Finding the largest value in an ArrayList

Question

I have two different functions for trying to find the largest value in an ArrayList. I have two since i was first seeing if they would return the same value and then performance time.

However they are reproducing the same value but it seems to be the last value of the ArrayList, regardless if its the largest or not. I think it might be taking the key instead of the value.

The code is below, and I think its just a simple mistake but can anyone point me in the right direction?

double highest = fitnessArray.get(0);

for (int s = 0; s <fitnessArray.size(); s++){
    if (fitnessArray.get(s)>highest)
        highest=fitnessArray.get(s);

}

System.out.println("highest fitness = " + highest 
                + " indoexOf = " + fitnessArray.indexOf(highest));

double highestFitness;

highestFitness = Collections.max(fitnessArray);
System.out.println("lowest fitness 2 = " + highestFitness );

The call to `indexOf()` is 100% redundant, as well as calling `get()` twice. — Niklas R, Oct 08 '12 at 19:31
your code as is seems fine, i tried it and it gave me the correct value. What's in `fitnessArray`? — Majid Laissi, Oct 08 '12 at 19:35
Also your for loop should start at index 1 since you have already checked index 0 before the for loop. — Nathan Villaescusa, Oct 08 '12 at 19:36

score 9 · Answer 1 · answered Dec 16 '14 at 14:03

Use already existing api

Collections.max(arrayList);

Example

import java.util.ArrayList;
import java.util.Collections;

public class Main {

  public static void main(String[] args) {

    ArrayList<Integer> arrayList = new ArrayList<Integer>();

    arrayList.add(new Integer("3"));
    arrayList.add(new Integer("1"));
    arrayList.add(new Integer("8"));
    arrayList.add(new Integer("3"));
    arrayList.add(new Integer("5"));

    Object obj = Collections.max(arrayList);
    System.out.println(obj);
  }
}

Documentation

You can also consider as a slightly worse solution

Collections.sort(arrayList); // Sort the arraylist
arrayList.get(arrayList.size() - 1); //gets the last item, largest for an ascending sort

Second approach might be useful if you'll need sorted list later.

"slightly worse solution" depends on the application. If the `n` is large enough, say, 1 billion elements and let's assume 1 unit of work to process each, O(n log(n)) is something like 30 billion units of work, potentially pretty awful. — ggorlen, Apr 04 '22 at 03:38

Nathan Villaescusa · Answer 2 · 2012-10-08T19:47:53.883

1

You might have better luck if you store the index that the largest number is at:

if (fitnessArray.size() > 0) {
    double highest = fitnessArray.get(0);
    int highestIndex = 0;

    for (int s = 1; s < fitnessArray.size(); s++){
        double curValue = fitnessArray.get(s);
        if (curValue > highest) {
            highest = curValue;
            highestIndex = s;
        }
    }

    System.out.println("highest fitness = " + highest + " indoexOf = " + highestIndex);
}

edited Oct 08 '12 at 19:47

answered Oct 08 '12 at 19:34

Nathan Villaescusa

17,331
4
53
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Ah yeah, I just copied that from the original example. I assumed they were checking beforehand whether the array was empty. – Nathan Villaescusa Oct 08 '12 at 19:44
1

You should store the size and not always call the method. A for each is also preferrable, regarding to effective java. – Christian Oct 08 '12 at 20:56

Finding the largest value in an ArrayList

2 Answers2