读取文本文件（使用ArrayList进行分隔，排序，查找数字之间）

Question

I'm really shy to say this problem. Because i couldn't understand how to do it. I've a text file. In this text file there is a 1 million numbers. These are between 0 to 999 and every line contains a number. I have to separate these numbers with (\\n), that means every line will contains one integer number. Then i have to sort these. After sorting i will take 2 inputs from user. Then i have to find and write numbers to a new text file between these inputs in the first text file.

I know most people will say "we're not fool. You don't know anything, this site is not like this." That's right, but i have to do this. I just want to know how can i do this. Which methods or strategies do i need? Because i've found similar things in here but i don't know are these true with my work.

Answer 1

For reading the file I suggest you use this, will make your task quite simple.
http://docs.oracle.com/javase/7/docs/api/java/io/BufferedReader.html

For sorting, look at this class, it has a sort method.
http://docs.oracle.com/javase/7/docs/api/java/util/Arrays.html

For converting String values (which you read from the file)
to int or Integer values use this class, it has parseInt method.
http://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html

Answer 2

Here's a simple example using Java 8:

public static void main(final String[] args) throws IOException {
    final Pair input = readInput();
    final Path sourceFile = Paths.get("path", "to", "input", "file");
    final Path destFile = Paths.get("path", "to", "output", "file");
    final IntStream parsed = Files.lines(sourceFile).
            mapToInt(Integer::parseInt).
            filter(i -> i >= input.getLow() && i <= input.getHigh()).
            parallel().
            sorted();
    try (final PrintWriter outputFile = new PrintWriter(Files.newBufferedWriter(destFile, StandardCharsets.UTF_8))) {
        parsed.forEach(outputFile::println);
    }
}

private static Pair readInput() {
    final Console console = System.console();
    final String low = console.readLine("Please enter the start number (inclusive): ");
    final String high = console.readLine("Please enter the end number (inclusive): ");
    return new Pair(Integer.parseInt(low), Integer.parseInt(high));
}

private static final class Pair {
    private final int low;
    private final int high;

    private Pair(int low, int high) {
        this.low = low;
        this.high = high;
    }

    public int getHigh() {
        return high;
    }

    public int getLow() {
        return low;
    }
}

The readInput() method uses a Console to read the high and low values into a class Pair that holds the user input.

The main method calls readInput() then reads lines from the input file, filters out numbers that are lower than low and higher than high and parallel sorts the resultant output.

Once we have the pipeline setup, we then call forEach on it and write the value to the output file.

Answer 3

For sorting numbers you can use counting sort algorithm: you have numbers of limited range, from 0 to 999. So, you can count how many times each number occurred in input file:

int[] count = new int[1000];
//depending of scope, maybe you need to fill array with zeroes

...
int number = ...;//read number from file;
count[number]++;

Now, in order to output it as sorted, you just need two loops like this:

for (int i = 0; i < 1000; ++i) {
    for (int j = 0; j < count[i]; ++j) {
        System.out.println(i);
    }
}

I wrote this just to make sure idea is clear, although your task doesn't require to output them all. Now, let's imagine you read these two numbers;

int start, finish = 0; //value read from input

We gonna modify the loops above like this:

int counter = 0;
for (int i = 0; i < 1000; ++i) {
    for (int j = 0; j < count[i]; ++j) {
        if ((counter >= start) && (counter < finish)) {//or maybe <=, depending on what exactly do you need
            System.out.println(i);
        }
        counter++;
        if (counter > finish) { //we finished, further iterations are waste of CPU time
            break(2);
        }  
    }
}

This method is not optimal, but easy to understand and implement. For more fast solutions, you may want to get rid of inner loop, replacing it with logic to add count[i] to counter at once and detecting moment when we need to start output.