如何在java中打印星形图案

Question

我试图找到以下星形图案的任何序列或公式，但我没有找到，所以只是在下面尝试

public class MyClass {

    public static void main(String args[]) {

        System.out.println("....*....");

        System.out.println("...***...");

        System.out.println("*********");

        System.out.println(".*******.");

        System.out.println("*********");

        System.out.println("...***...");

        System.out.println("....*....");

    }

}

这里的 “.” 表示 “” 空格字符。我们可以用循环打印这颗星星吗？

Answer 1

我使用的是 Java 11。您应该能够绘制星号的任何对映异形图案。有关详细说明，请阅读注释。import java.util.Arrays;class Star {&nbsp; &nbsp; public static void main(String[] args) {&nbsp; &nbsp; &nbsp; &nbsp; int[] stars = new int[] { 1, 3, 9, 7, 9, 3, 1 }; // asterisks per line&nbsp; &nbsp; &nbsp; &nbsp; int max = Arrays.stream(stars).max().getAsInt(); // max asterisks in any line&nbsp; &nbsp; &nbsp; &nbsp; for (int i = 0; i < stars.length; i++) { // prints the asterisks for a given row, and pads it left and right with spaces&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; System.out.print(" ".repeat((max - stars[i]) / 2) + "*".repeat(stars[i]) + " ".repeat((max - stars[i] + 1) / 2));&nbsp; &nbsp; }}

Answer 2

希望它有帮助！对于低年级学生来说，这很容易理解，而不是创建数组class starPattern {&nbsp; &nbsp; public static void main(String []args) {&nbsp; &nbsp; &nbsp; &nbsp; for(int i=1;i<=7;i++) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; for(int j=1;j<=9;j++) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if(i==3 || i==5 || j==5 || (i==4 && j>1 && j<9) || ((i==2 || i==6) && (j>3 && j<7))) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; System.out.print("*");&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }else{&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; System.out.print(" ");&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; System.out.println();&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }}我的电脑老师检查一下，这是正确的方法！关于此模式的简要说明1↓&nbsp; 1→&nbsp; 2&nbsp; &nbsp;3&nbsp; &nbsp;4&nbsp; &nbsp;5&nbsp; &nbsp;6&nbsp; &nbsp;7&nbsp; &nbsp;8&nbsp; &nbsp;92&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;*&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;3&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;4&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*5&nbsp; &nbsp; &nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;6&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*7&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp;*&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;8&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;*&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;

Answer 3

不知道你是什么意思，但如果你真的想要一个循环，这里有一个解决方案！public class Star {&nbsp; &nbsp; public static void main(String[] args) {&nbsp; &nbsp; &nbsp; &nbsp; int[] dots = new int[] {4, 3, 0, 1, 0, 3, 4};&nbsp; &nbsp; &nbsp; &nbsp; int width= 9;&nbsp; &nbsp; &nbsp; &nbsp; for (int i = 0; i < dots.length; i++) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; for (int j = 0; j < width; j++) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (j < dots[i] || j > width - dots[i] - 1) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; System.out.print(".");&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; System.out.print("*");&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; System.out.println();&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }}

Answer 4

是的，它可以使用循环来完成，不应该通过sout语句来完成。看起来你也是编程的初学者，所以我建议你自己尝试这类问题的逻辑。它将有助于你的成长。只需逐行尝试即可。这些问题通常需要2个循环，一个在垂直方向移动，另一个在水平方向移动。你可以把它想象成一个i*j矩阵（例如3X3，4X4）。因此，自己尝试一下，从一些基本的开始。如果您仍然需要解决方案，请先向我发送 ping，但请先自行尝试。