The title above sums up my question, to clarify things an example is:
array[0] = 1
array[1] = 3
array[2] = 7 // largest
array[3] = 5
so the result I would like is 2, since it contains the largest element 7.
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8What you have tried so far?Abimaran Kugathasan– Abimaran Kugathasan2014-04-07 11:56:25 +00:00Commented Apr 7, 2014 at 11:56
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2possible duplicate of How to find the largest int in an array using javarobertoia– robertoia2014-04-07 11:57:24 +00:00Commented Apr 7, 2014 at 11:57
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1This is a trivial thing to do actually. There are a lot of searching algorithms for this, have you even tried something?RB-Develop– RB-Develop2014-04-07 11:57:25 +00:00Commented Apr 7, 2014 at 11:57
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1there are simpler (and quicker) things than sorting...Betlista– Betlista2014-04-07 11:58:54 +00:00Commented Apr 7, 2014 at 11:58
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3Down-vote from me - I expect to see some effort on your part. This is not a difficult challenge.Duncan Jones– Duncan Jones2014-04-07 12:01:10 +00:00Commented Apr 7, 2014 at 12:01
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Show 4 more comments
9 Answers
int maxAt = 0;
for (int i = 0; i < array.length; i++) {
maxAt = array[i] > array[maxAt] ? i : maxAt;
}
5 Comments
ifloop
@Duncan Thanks for the hint, fixed this
Eric
@ifLoop, it's not fixed. The first iteration of the loop will crash with an exception (-1 out of range) ... edit: I see that you've corrected that now.
Eric
@Duncan, sorry I did get you and the author mixed up - my bad.
Boris Brodski
It might be neatest, if you would start your loop with
1 :) Also you can't differentiate between empty array and one element array.
Tarik
As suggested by Boris, starting at one makes sense.
public int getIndexOfLargest( int[] array )
{
if ( array == null || array.length == 0 ) return -1; // null or empty
int largest = 0;
for ( int i = 1; i < array.length; i++ )
{
if ( array[i] > array[largest] ) largest = i;
}
return largest; // position of the first largest found
}
7 Comments
Duncan Jones
You don't need to store
largest.Eric
@Duncan, true but it is a little more "readable" when you do ;)
Duncan Jones
Agree to disagree :-) I think
indexOfLargest would suffice as a single, self-explanatory variable.Duncan Jones
I wouldn't have thought so. For the avoidance of doubt, this is what I'm suggesting (copy into IDE to format nicely!):
int indexOfLargest = 0; for ( int i = 0; i < array.length; i++ ) { if ( array[i] > array[indexOfLargest] ) { indexOfLargest = i; }}. Basically the same as ifLoop's answer.Boris Brodski
@Duncan I would start the loop with
1, not 0. |
one way will be:
Integer[] array = new Integer[4];
array[0] = 1;
array[1] = 3;
array[2] = 7;
array[3] = 5;
List<Integer> iList = Arrays.asList(array);
System.out.println(iList.indexOf(Collections.max(iList)));
System.out.println(iList.indexOf(Collections.min(iList)));
4 Comments
Duncan Jones
This does not find the index.
Duncan Jones
This is a nice solution if the array was already of
Integers. Sadly in many cases it's not and there isn't a pleasant way to move between the two (without external libs).
Jafar Ali
what if the values are repeatable.
Shekhar Khairnar
If the values are repeatable the first occurence of index with repeated value will be considered.
Using Java 8 streams:
List<Integer> list = Arrays.asList(1, 3, 7, 5);
IntStream.range(0, list.size())
.reduce((i, j) -> list.get(i) > list.get(j) ? i : j)
.getAsInt();
Comments
public int getIndexOfMax(int array[]) {
if (array.length == 0) {
return -1; // array contains no elements
}
int max = array[0];
int pos = 0;
for(int i=1; i<array.length; i++) {
if (max < array[i]) {
pos = i;
max = array[i];
}
}
return pos;
}
Comments
Please find below code for the same
Integer array[] = new Integer[4];
array[0] = 1;
array[1] = 3;
array[2] = 7;
array[3] = 5;
List < Integer > numberList = Arrays.asList(array);
int index_maxNumber = numberList.indexOf(Collections.max(numberList));
System.out.println(index_maxNumber);
Comments
Two lines code will do that in efficient way
//find the maximum value using stream API of the java 8
Integer max =Arrays.stream(numbers) .max(Integer::compare).get();
// find the index of that value
int index = Arrays.asList(numbers).indexOf(max);
1 Comment
Tarik
The most elegant solution so far but due to Java streams implementation, not very efficient. C# does a better job with the IEnumerable interface implemented by arrays.
Another functional implementation
int array[] = new int[]{1,3,7,5};
int maxIndex =IntStream.range(0,array.length)
.boxed()
.max(Comparator.comparingInt(i -> array[i]))
.map(max->array[max])
.orElse(-1);
Comments
Would do it like this (as I don't know any predefined function to get the index of highest element, only the element itself, of course you could get the index with list.indexOf(element) then, but array needs to be converted to list and 2 iterations):
maxIndex = 0;
for (int i = 0; i < array.length; i++) {
if (array[i] > array[maxIndex]) {
maxIndex = i;
}
}
