我有一项任务,可以从7位数的电话号码中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]);
}
}
}
答案 0 :(得分:1)
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<Character, char[]> digitMap;
static {
digitMap = new HashMap<Character, char[]>();
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<String> 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<String> results = new ArrayList<String>();
// 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());
}
}