Java编写了2d数组的搜索算法，该算法通过在任何排序块中移动1步来查找所有路径

Question

我正在调试我的类。这个类正在通过一个boggle board并找到每一条可用的路径。路径包括水平、垂直和对角线。你不能重复一步。我现在有很多打印语句试图找出我的错误，但我似乎找不到。我把这些留在家里了。我想这与我的当前路径[]有关。任何提示都将不胜感激

public class FindWords {
String [][] board;
final int LAST_LETTER = 2;
final int BEEN_THERE = 1;
final int AVAILABLE = 0;
BoggleDictionary Dictionary;

private StackADT<SearchWords> searchStack = new ArrayStack<SearchWords>();
public StackADT<SearchWords> foundStack = new ArrayStack<SearchWords>();

public FindWords (String [][] board) throws Exception {
    this.board = board;
    this.Dictionary = new BoggleDictionary();
    //this.foundStack = null;
}

public void startSearch(){


    for (int i = 0; i < board.length; i++){
        for (int j = 0; j < board[0].length; j++){
            System.out.println(board[i][j]);
            String firstLetter = "";
            int [][] pathBoard = makeBlankBoard();

            pathBoard[i][j] = LAST_LETTER;

            System.out.println(">>" + Arrays.deepToString(pathBoard));
            firstLetter = board[i][j];
            System.out.println("first letters : " + firstLetter + " <====");
            SearchWords thisPath = new SearchWords(pathBoard, firstLetter);
            searchStack.push(thisPath);

        }
    }
    Search();
}

private boolean Search(){
    while (!searchStack.isEmpty())
    {

        SearchWords searchObj = searchStack.pop();
        int [][] currentPath = searchObj.getPath();

        int [][] array = new int [currentPath.length][currentPath[0].length];
        for (int i = 0; i <currentPath.length; i++){
            for (int j = 0; j <currentPath[0].length; j++){
                array[i][j] = currentPath[i][j];
            }
        }
        String makingString = searchObj.getString();


        if (makingString.length() > 2){
            if (Dictionary.contains(makingString)){
                foundStack.push(searchObj);
            }
        }

        for (int i = 0; i < array.length; i ++){
            for (int j = 0; j < array[0].length; j++){
                if (array[i][j] == LAST_LETTER){ //finding the last position in the string
                    int x = i;
                    int y = j;
                    //array[i][j] = BEEN_THERE;
                    pushPosition(searchObj, x+1, y+1, i, j); //lower left then going counter-clockwise
                    pushPosition(searchObj, x, y+1, i, j);
                    pushPosition(searchObj, x-1, y+1, i, j);
                    pushPosition(searchObj, x-1, y, i, j);
                    pushPosition(searchObj, x-1, y-1, i, j);
                    pushPosition(searchObj, x, y-1, i, j);
                    pushPosition(searchObj, x+1, y-1, i, j);
                    pushPosition(searchObj, x+1, y, i, j);

                }
            }
        }
    }

    return true;

}



private void pushPosition (SearchWords obj, int x, int y, int i, int j){
    int [][] currentPath = obj.getPath();
    String makingString = obj.getString();
    System.out.println("In push method: " + Arrays.deepToString(currentPath) +x+y+i+j);

    if (validPosition(x, y, currentPath)){

        currentPath[x][y] = LAST_LETTER;
        currentPath[i][j] = BEEN_THERE;

        System.out.println("after valid ch: "+ Arrays.deepToString(currentPath));

        makingString = makingString + board[x][y];
        SearchWords newPath = new SearchWords(currentPath, makingString);

        System.out.println("is string getting longer: " + makingString);
        System.out.println("Stack size: " + searchStack.size());


        System.out.println("pushing back on stack"+ Arrays.deepToString(currentPath));
        searchStack.push(newPath);
    }
}

private boolean validPosition (int x, int y, int [][] path){
    boolean result = false;

    if (x >= 0 && x < board.length && y >= 0 && y < board[x].length){
        if (path[x][y] == AVAILABLE){
            System.out.println("Checked position : " + x + y);
            result = true;
        }
    }
    return result;
}
private int [][] makeBlankBoard(){
    int row = board.length;
    int col = board[0].length;
    int [][] blankBoard = new int [row][col];

    for (int i = 0; i < board.length; i++){
        for (int j = 0; j < board[0].length; j++){
            blankBoard[i][j] = AVAILABLE;

        }
    }
    return blankBoard;
}

}

更新了有效的类。需要在调用push方法之前创建pathBoard的新副本。谢谢你，基尔斯克里恩

private void Search(){
    while (!searchStack.isEmpty())
    { 
        System.out.println("stack size in search: " + searchStack.size());

