java 我正在尝试解决'15难题',但我得到'OutOfMemoryError' [已关闭]

gmxoilav  于 2023-02-02  发布在  Java
关注(0)|答案(1)|浏览(79)

很难判断此处所问的问题。此问题模棱两可、模糊不清、不完整、过于宽泛或过于修辞,无法以其当前形式合理地回答。若要获得澄清此问题以便重新打开的帮助,请单击visit the help center
12年前就关闭了。
有没有一种方法,我可以优化这个代码,以不运行内存不足?

import java.util.HashMap;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Random;
import java.util.Stack;

public class TilePuzzle {

    private final static byte ROWS = 4;
    private final static byte COLUMNS = 4;
    private static String SOLUTION = "123456789ABCDEF0";
    private static byte RADIX = 16;

    private char[][] board = new char[ROWS][COLUMNS];
    private byte x; // Row of the space ('0')
    private byte y; // Column of the space ('0') private String representation;
    private boolean change = false; // Has the board changed after the last call to toString?

    private TilePuzzle() {
        this(SOLUTION);
        int times = 1000;
        Random rnd = new Random();
        while(times-- > 0) {
            try {
                move((byte)rnd.nextInt(4));
            }
            catch(RuntimeException e) {
            }
        }
        this.representation = asString();
    }

    public TilePuzzle(String representation) {
        this.representation = representation;
        final byte SIZE = (byte)SOLUTION.length();
        if (representation.length() != SIZE) {
            throw new IllegalArgumentException("The board must have " + SIZE + "numbers.");
        }

        boolean[] used = new boolean[SIZE];
        byte idx = 0;
        for (byte i = 0; i < ROWS; ++i) {
            for (byte j = 0; j < COLUMNS; ++j) {
                char digit = representation.charAt(idx++);
                byte number = (byte)Character.digit(digit, RADIX);
                if (number < 0 || number >= SIZE) {
                    throw new IllegalArgumentException("The character " + digit + " is not valid.");
                } else if(used[number]) {
                    throw new IllegalArgumentException("The character " + digit + " is repeated.");
                }
                used[number] = true;
                board[i][j] = digit;
                if (digit == '0') {
                    x = i;
                    y = j;
                }
            }
        }
    }

    /**
     * Swap position of the space ('0') with the number that's up to it.
     */
    public void moveUp() {
        try {
            move((byte)(x - 1), y);
        } catch(IllegalArgumentException e) {
            throw new RuntimeException("Move prohibited " + e.getMessage());
        }
    }

    /**
     * Swap position of the space ('0') with the number that's down to it.
     */
    public void moveDown() {
        try {
            move((byte)(x + 1), y);
        } catch(IllegalArgumentException e) {
            throw new RuntimeException("Move prohibited " + e.getMessage());
        }
    }

    /**
     * Swap position of the space ('0') with the number that's left to it.
     */
    public void moveLeft() {
        try {
            move(x, (byte)(y - 1));
        } catch(IllegalArgumentException e) {
            throw new RuntimeException("Move prohibited " + e.getMessage());
        }
    }

    /**
     * Swap position of the space ('0') with the number that's right to it.
     */
    public void moveRight() {
        try {
            move(x, (byte)(y + 1));
        } catch(IllegalArgumentException e) {
            throw new RuntimeException("Move prohibited " + e.getMessage());
        }
    }

    private void move(byte movement) {
        switch(movement) {
        case 0: moveUp(); break;
        case 1: moveRight(); break;
        case 2: moveDown(); break;
        case 3: moveLeft(); break;
        }
    }

    private boolean areValidCoordinates(byte x, byte y) {
        return (x >= 0 && x < ROWS && y >= 0 && y < COLUMNS);
    }

    private void move(byte nx, byte ny) {
        if (!areValidCoordinates(nx, ny)) {
            throw new IllegalArgumentException("(" + nx + ", " + ny + ")");
        }
        board[x][y] = board[nx][ny];
        board[nx][ny] = '0';
        x = nx;
        y = ny;
        change = true;
    }

    public String printableString() {
        StringBuilder sb = new StringBuilder();
        for (byte i = 0; i < ROWS; ++i) {
            for (byte j = 0; j < COLUMNS; ++j) {
                sb.append(board[i][j] + " ");
            }
            sb.append("\r\n");
        }
        return sb.toString();
    }

    private String asString() {
        StringBuilder sb = new StringBuilder();
        for (byte i = 0; i < ROWS; ++i) {
            for (byte j = 0; j < COLUMNS; ++j) {
                sb.append(board[i][j]);
            }
        }
        return sb.toString();
    }

    public String toString() {
        if (change) {
            representation = asString();
        }
        return representation;
    }

    private static byte[] whereShouldItBe(char digit) {
        byte idx = (byte)SOLUTION.indexOf(digit);
        return new byte[] { (byte)(idx / ROWS), (byte)(idx % ROWS) };
    }

    private static byte manhattanDistance(byte x, byte y, byte x2, byte y2) {
        byte dx = (byte)Math.abs(x - x2);
        byte dy = (byte)Math.abs(y - y2);
        return (byte)(dx + dy);
    }

    private byte heuristic() {
        byte total = 0;
        for (byte i = 0; i < ROWS; ++i) {
            for (byte j = 0; j < COLUMNS; ++j) {
                char digit = board[i][j];
                byte[] coordenates = whereShouldItBe(digit);
                byte distance = manhattanDistance(i, j, coordenates[0], coordenates[1]);
                total += distance;
            }
        }
        return total;
    }

    private class Node implements Comparable<Node> {
        private String puzzle;
        private byte moves; // Number of moves from original configuration
        private byte value; // The value of the heuristic for this configuration.
        public Node(String puzzle, byte moves, byte value) {
            this.puzzle = puzzle;
            this.moves = moves;
            this.value = value;
        }
        @Override
        public int compareTo(Node o) {
            return (value + moves) - (o.value + o.moves);
        }
    }

    private void print(Map<String, String> antecessor) {
        Stack toPrint = new Stack();
        toPrint.add(SOLUTION);
        String before = antecessor.get(SOLUTION);
        while (!before.equals("")) {
            toPrint.add(before);
            before = antecessor.get(before);
        }
        while (!toPrint.isEmpty()) {
            System.out.println(new TilePuzzle(toPrint.pop()).printableString());
        }
    }

    private byte solve() {
        if(toString().equals(SOLUTION)) {
            return 0;
        }

