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AP Computer Science 2.5 Standard Algorithms 176 Views


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Description:

In this computer science drill question, figure out which implementation will copy one array over to another.

Language:
English Language

Transcript

00:00

Thank you We sneak And here's your shmoop du jour

00:05

brought to you by people who can say the alphabet

00:07

backwards either they're extremely smart they know how to integrate

00:11

backwards through an array or they've just done a lot

00:13

of sobriety tests and you don't want to go there

00:17

All right Which of the following implementation successfully copies over

00:21

array my alf to a realist alphabet All right And

00:26

here your potential answers brought to you by the romans

00:32

Okay option one uses a for each loop which will

00:35

cycle once for each element in an array In this

00:39

case for each string in the array my alf will

00:42

add the content of that element to the array list

00:45

alphabet and the loop will in as soon as we've

00:48

reached the end of my alf that's about as simple

00:51

as it gets and it will totally work all right

00:54

Option two option two creates an iterated that begins at

00:57

zero And while the current element of my health is

01:01

not equal to z where it'll add that particular value

01:06

to alphabet add one to the generator and do it

01:08

all over again Do that to me one more time

01:13

Great hold when the wild statement finally does find a

01:16

value equal to z it'll end the loop and travel

01:19

down to the next statement adding the current value of

01:21

my alf meaning z we just found the alphabet so

01:25

option two fits the bill to woo hoo wiseguys will

01:29

have already noticed that none of the possible answers are

01:31

all three options work and think we're totally done here

01:33

but because we're also smart and uptight about things we're

01:37

going to not just assume the answer is d all

01:39

right so let's just check out option three and be

01:41

sure it doesn't work looks decent on first glance right

01:45

Well standard looking loop something niggling with the array list

01:47

Nothing obviously a miss Ah wait we can't use that

01:51

alphabet dot set call on indices that don't exist yet

01:55

remember our ray list alphabet is totally empty all set

01:58

replaces elements that already existe with other elements We need

02:01

to use an ad call instead and if we're being

02:04

nitpicky Well that four loop set up is saying well

02:08

i is less than alphabet dot size and that would

02:11

have to be changed to the size of alphabet at

02:13

this stage would be zero so the loop would never

02:16

run in all changing that terminator parameter to while i

02:20

is less than my health dot length it would allow

02:22

the loop to run the full twenty six Anyway we're 00:02:26.018 --> [endTime] done Wait Okay we're leaving we're leaving

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