        SearchWords searchObj = searchStack.pop();
        int lastLetterRow = searchObj.getRow();
        int lastLetterCol = searchObj.getCol();
        String stringSoFar = searchObj.getString();
        int [][] pathBoard = searchObj.getPath();

        System.out.println ("row then Col: " + lastLetterRow + lastLetterCol);
        System.out.println ("string so far: " + stringSoFar);
        System.out.println("Path board so far in search: " + Arrays.deepToString(pathBoard)+ "\n");

        if (stringSoFar.length() > 2){
            if (Dictionary.contains(stringSoFar)){
            foundStack.push(searchObj);
            System.out.println("Found word!! " + stringSoFar);
        }
            }
        System.out.println("made it past if dict");

        //lower left then going counter-clockwise
        pushPosition (pathBoard, stringSoFar, lastLetterRow+1, lastLetterCol+1, lastLetterRow, lastLetterCol);
        pathBoard = makeCopy(pathBoard);
        pushPosition (pathBoard, stringSoFar, lastLetterRow, lastLetterCol+1, lastLetterRow, lastLetterCol);
        pathBoard = makeCopy(pathBoard);
        pushPosition(pathBoard, stringSoFar, lastLetterRow, lastLetterCol+1, lastLetterRow, lastLetterCol);
        pathBoard = makeCopy(pathBoard);
        pushPosition(pathBoard, stringSoFar, lastLetterRow-1, lastLetterCol+1, lastLetterRow, lastLetterCol);
        pathBoard = makeCopy(pathBoard);
        pushPosition(pathBoard, stringSoFar, lastLetterRow-1, lastLetterCol, lastLetterRow, lastLetterCol);
        pathBoard = makeCopy(pathBoard);
        pushPosition(pathBoard, stringSoFar, lastLetterRow-1, lastLetterCol-1, lastLetterRow, lastLetterCol);
        pathBoard = makeCopy(pathBoard);
        pushPosition(pathBoard, stringSoFar, lastLetterRow, lastLetterCol-1, lastLetterRow, lastLetterCol);
        pathBoard = makeCopy(pathBoard);
        pushPosition(pathBoard, stringSoFar, lastLetterRow+1, lastLetterCol-1, lastLetterRow, lastLetterCol);
        pathBoard = makeCopy(pathBoard);
        pushPosition(pathBoard, stringSoFar, lastLetterRow+1, lastLetterCol, lastLetterRow, lastLetterCol);

    }
    System.out.println("FOUND WORDS:");
    while(!foundStack.isEmpty()){
        SearchWords foundWords = foundStack.pop();
        System.out.println(foundWords.getString());
    }
}   

Answer 1

如注释中所述，您遇到的具体问题似乎是多个路径共享同一个数组，该数组描述了您在其中搜索的路径。这会产生bug，其中它们的搜索历史记录会相互覆盖，在currentPath数组中留下一组不是特别有用的信息

为每个被推送的路径执行数组复制可能会解决这个问题（代码中可能还有其他问题，但我没有发现任何问题）

作为一个整体，我不需要每次都使用不同的方法来存储阵列。相反，您可以按以下方式跟踪每条路径：

• 路径中最后一个字母的位置（对于在路径中执行下一步非常有用，如在算法中）
• 指向该点的上一条路径（将是同一类型的对象）

代码示例如下所示（请注意，我还没有对此进行测试，因为它只是一个示例）：

class SearchPath {
    int x, y;
    String string;
    SearchPath prior;

    // x,y are most recent position; prior is the path up to this point, or null
    public SearchPath(int x, int y, String board, SearchPath prior) {
        this.x = x;
        this.y = y;
        this.string = (prior != null ? prior.string : "") + board[x][y];
        this.prior = prior;
    }

    // To check if an x,y location collides with this path
    public contains(int x, int y) {
        if (this.x == x && this.y == y) {
            return true;
        } else if (prior == null) {
            return false;
        } else {
            return prior.contains(x,y);
        }
    }
}

在初始循环（startSearch）中，您将推送没有优先路径的对象（new SearchPath(i, j, board, null)）。在search循环中，您将弹出其中一个，然后在检查新的x/y坐标是否未通过contains方法使用后，按下new SearchPath(searchPath.x - 1, searchPath.y, board, searchPath)。（请注意，使用堆栈存储搜索路径可能也不是必需的，因为您只需进行递归搜索调用即可获得相同的搜索模式。）