How to find array index of largest value?

Question

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 th

Answer 1

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
}

Answer 2

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);

Answer 3

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);

Answer 4

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);

Answer 5

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;
}

Answer 6

int maxAt = 0;

for (int i = 0; i < array.length; i++) {
    maxAt = array[i] > array[maxAt] ? i : maxAt;
}