Java arrays returning highest array index

Question

I am supposed to write a program that finds the largest index in an unsorted array of integers.

Use a method that takes an array of integers as a parameter. The method should search the array and return the index of the largest value.

So i wrote a program that returns the highest value, for example it returns 13 as the highest number instead of the index of 2. So how would i return the index instead of the highest number itself? If that is an easy fix, does the rest of my code look correct? Thanks!

public class LargestInteger 
{
    public static void main(String[] args)
    {
        int[] largeArray = {5,4,13,7,7,8,9,10,5};

        System.out.println(findLargest(largeArray));
    }

    public static int findLargest(int array[])
    {
        int largest = array[0];

        for(int i = 0; i < array.length; i++)
        {
            if(array[i] > largest)
                largest = array[i];   
        }

        return largest;
    }
}

if you want to return index only, simply change largest = array[i]; to largest = i; — Baby
– Baby, Commented Feb 11, 2014 at 3:47
Be careful with your wording. What you asked is NOT what the problem statement you pasted says. — Alex
– Alex, Commented Feb 11, 2014 at 3:51

Rakesh KR · Accepted Answer · 2014-02-11 04:04:52Z

3

Can also done with,

int[] largeArray     = {5,4,13,7,7,8,9,10,5};
int   largestElement = findLargest(largeArray);
int   index          = Arrays.asList(5,4,13,7,7,8,9,10,5).indexOf(largestElement);

answered Feb 11, 2014 at 4:04

Rakesh KR

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Comments

Gabriel · Accepted Answer · 2014-05-19 22:20:35Z

2

In java 8 you can do it simply:

To get the highest:

int[] largeArray = {5,4,13,7,7,8,9,10,5};
System.out.println(IntStream.of(largeArray).max().getAsInt());

result: 13

And to sum them:

System.out.println(IntStream.of(largeArray).sum());

result: 68

edited May 19, 2014 at 22:20

answered May 19, 2014 at 22:09

Gabriel

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1 Comment

max_max_mir Over a year ago

The OPs question was to find the index, not the max value (although the code posted by the OP is incorrect).

Jason · Accepted Answer · 2014-02-11 03:46:25Z

1

You need to keep and store the largest index. Try this:

public static int findLargest(int array[])
{
    int largest = array[0];
    int largestIndex = 0;

    for(int i = 0; i < array.length; i++)
    {
        if(array[i] > largest) {
            largest = array[i]; 
            largestIndex =i;
        }  
    }

    return largestIndex;
}

answered Feb 11, 2014 at 3:46

Jason

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Comments

Arbiter · Accepted Answer · 2014-02-11 03:49:07Z

0

Create a for loop at the end of the method, and a variable like index. Search through the list and find if the variable matches, remembering to increment the index variable every time.

Example:

int index = 0;
for (int i : array)
{
    if (i == largest)
    {
         break;
    }
    index++;
}
return index;

answered Feb 11, 2014 at 3:49

Arbiter

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Comments

msmolcic · Accepted Answer · 2014-02-11 03:52:16Z

You can add aditional method that gets largest number and an array then search for index of that number:

public class LargestInteger 
{
    public static void main(String[] args)
    {
        int[] largeArray = {5,4,13,7,7,8,9,10,5};

        System.out.println(indexOfLargest(largeArray, findLargest(largeArray)));
    }

    public static int findLargest(int array[])
    {
        int largest = array[0];

        for(int i = 0; i < array.length; i++)
        {
            if(array[i] > largest)
                largest = array[i];   
        }

        return largest;
    }

    public static int indexOfLargest(int array[], int number)
    {

        for (int i = 0; i < array.length; i++) {
            if (array[i] == number)
                return i;            
        }

        return -1;
    }

}

ccyid · Accepted Answer · 2022-10-11 17:12:07Z

0

Just thought we need a simple answer in the era of stream interface. If your array is int[] a, you can get argmax (the index of the largest value) with the following code:

IntStream.range(0, a.length).boxed().max(Comparator.comparing(i->a[i])).get();

If you want argmin, just change max() to min(). Because it compares elements in their natural order, any array with comparable elements will do, like double[] or long[].

answered Oct 11, 2022 at 17:12

ccyid

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