Java填字游戏:有没有办法计算打印单词前后的空格?
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Words:测试文件
我正在为我的学校设立一个课程,但我遇到了麻烦。你可以帮我找到一种方法来打印前面和后面的空格吗?
public class Main {
public static void main(String[] args) {
System.out.println("crossword generator ver. 1.0");
File wordlist = new File("words.txt");
try {
Scanner s = new Scanner(wordlist);
String words[] = new String[1000000];
int lineNr = 0;
while (s.hasNext() && lineNr < 1000000) {
words[lineNr] = s.nextLine();
lineNr++;
}
System.out.println("Wordlist succesfully loaded");
Random r = new Random();
String solution = words[r.nextInt(lineNr)];
System.out.println("Solution = " + solution);
for (int i = 0; i<solution.length(); i++){
char c = solution.charAt(i);
String word;
do{
word = words[r.nextInt(lineNr)];
} while(word.indexOf(c) == -1);
System.out.printf("(%c): %s
", c ,word);
}
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
// TODO Auto-generated method stub
}
}
答案
你已经拥有了关键成分:indexOf()
创建空间量有点棘手:创建与indexOf相同的数量与我们需要的完全相反。首先,我们必须计算最高的indexOf,这样我们就可以在每个单词前面减去当前单词中的indexOf。
我们必须记住这些词,因为我们经历了整个循环两次。
下面的解决方案有点脏 - 更好的方法是为随机单词的实例创建一个新类(使用它们的小写版本和indexOf),这也可以保存一个有效的indexOf位置列表,这样你就不会总是使用第一次出现的角色。
它只是意味着成为一条踏脚石。还有很多工作要做,例如你可以决定只使用小写字,然后在最终输出中使“热”字符大写。
此代码忽略大写/小写,因此如果您的解决方案单词以大写字符开头,则不会锁定某些随机单词。这里实现的方式也很脏。
加载列表,顺便说一句,可以大大简化,如下所示。这也将避免不必要的大单词列表数组(否则有时可能太小)。
public static void main(String[] args) {
System.out.println("
crossword generator ver. 1.0");
// Load word list.
final List<String> wordList;
try {
final File wordListFile = new File("words.txt");
wordList = Files.readAllLines(wordListFile.toPath(), StandardCharsets.UTF_8);
} catch (IOException e) {
e.printStackTrace();
return;
}
System.out.println("
Word list successfully loaded.");
// Pick solution word.
final int wordCount = wordList.size();
final Random rand = new Random();
final String solution = wordList.get(rand.nextInt(wordCount));
final String solutionLC = solution.toLowerCase(); // So that we won't depend on upper/lower case.
final int solutionLen = solution.length();
System.out.println("
Solution = " + solution + "
");
// Choose words whose characters are in the solution word.
final String[] chosenWords = new String[solutionLen];
int highestIndex = 0;
for (int i = 0; i < solutionLen; i++) {
final char c = solutionLC.charAt(i);
String word;
int indexOfChar;
do {
word = wordList.get(rand.nextInt(wordCount));
indexOfChar = word.toLowerCase().indexOf(c);
} while (indexOfChar < 0);
chosenWords[i] = word;
highestIndex = Math.max(highestIndex, indexOfChar);
}
// Print crossword excerpt.
for (int i = 0; i < solutionLen; i++) {
final char cLC = solutionLC.charAt(i);
final char c = solution.charAt(i);
final int indexOfChar = chosenWords[i].toLowerCase().indexOf(cLC);
System.out.println("(" + c + "): " + createStringOfIdenticalCharacters(highestIndex - indexOfChar,
' ') + chosenWords[i]);
}
}
public static String createStringOfIdenticalCharacters(final int count,
final char c) {
final char[] retPreliminary = new char[count];
Arrays.fill(retPreliminary, c);
return new String(retPreliminary);
}
示例输出:
crossword generator ver. 1.0
Word list successfully loaded.
Solution = councilor
(c): Corcyra
(o): Harbour
(u): nonillustrative
(n): unexiled
(c): sepulchering
(i): Torrington
(l): builtin
(o): nonnarcissistic
(r): Balsamodendron
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