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merging and sorting 2 sorted arrays with big o looking for clarification.

This is a school work. I am not looking for code help but since my teacher isn't helping I came here.
I am asked to merge and sort two sorted arrays following two cases:
When sizes of the two arrays are equal
When sizes of the two arrays are different
Now I have done case 2 which also does case 1 :/ I just don't get it how I could write a code for case 1 or how it could differ from case 2. array length doesn't connect with the problem or I am not understanding correctly.
Then I am asked to compute big(o).
I am not looking for code here. If anyone by any chance understands what my teacher is asking really please give me hints to solve it.

It is very good to learn instead of copying.
as you suggest, there is no difference between case 1 and 2 but the worst case of algorithms depend on your solution. So I describe my solution (no code) and give you its worst case.
You can in both case, the arrays must ends with infinity so add infinity to them. then iterate over all of elements of each array, at each time, pick the one which is smaller and put in in your result array (merge of tow arrays).
With this solution, you can calculate worst case easily. we must iterate both of arrays once, and we add a infinity to both of them, if their length is n and m so our worst and best case is O(m + n) (you do m + n + 2 - 1 comparison and -1 because you don't compare the end of both array, I mean infinity)
but why adding infinity add the end of array? because for that we must make a copy of array with one more space? it is one way and worst case of that is O(m + n) for copying arrays too. but there is another solution too. you can compare until you get at the end of array, then you must add the rest of array which is not compared completely to end of your result array. but with infinity, it is automatic.
I hope helped you. if there is something wrong, comment it.

Merging two sorted arrays is a linear complexity operation. This means in terms of Big-O notation it is O(m+n) where m and n are lengths of two sorted arrays.
So when you say the array length doesn't connect with the problem your understanding is correct. Irrespective of the lengths of two sorted arrays the merging of these arrays involves taking elements from each sorted array and comparing them and copying the one to new array(depending whether you want the merged sorted array in ascending or descending order) and incrementing the counter of the array from which you copied the element to new sorted array.

Another way to approach this question is to look at each array as having a head and a tail, and solving the problem recursively. This way, we can use a base case, two arrays of size 1, to sort through the entirety of the two arrays m and n. Since both arrays are already sorted, simply compare the two heads of each array and add the element that comes first to your newly-created merged array, and move to the next element in that array. Your function will call itself again after adding the element. This will keep happening until one of the two arrays is empty. Now, you can simply add what is left of the nonempty array to the end of your merged array, and you are done.
I'm not sure if your professor will allow you to use recursive calls, but this method could make the coding much easier. Runtime would still be O(m+n), as you are basically iterating through both arrays once.
Hope this helps.

A good data structure for storing and searching integers?

Edit: Typos fixed and ambiguity tried to fix.
I have a list of five digit integers in a text file. The expected amount can only be as large as what a 5-digit integer can store. Regardless of how many there are, the FIRST line in this file tells me how many integers are present, so resizing will never be necessary. Example:
3
11111
22222
33333
There are 4 lines. The first says there are three 5-digit integers in the file. The next three lines hold these integers.
I want to read this file and store the integers (not the first line). I then want to be able to search this data structure A LOT, nothing else. All I want to do, is read the data, put it in the structure, and then be able to determine if there is a specific integer in there. Deletions will never occur. The only things done on this structure will be insertions and searching.
What would you suggest as an appropriate data structure? My initial thought was a binary tree of sorts; however, upon thinking, a HashTable may be the best implementation. Thoughts and help please?

It seems like the requirements you have are
store a bunch of integers,
where insertions are fast,
where lookups are fast, and
where absolutely nothing else matters.
If you are dealing with a "sufficiently small" range of integers - say, integers up to around 16,000,000 or so, you could just use a bitvector for this. You'd store one bit per number, all initially zero, and then set the bits to active whenever a number is entered. This has extremely fast lookups and extremely fast setting, but is very memory-intensive and infeasible if the integers can be totally arbitrary. This would probably be modeled with by BitSet.
If you are dealing with arbitrary integers, a hash table is probably the best option here. With a good hash function you'll get a great distribution across the table slots and very, very fast lookups. You'd want a HashSet for this.
If you absolutely must guarantee worst-case performance at all costs and you're dealing with arbitrary integers, use a balanced BST. The indirection costs in BSTs make them a bit slower than other data structures, but balanced BSTs can guarantee worst-case efficiency that hash tables can't. This would be represented by TreeSet.

Given that
All numbers are <= 99,999
You only want to check for existence of a number
You can simply use some form of bitmap.
e.g. create a byte[12500] (it is 100,000 bits which means 100,000 booleans to store existence of 0-99,999 )
"Inserting" a number N means turning the N-th bit on. Searching a number N means checking if N-th bit is on.
Pseduo code of the insertion logic is:
bitmap[number / 8] |= (1>> (number %8) );
searching looks like:
bitmap[number/8] & (1 >> (number %8) );
If you understand the rationale, then a even better news for you: In Java we already have BitSet which is doing what I was describing above.
So code looks like this:
BitSet bitset = new BitSet(12500);
// inserting number
bitset.set(number);
// search if number exists
bitset.get(number); // true if exists

If the number of times each number occurs don't matter (as you said, only inserts and see if the number exists), then you'll only have a maximum of 100,000. Just create an array of booleans:
boolean numbers = new boolean[100000];
This should take only 100 kilobytes of memory.
Then instead of add a number, like 11111, 22222, 33333 do:
numbers[11111]=true;
numbers[22222]=true;
numbers[33333]=true;
To see if a number exists, just do:
int whichNumber = 11111;
numberExists = numbers[whichNumber];
There you are. Easy to read, easier to mantain.

