CODEWITHSUNDEEP
javaarrays
user3199501
user3199501

Reputation: 1

How to find the smallest and the biggest number of items in array in Java?

I was wondering if there is any method to display the highest and the smallest number of items stored in array/linkedlist. 

public Checkout()
{

   tillQueue = new LinkedList<String>(); 
   currentLen = 0;      
   lengthList = new ArrayList <Integer>();

   mean = 0; 
   r = new Random();
}


public void simulation()
{
    System.out.println("Time" + "\t" + "Rnd" + "\t" + "Queue Status");
    for (int t=2; t<10; t++) //number of time-steps in the simulation.
    {
        rndNumber = r.nextInt(6) +1; // generates random number between 1 and 6.          

        if (rndNumber==4 || rndNumber==2 || rndNumber==6) 
        {            
             // if rndNumber is 4, t is added to a queue.
                tillQueue.add(String.valueOf(t));
                currentLen++;

        }
        else if ((rndNumber==1 || rndNumber==3) && !tillQueue.isEmpty())
        {                              
            tillQueue.remove();
            currentLen--;
           //if rndNumber is either 1 or 3, person is removed from the queue
        }

and that's the outcome:

Queue number 1

Queue

[2]

[2, 3]

[2, 3, 4]

[3, 4]

[3, 4]

[4]

Queue is empty

Queue is empty

Mean length of queue 1= 1.375

Max length of queue 1= 0.0

Min length of queue 1= 0.0

As you can see from the output the highest number of people is 3 (2,3,4) and lowest 0 (empty) How can I calculate those values?

Upvotes: 0

Views: 307

Answers (3)

ram
ram

Reputation: 777

If you are intending to use lengthList,

Before the loop or business logic,

lengthList.add(tillQueue.size())

During the loop or any business logic, after any add or remove operations,

lengthList.add(tillQueue.size())

After your implementation is complete, as Sergio suggested,

Integer max_number = Collections.max(lengthList);
Integer min_number = Collections.min(lengthList);

Upvotes: 0

Destruktor
Destruktor

Reputation: 463

If your intent is to track the length of the list as it changes, you will need to calculate the number of members at each step. So, outside of the for loop, int count = 0; then, at the end of the loop add an conditional statement that updates if the current length is greater than the current max

public void simulation()
{
    System.out.println("Time" + "\t" + "Rnd" + "\t" + "Queue Status");
    int count = 0;
    for (int t=2; t<10; t++) //number of time-steps in the simulation.
    {
        //do all your stuff...

    if (count > tillQueue.size())
    {
        count = tillQueue.size();
    }
}

The same principle would apply to the least value, though I think in this case it is always 0?

Upvotes: 0

S. A.
S. A.

Reputation: 3754

You could use the Collections class.

Integer max_number = Collections.max(arrayList);
Integer min_number = Collections.min(arrayList);

See: http://docs.oracle.com/javase/7/docs/api/java/util/Collections.html

Upvotes: 5

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