Number To Words Java

I have a homework in my programming class that makes a program that can change numbers(Integer) to Words(String).
We can only use If else or Switch statement and my code is already working but My teacher said that my code is not correct because I use private static String numberToWord(int number) and String unitsArray he said that I may use only the things that we discuss in our class so I need to reconstruct my code.
Please help me to reconstruct my code. I'm so confused right now I can't think properly.

Integer

String

Switch

private static String numberToWord(int number)

String unitsArray

This is my code in java:

public static void main(String args)
int number = 0;
Scanner scanner = new Scanner(System.in);
System.out.print("Please type a number(max upto 9 digits)");

// read the number
number = scanner.nextInt();
if (number == 0)
System.out.print("Number in words: Zero");
else
System.out.print(numberToWord(number));



private static String numberToWord(int number)
// variable to hold string representation of number
String words = "";
String unitsArray = "zero", "one", "two", "three", "four", "five", "six",
"seven", "eight", "nine", "ten", "eleven", "twelve",
"thirteen", "fourteen", "fifteen", "sixteen", "seventeen",
"eighteen", "nineteen" ;
String tensArray = "zero", "ten", "twenty", "thirty", "forty", "fifty",
"sixty", "seventy", "eighty", "ninety" ;

if (number == 0)
return "zero";

// add minus before conversion if the number is less than 0
if (number < 0)
// convert the number to a string
String numberStr = "" + number;
// remove minus before the number
numberStr = numberStr.substring(1);
// add minus before the number and convert the rest of number
return "minus " + numberToWord(Integer.parseInt(numberStr));

// check if number is divisible by 1 million
if ((number / 1000000) > 0)
words += numberToWord(number / 1000000) + " million ";
number %= 1000000;

// check if number is divisible by 1 thousand
if ((number / 1000) > 0)
words += numberToWord(number / 1000) + " thousand ";
number %= 1000;

// check if number is divisible by 1 hundred
if ((number / 100) > 0)
words += numberToWord(number / 100) + " hundred ";
number %= 100;

if (number > 0)
// check if number is within teens
if (number < 20)
// fetch the appropriate value from unit array
words += unitsArray[number];
else
// fetch the appropriate value from tens array
words += tensArray[number / 10];
if ((number % 10) > 0)
words += "-" + unitsArray[number % 10];



return words;

}


2 Answers
2

you just need to check how the switch statement works and do not use the function, but put everything in the main function.
I'll give you a hint:

switch(number){
case 1: "one";
break;


static

It is good, that this code works fine, but I can see some problems in it:

You use recursion (i.e. loops). Therefore each time when you type str = str1 + str2 you have three strings in the JVM. You should use StringBuilder instead and at the end, convert it to the final string.

str = str1 + str2

In numberToWord method, you declare two arrays: unitsArray and tensArray. Every time, when you call this method, these array are been initialized again. If you use static method, just use static initialization block to declare these arrays only once and reuse it.

numberToWord

unitsArray

tensArray

static

Scanner scanner scanner should be closed (it contains close() method). It is not critical for the scanner, but anyway, this is must have: when you use resource, do release it before quit.

Scanner scanner

close()

try (Scanner scan = new Scanner(System.in))
// payload



1031 is one thousand and thirty one

Your client code looks like:

number = scanner.nextInt();

if (number == 0)
System.out.print("Number in words: Zero");
else
System.out.print(numberToWord(number));


This is very bad practice. This check should be incapsulated in your logic. Client should not worry about corner cases: System.out.print(numberToWord(scanner.nextInt())).

System.out.print(numberToWord(scanner.nextInt()))

Actually, I do not mind to use static in some cases. But in the given example, your teacher is right: you have a task convert number to the string representation and this task contains some logic and internal constants. It is better to incapsulate it into separate class. It could be static class with only one public method String convert(long val) (but this is old-style and not OOP). You could create separate class, move your static method into it and remove static. This is enough (plus I would initialize array once, in the constructor). In general this is enough.

static

convert number to the string representation

static

String convert(long val)

static

P.S.
Could I give you my solution to this problem. I split this task into two steps: first - convert the string into list of simple numbers (like one, ten, hundred, million) and then using predefined dictionary, convert these simple numbers into final string.

E.g.

value: 1001

1001

numbers: [1, 1000, 1]

[1, 1000, 1]

results: one thousand and one

one thousand and one

My solution:

public final class NumberToWordConverter

private final Long AND = null;
private final Map<Long, String> map = new HashMap<>();


map.put(AND, "and");
map.put(0L, "zero");
map.put(1L, "one");
map.put(2L, "two");
map.put(3L, "three");
map.put(4L, "four");
map.put(5L, "five");
map.put(6L, "six");
map.put(7L, "seven");
map.put(8L, "eight");
map.put(9L, "nine");
map.put(10L, "ten");
map.put(11L, "eleven");
map.put(12L, "twelve");
map.put(13L, "thirteen");
map.put(14L, "fourteen");
map.put(15L, "fifteen");
map.put(16L, "sixteen");
map.put(17L, "seventeen");
map.put(18L, "eighteen");
map.put(19L, "nineteen");
map.put(20L, "twenty");
map.put(30L, "thirty");
map.put(40L, "forty");
map.put(50L, "fifty");
map.put(60L, "sixty");
map.put(70L, "seventy");
map.put(80L, "eighty");
map.put(90L, "ninety");
map.put(100L, "hundred");
map.put(1000L, "thousand");
map.put(1000000L, "million");
map.put(1000000000L, "billion");


public String numberToWords(long val)
String str = parts(Math.abs(val), 1, new LinkedList<>()).stream()
.map(map::get)
.collect(Collectors.joining(" "));
return val < 0 ? "minus " + str : str;


private Deque<Long> parts(long val, long mul, Deque<Long> queue)
if (val >= 1000)
parts(val / 1000, mul * 1000, queue);

if (val % 1000 > 0)
parts(val % 1000, mul, queue);
else if (val <= 20)
if (mul == 1)
queue.add(AND);

queue.add(val);

if (mul > 1)
queue.add(mul);
else if (val < 100)
if (mul == 1)
queue.add(AND);

queue.add(val / 10 * 10);

if (val % 10 > 0)
queue.add(val % 10);

if (mul > 1)
queue.add(mul);

else
queue.add(val / 100);
queue.add(100L);

if (val % 100 > 0)
parts(val % 100, 1, queue);

if (mul > 1)
queue.add(mul);


return queue;




Thanks for contributing an answer to Stack Overflow!

But avoid …

To learn more, see our tips on writing great answers.

Required, but never shown

Required, but never shown

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy