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How to get number from string array in java

i have a string like String[] str = {"[5, 2, 3]","[2, 2, 3, 10, 6]"} and i need to take numbers to add into an integer list.
i tried to split first index into numbers to see if it will work, looks like:
String[] par = str[0].split("[, ?.#]+");
After the split i tried to see what array i get:
for(String a: par)
System.out.println(a);
But when i wrote that code i get an array like this:
[5
2
3]
So, how can i get rid of this square brackets?

Instead of your current pattern, I would use \\D+ which will split on one or more non-digits. Add a guard for the empty string too. Something like
String[] str = { "[5, 2, 3]", "[2, 2, 3, 10, 6]" };
for (String par : str) {
for (String t : par.split("\\D+")) {
if (t.isEmpty()) {
continue;
}
System.out.println(Integer.parseInt(t));
}
}
Outputs
5
2
3
2
2
3
10
6

Read data as arrays from a Text File in Java

I have a text file with a bunch of arrays with a specific number I have to find in the array. The text file looks like this:
(8) {1, 4, 6, 8, 12, 22}
(50) {2, 5, 6, 7, 10, 11, 24, 50, 65}
(1) {1}
(33) {1, 2, 5, 6, 11, 12, 13, 21, 25, 26, 30, 33, 60, 88, 99}
(1) {1, 2, 3, 4, 8, 9, 100}
(1) {2, 3, 5, 6, 11, 12, 13, 21, 25, 26, 30, 33, 60, 88, 99}
where the number inside the parenthesis is the number I have to find using binary search. and the rest is the actual array.
I do not know how I would get this array from the text file and be able to read it as an actual array.
[This is a question on a previous coding competition I took, and am going over the problems]
I already have a method to do the binary search, and I have used scanner to read the file like this:
Scanner sc = new Scanner(new File("search_race.dat"));
and used a while loop to be able to loop through the file and read it.
But I am stuck on how to make java know that the stuff in the curly braces is an array and the stuff in the parenthesis is what it must use binary search on said array to find.

You could simply parse each line (the number to find and the array) as follow :
while (sc.hasNext()) {
int numberToFind = Integer.parseInt(sc.next("\\(\\d+\\)").replaceAll("[()]", ""));
int[] arrayToFindIn = Arrays.stream(sc.nextLine().split("[ ,{}]"))
.filter(x->!x.isEmpty())
.mapToInt(Integer::parseInt)
.toArray();
// Apply your binary search ! Craft it by yourself or use a std one like below :
// int positionInArray = Arrays.binarySearch(arrayToFindIn, numberToFind);
}
If you don't like the replaceAll, you could replace the first line in the loop by the two below :
String toFindGroup = sc.next("\\(\\d+\\)");
int numberToFind = Integer.parseInt(toFindGroup.substring(1, toFindGroup.length()-1));
Cheers!

TL;DR: You have to check character by character and see if it's a curly brace or a parenthesis or a digit
Long Answer:
First, create a POJO (let's call this AlgoContainer, but use whatever name you like) with the fields int numberToFind and ArrayList<Integer> listOfNumbers.
Then, read the file like #ManojBanik has mentioned in the comments
Now create an ArrayList<AlgoContainer> (it's size should be the same as the ArrayList<String> that was gotten while reading the file line by line)
Then loop through the ArrayList<String> in the above step and perform the following operations:
Create and instantiate an AlgoContainer object instance (let's call this tempAlgoContainer).
check if the first character is an open parentheses -> yes? create an empty temp String -> check if the next character is a number -> yes? -> append it to the empty String and repeat this until you find the closing parenthesis.
Found the open parenthesis? parse the temp String to int and set the numberToFind field of tempAlgoContainer to that number.
Next up is the curly bracket stuff: found a curly bracket? create a new empty temp String -> check if the next character is digit -> yes? append then append it to the empty String just like in step #2 until you find a comma or a closing curly brace.
Found a comma? parse the temp String to int and then add it to the listOfNumbers (which is a field) of tempAlgoContainer -> make the temp String empty again.
Found a closing curly brace? repeat the above step and break out of the loop. You are now ready to process whatever you want to do. Your data is ready.
Also, it's a good idea to have a member function or instance method of AlgoContainer (call it whatever you want) to perform the binary search so that you can simply loop through the ArrayList<AlgoContainer> and call that BS function on it (no-pun-intended)

