[英]Working With Puzzle and figuring how to fix out of bound error
我的錯誤說明
我應該從文件中讀取單詞搜索拼圖我的問題我一直在越界錯誤。 我不知道如何根據文件的規范正確地設置單詞拼圖的格式。 拼圖的格式通常是這樣的:5 5
轉移
紅衛兵
知乎
農夫
瓦姆恩
qtfox
到目前為止,這是我的工作,我覺得我已經從文件中讀取,但將其轉換為單詞搜索拼圖。 特別是盡量不對單詞搜索拼圖的行和列的規范進行硬編碼。
public static char[][] fill(){
// Created 2 different scanner one for user input and one to read the file
Scanner file1 = new Scanner(System.in);
// created a count to add the keywords
int count = 0;
//System.out.print("Please enter a keyword to search for.");
// Asking user to input a valid file name
System.out.print("Please enter a valid puzzle file name\nYou will be asked for the same file name again later.");
String wordFile = file1.nextLine();
FileReader infile;
boolean validFile = false;
// Creating a while loop that will keep asking for a valid file name
while(!validFile) {
// Using a try and catch to obtain correct file
try {
infile = new FileReader(wordFile);
file1 = new Scanner(infile);
validFile = true;
}
//ask the user to put a valid file name if they are wrong
catch(IOException e) {
System.out.println("Not a valid file name, please enter again!");
wordFile = file1.nextLine();
}
}
String numbers = file1.nextLine();
String[] fileArray = numbers.trim().split(" ");
int rows = Integer.parseInt(fileArray[0]);
int columns = Integer.parseInt(fileArray[1]);
char[][] fillThemLetters = new char [rows][columns];
String letters = file1.nextLine().trim().replace(" ", "");
char [] condensed = letters.toCharArray();
for (int i = 0; i < condensed.length; i++) {
for(int row = 0; row < rows; row++){
for(int column = 0; column < columns; column++)
{
char index = condensed[i];
fillThemLetters[row][column] = index;
i++;
}
}
}
return fillThemLetters;
}
您的索引越界錯誤是由此處引起的:
for(int column = 0; column < columns; column++)
{
char index = condensed[i];
fillThemLetters[row][column] = index;
i++; // <--------- THIS IS WRONG!!!
}
看到i++
嗎? 您無緣無故地從最外層循環增加計數器。 讓循環處理它自己的增量,它已經內置了,像這樣:
for (int i = 0; i < condensed.length; i++) { // <--- You already have i++ here!
解決這個問題后,您將遇到更多問題——您的代碼沒有按照您認為的那樣做,但這些都是單獨的問題,應該單獨發布。
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