每个可能的单词，并使用PrintWriter将其保存为.txt文件。 我的代码在下面，但我的输出（目前只是打印到控制台）是相同的3“字”2187次。
package ks2_Lab19; import java.util.Scanner; import java.io.PrintWriter; import java.io.FileNotFoundException; public class WordGenerator { private static String[] two = {"a", "b", "c"}; private static String[] three = {"d", "e", "f"}; private static String[] four = {"g", "h", "i"}; private static String[] five = {"j", "k", "l"}; private static String[] six = {"m", "n", "o"}; private static String[] seven = {"p", "r", "s"}; private static String[] eight = {"t", "u", "v"}; private static String[] nine = {"w", "x", "y"}; private static char[] numArray; private static String[] wordList = new String[2187]; public static void convert(char[] input){ for (int i = 0; i < 2184; i = i + 3){ for (int a = 0; a < 7; a++){ for (int b = 0; b < 3; b++){ if (input[a] == '1' || input[a] == '0') { wordList[i+b] = wordList[i+b] + " "; }//if 0 or 1 if (input[a] == '2'){ wordList[i+b] = wordList[i+b] + two[b]; }//if 2 if (input[a] == '3'){ wordList[i+b] = wordList[i+b] + three[b]; }//if 3 if (input[a] == '4'){ wordList[i+b] = wordList[i+b] + four[b]; }//if 4 if (input[a] == '5'){ wordList[i+b] = wordList[i+b] + five[b]; }//if 5 if (input[a] == '6'){ wordList[i+b] = wordList[i+b] + six[b]; }//if 6 if (input[a] == '7'){ wordList[i+b] = wordList[i+b] + seven[b]; }//if 7 if (input[a] == '8'){ wordList[i+b] = wordList[i+b] + eight[b]; }//if 8 if (input[a] == '9'){ wordList[i+b] = wordList[i+b] + nine[b]; }//if 9 }//possible output for loop }//input array for loop }//write to wordList for loop } public static void main(String[] args) { //initialize output file name and PrintWriter object String fileName = "output.txt"; PrintWriter outputStream = null; String output = ""; //try and catch exception try { outputStream = new PrintWriter(fileName); } catch (FileNotFoundException e){ System.out.println("Error opening file " + fileName + "."); System.exit(0); } //initialize scanner and wordList array Scanner kb = new Scanner(System.in); for (int i=0; i < 2187; i++){ wordList[i] = ""; } //announce and accept input System.out.println("Please input a 7 digit phone number without special characters."); String num = kb.next(); numArray = num.toCharArray(); convert(numArray); for (int p = 0; p < 2187; p++){ System.out.println(wordList[p]); } } } 我相信下面的代码正确地解决了问题，并演示了尾递归的使用，这听起来像练习的目标。 另请注意，我正在使用Collection库中的各种项目，而不是仅使用原始数组和大型if / then / else块。
import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map; public class WordGenerator { private static Map digitMap; static { digitMap = new HashMap(); digitMap.put(Character.valueOf('0'), new char[] { ' ' }); digitMap.put(Character.valueOf('1'), new char[] { ' ' }); digitMap.put(Character.valueOf('2'), new char[] { 'a', 'b', 'c' }); digitMap.put(Character.valueOf('3'), new char[] { 'd', 'e', 'f' }); digitMap.put(Character.valueOf('4'), new char[] { 'g', 'h', 'i' }); digitMap.put(Character.valueOf('5'), new char[] { 'j', 'k', 'l' }); digitMap.put(Character.valueOf('6'), new char[] { 'm', 'n', 'o' }); digitMap.put(Character.valueOf('7'), new char[] { 'p', 'r', 's' }); digitMap.put(Character.valueOf('8'), new char[] { 't', 'u', 'v' }); digitMap.put(Character.valueOf('9'), new char[] { 'w', 'x', 'y' }); } public static void convert(String input, String resultSoFar, List allResults) { if (input.length() == 0) { // We have hit the end of the input phone number and thus the end of // recursion allResults.add(resultSoFar); } else { // Strip the next character off the front of the phone number Character nextDigit = Character.valueOf(input.charAt(0)); // Look up the list of mappings from that digit to all letters char[] mappingArray = digitMap.get(nextDigit); // More robust error handling would throw an exception or do // something else when an unknown character was encountered in the // phone number. if (mappingArray != null) { // We have processed the first digit in the rest of the number, // so recurse with the rest of the number String inputTail = input.substring(1); // By iterating through the array the mapping lists do not all // have to be the same size. for (char nextLetter : mappingArray) { // Put the next mapped letter on the end of the result being // built and recurse convert(inputTail, resultSoFar + nextLetter, allResults); } } } } public static void main(String[] args) { // Simplified version that does not ask for input String num = "8675309"; List results = new ArrayList (); // Starting condition is that the entire input needs to be processed, // the result so far is empty, and we have nothing in the list of final // answers convert(num, "", results); for (String nextResult : results) { System.out.println(nextResult); } System.out.println("End of results list. Total words generated: " + results.size()); } }