Q:

Finding the second largest number of given integers without using array

Using if loops, I am tasked with putting the largest & 2nd largest integers into a pair, and the smallest & 2nd largest integers into a pair, from user provided numbers.
I have tried several different if conditions, and while my program can find the 2nd smallest integer correctly, if i apply the same logic (with reversed greater than/smaller than signs), I don't get the correct answer. 
           numN = keyboard.nextInt();
           if (numN > numL1){
              numL1 = numN;
           }

           if (numN < numS1){
              numS1 = numN; 
           }
           else if (numN < numS2 && numS2 > numS1){
              numS2 = numN;
           }
           else if (numN > numL2 && numL2 < numL1){
              numL2 = numN;
           }

If User inputs the four numbers 1,2,3,4 
Actual results: Largest and Smallest Pairs:  (4,4) (1,2)
Needed results: Largest and Smallest Pairs:  (4,3) (1,2)

A:

you can do this like below:
int max = Integer.MIN_VALUE, secondMax = Integer.MIN_VALUE, min = Integer.MAX_VALUE, secondMin = Integer.MAX_VALUE;

    int input = keyboard.nextInt();
    if (input > max) {
        secondMax = max;
        max = input;
    } else if (input > secondMax) {
        secondMax = input;
    }
    if (input < min) {
        secondMin = min;
        min = input;
    } else if (input < secondMin) {
        secondMin = input;
    }

