java - 0,1矩阵中的递归路径查找(并保存所有可能的路径)java

Question

我正在尝试编写一个递归方法，该方法将找到一条路径，而无需回溯到 int 矩阵中包含值 0,1 的位置。 0 可以踩，1 不行。我还限制了一条超过 50 次移动的路径。

Location 是一个具有行值和列值 (x,y) 的对象。locationEquals 是一个函数，如果两个位置相同则返回 true，否则返回 false。 map 变量是我试图在其中查找路径的矩阵。

private static List<List<Location>> options = new ArrayList<List<Location>>();
public static void PathFind(List<Location> path)
{
 Location current = path.get(path.size() - 1);
    boolean done = false;
    if(locationEquals(current,new Location(24,38)))
    {
        options.add(path);
        return;
    }
    if(path.size() > 50) done = true;
    if(!done)
    {
    try
    {
    if(map[current.row][current.col + 1] == 0)
    {
    if(!path.contains(new Location(current.row, current.col + 1)))
        {
            List<Location> temp = path;
            temp.add(new Location(current.row, current.col + 1));
            PathFind(temp);
        }
    }
    }
    catch (Exception e){}
            try
    {
    if(map[current.row - 1][current.col] == 0)
    {
        if(!path.contains(new Location(current.row - 1, current.col)))
        {
            List<Location> temp = path;
            temp.add(new Location(current.row - 1, current.col));
            PathFind(temp);
        }
    }
    }
    catch (Exception e){}
    try
    {
    if(map[current.row][current.col - 1] == 0)
    {
        if(!path.contains(new Location(current.row, current.col - 1)))
        {
            List<Location> temp = path;
            temp.add(new Location(current.row, current.col - 1));
            PathFind(temp);
        }
    }
    }
    catch (Exception e){}
    try
    {
    if(map[current.row + 1][current.col] == 0)
    {
        if(!path.contains(new Location(current.row + 1, current.col)))
        {
            List<Location> temp = path;
            temp.add(new Location(current.row + 1, current.col));
            PathFind(temp);
        }
    }
    }
    catch (Exception e){}
    }

执行以下代码后'options'为空，这意味着它没有找到方法。但这个矩阵中肯定有一种方法，所以这是我的代码中我找不到的错误。

Answer 1

问题是您每次进入递归的下一步时都不会创建一个新列表(您的 temp 变量并不是真正临时的，因为它只是对您的路径的引用，而不是它的副本)。

为了解决这个问题，我替换了 List<Location> temp = path;与 List<Location> temp = new ArrayList<>(path);

所以代码是:

private static List<List<Location>> options = new ArrayList<List<Location>>();
public static void PathFind(List<Location> path) {
    Location current = path.get(path.size() - 1);
    boolean done = false;
    if (locationEquals(current, new Location(24, 38))) {
        options.add(path);
        return;
    }
    if (path.size() > 50) done = true;
    if (!done) {
        try {
            if (map[current.row][current.col + 1] == 0) {
                if (!path.contains(new Location(current.row, current.col + 1))) {
                    List<Location> temp = new ArrayList<>(path);
                    temp.add(new Location(current.row, current.col + 1));
                    PathFind(temp);
                }
            }
        } catch (Exception e) {
        }
        try {
            if (map[current.row - 1][current.col] == 0) {
                if (!path.contains(new Location(current.row - 1, current.col))) {
                    List<Location> temp = new ArrayList<>(path);
                    temp.add(new Location(current.row - 1, current.col));
                    PathFind(temp);
                }
            }
        } catch (Exception e) {
        }
        try {
            if (map[current.row][current.col - 1] == 0) {
                if (!path.contains(new Location(current.row, current.col - 1))) {
                    List<Location> temp = new ArrayList<>(path);
                    temp.add(new Location(current.row, current.col - 1));
                    PathFind(temp);
                }
            }
        } catch (Exception e) {
        }
        try {
            if (map[current.row + 1][current.col] == 0) {
                if (!path.contains(new Location(current.row + 1, current.col))) {
                    List<Location> temp = new ArrayList<>(path);
                    temp.add(new Location(current.row + 1, current.col));
                    PathFind(temp);
                }
            }
        } catch (Exception e) {
        }
    }
}