查找数组中的最高索引值

Question

Hello I want find the highest index n array. For example I have two arrays : [1,2,3,0,-1] - should return 2 - index with max values [] - should return -1 because array is empty

I have method:

   public int findMax(int[] array){
    int values =0;
    for(int i=0; i<array.length; i++){
    if(array == null)
    {return -1;}
    else(array{i}>values){
    return i;
    }
 }

What i'm doing wrong?

Answer 1

Your program has logical as well as compilation errors. I guess you want to write a code like this,

public static void main(String[] args) throws IOException {
    System.out.println(findMax(new int[] { 1, 2, 3, 0, -1 }));
    System.out.println(findMax(new int[] {}));
}

public static int findMax(int[] array) {
    if (array == null || array.length == 0) {
        return -1;
    }

    int maxIndex = 0;
    int maxNum = array[0];
    for (int i = 1; i < array.length; i++) {
        if (array[i] > maxNum) {
            maxNum = array[i];
            maxIndex = i;
        }
    }

    return maxIndex;
}

Like you expected, this gives following output,

2
-1

Answer 2

Suppose you want to find both maximum element and first index of maximum value.

Code for this:

public void findMax(int[] array){
    int values =-99999999; //values is any large negative number
    int index=-1;
    if(array.length == 0)
    {System.out.println("-1");}
    for(int i=0; i<array.length; i++){
    if(array[i]>values){ 
       values=array[i];
       index=i;
      }
    System.out.println("Max element is"+values +"and index is"+index);
    }
 }

Answer 3

Try this one:

public int findMax(int[] array) {
    if (array == null || array.length == 0)
        return -1;

    int index = 0;
    int max = -Integer.MAX_VALUE;
    for (int i=0; i< array.length; i++) {
        if (array[i] > max) {
            index = i;
            max = array[i];
        }
    }

    return index;
}

Answer 4

public int findMax(int[] array){
    return IntStream.range(0, array.length).boxed().max(Comparator.comparing(i -> array[i])).orElse(-1));
}