Not all programs are created equal. In this course, we'll focus on writing quality code that runs correctly and efficiently. We'll design, code and validate our programs and learn how to compare programs that are addressing the same task.

Loading...

来自 University of Toronto 的课程

学习编程：撰写高质量代码

296 个评分

Not all programs are created equal. In this course, we'll focus on writing quality code that runs correctly and efficiently. We'll design, code and validate our programs and learn how to compare programs that are addressing the same task.

从本节课中

Week 1

- Jennifer CampbellAssociate Professor, Teaching Stream

Department of Computer Science - Paul GriesAssociate Professor, Teaching Stream

Department of Computer Science

In this lecture we'll implement our second algorithm for determining whether a string

is a Palindrome. As before we start with the same results

we follow in our design recipe as we did for the last lecture.

So our second algorithm we split this string into two halves.

Reverse the second half. And then compare that reversed second half

to the first half. Again, we need to reverse a string, so

let's go get that code from the previous version.

As a quick reminder, if I have an even length string, then I'm going to divide

this in half, reverse the second half, and then compare it to the first half.

I'm going to write some indices at the top, 0, 1, 2 and 3.

If I have an odd length string on the other hand, such as a race car, my

approach is to not include the E and just reverse everything after the E.

That should give me RAC, which I then compare to the first half, and again I'm

going to write the indices of each letter above.

The last example that we have is, dented, split that into half, reverse the last

half and then unsuccessfully compare the, the reversed second half to our first

half. And now it's time to play with indices

because we need to know exactly where we're going to slice in each string.

The length of noon is 4, 4 divided by 2 is 2, so that tells us the index to, to slice

from for, to get the second half of this strip, 4 divided by 2.

The length of race car is 7. 7 divided by 2, well, if I just do

straight division, then that gives me 3 and a half.

But, if I do integer division, then that gives me 3.

So 3 is the number of characters in each half, but I don't want to include the e

there. So, I would, I'm going to have to figure

out how to work around this, with my slicing perhaps.

I can take some and subtract 7 divided by 2, 7 divided by 2 is 3, thus we got a

slash there, sorry, 7 minus 7 divided by 2 that gives me 4, which is the first index

of the character that I want. Now is that going to work over here, 4

minus 4 integer division 2. Well 4 integer division 2 is 2, 4 minus 2

is 2 and indeed, that gives me the index that I need in order to start the slice to

grab the second half of the string. In general then the links of the string

divided by two, that gives me the number of characters for each slice.

I can use this result as the, the last index in a slice, then, in order to grab

the first half of the string, and I can take the, the length of s minus this

value. In order to find the index of the first

character in the second half to graph. So over here, 7 minus 7 divided 2, gave me

4, and that is indeed the index over the first character in the second half that I

want to reverse. We're done thinking hard.

We know that the first half that we want to grab, is everything up to the length of

s divided by 2, and we know that we want to grab the, everything from the length of

s minus the length of s, divided by 2, for the second half.

And that we want to reverse the second half and check to see if it's equal to the

first half. That's a little bit convoluted because we

have a whole bunch of repetitions of length of s.

So I am going to have one separate step where I grab that length, and make it so

that we only have to do that calculation once.

I think the code is a little bit clearer this way.

Your opinion may differ. Getting rid of the scribbles.

We're just going to add a couple of comments we're so that other people who

might be reading this code later can understand what we were doing, n is the

number of characters in s. And here what we're doing is we are

comparing, the first half of s, to the reverse of the second path, because we so

painfully figured out that we were omitting the middle character in an odd

length string. Let's actually mention that.

All that's left now is for us to test. Try noon, and looks good so far.

How about racecar, yup, that's also good. And last, dented.

Good. We're done.