我正在制作一个应用程序，它将找到所有可以用 4x4 网格（Boggle）上的相邻图块组成的单词。我已经到了可以输入一串字母的位置，该算法将在最佳时间内找到所有可用的单词。但是，我不仅需要知道有效的单词，还需要知道它们在棋盘上的方格位置。
这是主要的游戏类。该算法递归地从一个字母开始，然后检查是否存在以该字母及其邻居开头的单词。如果不存在单词，则路径被封锁，算法继续前进到下一个字母。如果存在带有该前缀的单词，则对该邻居执行相同的操作。
import java.io.IOException;
import java.util.ArrayList;
public class Game {
private final int boardSize = 4;
private BoardSquare[][] squares;
private ArrayList<String> validWords;
private static WordTree trie;
private static int counter = 0;
private DevRunner runner;
public Game(String letters) throws IOException{
validWords = new ArrayList<String>();
runner = new DevRunner();
trie = new WordTree();
squares = new BoardSquare[boardSize][boardSize];
for (int y=0; y<boardSize; y++){
for (int x=0; x<boardSize; x++){
squares[x][y] = new BoardSquare(letters.charAt(y*boardSize + x), y*boardSize + x);
}
}
for (int y=0; y<boardSize; y++){
for (int x=0; x<boardSize; x++){
for (int a=-1; a<2; a++){
for (int b=-1; b<2; b++){
if (a == 0 && b == 0) continue;
if (x+b < 0 || y+a < 0) continue;
if (x+b > boardSize-1 || y+a > boardSize-1) continue;
squares[x][y].addNeighbor(squares[x+b][y+a]);
}
}
}
}
getPossibleCombinations();
System.out.println(counter + " words found.");
}
public void getPossibleCombinations(){
for (int y=0; y<boardSize; y++){
for (int x=0; x<boardSize; x++){
doNeigh(squares[x][y], "", new ArrayList<Integer>());
}
}
}
public void doNeigh(BoardSquare square, String path, ArrayList<Integer> locations) {
square.setIsActive(true);
path += square.getData();
locations.add(square.getPosition());
if (trie.has(path) != 0){
for (BoardSquare neighbor : square.getNeighbors()){
if (!neighbor.getIsActive()){
doNeigh(neighbor, path, locations);
};
}
}
if (trie.has(path) == 1 && !validWords.contains(path)){
System.out.print(path + " is a word! (");
validWords.add(path);
for (int i : locations){
System.out.print(i + " -> ");
}
System.out.print("\n");
sendWord(path);
counter++;
}
square.setIsActive(false);
}
public void sendWord(String s){
}
public static void main(String[] args){
try {
long t1 = System.currentTimeMillis();
Game g = new Game("SIOZTRTEBAINERLA");
long t2 = System.currentTimeMillis();
System.out.println("The algorithm took " + Long.toString(t2-t1) + " milliseconds to complete.");
} catch (IOException e) {
e.printStackTrace();
}
}
}
实际输出：
SITAR is a word! (0 -> 1 -> 2 -> 4 -> 5 -> 8 -> 9 -> 5 -> 2 -> 6 -> 8 -> 10 -> 6 -> 7 -> 11 -> 13 -> 14 -> 15 ->
SIT is a word! (0 -> 1 -> 2 -> 4 -> 5 -> 8 -> 9 -> 5 -> 2 -> 6 -> 8 -> 10 -> 6 -> 7 -> 11 -> 13 -> 14 -> 15 -> 6 -> 8 -> 10 -> 12 -> 13 -> 8 -> 10 -> 5 -> 6 -> 7 -> 11 -> 14 -> 15 -> 12 -> 14 -> 14 ->
SIR is a word! (0 -> 1 -> 2 -> 4 -> 5 -> 8 -> 9 -> 5 -> 2 -> 6 -> 8 -> 10 -> 6 -> 7 -> 11 -> 13 -> 14 -> 15 -> 6 -> 8 -> 10 -> 12 -> 13 -> 8 -> 10 -> 5 -> 6 -> 7 -> 11 -> 14 -> 15 -> 12 -> 14 -> 14 -> 5 -> 2 -> 3 -> 6 -> 7 -> 4 -> 6 -> 8 -> 9 -> 10 -> 6 -> 7 -> 9 -> 11 -> 6 -> 7 -> 14 -> 15 -> 13 -> 14 -> 15 ->
预期输出：
SITAR is a word! (0 -> 1 -> 4 -> 9 -> 5 ->
SIT is a word! (0 -> 1 -> 4 ->
SIR is a word! (0 -> 1 -> 5 ->
...
我不明白为什么我可以使用 doNeigh() 方法并String path通过递归构建它，但是当我尝试以相同的方式构建正方形位置的数组列表时，它包括一堆不组成的正方形这个单词。
任何帮助表示赞赏，谢谢。