        PriorityQueue<Node> toProcess = new PriorityQueue();
        Node initial = new Node(toString(), (byte)0, heuristic());
        toProcess.add(initial);

        Map<String, String> antecessor = new HashMap<String, String>();
        antecessor.put(toString(), "");

        while(!toProcess.isEmpty()) {
            Node actual = toProcess.poll();
            for (byte i = 0; i < 4; ++i) {
                TilePuzzle t = new TilePuzzle(actual.puzzle);
                try {
                    t.move(i);
                } catch(RuntimeException e) {
                    continue;
                }
                if (t.toString().equals(SOLUTION)) {
                    antecessor.put(SOLUTION, actual.puzzle);
                    print(antecessor);
                    return (byte)(actual.moves + 1);
                } else if (!antecessor.containsKey(t.toString())) {
                    byte v = t.heuristic();
                    Node neighbor = new Node(t.toString(), (byte)(actual.moves + 1), v);
                    toProcess.add(neighbor);
                    antecessor.put(t.toString(), actual.puzzle);
                }
            }
        }
        return -1;
    }

    public static void main(String... args) {
        TilePuzzle puzzle = new TilePuzzle();
        System.out.println(puzzle.solve());
    }
}
hpcdzsge

hpcdzsge1#

∮问题是
根本原因是您在toProcess队列和antecessorMap中创建和存储了大量String对象。
看看你的算法,看看你是否真的需要存储超过200万个节点和500万个字符串。
∮调查
这很难发现,因为程序很复杂。实际上,我甚至没有试图理解所有的代码。相反,我使用了**VisualVM**-一个Java分析器、采样器和CPU/内存使用监视器。
我启动了它:

然后看了一下内存的使用情况,我注意到的第一件事是(显而易见的)你正在创建大量的对象。
这是应用程序的截图:

正如您所看到的,使用的内存量是巨大的,在短短的40秒内,就消耗了2GB,并且填满了整个堆。

死胡同

我最初认为这个问题与Node类有关,因为即使它实现了Comparable,但它没有实现equals

public boolean equals( Object o ) {
    if( o instanceof Node ) {
        Node other = ( Node ) o;
        return this.value == other.value && this.moves == other.moves;
    }
    return false;
}

但这不是问题所在。
实际的问题原来是上面说的那个。

变通方案

如前所述,真正的解决方案是重新思考你的算法,在此期间,无论做什么,都只会拖延问题的解决。
但是变通方法可能是有用的,一个是重用您生成的字符串,您非常密集地使用TilePuzzle.toString()方法;这经常导致创建重复的字符串。
因为你要生成字符串排列,所以你可以在几秒钟内创建许多12345ABCD字符串,如果它们是同一个字符串,那么用相同的值创建数百万个示例是没有意义的。

    • String.intern()**方法允许字符串被重用,文档中说:

返回字符串对象的规范表示形式。
字符串池最初为空,由String类私有地维护。
调用intern方法时,如果池中已经包含了等于该String对象的字符串(由equals()方法确定),则返回池中的字符串;否则,将该String对象添加到池中,并返回对该String对象的引用。
对于常规应用程序,使用String.intern()可能不是一个好主意,因为它不允许示例被GC回收,但在本例中,由于您在Map和Queue中保存了引用,因此它是有意义的。
所以做出这个改变:

public String toString() {
    if (change) {
        representation = asString();
    }
    return representation.intern(); // <-- Use intern
}

差不多解决了记忆问题。
这是修改后的截图:

现在,即使在几分钟之后,堆使用量也不会达到100 MB。

额外备注

备注#1

您使用异常来验证移动是否有效,这是正常的;但当你抓住他们时,你却忽略了他们

try {
    t.move(i);
} catch(RuntimeException e) {
    continue;
}

如果你根本不使用它们,你可以通过不创建异常来节省大量的计算,否则你会创建数百万个未使用的异常。
进行此更改:

if (!areValidCoordinates(nx, ny)) {
    // REMOVE THIS LINE:
    // throw new IllegalArgumentException("(" + nx + ", " + ny + ")");

    // ADD THIS LINE:
    return;
}

并使用验证代替:

// REMOVE THESE LINES:
// try {
//     t.move(i);
// } catch(RuntimeException e) {
//     continue;
// }

// ADD THESE LINES:
if(t.isValidMovement(i)){
    t.move(i);
} else {
    continue;
}

备注2

你要为每个新的TilePuzzle示例创建一个新的Random对象,如果你在整个程序中只使用一个对象会更好,毕竟你只使用了一个线程。

备注3

这个解决方法解决了堆内存问题,但是创建了另一个涉及PermGen的问题。我只是增加了PermGen的大小,如下所示:

java -Xmx1g -Xms1g -XX:MaxPermSize=1g TilePuzzle

备注4

输出有时为49,有时为50。矩阵的打印方式如下:
五十次

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