To read the file, you can use Files.readAllLines()
To actually parse each line, you can use something like this.
First, to make things easier, remove any whitespace from the line.
line.replaceAll("\\s+", "");
This will essentially transform (8) {1, 4, 6, 8, 12, 22} into (8){1,4,6,8,12,22}.
Next, use a regular expression to validate the line. If the line does not match no further actions are required.
Expression: \([0-9]*\)\{[0-9]*(,[0-9]*)*}
\([0-9]*\) relates to (8) (above example)
\{[0-9]*(,[0-9]*)*} relates to {1,4,6,8,12,22}
If you don´t understand the expression, head over here.
Finally, we can parse the string into its two components: The number to search for and the int[] with the actual values.
// start from index one to skip the first bracket
int targetEnd = trimmed.indexOf(')', 1);
String searchString = trimmed.substring(1, targetEnd);
// parsing wont throw an exception, since we checked with the regex its a number
int numberToFind = Integer.parseInt(searchString);
// skip ')' and '{', align to the first value, skip the last '}'
String valuesString = trimmed.substring(targetEnd + 2, trimmed.length() - 1);
// split the array at ',' to get each value as string
int[] values = Arrays.stream(valuesString.split(","))
.mapToInt(Integer::parseInt).toArray();
With both of these components parsed, you can do the binary search yourself.
Example code as Gist on GitHub

You could read strings in file line by line and then use regex on each line to separate the string to groups.
Below regex should fit to match the line
\((\d+)\) \{([\d, ]+)\}
Then group(1) will give the digit inside the parentheses (as a String) and group(2) will give the String inside curly braces, which you can split using , and space(assuming every comma follows a space) and get an array of numbers (as Strings again).

Replacing more than one character in a String

I am printing an array, but I only want to display the numbers. I want to remove the brackets and commas from the String (leaving only the digits). So far, I have been able to remove the commas but I was looking for a way add more arguments to the replaceAll method.
How can I remove the brackets as well as the commas?
cubeToString = Arrays.deepToString(cube);
System.out.println(cubeToString);
String cleanLine = "";
cleanLine = cubeToString.replaceAll(", ", ""); //I want to put all braces in this statement too
System.out.println(cleanLine);
The output is:
[[0, 0, 0, 0], [1, 1, 1, 1], [2, 2, 2, 2], [3, 3, 3, 3], [4, 4, 4, 4], [5, 5, 5, 5]]
[[0000][1111][2222][3333][4444][5555]]

You can use the special characters [ and ] to form a pattern, and then \\ to escape the [ and ] (from your input) like,
cleanLine = cubeToString.replaceAll("[\\[\\]\\s,]", "");
or replace everything not a digit. Like,
cleanLine = cubeToString.replaceAll("\\D", "");

What you are doing is effectively using Java like you use a scripting language.
In this case, it happens to work well because your arrays only contain numbers and you don't have to worry about escaping characters that may also appear inside your array elements.
But it's still not efficient or Java-like to be converting strings several times, one of them with regular expressions (replaceAll), to get to your end-result.
A nicer and more efficient approach is to directly build the string that you need without any comma's or square brackets:
public static void main(String[] args) throws Exception {
int[][] cube = { { 0, 0, 0, 0 }, { 1, 1, 1, 1 }, { 2, 2, 2, 2 }, { 3, 3, 3, 3 }, { 4, 4, 4, 4 },
{ 5, 5, 5, 5 } };
StringBuilder builder = new StringBuilder();
for (int[] r : cube) {
for (int c : r) {
builder.append(c);
}
}
String cleanLine = builder.toString();
System.out.println(cleanLine);
}
Output:
000011112222333344445555

How do I convert a String to an String Array? [duplicate]

