[英]Working With Puzzle and figuring how to fix out of bound error
我的错误说明
我应该从文件中读取单词搜索拼图我的问题我一直在越界错误。 我不知道如何根据文件的规范正确地设置单词拼图的格式。 拼图的格式通常是这样的:5 5
转移
红卫兵
知乎
农夫
瓦姆恩
qtfox
到目前为止,这是我的工作,我觉得我已经从文件中读取,但将其转换为单词搜索拼图。 特别是尽量不对单词搜索拼图的行和列的规范进行硬编码。
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;
}
您的索引越界错误是由此处引起的:
for(int column = 0; column < columns; column++)
{
char index = condensed[i];
fillThemLetters[row][column] = index;
i++; // <--------- THIS IS WRONG!!!
}
看到i++
吗? 您无缘无故地从最外层循环增加计数器。 让循环处理它自己的增量,它已经内置了,像这样:
for (int i = 0; i < condensed.length; i++) { // <--- You already have i++ here!
解决这个问题后,您将遇到更多问题——您的代码没有按照您认为的那样做,但这些都是单独的问题,应该单独发布。
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