A Set is the go-to data structure to "find", and here's a tiny amount of code you need to make it happen:
Scanner scanner = new Scanner(new FileInputStream("myfile.txt"));
Set<Integer> numbers = Stream.generate(scanner::nextInt)
.limit(scanner.nextInt())
.collect(Collectors.toSet());

Getting the 5 lowest values with their index from a 2D Array

Any ideas how to get the 5 minimum numbers from a 2D Array. I would like to know their index as well. I'm using Processing but I'm interested to find the correct way to do that.
For example: I have a 4x4 array with the following values:
3-72-64-4
12-45-9-7
86-34-81-55
31-19-18-21
I want to get the five lowest number in my Array which are 3,4,7,9,12. The problem is that I want to know their original index as well.
Example:
Array[0,0] = 3
Array[0,3] = 4
Array[1,3] = 7
Array[1,2] = 9
Is there any formula or good programming way to do that?

There is actually a very good practice that is suited for your case. It's called the 'merge sort algorithm'. It will sort your values and then you just need to output the first 5 values. Here's a link specifically for java. Have fun coding and testing it! I did :D

Well obviously you can just cycle through it and brute force with 2 for loops. Getting the original index makes it harder, as then you cant use sorts, which are faster. If it is sorted or if there is some kind of pattern, you can use a search (binary search) but from what you've given, as it looks as if the data is random, you can't really do much.
If you don't care about indexes, you can try sorts, such as merge sort mentioned by ERed or other types of sorts (I prefer quickSort). Basically you treat the 2D array as a 1D array and assume each subsequent level is just a continuation of the previous level (basically its all just one giant row broken into pieces).

Fastest way to add a value in the middle of a sorted array - Java

I have a sorted array, lets say D={1,2,3,4,5,6} and I want to add the number 5 in the middle. I can do that by adding the value 5 in the middle and move the other values one step to the right.
The problem is that I have an array with 1000 length and I need to do that operation 10.000 times, so I need a faster way.
What options do I have? Can I use LinkedLists for better performance?

That depends on how you add said numbers. If only in ascending or descending order - then yes, LinkedList will do the trick, but only if you keep the node reference in between inserts.
If you're adding numbers in arbitrary order, you may want to deconstruct your array, add the new entries and reconstruct it again. This way you can use a data structure that's good at adding and removing entries while maintaining "sortedness". You have to relax one of your assumptions however.
Option 1
Assuming you don't need constant time random access while adding numbers:
Use a binary sorted tree.
The downside - while you're adding, you cannot read or reference an element by their position, not easily at least. Best case scenario - you're using a tree that keeps track of how many elements the left node has and can get the ith element in log(n) time. You can still get pretty good performance if you're just iterating through the elements though.
Total runtime is down to n * log(n) from n^2. Random access is log(n).
Option 2
Assuming you don't need the elements sorted while you're adding them.
Use a normal array, but add elements to the end of it, then sort it all when you're done.
Total runtime: n * log(n). Random access is O(1), however elements are not sorted.
Option 3
(This is kinda cheating, but...)
If you have a limited number of values, then employing the idea of BucketSort will help you achieve great performance. Essentially - you would replace your array with a sorted map.
Runtime is O(n), random access is O(1), but it's only applicable to a very small number of situations.
TL;DR
Getting arbitrary values, quick adding and constant-time positional access, while maintaining sortedness is difficult. I don't know any such structure. You have to relax some assumption to have room for optimizations.

A LinkedList will probably not help you very much, if at all. Basically you are exchanging the cost of shifting every value on insert with the cost of having to traverse each node in order to reach the insertion point.
This traversal cost will also need to be paid whenever accessing each node. A LinkedList shines as a queue, but if you need to access the internal nodes individually it's not a great choice.
In your case, you want a sorted Tree of some sort. A BST (Balanced Search Tree, also referred to as a Sorted Binary Tree) is one of the simplest types and is probably a good place to start.
A good option is a TreeSet, which is likely functionally equivalent to how you were using an array, if you simply need to keep track of a set of sorted numbers.

Iterating over Permutations of an Array

I'm working on some java code for some research I'm working on, and need to have a way to iterate over all permutations of an ArrayList. I've looked over some previous questions asked here, but most were not quite what I want to do, and the ones that were close had answers dealing with strings and example code written in Perl, or in the case of the one implementation that seemed like it would work ... do not actually work.
Ideally I'm looking for tips/code snippets to help me write a function permute(list, i) that as i goes from 0 to list.size()! gives me every permutation of my ArrayList.

There is a way of counting from 0 to (n! - 1) that will list off all permutations of a list of n elements. The idea is to rewrite the numbers as you go using the factorial number system and interpreting the number as an encoded way of determining which permutation to use. If you're curious about this, I have a C++ implementation of this algorithm. I also once gave a talk about this, in case you'd like some visuals on the topic.
Hope this helps!

If iterating over all permutations is enough for you, see this answer: Stepping through all permutations one swap at a time .
For a given n the iterator produces all permutations of numbers 0 to (n-1).
You can simply wrap it into another iterator that converts the permutation of numbers into a permutation of your array elements. (Note that you cannot just replace int[] within the iterator with an arbitrary array/list. The algorithm needs to work with numbers.)