Sti - 4 months ago 25

Java Question

I am trying to create a method which returns an int - the value of the largest integer in the sent array.

The way I want this method to work, is to check the first **and** the last element of the array in a for-loop, and work their way to the middle. So i = first integer, k = last integer. When

`i = 0, k = n-1`

`i = 1, k = n-2`

`if a[i]>a[k]`

I tried like this:

`public static int maxOfArray(int[] a)`

{

int length = a.length;

if(length<1)

throw new NoSuchElementException("Not at least one integer in array");

while (length > 1)

{

int k = length;

for(int i = 0; i < length/2; i++)

{

k--;

if(a[i]<a[k])

{

int j = a[i];

a[i] = a[k];

a[k] = j;

}

}

length /=2;

}

return a[0];

}

..but I don't really get it.. I'm having a hard time "picturing" what's happening here.. But it's not always working.. (though sometimes).

Also: The array {6,15,2,5,8,14,10,16,11,17,13,7,1,18,3,4,9,12}; will spit out 17 as the largest number. I realize I have to fix the odd-length bug, but I would like to solve this even-length array first..

Answer

**A bug is when encountering length is odd.**

In these cases, you "miss" the middle element.

**Example**: for input `int[] arr = { 8, 1, 5, 4, 9, 4, 3, 7, 2 };`

- the element `9`

will be compared and checked against itself, but then you reduce the size of `length`

, you exclude `9`

from the array you are going to iterate next.

I believe it can be solved by reducing the problem to `ceil(length/2)`

instead of `length/2`

(and handling special case of `length==1`

)

The other issue as was mentioned in comments is: you need to **iterate up to length/2** rather then up to

`length`

, otherwise you are overriding yourself.Lastly - **the sign is wrong**.

```
if(a[i]>a[k])
```

should be

```
if(a[i]<a[k])
```

Remember - you are trying to swap the elements if the first is smaller the the second in order to push the larger elements to the head of your array.