如何减少碰撞机会??
如实现100% 不会碰撞??
import java.util.*;class t1
{
public static void main(String[] args)
{
Object[][][] ojb1 = new Object[10][5][9];
Object[][][] ojb2 = new Object[7][5][9];
Object[][][] ojb3 = new Object[6][5][9];
int total = 0;
int count = 0; for(int i=0; i<104; i++ ){ while(count < 100){
int x = (int) (Math.random() * 5);
int y = (int) (Math.random() * 9);
int z = (int) (Math.random() * 10);
int m = (int) (Math.random() * 7);
int n = (int) (Math.random() * 6);
if(ojb1[z][x][y] == null && ojb1[m][x][y] == null && ojb3[n][x][y] == null){
ojb1[z][x][y] = new String("No Empty");
ojb1[m][x][y] = new String("No Empty");
ojb1[n][x][y] = new String("No Empty");
System.out.println("Added " + i);
break;
}else{
count+=1;
}
}
if(count >=100){
System.out.println("Conflict " + i);
total +=1;
} } System.out.println("Total of conflict: " + total);
}
}
如实现100% 不会碰撞??
import java.util.*;class t1
{
public static void main(String[] args)
{
Object[][][] ojb1 = new Object[10][5][9];
Object[][][] ojb2 = new Object[7][5][9];
Object[][][] ojb3 = new Object[6][5][9];
int total = 0;
int count = 0; for(int i=0; i<104; i++ ){ while(count < 100){
int x = (int) (Math.random() * 5);
int y = (int) (Math.random() * 9);
int z = (int) (Math.random() * 10);
int m = (int) (Math.random() * 7);
int n = (int) (Math.random() * 6);
if(ojb1[z][x][y] == null && ojb1[m][x][y] == null && ojb3[n][x][y] == null){
ojb1[z][x][y] = new String("No Empty");
ojb1[m][x][y] = new String("No Empty");
ojb1[n][x][y] = new String("No Empty");
System.out.println("Added " + i);
break;
}else{
count+=1;
}
}
if(count >=100){
System.out.println("Conflict " + i);
total +=1;
} } System.out.println("Total of conflict: " + total);
}
}
Object[][][] ojb1 = new Object[n][5][9];
Object[][][] ojb1 = new Object[x][5][9];
Object[][][] ojb1 = new Object[y][5][9]; 3 者位置[5][9]同时都是 null..就return true.. 例如:
ojb1 [1][2][3] = null
obj2 [5][2][3] = null
obj3 [3][2][3] = null
这时就 return true; 当3个ojbect 一边存入数据..一边计算剩下多少个同时共有的空位..(希望这个)
if(ojb1[z][x][y] == null && ojb1[m][x][y] == null && ojb3[n][x][y] == null)
的第2个判断应该是ojb2[m][x][y] == null
在a, b, c, d的范围内随即取值,不能重复(碰撞?)。
x[0]=a
x[1]=b
x[2]=c
x[3]=dRandom rand = new Random();
int temp;
for (int j = 4; j > 0; j--) {
int i = rand.nextInt(j);
System.out.println(x[i]);
temp = x[i];
x[i] = x[4 - j];
x[4 - j] = temp;
}大意就是这样
应该将x[i]与x[j]互换。以保证取过的值在数组的最后,再限制随机数的取值范围,则避免重复。