Swap misuse absurdity?

the following is a reverse array implementation in java. I walked through it logically and I feel like it should work.

public static int[] reverseArray(int[] arr){
int temp = 0;
int[] newArr;
newArr = Arrays.copyOf(arr, arr.length);
for (int i = 0; i < arr.length - 1; i++){
temp = newArr[i];
newArr[i] = newArr[arr.length - 1 - i];
newArr[arr.length - 1 - i] = temp;
}
return newArr;
}

I pass in an array [1, 2, 3, 4, 5]
It returns to me only the first the the last number swapped.
Returns [5, 2, 3, 4, 1]

You are looping till the end of Array which causes your logic to fail.

You must loop till arr.length/2 instead of looping the whole array. Looping the whole array will cause your program to do the following -

• swap elements and reverse the array in first half of loop : Now when the loop reaches the index - arr.length/2, all elements have already been swapped with each other and your Array has been reversed.
• Reswapping back of all the elements starts in the second half of loop : This effectively starts putting the Array elements back into their original place. When the loop reaches the index arr.length - 1 , your Array is reversed back to the original form it was.

Solution :

Just stop the loop when you reach arr.length/2 and your program will run fine. So your for-loop in reverseArray function should be as follows :

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

You are swapping elements twice.

In your example, when i is 1, you swap elements 1 and 3. Then when i is 3, you swap elements 3 and 1, effectively putting them back where they were originally.