我正在努力实现一种解决方案,以在5x5的随机字母板上找到所有单词。目前,它返回的只是几句话,而并非完整列表。我很确定我的问题存在于for循环的findWords方法中,但是我无法弄清楚如何使if语句继续在所有8个方向上遍历。
import java.io.File;
import java.util.*;
public class RandomWordGame {
private static char[][] board = new char[5][5];
private static Random r = new Random();
private static ArrayList<String> dictionary = new ArrayList<String>();
private static char[][] createBoard()
{
for (int i=0; i<board.length; i++)
{
for (int j=0; j<board.length; j++)
{
board[i][j] = (char) (r.nextInt(26) + 'a');
System.out.print(board[i][j]);
}
System.out.println("");
}
System.out.println();
return board;
}
public static ArrayList<String> solver(char[][] board)
{
if(board == null)
System.out.println("Board cannot be empty");
ArrayList<String> words = new ArrayList<String>();
for(int i=0; i<board.length; i++)
{
for(int j=0; j<board[0].length; j++)
{
findWords(i, j, board[i][j] + "");
}
}
return words;
}
public static void findWords(int i, int j, String currWord)
{
try
{
Scanner inputStream = new Scanner(new File("./dictionary.txt"));
while(inputStream.hasNext())
{
dictionary.add(inputStream.nextLine());
}
inputStream.close();
}catch(Exception e){
e.printStackTrace();
}
for(i=0; i>=0 && i<board.length; i++)
{
for(j=0; j>=0; j++)
{
currWord += board[i][j];
if(currWord.length()>5)
return;
if(dictionary.contains(currWord))
System.out.println(currWord);
}
}
}
public static void main(String[] args)
{
board = createBoard();
ArrayList<String> validWords = RandomWordGame.solver(board);
for(String word : validWords)
System.out.println(word);
}
}
最佳答案
此代码有些有趣的地方。首先,您总是从求解器方法返回一个空的ArrayList,但这并不是杀死您的原因,因为您正在打印findWords中的每个结果。
问题是findWords方法只是从拼图的左上方连续添加字母。
//i=0 and j=0 means it will always start at the top left tile
for(i=0; i>=0 && i<board.length; i++)
{
for(j=0; j>=0; j++)
{
//currWord is never reset, so it just keeps getting longer
currWord += board[i][j];
if(currWord.length()>5)
return;
if(dictionary.contains(currWord))
System.out.println(currWord);
}
}
现在,您只会找到以所选图块开头的单词,其余字母的选择顺序与从左上角开始添加到拼图的顺序相同。
我建议您花些时间在铅笔和纸上,并深入了解二维数组索引与其在网格中的位置之间的关系。