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AP Computer Science 4.5 Standard Algorithms 171 Views


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Description:

AP Computer Science 4.5 Standard Algorithms. Which of the following implementations successfully copies over array myalph to ArrayList alphabet?

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Transcript

00:00

Sorry And here's your shmoop du jour brought to you

00:05

by people who can say the alphabet backwards either they're

00:08

extremely smart they know how to reiterate backwards through an

00:11

array or they've just done a lot of sobriety tests

00:14

and you don't want to go there All right Which

00:18

of the following implementation successfully copies over array my alf

00:22

to a realist alphabet All right And here your potential

00:26

answers brought to you by the romans Okay option one

00:33

uses a for each loop which will cycle once for

00:36

each element in an array In this case for each

00:39

string in the array my alf we'll add the content

00:42

of that element to the array list alphabet and the

00:46

loop will in as soon as we've reached the end

00:48

of my al that's about as simple as it gets

00:51

and it will totally work all right Option two option

00:54

two creates an iterated that begins at zero And while

00:58

the current element of my health is not equal to

01:01

z where it'll add that particular value to alphabet add

01:06

one to the generator and do it all over again

01:09

Do that to me one more time Great hold when

01:13

the wild statement finally does find a value equal to

01:16

z it'll end the loop and travel down to the

01:18

next statement adding the current value of my alf meaning

01:22

z we just found the alphabet so option two fits

01:25

the bill to woo hoo wiseguys will have already noticed

01:29

that none of the possible answers are all three options

01:31

work and think we're totally done here but because we're

01:33

also smart and uptight about things we're going to not

01:37

just assume the answer is d all right so let's

01:39

just check out option three and be sure it doesn't

01:41

work looks decent on first glance right Well standard looking

01:45

loop something niggling with the array list Nothing obviously a

01:48

miss Ah wait we can't use that alphabet dot set

01:51

call on indices that don't exist yet remember our ray

01:55

list alphabet is totally empty all set replaces elements that

01:58

already exist with other elements We need to use an

02:01

ad call instead and if we're being nitpicky Well that

02:05

four loop set up is saying well i is less

02:08

than alphabet dot size and that would have to be

02:11

changed to the size of alphabet at this stage would

02:13

be zero so the loop would never run it all

02:16

Changing that terminator parameter too While i is less than

02:20

my health dot length would allow the loop to run

02:22

the full twenty six Anyway we're done Wait okay we're 00:02:27.198 --> [endTime] leaving we're leaving

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