The main thing to look at for this question is the main loop for the code. This loop first goes through each of the characters in the Strings:

int i = 0; i < s.length(); i++

Next, notice the condition on the inner loop:

vals[i%vals.length]

The modulus oniwill yield values that are between 0 and 2 (given the length of vals). This means that you will loop in the inner loop the following sequence of times:

5,4,2,5,4,2,5,4

This will replicate the given letter at the index (i) in the initial string, using 5,4,2,5, etc as the replication count. Thus, you will replicate "H" 5 times, "e" 4, etc. This gives you an output:

HHHHHeeeerrvvvvvaaaaeeuuuuussss

It is easiest to think of thevaluesarray as a String: "I love you!!".

Now, the loop is going to run forortimes. Notice what it does on each iteration. It swaps the values atiandvalues.length - i -1.Thus, it will do the following swaps:

0 and 11

1 and 10

等等……

This sequence of swaps will eventually reverse the array. Thus, your output is:

!!uoy evol I

Given that strings cannot be internally modified, you will have to store your result in a new string, namelys2. Now, for each character ins, you will have to make that charater lower case to begin with:

char c = Character.toLowerCase(s.charAt(i));

Next, for the odd values ofi, you will need to make your value to be upper case. The modulus operator is great for this! You can use % 2 to find the odd values. When the remainder of a division by 2 is equal to 1, you know you have an odd value. Hence, you have the condition:

if(i % 2 == 1) {...

Then, once you appropriately capitalize, you can place your character ons2.

The easiest way to consider this is by commenting on the correct answer. You must begin by defining the character array:

char[] vals = {'A','t', ' ', '6',' ','a','m','!'};

Next, you mustinitializethe string valuesto be an empty string. This is critical. Otherwise, you can't build your string!

String s = "";

Next, you have the loop. This goes through the characters and concatenates the values to the variables. The operation to concatenate the characters is the "+=". This will give you the string value of the array of characters.

The only major issue with this code is the use of the==operator to compare the two strings. Since the user has input this value, will not have an equality on this test. Youmustuse the method.equalsin order to check whether two strings are equal. The code should be:

if(books[i].equals(userInput)){

...

(There are some cases in which == will work for string comparison, namely when literals are involved. However, you should avoid relying on this and always use.equals().)

ThecompareTomethod for strings compares to string objects and returns:

• 0 when they are equal
• Something positive when the given string is alphabetically (really, lexicographically) later than the argument to the method.
• Something negative when the given string is alphabetically (really, lexicographically) before the argument to the method.

Our strings could be put in this order:

"Cursus Philosophicus Thomisticus"

"Happy words for happy people"

"Insane words for an insane world"

"Kritik der reinen Vernunft"

"Logica sive Ars Rationalis"

For each of these, we are asking, "Is it later in order than 'Kittens in a cart'?" This is true for "Kritik der reinen Vernunft" and "Logica sive Ars Rationalis". Thus, our output is:

[Logica sive Ars Rationalis, Kritik der reinen Vernunft]

Notice the order—this is due to the order in the original array.

The length of greet is 6 characters including the space at the end.

问候。substring (0, (len / 2))方程l to greet.substring(0, 3)

The substring of greet from the zeroth position to second position, not to the third position.