How to reverse an ArrayList without the reverse method?

Question

Hello I am trying to reverse an ArrayList without the reverse method. I just wanted to make it work without the method. I can't seem to get it right. This is what i have so far:

for (int x = nums.size()-1; x>=0; x--)
    {

        for(int z =0; z<nums.size();z++)
        {
            nums.set(z, x);

        }
    }

This is my output: run: 0 1 2 3

1 1 1 1 BUILD SUCCESSFUL (total time: 0 seconds)

you are doing it twice and rearranging it back to the way it was — user1064590
– user1064590, Commented Oct 6, 2013 at 23:35
@VirtualBaseClass - Actually, he's not. If he was, the output would be "0 1 2 3" not "1 1 1 1". — Stephen C
– Stephen C, Commented Oct 7, 2013 at 0:10
@StephenC yeah.. another problem is that he is setting the value x, not getting it from the array. Dunno why its 1111 though, shouldn't it be the last value that x takes which is 0. — user1064590
– user1064590, Commented Oct 7, 2013 at 0:14
@VirtualBaseClass - "Dunno why its 1111 though". I suggest that you get a pencil and paper and "hand execute" the code. Then you'll understand. — Stephen C
– Stephen C, Commented Oct 7, 2013 at 0:50
@StephenC Don't really need a pen and paper for it, thanks for the remark though. And I still think its 0000. Would be grateful if you point out what I am missing. — user1064590
– user1064590, Commented Oct 7, 2013 at 0:57

Thomas W · Accepted Answer · 2013-10-06 23:41:39Z

1

You can ascend from the bottom & simultaneously descend from the top (size() - 1) swapping elements, and stop when you meet in the middle.

int i = 0;
int j = nums.size()-1;
while (i < j) {
    int temp = nums.get(i);
    nums.set( i, nums.get(j));
    nums.set( j, temp);
    i++; j--;
}

answered Oct 6, 2013 at 23:41

Thomas W

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3 Comments

Beto Over a year ago

Do you mind putting it in a for loop? i wanna understand it better.

Stephen C Over a year ago

@JB - You should attempt to understand it in its current form. It is easier to understand this way than as a for loop.

Thomas W Over a year ago

Since there are two loop indexes you need two declarations, two initializations and two advancements. Putting the syntax in a "for" loop can be done (except perhaps for the declarations), but it would reduce clarity.

SimonT · Accepted Answer · 2013-10-06 23:44:54Z

0

You can swap in pairs at a time, coming in from both ends toward the center:

for (int i = 0; i < nums.size()/2; i++) {
    Integer left = nums.get(i);
    Integer right = nums.get(nums.size()-1-i);
    nums.set(i, right);
    nums.set(nums.size()-1-i, left);
}

By using Integer instead of int for left and right, Java doesn't have to keep converting between int and Integer values (the first being a primitive, the second being an actual object). This improves performance.

answered Oct 6, 2013 at 23:44

SimonT

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2 Comments

user1064590 Over a year ago

You don't know if autoboxing is happening here

SimonT Over a year ago

ArrayList can only store objects, so in this case Integer instances would be stored and not int. When you read the documentation at docs.oracle.com/javase/7/docs/api/java/util/… you will notice that the return type is given as E, which in this case refers to Integer. If this answer said int left = nums.get(i), then the Integer returned would be converted to int and then stored in left. Then it would be converted into Integer in nums.set(... again.

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