Java Arrays: Finding Unique Numbers In A Group of 10 Inputted Numbers

Question

I\'m at a loss here.

I have this homework assignment where I have to enable the user to input 10 numbers, place them in an array, and figure out which inputted numbe

Answer 1

Store all numbers in an array. For each stored number: check if number was inserted before and save that in a boolean array. Print all numbers that are not marked in the boolean array.

java.util.Scanner input = new java.util.Scanner(System.in);

int[] numbers = new int[10];
boolean[] usedBefore = new boolean[10];

// Insert all numbers
for (int i = 0; i < numbers.length; i++) {
    // Read number from console
    numbers[i] = input.nextInt();

    // Check if number was inserted before
    usedBefore[i] = false;
    for(int k = 0; k < i; k++) {
        if(numbers[k] == numbers[i]) {
            usedBefore[i] = true;
            break;
        }
    }
}

// Print all numbers that were not inserted before
for(int j = 0; j < numbers.length; j++) {
    if(!usedBefore[i]) {
        System.out.print(String.valueOf(numbers[j])+" ");
    }
}

Answer 2

Check the numbers at they are entered, then keep track of which ones are unique by marking the same positions in a second (boolean) array with true if they are unique and false otherwise.

Then, when you print out the unique values, only print the value from each position in numbers[] if that position in uniques[] contains true.

Scanner input = new Scanner(System.in);
int[] numbers = new int[10];
boolean[] uniques = new boolean[10];

for(int i = 0; i < 10; i++) {
    System.out.println("Please enter a value: \n" + "[" + (i + 1) + "]: ");
    numbers[i] = input.nextInt();
    uniques[i] = true;
    for(int j = 0; j < 10; j++) {
        if(numbers[i] == numbers[j] && i != j) {
            uniques[i] = false;
        }
    }
}

System.out.println("\nThe numbers you entered were: \n");
for(int i = 0; i < 10; i++) {
    System.out.println(numbers[i] + ", ");
}
System.out.println("done.\n\n");

System.out.println("\nThe uniqe numbers you entered were: \n");
for(int i = 0; i < 10; i++) {
    if(uniques[i]) {
        System.out.println(numbers[i] + ", ");
    }
}
System.out.println("done.\n\n");

Answer 3

Instead of an array of input values, put them in a Set of Integers. Sets, by definition, store only unique values. If you add 3 'foos', there will be only one 'foo' in the set.

// Add this to your top-level loop
Set<Integer> uniqueValues = new TreeSet<Integer>;
uniqueValues.add(number);

// Add this after the loop to write all unique values on one line
for (Integer value : uniqueValues) {
  System.out.print(value.toString() + " ");
}

// Now end the line.
System.out.println();

Answer 4

The fastest and most concise and efficient way is to destructively parse the array for uniques using its first number as a magic value, after all other operations on it are complete:

Scanner input = new Scanner(System.in);
int magic = 0;
int[] numbers = new int[10];

for(int i = 0; i < 10; i++) {
    System.out.println("Please enter a value: \n" + "[" + (i + 1) + "]: ");
    numbers[i] = input.nextInt();
}

System.out.println("\nThe numbers you entered were: \n");
for(int i = 0; i < 10; i++) {
    System.out.println(numbers[i] + ", ");
}
System.out.println("done.\n\n");

System.out.println("The unique numbers are: ");
magic = numbers[0];
System.out.println(magic + ", ");
for(int i = 0; i < 10; i++) {
    for(int j = 0; j < 10; j++) {
        if(numbers[i] == numbers[j] && j != i) {
            numbers[j] = magic;
        }
    }
    if(numbers[i] != magic) {
        System.out.println(numbers[i] + ", ");
    }
}
System.out.println("done.\n\n");

Yes, I have two answers - this one is significantly different from the other one, and is better, though it is much more difficult for beginners to understand. Both solutions are valid, however.