Question

我有一个数组来存储一组用于绘制一条线的坐标。所以这里有一些示例坐标

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。

enter image description here

首先，我计算左列中的点后跟顶部列中的点的时间。因为我想要一条线，每个点后跟另一个单点。这意味着如果我们输入{10,20}，{55,80}作为下一个点的可行性是100％。

我不是很确定这一切所以请纠正我！

所以这是我的矩阵

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：

    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，当我尝试访问plotMatrix和markovMatrix时，我得到ArrayOutOfBoundsException:7。但是，在每个循环结束时输出设置为0。任何想法如何解决这个问题？

Answer 1

我不太确定它的答案是否正确，

但是（;输出＆lt; markovMatrix.length;输出++）将从0步进到7，而在markovMatrix中只有0到6个条目。

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

但是我怀疑你真的想从0步到6步。这就是你的问题。

Answer 2

何时完成内循环输出= 7循环，这是数组的长度。您应该跳过最后一次迭代，因为您的数组索引是从0到6。

使用markov链时获取ArrayIndexOutOfBoundsException

2 个答案: