在使用马尔可夫链时获取ArrayIndexOutOfBoundsException

Question

我有一个数组来存储一组坐标，用于绘制一段线。下面是一些示例坐标

double[][] plotMatrix = {{10,20},{55,80},
                         {120,40},{225,30},
                         {327.5,100},
                         {427.5,30},
                         {529,60}};

下一步是创建一个二维的马尔可夫矩阵。

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首先，我计算左列中的点后面跟着顶列中的点的次数。因为我想要一条线，所以每个点后面跟着一个点。这意味着如果我们有{10,20}作为输入，{55,80}是下一个点的概率是100%。

我对所有这些都不是很确定，所以请纠正我！

这是我的矩阵

double[][] markovMatrix = { {0.0,1.0,0.0,0.0,0.0,0.0,0.0},
                                    {0.0,0.0,1.0,0.0,0.0,0.0,0.0},
                                    {0.0,0.0,0.0,1.0,0.0,0.0,0.0},
                                    {0.0,0.0,0.0,0.0,1.0,0.0,0.0},
                                    {0.0,0.0,0.0,0.0,0.0,1.0,0.0},
                                    {0.0,0.0,0.0,0.0,0.0,0.0,1.0},
                                    {0.0,0.0,0.0,0.0,0.0,0.0,0.0}};

我的算法是：

    int seed = 0;
    int output = 0;

    for(int i = 0; i < 40;i++){
        double choice = r.nextDouble();

        double currentSum = 0.0;

        for(;output < markovMatrix.length;output++){

            currentSum += markovMatrix[seed][output];

            if(choice <= currentSum){
                break;
            }
        }

        System.out.println(output);
        polygon.lineTo(plotMatrix[output][0], plotMatrix[output][1]);

        seed = output;

        output = 0;
    }

我的问题是，当我试图同时访问plotMatrix和markovMatrix时，我得到了一个ArrayOutOfBoundsException:7。但是，在每个循环结束时，输出被设置为0。有什么办法可以解决这个问题吗？

Answer 1

我不太确定这是不是正确的答案，

但是for(;output < markovMatrix.length;output++)将从0步进到7，而您在markovMatrix中只有0到6个条目。

使用for(;output < markovMatrix.length-1;output++)通过从1步进到6步来修复ArrayIndexOutOfBoundsException。

然而，我怀疑你真的想从0到6，这就是你的问题所在。

Answer 2

当使用内部循环output=7完成循环时，它是数组的长度。你应该跳过最后一次迭代，因为你的数组索引是从0到6。