以下是形状的外观like:
到目前为止,这是我的代码:
public class Diamonds {
public static void main(String[] args) {
for (int i = 1; i < 10; i += 2) {
for (int j = 0; j < 9 - i / 2; j++) {
System.out.print(" ");
}
for (int j = 0; j < i; j++) {
System.out.print("*");
}
System.out.print("\n");
}
for (int i = 7; i > 0; i -= 2) {
for (int j = 0; j < 9 - i / 2; j++) {
System.out.print(" ");
}
for (int j = 0; j < i; j++) {
System.out.print("*");
}
System.out.print("\n");
}
}
}
我无法获得第二个形状
答案 0 :(得分:1)
为了不破坏这个问题的乐趣,我将在不编写任何代码的情况下解释您需要做什么。
要获得第二个形状,您需要在已有的两个“外部”循环中的每一个中添加两个额外的嵌套for
循环。
第三个循环将产生固定数量的空格。请注意,第一个形状的右边缘和第二个形状的左边缘之间的距离是恒定的,因此您的第三个循环将很容易编码。
第四个循环将像第一个循环一样循环,但它们会改变位置:第一个外循环的第一个内循环将是第二个外循环中的第四个内循环,反之亦然。
答案 1 :(得分:0)
通过检查右侧的形状,我们可以注意到,对于左侧形状的线上的每个N
星号,右侧的星号为10 - N
,因此,接受您的代码并扩展它,我们可以得到:
public class Diamonds {
public static final String SPACE = " ";
public static void main(String[] args) {
for (int i = 1; i < 10; i += 2) {
for (int j = 0; j < 9 - i / 2; j++)
System.out.print(" ");
for (int j = 0; j < i; j++)
System.out.print("*");
System.out.print(SPACE);
for (int j = 0; j < 10 - i; j++)
System.out.print("*");
System.out.print("\n");
}
for (int i = 7; i > 0; i -= 2) {
for (int j = 0; j < 9 - i / 2; j++)
System.out.print(" ");
for (int j = 0; j < i; j++)
System.out.print("*");
System.out.print(SPACE);
for (int j = 0; j < 10 - i; j++)
System.out.print("*");
System.out.print("\n");
}
}
}
如果我们提取一些常用代码:
public class Diamonds {
public static final String SPACE = " ";
public static void main(String[] args) {
for (int i = 1; i < 10; i += 2) {
drawLine(i);
System.out.print("\n");
}
for (int i = 7; i > 0; i -= 2) {
drawLine(i);
System.out.print("\n");
}
}
private static void drawLine(int i) {
for (int j = 0; j < 9 - i / 2; j++)
System.out.print(" ");
for (int j = 0; j < i; j++)
System.out.print("*");
System.out.print(SPACE);
for (int j = 0; j < 10 - i; j++)
System.out.print("*");
}
}
答案 2 :(得分:0)
试试这个:
public static void main(String[] args) {
for (int i = 9; i > 0; i -= 2) {
for (int j = 0; j < 9 - i / 2; j++)
System.out.print(" ");
for (int j = 0; j < i; j++)
System.out.print("*");
System.out.print("\n");
}
for (int i = 2; i < 10; i += 2) {
for (int j = 0; j < 9 - i / 2; j++)
System.out.print(" ");
for (int j = 0; j <= i; j++)
System.out.print("*");
System.out.print("\n");
}
}
输出:
*********
*******
*****
***
*
***
*****
*******
*********
答案 3 :(得分:0)
public class ReverseDiamond {
public static void main(String[] ar) {
int rows = 10;
ReverseDiamond diamond = new ReverseDiamond();
for(int i = 0; i < rows; i++)
diamond.printLine(rows, i);
for(int i = rows - 2; i >= 0; i--)
diamond.printLine(rows, i);
}
private void printLine(int rows, int currRow) {
for(int space = 1; space <= currRow; space++)
System.out.print(" ");
for(int star = 1; star < 2 * (rows - currRow); star++)
System.out.print("*");
System.out.println();
}
}
答案 4 :(得分:0)
您可能喜欢这样:
public class Diamonds {
public static void main(String[] args) {
int totalStars = 9;
int rows = 9;
for (int r = 0,stars=-1,gap=totalStars; r < rows; r++ ) {
stars+= (r<=rows/2) ?2:-2;
gap=totalStars-stars;
printChars(' ', gap);
printChars('*', stars);
printChars(' ', gap);
int gap2=stars+1;
int stars2=gap+1;
printChars(' ', gap2);
printChars('*', stars2);
printChars(' ', gap2);
System.out.println();
}
}
private static void printChars(char c ,int times) {
for (int i = 0; i < times; i++) {
System.out.print(c);
}
}
}