在Java中的子字符串中搜索字符串的最快方法？

Question

Currently I am working on a Kattis problem, "String Matching"( https://open.kattis.com/problems/stringmatching ). I am getting correct output for my program, however since the file is so big, and the time limit for completing the problem is 2 seconds I keep getting the "Time Limit Exceeded" error on Kattis. I've attempted two ways to solve the problem and the second test case exceeds my time limit on both. Here is what I have done:

    while (sc.hasNext()) {

        String pattern = sc.nextLine();
        String text = sc.nextLine();

        for(int i = 0; i < text.length()-pattern.length()+1; i++) {
            if(text.regionMatches(i,  pattern, 0, pattern.length())) {
                System.out.print(i + " ");
            }
        }
        System.out.println();
    }

I have also tried it this way:

    while(sc.hasNext()) {

        String pattern = sc.nextLine();
        String text = sc.nextLine();

        for(int i = 0; i < text.length()-pattern.length()+1; i++) {
            if(pattern.equals(text.substring(i, i+pattern.length()))) {
                System.out.print(i + " ");
            }
        }   
    System.out.println();
}

What is a faster way to take a String and compare it to see if it exists in a larger String?

Answer 1

As per your comment on getting the positions, it's pretty straightforward with a regex matcher. See below:

Pattern p = Pattern.compile("your regex pattern here");
Matcher m = p.matcher("the string you're testing");
if (m.find()) {
    int start = m.start();
    int end = m.end();

    // do something
}

EDIT: A working example below.

public static void main(String...args) throws Exception {
    printMatchInfo("\\w+", "The quick brown fox jumps over the lazy dog");
}

private static void printMatchInfo(String regex, String input) {
    Pattern p = Pattern.compile(regex);
    Matcher m = p.matcher(input);
    System.out.println(String.format("Checking %s against %s", input, regex)); 
    while (m.find()) {
        System.out.println(String.format("Found %s with (start, end) (%d, %d)", m.group(), m.start(), m.end()));
    }
}

Output

Checking The quick brown fox jumps over the lazy dog against \w+
Found The with (start, end) (0, 3)
Found quick with (start, end) (4, 9)
Found brown with (start, end) (10, 15)

...

Answer 2

The fastest solution for case of massive data would give Suffix Tree or Suffix Array

Naive implementation of suffix tree is pretty simple, you can check my solution here: https://github.com/sergpank/lcs/blob/master/src/MonoMain.java

If you need to build Suffix Tree faster, try Ukkonen Algorythm or any other advanced data structure.