Split function while reading multiple lines

The input begins with the number \$t\$ of test cases in a single line
(\$t \le 10\$). In each of the next t lines there are two or more
numbers \$m\$ and \$n\$ (\$1 \le m \le n \le 1000000000\$, \$n-m \le 100000\$) separated by a space.


Print each number in a separate line which can be used further for
summation.


Input

2

50 100

100 50 105

Output

50

100

100

50

105

This is the code that I've written that is giving me output:

```
import java.util.Scanner;
import java.util.StringTokenizer;

public class Generation {

public static void main(String[] str) {

Scanner keyboard = new Scanner(System.in);
int inputSize;
do {
System.out.println("Enter the value of T Size");
inputSize = keyboard.nextInt();
keyboard.nextLine();
if (inputSize 10) {
System.out.println("Not a Valid Input Size");
}
} while (inputSize 10);

String[] inputValue = new String[inputSize];
int tokenCount = 0;
for (int i = 0; i < inputSize; i++) {
System.out.println("Enter the inputs");
inputValue[i] = keyboard.nextLine();
StringTokenizer strToken = new StringTokenizer(inputValue[i], " ");
tokenCount += strToken.countTokens();
}
keyboard.close();

//suppose this is 2nd part
int[] splitedString = new int[tokenCount];
int tempTokenCount = 0;
for (int i = 0; i < inputSize; i++) {
String[] tempSplitArray = inputValue[i].split(" ");
for (int j = 0; j < tempSplitArray.length; j++) {
splitedString[tempTokenCount] = Integer
.parseInt(tempSplitArray[j]);
tempTokenCount++;
}

}
/*for (String s : inputValue) {
System.out.println(s);
}*/
for (Integer s : splitedString) {
System.out.println(s

try-with-resources

If you're on Java 7, you should use try-with-resource on your Scanner:

try (Scanner scanner = new Scanner(System.in)) {
    // your code here
}

This takes care of closing your Scanner at the end, so you do not have do perform an explicit close().

First input

I am not too sure why are you ensuring inputSize cannot be less than 2, since the problem statement only mentioned

-
Read in a line for x times

Stream.generate(scanner::nextLine)

This creates a new stream where each element is read from the scanner, i.e. System.in, i.e. the user input. This is accessed via a method reference.

-
On each line, split by a ' ' character

stream.flatMap(Pattern.compile(" ")::splitAsStream)

Pattern.splitAsStream(CharSequence) is a convenient way of tokenizing a line as a stream, which in turn allows us to perform a flatMap() by replacing each element of the source stream with the contents of the resulting stream.

-
Convert each token as an integer within a certain range

stream.map(Generation::convert)

The sample implementation shown below for convert(String) does not perform any validation. Actually, as a result of the flatMap() operation in the previous step, it will be quite tricky to make sure that \$1 \le m \le n \le 1000000000\$, so I'll leave this as an exercise to the reader... (hint: map each String input as a List to facilitate the validation first).

-
Collect all results into a Collection for further usage

stream.filter(Objects::nonNull).collect(Collectors.toList())

Since convert(String) may give us null values for invalid integers, we need to exclude them first. Collecting as toList() is just for illustration. If you want to sum the results together immediately, for example, you can change that to .mapToInt(Integer::intValue).sum()`.

public class Generation { // you probably need a better name too
    public static void main(String[] args) {
        try (Scanner scanner = new Scanner(System.in)) {
            List result = Stream.generate(scanner::nextLine)
                    .limit(getCount(scanner))
                    .flatMap(Pattern.compile(" ")::splitAsStream)
                    .map(Generation::convert)
                    .filter(Objects::nonNull)
                    .collect(Collectors.toList());
            result.forEach(System.out::println);
        }
    }

    private static int getCount(Scanner scanner) {
        // see implementation above
    }

    private static Integer convert(String value) {
        try {
            return Integer.valueOf(value);
        } catch (NumberFormatException e) {
            return null;
        }
    }
}

StackExchange Code Review Q#90794, answer score: 6