从 Java 中的文本文件中读取数据为 arrays

Question

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.

Answer 1

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)

Hope this helps!

Answer 2

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:

1. Create and instantiate an AlgoContainer object instance (let's call this tempAlgoContainer ).

2. 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.

3. Found the open parenthesis? parse the temp String to int and set the numberToFind field of tempAlgoContainer to that number.

4. 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.

5. 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.
6. 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)

Answer 3

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 asGist on GitHub

Answer 4

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!