[英]Placing numbers in a 2D array by a chess knight move pattern in Java using recursion
[英]How to solve this chess knight problem in Java?
我想使用Java解決國際象棋難題。 我編碼為Knight片段從開始字段(1; 1)移動到任何地方,除了負的x和y之外,如果所有內容都有效,則將此訪問字段放入列表中,否則返回到上一個字段。 但是它根本不起作用,這種情況永遠不會成立,它一直都是負數,並且不會返回到先前的領域,可能是什么引起了問題?
import java.util.ArrayList;
import java.util.List;
import java.util.Random;
public class Main
{
static Vector now;
static Vector prev;
static List<Vector> visited = new ArrayList<>();
public static void main(String[] args)
{
now = new Vector(); // sets x = 1, y = 1
prev = now; // also x = 1, y = 1
visited.add(now); // we are already on (1;1)
generate();
}
static void generate()
{
Random rand = new Random();
for (int i = 0; i < 50; i++)
{
int a = rand.nextInt(8);
move(a);
if((isValid()) && hasNotVisited()) // if x and y > 0 , because the smallest coord is (1;1), and check is we haven't visited that field
{
visited.add(now);
prev = now; // previous coord is now, then make a new step
}
else
{
now = prev; // else, return to previous coord
// For example, we are one (3;2), we move to (1;0), that's not valid, we should move back to (3;2)
}
}
}
static void move(int a)
{
switch (a){
case 0:
now.x += 2;
now.y++;
break;
case 1:
now.x += 2;
now.y--;
break;
case 2:
now.x -= 2;
now.y++;
break;
case 3:
now.x -= 2;
now.y--;
break;
case 4:
now.y += 2;
now.x++;
break;
case 5:
now.y += 2;
now.x--;
break;
case 6:
now.y -= 2;
now.y++;
break;
case 7:
now.y -= 2;
now.y--;
break;
}
}
static boolean hasNotVisited()
{
for (Vector aVisited : visited) {
if (aVisited == now)
return false;
}
return true;
}
static boolean isValid()
{
return (0 < now.x && now.x <= 10) && (0 < now.y && now.y <= 10);
}
}
謝謝!
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