如何使Java从命令行参数输出唯一整数？

Question

The code below checks to ensure the command-line arguments are integers? However, I need it to only print each integer in the command line once, regardless of how many times that integer appears in the command line. How do I do that?

public class Unique {
    public static void main(String[] args) {
        for (int i = 0; i<args.length; i++) {
            try {
                int j = Integer.parseInt(args[i]);
                System.out.println(j);
            }
            catch (NumberFormatException e) {
                System.out.println("Your command line argument "+ args[i] +" is a non-integer!");
            }
        }
    }
}

Answer 1

What you need is a Set. A set does not contain duplicates, as it is defined mathematically. Java has a Set class that does that job.

Set<Integer> set = new HashSet<Integer>();
for (int i = 0; i < args.length; i++)
{
    try
    {
        int j = Integer.parseInt(args[i]);
        set.add(j);
    }
    catch (NumberFormatException e)
    {
        System.out.println("Your command line argument "+ args[i] +" is a non-integer!");
    }
}
System.out.println(set);

Or iterate over all the elements like this:

for (Integer i : set)
{
    System.out.println(i);
}

Answer 2

public class Unique {

public static void main(String[] args) {

    final Set<Integer> uniqueIntegers = new HashSet<>();        

    for (int i = 0; i<args.length; i++) {
        try {
            int j = Integer.parseInt(args[i]);
            uniqueIntegers.add(j);
        }
        catch (NumberFormatException e) {
            System.out.println("Your command line argument "+ args[i] +" is a non-integer!");
        }
    }

    System.out.println(uniqueIntegers);
}

Answer 3

public class UniqueNumbers {
  public static void main(String[] args) {
    args = new String[6];
    args[0] = "5";
    args[1] = "6";
    args[2] = "2";
    args[3] = "5";
    args[4] = "1";
    args[5] = "1";

    Arrays.sort(args); 
   // Sort the array. This will use natural order:
   // low -> high for integers, a -> z and low to high for strings
   // Essentially this causes our array to sort from low to high, despite not being integers

    for (int i = 0; i < args.length; i++) { // Loop over the entire array
        if (i == 0) { // (1)
            System.out.println(args[0]);
        } else if (!(args[i - 1].equals(args[i]))) { // (2)
            System.out.println(args[i]);
        }
    }
  }
}

Output:

1
2
5
6

Clarification:

To make it easy to demonstrate, I've manually placed the values into the array. This makes no difference, you can just keep using your input method.

(1): Read (2) first. Because of the method below, we would essentially skip the first element. The second part of the else if essentially means if i > 0 . If we wouldn't do this, we would get an ArrayIndexOutOfBoundsException when we do args[i - 1] because that would try to access args[-1] . That's why we skip the first element: to avoid this. However, this would also mean that our first value ( 1 ) would be ignored. That's why we just want to make sure the first value is always printed.

(2): Now we check if the current element ( args[i] ) is equal ( .equals() ) to the previous elements ( args[i -1] ). If this is not the case ( ! inverts a statement), we print the current value.

With this method, you can solve your assignment without any datastructures apart from the the standard one.

More visualization:

Start:

5 6 2 5 1 1

Sort:

1 1 2 5 5 6

Loop:

!  
1 1 2 5 5 6 -> Output: 1

  !
1 1 2 5 5 6 -> Not different from previous one, no output

    !
1 1 2 5 5 6 -> Different from previous one, output: 1 2

      !
1 1 2 5 5 6 -> Different from previous one, output 1 2 5

        !
1 1 2 5 5 6 -> Not different from previous one, no output

          !
1 1 2 5 5 6 -> Different from previous one, output 1 2 5 6