是否可以使此代码更高效（java8，解析文件）？

Question

我使用字节比较分析（查找到的模板进行了替换和删除）大型文本文件。在您看来，是否有可能加快此代码的速度（停留在java.io上并在一个流中）？

import org.apache.commons.io.IOUtils;
import java.io.*;
import java.time.Instant;
import java.util.Arrays;

public class ParserSingle {
private static final String RESULT = "static/result.txt";
private static final String PATTERN_OLD = "static/pattern.txt";
private static final String PATTERN_NEW = "static/newPattern.txt";
private static final String ORIGIN = "static/big.txt";
public static void main(String[] args) throws IOException {
    File originFile = new File(ORIGIN);
    try (FileInputStream streamPatternOld = new FileInputStream(new File(PATTERN_OLD));
         FileInputStream streamPatternNew = new FileInputStream(new File(PATTERN_NEW));
         BufferedReader origin = new BufferedReader(new FileReader(originFile));
         BufferedWriter result = new BufferedWriter(new FileWriter(new File(RESULT)))) {
        Instant start = Instant.now();
        new ParserSingle().process(IOUtils.toByteArray(streamPatternOld),
                IOUtils.toByteArray(streamPatternNew),
                0,
                originFile.length(),
                origin,
                result);
        Instant end = Instant.now();
        System.out.println("seconds running: " + (end.getEpochSecond() - start.getEpochSecond()));
    }
}

public void process(byte[] patternOld,
                    byte[] patternNew,
                    long start,
                    long limit,
                    BufferedReader origin,
                    BufferedWriter result) throws IOException {
    long originCurrent = 0;
    int posCurrent = 0, value;
    byte[] values = new byte[patternOld.length];
    origin.skip(start);
    while ((value = origin.read()) != -1) {
        values[posCurrent++] = (byte) value;
        originCurrent++;
        if (!Arrays.equals(Arrays.copyOf(patternOld, posCurrent), Arrays.copyOf(values, posCurrent))) {
            result.write(new String(Arrays.copyOf(values, posCurrent)));
            values = new byte[patternOld.length];
            posCurrent = 0;
        } else if (posCurrent == patternOld.length) {
            posCurrent = 0;
            //result.write(new String(values)); //one to one
            result.write(new String(patternNew)); //replace //comment out for removal
        }
    }
}
}

感谢您的帮助！