Как найти наибольшее четное число массива

Explanation

Your code has two problems. The first is, as others have already pointed out, that you start with Integer.MAX_VALUE as initial guess, which is the wrong logic.

You need to use the worst possible largest value as an initial guess. Otherwise your elements will always be smaller, and thus your initial guess is the biggest element. That’s why we start by guessing MIN_VALUE as largest element. Since the elements of the array can then only get larger. Just play the algorithm on paper for a small example like {1, 2} and you see why that makes sense.

The second problem is that you are actually considering all values, but you only wanted to consider the even values. We easily fix that by skipping all odd values.

Code

Here is your code with both fixes:

public static void main(String[] args) {
    int[] a = new int[]{10, 46, 78, 32, 3, 80, 92, 11, 39, 57};
    System.out.println(Arrays.toString(a));

    // Start with lowest value as initial guess
    int largest = Integer.MIN_VALUE;
    for (int number : a) {
        // Skip number if odd
        if (number % 2 == 1) {
            continue;
        }

        // Now we only consider and collect even numbers
        if (number > largest) {
            // Update the current guess
            largest = number;
        }
    }
    // We now considered all elements, the guess is
    // final and correct.
    // And also even since we skipped odd values.
    System.out.println(largest);
}

Notes

If the array does not contain any even number, then the output will be Integer.MIN_VALUE, you might consider this special case and catch it with some if clause.

Others suggest using an element of the array as initial guess. Since you only want even values, you may only consider using even values of the array for this initial guess. Otherwise, if the array does not contain even values, you would output an odd number again.

You could use a general and compact Stream solution as alternative to a custom method:

int maxEven = Arrays.stream(a)
    .filter(a -> a % 2 == 0) // Only even values
    .max()  // OptionalInt
    .orElse(-1); // Gets the value or uses -1 if not present

Перейти к содержанию

const N = 10;

var

    a: array[1..N] of integer;

    i, max: byte;


begin

    randomize;

    for i:=1 to N do a[i]:=random(100);

    for i:=1 to N do write(a[i]:3);

    writeln;

    max := 2;

    i := 4;

    while i <= N do begin

        if a[i] > a[max] then

            max := i;

        i := i + 2;

    end;

    writeln('№ ',max,' = ',a[max]);

end.

 28 45 48 10  1 85 47 53 32 36

№ 6 = 85

#include 

#define N 10


main() {

    int a[N];

    int i, max;

    for (i=0; i
        printf("%3d", i);

    }

    printf("n");

    srand(time(NULL));

    for (i=0; i
        a[i] = rand() % 100;

        printf("%3d", a[i]);

    }

    printf("n");

    
    max = 0;

    for (i=2; i
        if (a[i] > a[max]) max = i;

    printf("%d: %dn",max,a[max]);

}

  0  1  2  3  4  5  6  7  8  9

 37 81 23 91 66 36 59  6 25 57

4: 66

from random import random

N = 10

arr = [0] * N

for i in range(N):

    arr[i] = int(random() * 100)

print(arr)

m = 0

for i in range(2,N,2):

    if arr[i] > arr[m]:

        m = i

print(m,'-',arr[m])

[42, 25, 94, 79, 86, 51, 99, 70, 74, 25]

6 - 99

алг макс с чет индексами

нач

  цел N = 10

  цел таб arr[1:N]

  цел i,m

  нц для i от 1 до N

    arr[i] := irand(10,99)

    вывод arr[i]:3

  кц

  вывод нс


  m := 2

  нц для i от 4 до N шаг 2

    если arr[i] > arr[m] то

      m := i

    все

  кц

  вывод m,'-', arr[m]

кон

 69 30 80 54 55 82 64 85 34 77

8-85

N = 10

dim arr(N)

for i=0 to N-1

    arr[i] = int(rand*100)

    print arr[i] + " ";

next i

print


m = 0

for i=2 to N-1 step 2

        if arr[i] > arr[m] then

                m = i

        endif

next i

print m+"-"+arr[m]

49 12 1 37 96 47 70 1 5 66 

4-96