Question

我试图找出二维数组中的行程编码。我随机填充了我的尺寸x尺寸的板，0和1。然后我的程序运行曲折行走（起始位置=右上角）以读取该模式中的0和1。这对我有用，如下所示。但是，我需要帮助读取连续的0或1并计算它们每次出现的次数。这是一个例子：

/* n = 4 (i.e. 4 x 4 board)
   1010
   1010
   0001
   1000
 Run-length coding on the zigzag path: 
 (0,2)
 (1,1)
 (0,1)
 (1,2)
 (0,3)
 (1,2)
 (0,4)
 (1,1) */

这是我到目前为止所拥有的。

   void runLengthCoding()
{
    int flag = 1; // alternate between one and negative one depending on direction.

    //2*maxsize-1 is the number of segments.
    for(int i = 2 * maxSize - 1; i >= 0; i--)   //outer for loop goes through the segments. #of segments
    {    
        //determine the starting element.
        int r, c; 

        if (flag == -1)// if(i%2==1)
        {
            if(i > maxSize)
                r = i - maxSize - 1;
            else 
                r = maxSize - 1; 
        }
        else
        {
            if(i >= maxSize)
                r = 0;
            else 
                r = maxSize - i; 
        }
        c = i - maxSize + r;

        while(r >=0 && r <= maxSize -1 && c >= 0 && c <= maxSize - 1)
        {
            System.out.print(A[r][c] + " ");

            int cnt = 0;

            if (A[r][c] == 0)
            if(flag == 1)
            {
                r++;
                c++;
            }
            else
            {
                r--;
                c--;
            }
        }
        //change the moving direction
        flag = -flag;

        System.out.println();

    }
}

// print the run-length coding result, i.e., content of rlc[][] 
void printCodingResult()
{
    System.out.println("Run-length coding on the zigzag path: ");

}

为了执行rlc [] []，我想每次A [r] [c]从0变为1或反之，记录并重置计数。但是我该如何融入呢。 rlc [] []将如何记住这一点？从示例中可以看出，rlc [] []以两列的格式显示（一个用于0或1，第二个用于计数）。欣赏任何想法。感谢。

Answer 1

void runLengthCoding()
{
    int flag = -1;// 1: up      -1: down
    System.out.println("runLengthCoding: ");
    for (int i=0; i<=(maxSize-1)*2; i++)//i=pass variable
    {
        int r, c, t;
        if (flag == 1)
        {
            c = Math.min(maxSize-1, (2*(maxSize-1))-i);
            t=(maxSize-1)-c;
            r = i-t;
        }
        else
        {
            r = Math.max(0, i-(maxSize-1));
            t=i-r;
            c = (maxSize-1)-t;
        }


        while (r >=0 && c >=0 && r < maxSize && c < maxSize)
        {
            System.out.print(A[r][c] + " ");
            r -= flag;
            c -= flag;
        }

        flag = -flag;//changes flag in order to change direction of scan
        System.out.println();
    }
}

运行长度编码二维数组

1 个答案: