This week's book giveaway is in the OCAJP 8 forum. We're giving away four copies of OCA Java SE 8 Programmer I Study Guide and have Edward Finegan & Robert Liguori on-line! See this thread for details.

I want to increment the value of the variable **x** of object each time a new element is added to array **with the faster way possible** . The new element can be added in any position and we increment all **x** element after the insertion position (position +1)

Before add (6,6) :

`T = (1,1) (2,2) (3,3) (4,4)`

After add (6,6) in different positions:

1) T = (6,6) (2,1) (3,2) (4,3) (5,4)

or

2) T = (1,1) (2,2) (6,6) (4,3) (5,4)

or

3) T = (1,1) (2,2) (3,3) (6,6) (5,4)

I used the method arraycopy to add the new element, and loop for to increment the variable x for each element, as follow:

1. increment all `x` of object elements with loop for

2. `Ta[0] = (6,6)`

3. `araycopy(T, 0, Ta, 1, T.size-1 );`

because it is faster than

While (i< T.length){

T[i] = T[i+1]

T[i].x ++;

i++;
}

I need to add the new element and increment the other objects of array simultaneously with a faster time.

//-------------------

//----------------

//------- The results:

Time without Incrementation : 15 ms

Time with Incrementation: 156 ms

My objective is to minimize as possible the time of incrementation process

(Time with Incrementation < Time without Incrementation * 2 )

because actually

Time with Incrementation (156 ms) = Time without Incrementation (15 ms )* 10

I notes that i can added a new element in any position, but i chose the worst case (adding an element in the first position that requires the incrementation of all x element of the arraylist)

you'll usually server yourself better by writing code that is easy to follow than uses every possible trick to save time. If you use enough tricks and save yourself a microsecond each time, but spend and extra eight hours debugging this at some point int the future, your code is going to have to run a LOT to make up that lost time.

There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

Moulay Driss
Greenhorn

Joined: May 05, 2013
Posts: 5

posted

0

suppose an array of object: Element T[ ] , each object contains two integer variables (x,y).

T = (1,1) (2,2) (3,3) (4,4)

I want to increment the value of the variable **x** of object each time a new element is added to array **with the faster way possible** . The new element can be added in any position and we increment all **x** element after the insertion position (position +1)

Before add (6,6) :

`T = (1,1) (2,2) (3,3) (4,4)`

After add (6,6) in different positions:

1) T = (6,6) (2,1) (3,2) (4,3) (5,4)

or

2) T = (1,1) (2,2) (6,6) (4,3) (5,4)

or

3) T = (1,1) (2,2) (3,3) (6,6) (5,4)

I used the method arraycopy to add the new element, and loop for to increment the variable x for each element, as follow:

1. increment all `x` of object elements with loop for

2. `Ta[0] = (6,6)`

3. `araycopy(T, 0, Ta, 1, T.size-1 );`

because it is faster than

While (i< T.length){

T[i] = T[i+1]

T[i].x ++;

i++;
}

I need to add the new element and increment the other objects of array simultaneously with a faster time.

//-------------------

//----------------

//------- The results:

Time without Incrementation : 15 ms

Time with Incrementation: 156 ms

My objective is to minimize as possible the time of incrementation process

(Time with Incrementation < Time without Incrementation * 2 )

because actually

Time with Incrementation (156 ms) = Time without Incrementation (15 ms )* 10

I notes that i can added a new element in any position, but i chose the worst case (adding an element in the first position that requires the incrementation of all x element of the arraylist)

Moulay Driss wrote:My objective is to minimize as possible the time of incrementation process

That's not a very well-defined spec, but if it is true, you will quite probably get the most bang for your buck if you buy a faster server. That would have the added bonus of speeding up EVERYTHING on the box, not just this one tiny piece.