Java的创建螺旋 [英] Java creation of a spiral
问题描述
可能重复：
在螺旋循环
Possible Duplicate:
Looping in a spiral
我要创建一个程序一个3×3的矩阵来填充。我想导致一些看起来像这样
I'm creating a program to populate a 3 by 3 matrix. I want to result in something looking like this
5 4 3
6 1 2
7 8 9
正如你可能已经注意到这是一个恶性循环。 现在，我使用的算法是这样的：我有一个二维数组，其中值重新present数量的坐标。首先，我分配的每个数字在这个数组中的坐标将有10数值，然后开始在9我缩小x坐标，并指定的坐标currentnum的价值 - 1，直到达到结束或者其值不为10;然后，我做同样的事情，除非我增加Y的值;然后减少x的值;然后，ÿ;
As you have probably noticed it is a spiral. Now the algorithm I'm using is this: I have a 2-d array where the values represent the coordinates of the number. First I assign that every number coordinate in this array will have a value of 10. Then starting at 9 I decrease my x coordinate and assign the value of the coordinate to currentnum - 1 until it reaches the end or its value is not 10; Then I do the same thing except I increase the value of Y; Then decrease the value of x; Then of Y;
我给你10的原因，每一个号码是太喜欢它作为一个道路为我的计划。由于当前NUM将不会超过9。如果一个正方形的值是10它就像一个绿灯。如果不是10含义值已被分配给该方它打破了它。
The reason I assign 10 to every number is so like it acts as a road for my program. Since current num will never exceed 9. If the value of a square is 10 it is like a green light. If it is not 10 meaning a value has been assigned to that square it breaks out of it.
下面是我的code，请注意是用Java编写的
Here is my code, please note it is written in Java
public class spiral {
/**
* @param args
*/
public static void main(String[] args) {
int spiral [] [] = new int[3][3];
for(int i = 0; i <= 2; i++){
for(int j = 0; j <= 2; j++){
spiral[i][j] = 10;
}
}
//0 is x value, 1 is y value
spiral[0][0] = 9;
int x = 1;
int y = 1;
int counter = 1;
int currentnum = 9;
int gridsquare = 3;
for(int i = 0; i <= 8; i++){
if(counter == 5){
counter = 1;
}
if(counter == 1){
System.out.println(x + " " + y);
for(int j = 0;j <= 1;j++){
if(spiral[x][y] == 10){
spiral[x][y] = currentnum;
currentnum--;
x += 1;
}
else{
y += 1;
break;
}
}
}
if(counter == 2){
for(int k = 0; k <= 0; k++){
System.out.print(x + " " + y);
if(spiral[x][y] == 10){
spiral[x][y] = currentnum;
currentnum--;
y += 1;
}
else{
x -= 1;
break;
}
}
}
if(counter == 3){
for(int z = 0; z <= 0; z++){
if(spiral[x][y] == 10){
spiral[x][y] = currentnum;
currentnum--;
x -= 1;
}
else{
y -= 1;
break;
}
}
}
if(counter == 4){
for(int b = 0; b <= 0; b++){
if(spiral[x][y] == 10){
spiral[x][y] = currentnum;
currentnum--;
y -= 1;
}
else{
x += 1;
break;
}
}
}
counter++;
}
System.out.print(currentnum);
}
}
我得到这个错误
I'm getting this error
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 3
at spiral.main(spiral.java:44)
由于我是新来的Java会有人提出此一更多钞票修复。此外，如果你看到我的算法任何问题，请不要告诉我。
Since I'm new to Java would someone please suggest a posible fix for this. Also if you see any problems with my algorithm please do inform me.
推荐答案
您不必pre-充满10：零的作品一样好
You do not need to pre-fill with 10: zero works just as well.
我觉得解决螺旋最好的办法是想你怎么做手工：开始在一个角落里，去水平，直到你打不为零或阵列的边缘。然后右转。当停止当前一些去通过N *ñ。
I think the best approach to solving the spiral is to think of how you do it manually: start in a corner, and go horizontally until you hit non-zero or an edge of the array. Then you turn right. Stop when the current number goes past N*N.
现在让我们来看看什么是算法的每个部分的含义：
Now let's look at what each part of the algorithm means:
- 开始在角落机构将x = 0和Y = 0。
- 去在一条直线上装置X = X + DX，Y = Y + DY，其中DX任或镝是零，和dy或dx为1或-1。
- 在谈到DX分配到Dy和-dy到DX正确的方式。
下面是如何看起来在code：
Here is how it looks in the code:
int current = 1;
// Start in the corner
int x = 0, y = 0, dx = 1, dy = 0;
while (current <= N*N) {
// Go in a straight line
spiral[x][y] = current++;
int nx = x + dx, ny = y + dy;
// When you hit the edge...
if (nx < 0 || nx == N || ny < 0 || ny == N || spiral[nx][ny] != 0) {
// ...turn right
int t = dy;
dy = dx;
dx = -t;
}
x += dx;
y += dy;
}
这篇关于Java的创建螺旋的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
