一道Java编程题目

bobo415 2009-12-22 10:36:01

编程输出下列矩阵
例如：
输入： 5
输出：
1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9

...全文

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gigimyl520 2010-07-26

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原來可以這樣學習了

gao11811 2009-12-23

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mark

sweetBug 2009-12-23

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学习

yegaofei 2009-12-23

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从中心点开始往回走，每走一部数值减一。其中走的路线也是有规律的，横的走几格，必定伴随着纵向的也走几格。

ublic class FunClass {

/**
* @param args
*/
public static void main(String[] args) {
int arg = 5; //设置你要得数

int[][] arry = new int[arg][arg];
int centerColumn = 0;
int centerRow = 0;
if(arg % 2 == 0){
centerColumn = arg / 2 - 1;
centerRow = arg / 2 ;
} else {
centerColumn = arg / 2;
centerRow = arg / 2;
}

int value = arg * arg;
arry[centerRow][centerColumn] = value--;

boolean isMoveVertical = false;
boolean isToLeft = false;
boolean isUp = true;
if(arg % 2 == 0){
isToLeft = false;
isUp = true;
} else {
isToLeft = true;
isUp = false;
}
int currCol = centerColumn;
int currRow = centerRow;
for(int i = 1; i < arg; i++){
for(int j = 0 ; j < 2; j++){
for(int k = 0; k < i; k++){
if(isMoveVertical){
if(isUp){
currRow--;
} else {
currRow++;
}
} else {
if(isToLeft){
currCol--;
} else {
currCol++;
}
}
arry[currRow][currCol] = value--;
}
isMoveVertical = !isMoveVertical;
}
isToLeft = !isToLeft;
isUp = !isUp;
}

for(int i = 0; i < arg - 1; i++){
arry[0][i] = i+1;
}

//output the array
for(int i = 0; i < arg; i++){
for(int j = 0; j < arg; j++){
System.out.print(arry[i][j]);
System.out.print(" ");
}
System.out.println();
}

}

}

huosidun0302 2009-12-23

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顶顶顶

xuhailang00 2009-12-23

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很强大。

商科程序员 2009-12-23

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这是我06年5月写的程序:
http://blog.csdn.net/chouy/archive/2006/06/15/800188.aspx

sweatcoffee 2009-12-23

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学习

huanglinhao 2009-12-22

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来顶一下还有来学习学习

NetMatrix 2009-12-22

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也过来学习一下

miguboy 2009-12-22

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去wwww.chinajavaworld.com。你搜一下螺旋矩阵，里面有个很好的例子，讲解的和透彻，个人觉得很好。推荐。

nehnre 2009-12-22

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	public static void main(String args[]){

		int constNum = 11;

		int n=1;

		//0 1 2 3 4 / 9 14 19 24 / 23 22 21 20 / 15 10 5 / 6 7 8 / 13 18 / 17 16 / 11 / 12

		int [] test = new int[constNum*constNum];

		

		int count = constNum;

		boolean hasSub = true;

		int []operator = new int[4];

		operator[0] = 1;

		operator[1] = constNum;

		operator[2] = -1;

		operator[3] = -constNum;

		int index = 0;

		int last = -1;

		while (count > 0) {

			for (int i = 0; i < count; i++) {

				test[last + operator[index]] = n++;

				last = last + operator[index];



			}

			index = (index +1)%4;

			if(hasSub){

				count--;

				hasSub = false;

			}

			else{

				hasSub = true;

			}

		}

		for(int i=0;i<constNum*constNum;i++){

			if(i%constNum==0){

				System.out.println("");

			}

			System.out.print(test[i] + "\t");

		}

	}

nehnre 2009-12-22

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public static void main(String args[]){
int constNum = 11;
int n=1;
//0 1 2 3 4 / 9 14 19 24 / 23 22 21 20 / 15 10 5 / 6 7 8 / 13 18 / 17 16 / 11 / 12
int [] test = new int[constNum*constNum];

int count = constNum;
boolean hasSub = true;
int []operator = new int[4];
operator[0] = 1;
operator[1] = constNum;
operator[2] = -1;
operator[3] = -constNum;
int index = 0;
int last = -1;
while (count > 0) {
for (int i = 0; i < count; i++) {
test[last + operator[index]] = n++;
last = last + operator[index];

}
index = (index +1)%4;
if(hasSub){
count--;
hasSub = false;
}
else{
hasSub = true;
}
}
for(int i=0;i<constNum*constNum;i++){
if(i%constNum==0){
System.out.println("");
}
System.out.print(test[i] + "\t");
}
}

zhuzeitou 2009-12-22

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public class Matrix {

	private int[][] matrix;

	private int n;



	public Matrix(int n) {

		this.n = n;

		matrix = new int[n][n];

		int[] direct = new int[] { 1, 0, -1, 0 };

		int directX = 0, directY = 3, currentX = -1, currentY = 0;

		for (int count = 0; count < n * n;) {

			int nextX = currentX + direct[directX], nextY = currentY

					+ direct[directY];

			if (nextX >= n || nextY >= n || nextX < 0 || nextY < 0

					|| matrix[nextY][nextX] != 0) {

				directX = (directX + 1) % 4;

				directY = (directY + 1) % 4;

				continue;

			}

			matrix[nextY][nextX] = ++count;

			currentX = nextX;

			currentY = nextY;

		}

	}



	public String toString() {

		StringBuffer sb = new StringBuffer();

		if (matrix != null) {

			int length = new String(n * n + "").length();

			String format = "%-" + length + "d ";

			for (int[] array : matrix) {

				for (int i : array) {

					sb.append(String.format(format, i));

				}

				sb.append('\n');

			}

		}	

		sb.trimToSize();

		return sb.toString();

	}

}

bobo415 2009-12-22

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大家多顶下
希望能写个程序粘出来
欢迎不同见解

wanghuailong 2009-12-22

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关注一下@

ComputerHeart 2009-12-22

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链表？同意楼上的观点！分解成正方框。

zhuzeitou 2009-12-22

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转向条件就是超过边界(小于0或者大于等于参数)或者即将到达的数组元素不为0

dlnu05610 2009-12-22

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public static int[][]  aa(int k){

		int[][] hui = new int[k][k];

		int row=0;

		int fx=0;

		for(int i=1;i<k*k;){

			if(fx%4==0){

				for(int j=row;j<k-row-1&&i<=k*k;j++){

					hui[row][j]=i;

					i++;

				}

				fx++;

			}else if(fx%4==1){

				for(int j=row;j<k-row-1&&i<=k*k;j++){

					hui[j][k-row-1]=i;

					i++;

				}

				fx++;

			}else if(fx%4==2){

				for(int j=row;j<k-row-1&&i<=k*k;j++){

					hui[k-1-row][k-j-1]=i;

					i++;

				}

				fx++;

			}else if(fx%4==3){

				for(int j=row;j<k-row-1&&i<=k*k;j++){

					hui[k-j-1][row]=i;

					i++;

				}

				fx++;

				row++;

			}

		}

		

			if(k%2==1){

				hui[(int)(k/2)][(int)(k/2)]=k*k;

			}else{

				hui[k/2][k/2-1]=k*k;

			}

		

		return hui;

	}