This question already has answers here:
How do I convert a String to an int in Java?
(47 answers)
Closed 7 years ago.
I'm reading from a file using Scanner, and the text contains the following.
[8, 3, 8, 2, 3, 4, 4, 4, 5, 8]
This was originally an integer Array that I had to convert to a String to be able to write in the file. Now, I need to be able to read the file back into java, but I need to be able to add the individual numbers together, so I need to get this String back into an array. Any help? Here's what I have:
File f = new File("testfile.txt");
try{
FileWriter fw = new FileWriter(f);
fw.write(Arrays.toString(array1));
fw.close();
} catch(Exception ex){
//Exception Ignored
}
Scanner file = new Scanner(f);
System.out.println(file.nextLine());
This prints out the list of numbers, but in a string. I need to access the integers in an array in order to add them up. This is my first time posting, let me know if I messed anything up.

You can use String#substring to remove the square brackets, String#split to split the String into an array, String#trim to remove the whitespace, and Integer#parseInt to convert the Strings into int values.
In Java 8 you can use the Stream API for this:
int[] values = Arrays.stream(string.substring(1, string.length() - 1)
.split(","))
.mapToInt(string -> Integer.parseInt(string.trim()))
.toArray();
For summing it, you can use the IntStream#sum method instead of converting it to an array at the end.

You don't need to read the String back in an Array, just use Regex
public static void main(String[] args) throws Exception {
String data = "[8, 3, 8, 2, 3, 4, 41, 4, 5, 8]";
// The "\\d+" gets the digits out of the String
Matcher matcher = Pattern.compile("\\d+").matcher(data);
int sum = 0;
while(matcher.find()) {
sum += Integer.parseInt(matcher.group());
}
System.out.println(sum);
}
Results:
86

List<Integer> ints = new ArrayList<>();
String original = "[8, 3, 8, 2, 3, 4, 4, 4, 5, 8]";
String[] splitted = original.replaceAll("[\\[\\] ]", "").split(",");
for(String s : splitted) {
ints.add(Integer.valueOf(s));
}

Input from text file to array

The input will be a text file with an arbitrary amount of integers from 0-9 with NO spaces. How do I populate an array with these integers so I can sort them later?
What I have so far is as follows:
BufferedReader numInput = null;
int[] theList;
try {
numInput = new BufferedReader(new FileReader(fileName));
} catch (FileNotFoundException e) {
System.out.println("File not found");
e.printStackTrace();
}
int i = 0;
while(numInput.ready()){
theList[i] = numInput.read();
i++;
Obviously theList isn't initialized, but I don't know what the length will be. Also I'm not too sure about how to do this in general. Thanks for any help I receive.
To clarify the input, it will look like:
1236654987432165498732165498756484654651321
I won't know the length, and I only want the single integer characters, not multiple. So 0-9, not 0-10 like I accidentally said earlier.

Going for Collection API i.e. ArrayList
ArrayList a=new Arraylist();
while(numInput.ready()){
a.add(numInput.read());
}

You could use a List<Integer> instead of a int[]. Using a List<Integer>, you can add items as desired, the List will grow along. If you are done, you can use the toArray(int[]) method to transform the List into an int[].

1 . Use guava to nicely read file's 1st line into 1 String
readFirstLine
2 . convert that String to char array - because all of your numbers are one digit lengh, so they are in fact chars
3 . convert chars to integers.
4 . add them to list.
public static void main(String[] args) {
String s = "1236654987432165498732165498756484654651321";
char[] charArray = s.toCharArray();
List<Integer> numbers = new ArrayList<Integer>(charArray.length);
for (char c : charArray) {
Integer integer = Integer.parseInt(String.valueOf(c));
numbers.add(integer);
}
System.out.println(numbers);
}
prints: [1, 2, 3, 6, 6, 5, 4, 9, 8, 7, 4, 3, 2, 1, 6, 5, 4, 9, 8, 7, 3, 2, 1, 6, 5, 4, 9, 8, 7, 5, 6, 4, 8, 4, 6, 5, 4, 6, 5, 1, 3, 2, 1]