Sti - 17 days ago 5

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.

Source (Stackoverflow)

Comments