Java 随机移动'；a'；在二维阵列中，在跟踪通过每个点的时间数的同时，将其移动到指定位置

我有一个2D数组（比如double[10][10]），其中包含一些1.0和10.0，其余的都是0.0。我试图通过循环这个数组来找到1.0（起点），从那里随机地（使用random.nextInt（4））向上、向下、向左或向右移动它，直到它达到10.0。我创建了一个emptyArray来跟踪它经过每个点的时间（或者至少我认为我做到了）。编译时没有出现任何问题，但在尝试将其显示到框架中时没有得到任何结果。你知道我哪里做错了或是失踪了吗

{0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0}
{0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0}
{0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0}
{0.0,0.0,0.0,0.0,1.0,1.0,0.0,0.0,0.0,0.0}
{0.0,0.0,0.0,0.0,1.0,1.0,0.0,0.0,0.0,0.0}
{0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0}
{0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0}
{0.0,10.0,10.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0}
{0.0,10.0,10.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0}
{0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0}

二维阵列的示例

    double[][] getPath(double[][] dataIn) {
    double[][] emptyArray = new double[dataIn.length][dataIn[0].length];
    double[][] drunkLoc = new double[dataIn.length][dataIn[0].length];
    for (int i = 0; i < dataIn.length; i++) {
        for (int j = 0; j < dataIn[i].length; j++) {
            if (dataIn[i][j] == 1.0) {

                double drunkHome = 10.0;
                drunkLoc[i][j] = dataIn[i][j];
                do {
                    int dir = getDirection();
                    switch(dir) {
                        case 0: 
                            if ((i > 0) && (j > 0)) {
                                drunkLoc[i][j] = drunkLoc[i-1][j];
                                double value = emptyArray[i][j];
                                emptyArray[i][j] = value + 1;
                                emptyArray[i][j] = (255<<24)  | (255<<16) | (255<<8) | 255;
                            } else {
                                break;
                            }
                            break;
                        case 1: 
                            if ((i > 0) && (j > 0)) {
                                drunkLoc[i][j] = drunkLoc[i][j-1];
                                double value = emptyArray[i][j];
                                emptyArray[i][j] = value + 1;
                                emptyArray[i][j] = (255<<24)  | (255<<16) | (255<<8) | 255;
                            } else {
                                break;
                            }
                            break;
                        case 2: 
                            if ((i > 0) && (j > 0)) {
                                drunkLoc[i][j] = drunkLoc[i+1][j];
                                double value = emptyArray[i][j];
                                emptyArray[i][j] = value + 1;
                                emptyArray[i][j] = (255<<24)  | (255<<16) | (255<<8) | 255;
                            } else {
                                break;
                            }
                            break;
                        case 3: 
                            if ((i > 0) && (j > 0)) {
                                drunkLoc[i][j] = drunkLoc[i][j+1];
                                double value = emptyArray[i][j];
                                emptyArray[i][j] = value + 1;
                                emptyArray[i][j] = (255<<24)  | (255<<16) | (255<<8) | 255;
                            } else {
                                break;
                            }
                            break;
                        default:
                    }
                } while (drunkLoc[i][j] != drunkHome);
            }
        }
    }
    return emptyArray;
}

double[]getPath（double[]dataIn）{
double[]emptyArray=new double[dataIn.length][dataIn[0.length]；
double[]loc=新的double[dataIn.length][dataIn[0.length]；
for（int i=0；i0）和&（j>0））{
醉鬼[i][j]=醉鬼[i-1][j]；
双值=空数组[i][j]；
空数组[i][j]=值+1；
emptyArray[i][j]=（255好的，供您学习，以下是一些示例代码，它将为您提供您正在寻找的答案（从（XPO，YPO）到（destX，destY）在一个尺寸为PendDimension x PendDimension的“板”上的移动次数：

        int penDimension = 10;
        int destX = 2;
        int destY = 2;
        int xpos = 5;
        int ypos = 5;

        // Add this to keep track of no moves through each square
        int[][] moveCounts = new int[penDimension][penDimension];

        Random r = new SecureRandom();
        long noMoves = 0;
        while (xpos != destX || ypos != destY) {
            switch (r.nextInt(4)) {
            case 0 : xpos++; break;
            case 1 : xpos--; break;
            case 2 : ypos++; break;
            case 3 : ypos--; break;
            }
            if (xpos < 0) xpos = 0;
            if (ypos < 0) ypos = 0;
            if (xpos > penDimension) xpos = penDimension;
            if (ypos > penDimension) ypos = penDimension;
            noMoves++;

            // Add this to keep track of no moves through each square
            moveCounts[ypos][xpos]++;

        }
        System.out.println("Number of moves: " + noMoves);

我只是像上面那样写，因为我觉得这样更容易理解

您也可以编写“new-SecureRandom（）”，而不是编写“new-Random（）”，这可能是您所学的内容。但是SecureRandom的质量要高得多（但速度较慢）随机数生成器。通常，在编写重复生成大量随机数的“模拟”时，最好避免使用标准的随机类，并使用更高质量的生成器。
我还不太清楚您在这里实际要做什么，但您是否在每次迭代时都登录了数组ion并将输出与预期结果进行比较？顺便说一句，您的条件“（i>0）和（j>0）”对于dirs 1-3是错误的。加上带有移位的奇异行（为什么不写“-1”？！？）将始终覆盖“值+1”我想这不是你想要的。嗨@NeilCoffey，我正在尝试获取数组的返回值，其中包含从“1.0”到“10.0”的随机过程中“a”经过的时间数。我是java新手，所以我不太明白你所说的“在每次迭代中记录数组的if”是什么意思，你能解释一下吗。谢谢。（I>0）&&（j>0）4种可能情况中的每种情况的一部分都是为了防止数组中的“a”用完。这是正确的方法吗？要注销数组的内容，请执行System.out.println（Arrays.deepToString（array））；在使用“while”命令行之前，在每个数组上调用此命令，看看它是什么样子。很好，我会尝试一下，非常感谢@Neil Coffey。但我很好奇，如果我使用上面的方法，是否有一种方法可以计算“a”通过每个点（而不是目的地）的时间数？这意味着我必须记录麻木时间“a”在到达目的地之前经过每个空间。有什么建议吗？我如何获得数组中指定单元格的x，y？例如，如果我想获得第一个“1.0”的x，y，我该怎么做？我想使用For循环遍历行和列，设置一个if条件来匹配x，y='1.0'。然后我被卡住了，如何返回x和y，但不返回x，y中的值“1.0”？如果确实需要，请创建一个数组，移动位置后，在该位置增加值。我将更新代码。
xpos = Math.max(0, xpos - 1);