A quesiton like this is most easily understood by doing a careful reading of the code in question. Let's consider each major section of code:

int[] ret = new int[arr.length + 1];

这行代码creates a new array, one that is 1 longer than the arrayarr.

for(int i = 0; i < index; i++) . . .

This loop copies intoretthe values ofarrup toindex - 1.

(This is because of thei < indexcondition)

Then, the code stores the valuevalinret[index]

for(int i = index + 1; i < ret.length; i++) . . .

It then finishes copying the values intoret.

Thus, what this does is insert a value into the original arrayarr, returning the new array that is one size larger. (This is necessary because of the static sizing of arrays in Java.)

The most obvious possible error is that the arrayarr might be a null value. You need to check for these kinds of values before using the variables. (If you do arr[0]on a null value, an exception will be thrown.) In addition, it is possible that a value for index might be given that is too large. Consider ifindex = 100butarris only 4 elements long. Then, you will haveretbe a 5 value array. When the first loop starts to run, you will go all the way to 99 (or at least attempt to do so) for the index valuei; however, once you get toret[4] = arr[4], there will be an out of bounds error onarr, which only has indices 0, 1, 2, and 3. Of course, there could be other problems later on withret, but the question only asks about this first loop.

Let us presume that the arrayarrlooks like the following list of integers:

{4,51,41,0,0,0,0,0,0,0}

Presume that our new value is77.

A stack couldeitherpush on to the beginning and move everything back one, giving you something like:

{77,4,51,41,0,0,0,0,0,0}

However, none of the options do this. (No, not even the one that uses the index0. This one does not add on to the array so much as replace the first element of it.)

So, the other option is to put it on the "end" of the list. The variablearrFillis being used for this purpose. If it is3, this means that it is the value of the fourth element. Thus, you can set arr[4] = 77 (where 4 really isarrFill).

This will give you:

{4,51,41,77,0,0,0,0,0,0}

You also need to add one to the value ofarrFill.

The other options do not correctly address the array. They either are too large or too small by one.

Insertion into an ArrayList is typically O(1). However, since an ArrayList uses an array as its underlying data structure, there can be a need to resize the underlying array. Resizing an array requires creating a new array and copying all the data from the old array to the new array which takes O(N) time.