我意识到之前已经问过这个问题,但是我无法理解我读到的很多答案。我一直在研究这段代码:
static ArrayList<String> psalmsTitlesArray = new ArrayList<>(47);
static ArrayList<String> psalmsNumbersArray = new ArrayList<>(47);
static String psalmTextFileArray[] = new String[47];
static int index = 0;
public static void main(String[] args) throws IOException {
//this program demonstrates a binary array search. It is going to search
//the text file "Psalms.txt" for a number list of the pslams.
// eg. the user wants to see psalm 7
// the program will display "prayer of the virtuous under persecution".
BufferedReader fileRead = new BufferedReader(new FileReader("Psalms.txt")); //create a BufferedReader to read the file
String fileLine = "";
int lineNumber = 1;
for (int i = 0; i < 47; i++) {
fileLine = fileRead.readLine(); // stores each line of text in a String
if (fileLine == null) { //if the line is blank
break; // don't read it
}
psalmTextFileArray[i] += fileLine;
if (lineNumber % 2 == 0) { //if the line is not an even number
psalmsTitlesArray.add(fileLine); //add the line to the Titles array
lineNumber++;
} else { //otherwise,
psalmsNumbersArray.add(fileLine); //add it to the numbers array
lineNumber++;
}
}
String userInput = JOptionPane.showInputDialog(null, "What psalm would you like me to search for?", "Psalm Finder",
JOptionPane.QUESTION_MESSAGE);
if (userInput == null) {
System.exit(0);
}
int numberInput = Integer.parseInt(userInput);
binarySearch(psalmsNumbersArray, 0, (psalmsNumbersArray.size() - 1), userInput);
for (int i = 0; i < psalmTextFileArray.length; i++) {
index = psalmTextFileArray[i].indexOf(numberInput);
}
JOptionPane.showMessageDialog(null, "I found Psalm #" + userInput + ". It is: \n" + psalmTextFileArray[index]);
}
public static boolean binarySearch(ArrayList<String> myPsalmsArray, int left, int right, String searchForPsalm) {
int middle;
if (left > right) {
return false;
}
middle = (left + right) / 2;
if (myPsalmsArray.get(middle).equals(searchForPsalm)) {
return true;
}
if (searchForPsalm.compareTo(myPsalmsArray.get(middle)) < 0) {
return binarySearch(myPsalmsArray, left, middle - 1,
searchForPsalm);
} else {
return binarySearch(myPsalmsArray, middle + 1, right,
searchForPsalm);
}
}
}
此代码从文件&#34; Psalm.txt&#34;中读取。从本质上讲,Pslams按顺序从1到99(但不包括每个Psalm,例如,文本文件中不存在pslam 4)
我试图使用indexOf()来查找字符首次出现的索引。然后,我将能够将它向前移动到数组中的一个索引并找到标题,因为数组中的信息列为:
[2,诗篇2标题,3,诗篇3标题,4 ...](等)
这是导致问题的一大块代码:
for (int i = 0; i < psalmTextFileArray.length; i++) {
index = psalmTextFileArray[i].indexOf(numberInput);
}
JOptionPane.showMessageDialog(null, "I found Psalm #" + userInput + ". It is: \n" + psalmTextFileArray[index]);
所以,如果我能在数组中找到2的索引,将它向前移动一个索引,我会得到标题。 此外,索引始终抛出-1异常。这是什么意思?
道歉,如果这令人困惑,我可以澄清任何不清楚的事情(我会尽我所能!)
答案 0 :(得分:1)
由于您的代码无效,您将获得异常
int numberInput = Integer.parseInt(userInput);
binarySearch(psalmsNumbersArray, 0, (psalmsNumbersArray.size() - 1), userInput);
for (int i = 0; i < psalmTextFileArray.length; i++) {
index = psalmTextFileArray[i].indexOf(numberInput);
}
JOptionPane.showMessageDialog(null, "I found Psalm #" + userInput + ". It is: \n" + psalmTextFileArray[index]);
上面的代码psalmTextFileArray[i]
中的将返回一个String
对象。然后你正在做indexOf(numberInput)
如JavaDoc
此对象表示的字符序列中第一次出现的字符的索引,如果未出现该字符,则为-1。
因此,如果您的numberInput
不在String
对象中,则index
将变为-1
index = -1
在下一行,当您致电showMessageDialog
时,您正在使用-1
访问数组的位置,因为您正在使用psalmTextFileArray[index]