Find how many equal pairs there is in input

Input is a series of positive integers with a 0 at the end. My task is to print out how many pairs of equal numbers this series contains.

For example:

Input: 2 34 34 4 6 2 27 27 8 8 8 5 0

Pairs: 4 (34 & 34, 27 & 27, 8 & 8, 8 & 8)

I have come up with this so far but I think I'm really lost.

import java.util.Scanner;
public class P3_5 {
    public static void main(String[]args) {
        int k1 = 0;
        int k2 = 0;
        int pair = 0;
        Scanner scan = new Scanner(System.in);
        System.out.print("Input series of positive integers: ");
        
        while (scan.nextInt() != 0) {
            k1 = scan.nextInt();
            k2 = scan.nextInt();
            if (k1 == k2) {
                pair++;
            }
        }
        System.out.print(pair);
	}
}

You should only ask for one new int within your loop. Ask for the initial int outside the loop and then ask for one within the loop. After you get a new one into k2 inside the loop compare it against k1 and maybe increment pair. Then set k1 = k2 and continue with the next loop iteration.

You may have a look onto HashMaps in order to remember which numbers you have already seen to avoid to count them multiple times

You should not ask for two inputs at once, since a number can form a pair with two other numbers, like maloomeister already said in the comments.

From your question I assume a pair is two consecutive equal numbers. Also, a number can form a pair with two other numbers (in the case of more than two equal consecutive numbers), 8 in your example.

You have to somehow remember the previous number, in order to compare it with the current one, in each iteration. Since a certain number can occur in two pairs, it can be deduced that for k consecutive numbers, the number of pairs is k − 1. So we need to record the current length of the sequence of consecutive equal numbers, and if it resets to zero (i.e. when an other number occurs), then we add it to the total number of pairs:

// The total number of pairs found
int pairs = 0;

// The current number of consecutive equal numbers (always >= 1)
int sequence = 1;

// Record the previous number encountered. Since the first element of an
// array cannot be compared with the previous one (since there is none!),
// we will need to start at the second element.
int last = ints[0];
// It is assumed that the length of the array is at least one. You should,
// however, add code to check if the length is not zero.

for (int i = 1; i < ints.length; i++) {
    // Check if the current number matches the previous one. If true, then
    // our sequence of equal values increments
    if (ints[i] == last) {
        sequence++;
    }
    else {
        // If the current element is not equal to the previous one, then
        // our sequence of equal numbers has ended. In that case, add the
        // length of the sequence minus one to the total number of pairs.
        pairs += (sequence - 1);

        // ...and reset the sequence length
        sequence = 1;
    }

    // Update the previously seen number
    last = ints[i];
}

I used an array here, but instead of using ints[], you can also ask the user to input an int.

Since you apparently only need to count consecutive pairs. This means that 8,8,8 yields [8,8] and [8,8] since the first and third 8's are not consecutive. You can do it with a simple loop.

int k = 0;
int count = 0;
int previousK = 0;
String s = "2 34 34 4 6 2 27 27 8 8 8 5 0";

Scanner input = new Scanner(s);
String pairs = "";
while ((k = input.nextInt()) != 0) {

	// if current number equals previous, a pair has been found.
    if (k == previousK) {
 
       // update output string and count
		pairs +=  k + " & " + k + ", ";
		count++;
	}
 
   // update previous value with this value
	previousK = k;
}
// get rid of lingering ","
int len = pairs.length()-2;
System.out.println("Pairs: =  " + count + " " + pairs.substring(0,len)+")");

Prints

Pairs: 4 (34 & 34, 27 & 27, 8 & 8, 8 & 8)