以蛇状图案遍历网格

问题描述

我有一个由直角坐标系上四个角定义的矩形。
矩形由大小相同的单元格组成。

I have a rectangle defined by its four corners on a Cartesian coordinate system.The rectangle consist of identically sized cells.

给出一个角单元作为起点和方向（沿x轴或y轴）如何以蛇状图案遍历此网格？

Given one of the corner cells as a starting point and a direction (either along x or y axis) how can I traverse this grid in snake like pattern?

例如，此处（x4，y4）是起始单元格，方向定义为沿x轴。蓝线描述了所需的路径。

For example, here (x4, y4) is the starting cell and direction is defined to be along the x axis. Blue line depicts the desired path.

谢谢！

推荐答案

尝试4对点，以找到定义该框的相对点。然后，您将获得一个起点和一个对点。在您的情况下，followAlongX将作为输入，因此不需要测试方向，但是我添加了经典的决策文本。

Try the 4 pair of points to find the opposite point that defines the box. Then you have a Start and an Opposite pair of points. In your case, followAlongX will be an input so it's not needed to test the direction but I added a classic decision text otherwise.

最后，尝试类似（Java ，用记事本编写，未经测试）：

In the end, try something like (Java, written in Notepad, not tested):

int stepY, stepX, yMin, yMax, yOpposite, yStart, xMin, xMax, xOpposite, xStart;
if (yOpposite > yStart) {
    stepY = 1;
    yMin = yStart;
    yMax = yOpposite;
}
else {
    stepY = -1;
    yMax = yStart;
    yMin = yOpposite;
}

if (xOpposite > xStart) {
    stepX = -1;
    xMin = xStart;
    xMax = xOpposite;
}
else {
    stepX = 1;
    xMin = xOpposite;
    xMax = xStart;
}

// boolean followAlongX = false;
// if (xMax-xMin>yMax-yMin) {
//     loopOnX = true;
// }

List<Points> path = new ArrayList<>();
if (followAlongX) {
    for (int i=yMin; i!=yMax; i+=stepY) {
        for (int j=xmin; j!=xmax; j+=stepX) {
            path.add(new Point(i,j));
        }
        stepX = -stepX;
        int temp = xMin;
        xMin = xMax;
        xMax = temp;
    }
}
else {
    for (int j=xmin; j!=xmax; j+=stepX) {
        for (int i=yMin; i!=yMax; i+=stepY) {
            path.add(new Point(i,j));
        }
        stepY = -stepY;
        int temp = yMin;
        yMin = yMax;
        yMax = temp;
    }
}
return path.toArray(new Point[path.size()]);

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