我正在使用四叉树系统渲染地形。我需要使用方法splitHeightmap(float[] originalMap, int quadrant)
将高度图分为四个部分,象限为0-3之间的数字。映射需要拆分为四分之一,因此如果将0作为象限传递,则数组的左下四分之一将作为新的float数组返回。我有一些基本代码,但是我不确定如何根据所需象限实际采样地图:
protected float[] splitHeightMap(float[] heightMap, TerrainQuadrant quadrant) {
float[] newHeightMap = new float[size >> 1];
int newSize = size >> 1;
for (int i = 0; i < newSize; i++)
for (int j = 0; j < newSize; j++)
newHeightMap[i * newSize + j] = sampleHeightAt(heightMap, i, j);
return newHeightMap;
}
protected float sampleHeightAt(float[] heightMap, int x, int z) {
return heightMap[z + x * size];
}
编辑:我写了我认为应该工作的内容,但正在获取索引65792(原始高度图为512x512)的ArrayIndexOutOfBoundsException:
protected float[] splitHeightMap(float[] heightMap, TerrainQuadrant quadrant) {
float[] newHeightMap = new float[(size >> 1) * (size >> 1)];
int newSize = size >> 1;
int xOffset = 0, zOffset = 0;
int xCount = 0, yCount = 0;
switch (quadrant) {
case BottomRight:
xOffset = newSize;
break;
case TopLeft:
zOffset = newSize;
break;
case TopRight:
xOffset = newSize;
zOffset = newSize;
break;
default:
break;
}
for (int x = xOffset; x < xOffset + newSize; x++)
for (int z = zOffset; z < zOffset + newSize; z++) {
newHeightMap[xCount + yCount * newSize] = heightMap[z + x * size]; // should this be 'z + x * size' or 'x + z * size'?
xCount++;
yCount++;
}
return newHeightMap;
}
最佳答案
如果我对您的理解正确,则必须使用二维数组。使用这种类型的数组,可以轻松获取数组的任何区域。
现在,您的heightMap
类型为float[][]
,因此必须在填充此数组的位置更正代码。
我为您实现了一个示例,并使用了4x4阵列:
| 2 2 3 3 |
| 2 2 3 3 |
| 0 0 1 1 |
| 0 0 1 1 |
据我了解,您想获取所有“ 0”,所有“ 1”,所有“ 2”和所有“ 3”之类的区域。
public static void main ( String[] args )
{
//setting up initial array 'heightMap' (you can name it like you want)
float[][] f = { { 2, 2, 3, 3 }, { 2, 2, 3, 3 }, { 0, 0, 1, 1 }, { 0, 0, 1, 1 } };
float[][] f2 = splitHeightMap ( f, TerrainQuadrant.BotttomRight );
for ( float[] floats : f2 )
{
System.out.println ( Arrays.toString ( floats ) );
}
}
protected static float[][] splitHeightMap ( float[][] heightMap, TerrainQuadrant quadrant )
{
//this gives you half of the 'heightMap' length
int newSize = heightMap.length >> 1;
float[][] newHeightMap = new float[ newSize ][ newSize ];
//its your offsets, indicating from what place to start iteration
int xOffset = 0;
int yOffset = newSize;
//its max values to reach while iterating
int xRestriction = newSize;
int yRestriction = heightMap.length;
//setting up params according to 'quadrant'
switch ( quadrant )
{
case BottomRight:
xOffset = newSize;
yOffset = newSize;
xRestriction = heightMap.length;
break;
case TopLeft:
yOffset = 0;
yRestriction = newSize;
break;
case TopRight:
yOffset = 0;
xOffset = newSize;
xRestriction = heightMap.length;
yRestriction = newSize;
break;
default:
break;
}
//counters not to reach new array bounds
int xCount = 0, yCount = 0;
for ( int y = yOffset; y < yRestriction; y++ )
{
//taking row at 'y' position
float[] row = heightMap[ y ];
for ( int x = xOffset; x < xRestriction; x++ )
{
//taking value from 'y' row at 'x' position.
float value = row[ x ];
//set fetched value to new map.
newHeightMap[ yCount ][ xCount ] = value;
//increase x position, but do not touch row
xCount++;
}
//new row - new 'x' position
xCount = 0;
yCount++;
}
return newHeightMap;
}
此实现向您显示:
| 1 1 |
| 1 1 |
要更改它-更改主要方法。
关于java - 将高度图划分为多个象限,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/25325127/