我正在尝试编写程序时遇到问题。基本上它从两个文件中读取,将它们转换为数组,并比较相似之处。到目前为止它已经工作了,但是当第一个.txt文件读取“a b c d e f g h i j a a a a a a a”而第二个.txt文件读取“a b c d e f g h i k”时,输入文件名后出现以下错误:
java.lang.ArrayIndexOutOfBoundsException: 10
at PlagiarismDetector.compareStrings(PlagiarismDetector.java:77)
at PlagiarismDetector.main(PlagiarismDetector.java:25)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(Unknown Source)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source)
at java.lang.reflect.Method.invoke(Unknown Source)
at edu.rice.cs.drjava.model.compiler.JavacCompiler.runCommand(JavacCompiler.java:272)
我的代码如下。感谢任何可以帮助的人=)
import java.util.Scanner;
import java.io.*;
public class PlagiarismDetector
{
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
System.out.println("What file is the first file?");
String fileOne = reader.next();
String stringOne = readStringFromFile(fileOne);
System.out.println("What file is the second file?");
String fileTwo = reader.next();
String stringTwo = readStringFromFile(fileTwo);
if (stringOne == null || stringTwo == null)
{
return;
}
System.out.println("Comparing the 2 files......");
System.out.println("The result of the 2 files is ....");
if (compareStrings(stringOne, stringTwo))
{
System.out.println("Plagiarism detected. Cheaters!!!!");
}
else
{
System.out.println("No plagiarism detected");
}
}
public static String readStringFromFile(String filename)
{enter code here
String builder = "";
try
{
Scanner fileReader = new Scanner(new File(filename));
while (fileReader.hasNextLine())
{
builder = builder + fileReader.nextLine() + "\n";
}
return builder;
}
catch (Exception e)
{
System.out.println("An error occurred while trying to open the file " + filename + ". Is the file located inside the same folder as the .class file and with the identical name?");
return null;
}
}
public static boolean compareStrings (String a, String b)
{
boolean checkForPlagiarism = true;
String[] piecesA = a.split("\\s");
String[] piecesB = b.split("\\s");
int count1 = 0;
int count2 = 0;
for (int counter = 0; counter <= piecesA.length - 1; counter++)
{
for(int counter2 = 0; counter<= piecesB.length - 1; counter++)
{
if(piecesA[counter].equals(piecesB[counter2]))
{
count1++;
}
}
}
for (int counter = 0; counter <= piecesB.length - 1; counter++)
{
for(int counter2 = 0; counter <= piecesA.length - 1; counter++)
{
if(piecesA[counter].equals(piecesB[counter]))
{
count2++;
}
}
}
if((count1/(int)piecesA.length)*100 >= 90 && (count2/(int)piecesB.length)*100 >= 90)
{
checkForPlagiarism = false;
}
return checkForPlagiarism;
}
}
答案 0 :(得分:2)
if(piecesA[counter].equals(piecesB[counter]))
应该是
if(piecesA[counter2].equals(piecesB[counter]))
答案 1 :(得分:1)
这看起来很可疑:
for (int counter = 0; counter <= piecesB.length - 1; counter++)
{
for(int counter2 = 0; counter <= piecesA.length - 1; counter++)
{
if(piecesA[counter].equals(piecesB[counter]))
{
count2++;
}
}
}
计数器受错误数组长度的限制
你可能想这样做:
for (int counter = 0; counter <= piecesA.length - 1; counter++)
{
for(int counter2 = 0; counter2 <= piecesB.length - 1; counter2++)
{
if(piecesA[counter].equals(piecesB[counter2]))
{
count2++;
}
}
}
此外,惯例是编写如下循环:
for(int counter2 = 0; counter2 < piecesB.length; counter2++)