這是我的主要方法。我有一個ArrayList爲食草動物和我的植物在(1000 x 1000)的畫布大小我有他們做的是: 所有的食草動物吃一種植物,是最接近他們。我知道需要做的是一旦所有的植物都被吃掉了,他們的主要方法應該退出並打印一些統計數據。我的理解是,在我的食草動物開始尋找下一個植物之前,我需要退出,因爲它會給我一個錯誤,因爲arrayList現在是空的。但是,如果我使用「System.exit(0);」在我的findNearestWithinFiftyMeters方法之前,它會執行我的代碼的其餘部分。如何讓它退出但仍然打印我的統計信息?如何退出我的主要方法?
import java.util.ArrayList;
public class TestDriver {
public static void main(String[] args){
//Set the canvas size to (1000,1000) and x and y scale to (0,1000)
StdDraw.setCanvasSize(1000,1000);
StdDraw.setXscale(0.0 , 1000.0);
StdDraw.setYscale(0.0 , 1000.0);
//ArrayList for the Plants
ArrayList<Plant> thePlants = new ArrayList<Plant>();
//Create 300 Plants
for (int i=0; i<300; i++){
thePlants.add(new Plant (1000 * Math.random(),1000 * Math.random()));
}
//ArrayList for the Herbivores
ArrayList<Herbivore> theHerbivores = new ArrayList<Herbivore>();
//Create 20 Herbivores
for (int i=0; i<20; i++){
theHerbivores.add(new Herbivore (1000 * Math.random(), 1000 * Math.random()));
}
//Draw Graphics:
while(theHerbivores.size() > 0){
//Clears the board
StdDraw.clear();
//created Herbivores of size
for(int i=0; i<theHerbivores.size(); i++){
//Draws the herbivore at position i
theHerbivores.get(i).draw();
if(thePlants.size() == 0){
StdDraw.clear();
}
//Finds the plant that is closest to the herbivore
Plant closest = findNearestWithinFiftyMeters(thePlants , theHerbivores.get(i));
if(closest != null){
//IF the closest plant is NOT null then move towards closest plant
theHerbivores.get(i).moveToward(closest.getX(), closest.getY());
//now that we have the closest plant in regards to this herbivore
//we want it to move to the plant
theHerbivores.get(i).createRandomTarget();
//Reset the target each time it finds a new plant to chew
}
else{
//if it IS null
//Walk in a random direction (towards the imaginary plant)
theHerbivores.get(i).move();
}
}
//Draws plants
for(Plant p: thePlants){
p.draw();
}
//Check for any herbivores that have moved off screen
for(int i=0; i < theHerbivores.size(); i++){
//if an herbivore moved too far left or right move to other side of screen
if(theHerbivores.get(i).getX()>1000){
theHerbivores.get(i).setX(0);
}
else if(theHerbivores.get(i).getX()<0){
theHerbivores.get(i).setX(1000);
}
//if an herbivore moved too far up or down
if(theHerbivores.get(i).getY()>1000){
theHerbivores.get(i).setY(0);
}
else if(theHerbivores.get(i).getY()<0){
theHerbivores.get(i).setY(1000);
}
}
//looping through all the plants to remove plants that have been eaten
for(int i=0; i< theHerbivores.size(); i++){
for(int j = 0; j < thePlants.size(); j++){
if(thePlants.get(j).distanceTo(theHerbivores.get(i).getX(),theHerbivores.get(i).getY()) < 3){
thePlants.remove(j);
theHerbivores.get(i).eat();
//INCREMENT HERBIVORE EATEN COUNT
}
}
StdDraw.show(1);
}
stepCounter++;
}//end while loop
System.out.println(stepCounter + " steps done in this simulation");
long estimatedTime = System.nanoTime() - startTime;
System.out.println(estimatedTime + "Length of time simulation used");
for(int i=0; i<theHerbivores.size(); i++){
System.out.println("Herbivore # " + i + "X: " + theHerbivores.get(i).getX() + "Y: " + theHerbivores.get(i).getY() + " EATEN: "+ theHerbivores.get(i).getEatCount());
}
return;
} // end of main method
static long startTime = System.nanoTime();
static int stepCounter = 0;
public static Plant findNearestWithinFiftyMeters(ArrayList<Plant> thePlants , Herbivore eater){
//plant variable for storage to find closest plant to that herbivore
Plant closest = thePlants.get(0);
for(int i=0; i<thePlants.size(); i++){
if(eater.distanceTo(closest.getX(), closest.getY()) > eater.distanceTo(thePlants.get(i).getX(),thePlants.get(i).getY())){
//if the plant in closest variable is farther away than the
//plant in index 'i' then replace the plant in closest with the plant in index 'i'
closest = thePlants.get(i);
}
}
if(eater.distanceTo(closest.getX(),closest.getY()) > 50){
//if distance is greater than 50(herbivore sight range) then set closest equal to null
closest=null;
}
return closest;
}
} // end of class
謝謝!這很有道理! – Michelle 2015-03-03 02:19:21
不客氣。如果你認爲這個答案是最好的,那麼考慮投票答案最好,或者儘可能提高答案。 – 2015-03-03 02:21:47