所以我被分配用Java制作带有星号的钻石，我真的很难过。到目前为止，这是我想出的：
public class Lab1 {
public static void main(String[] args) {
for (int i = 5; i > -5; i--) {
for (int j = 0; j < i; j++) {
System.out.print(" ");
}
for (int j = 0; j >= i; j--) {
System.out.print(" ");
}
System.out.println("*");
}
}
}
为了制作钻石，您需要设置空间和星星的形状。因为我是初学者，所以我只使用嵌套循环制作了这个简单的程序。
public class Diamond {
public static void main(String[] args) {
int size = 9,odd = 1, nos = size/2; // nos =number of spaces
for (int i = 1; i <= size; i++) { // for number of rows i.e n rows
for (int k = nos; k >= 1; k--) { // for number of spaces i.e
// 3,2,1,0,1,2,3 and so on
System.out.print(" ");
}
for (int j = 1; j <= odd; j++) { // for number of columns i.e
// 1,3,5,7,5,3,1
System.out.print("*");
}
System.out.println();
if (i < size/2+1) {
odd += 2; // columns increasing till center row
nos -= 1; // spaces decreasing till center row
} else {
odd -= 2; // columns decreasing
nos += 1; // spaces increasing
}
}
}
}
如您所见nos，是空格数。直到中间行需要减少，并且需要增加星星的数量，但在中间行之后则相反，即空间增加，星星减少。
size可以是任何数字。我在这里将它设置为 9，所以我将有一个 9 号星形，最大 9 行和 9 列...空间数（nos）将是9/2 = 4.5。但是java会把它当作4，因为int不能存储十进制数，并且中心行会是9/2 + 1 = 5.5，这将导致5 as int。
所以首先你会做行......因此有 9 行
(int i=1;i<=size;i++) //size=9
然后像我一样打印空格
(int k =nos; k>=1; k--) //nos being size/2
然后终于星星
(int j=1; j<= odd;j++)
一旦线路结束......
您可以使用 if 条件调整星号和空格。
for (int i = 0; i < 5; i++)
System.out.println(" *********".substring(i, 5 + 2 * i));
for (int i = 5; i > 0; i--)
System.out.println(" **********".substring(i - 1, 5 + (2 * i) - 3));
public class MyDiamond {
public static void main(String[] args) {
//Length of the pyramid that we want.151 is just an example
int numRows = 151;
//midrow is the middle row and has numRows number of *
int midrow = (numRows + 1) / 2;
int diff = 0;
for (int i = 1; i < numRows + 1; i++) {
for (int j = 1; j < numRows + 1; j++) {
if (((midrow - diff) <= j && (j <= midrow + diff))) {
System.out.print("*");
} else {
System.out.print(" ");
}
}
System.out.println();
if (i < midrow) {
diff++;
} else {
diff--;
}
}
}
}
public class Diamond {
//Size of the diamond
private int diagonal;
public Diamond(int diagonal) {
this.diagonal = diagonal;
}
public void drawDiamond() {
int n = diagonal;
for (int i = n / 2; i >= -n / 2; i--) {
for (int k = 0; k < i; k++) {
System.out.print(" ");
}
for (int j = 1; j <= (n - i * 2) && i >= 0; j++) {
System.out.print("*");
}
for (int k = 1; k <= -i && i < 0; k++) {
System.out.print(" ");
}
for (int j = (n / 2) * 2 + 2 * i; j >= -(n % 2 - 1) && i < 0; j--) {
System.out.print("*");
}
System.out.println();
}
}
public static void main(String[] args) {
//You pass diamond size here in the constructor
Diamond a = new Diamond(21);
a.drawDiamond();
}
}
主要问题是对角线的奇偶性。如果它甚至你不能正确绘制顶部星号。所以有 2 种类型的钻石 - 偶数和奇数对角线（顶部有 2 和 1 个星号）。
注意：使用 Count 全局变量，我们可以管理空间以及星号的递增和递减。
import java.util.*;
public class ParamidExample {
public static void main(String args[]) {
System.out.println("Enter a number");
Scanner sc = new Scanner(System.in);
int no = sc.nextInt();
int count = 1;
for (int i = 1; i <= 2 * no - 1; i++) {
for (int j = count; j <= no; j++) {
System.out.print(" ");
}
for (int k = 1; k <= count * 2 - 1; k++) {
System.out.print("* ");
}
if (i < no)
count++;
else
count--;
System.out.println("");
}
}
}
import static java.lang.System.out;
import java.util.Scanner;
public class Diamond {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int row = sc.nextInt();
sc.close();
Diamond d = new Diamond();
d.upperDiamond(row);
d.lowerDiamond(row - 2);
}
public void upperDiamond(int a) {
for (int i = 0; i < a; i++) {
for (int j = a - 1; j > i; j--)
out.print(" ");
for (int k = 0; k < 2 * i - 1; k++)
out.print("*");
out.print("\n");
}
}
public void lowerDiamond(int b) {
for (int i = 0; i < b; i++) {
for (int j = 0; j <= i; j++)
out.print(" ");
for (int k = 0; k < 2 * (b - i) - 1; k++)
out.print("*");
out.print("\n");
}
}
}
import java.util.Scanner;
public class Diamond {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in = new Scanner(System.in);
int input = in.nextInt();
int min = 1;
for (int i = 0; i < input; i++) {
for (int j = input - 1; j > i; j--) {
System.out.print(" ");
}
for (int k = 0; k < min; k++) {
if (k % 2 == 0) {
System.out.print("*");
} else {
System.out.print(".");
}
}
min += 2;
System.out.println();
}
int z = input + input - 3;
for (int i = 1; i < input; i++) {
for (int j = 0; j < i; j++) {
System.out.print(" ");
}
for (int k = 0; k < z; k++) {
if (k % 2 == 0) {
System.out.print("*");
} else {
System.out.print(".");
}
}
z -= 2;
System.out.println();
}
}
}
我可以看到您正在尝试做什么，这是思考钻石的一种非常巧妙的方式。
当我变为负数时，你会遇到 j 计数器的一些问题。看看如何使用 Math.abs()
还可以尝试在带有注释的基本步骤中编写一些伪代码，以使模式清晰：
//print 5 spaces + 1 star
//print 4 spaces + 2 stars
//print 3 spaces + 3 stars
//print 2 spaces+ 4 stars
.
.
.
//print 5 spaces + 1 star
然后，从字面上用变量（j 和 i）替换数字。
你现在有一个模型。这通常是编程中最难的部分……正确地建立模型。仅当您对模型的工作原理有很好的了解时才开始编码。
一旦你替换了变量，你可以尝试将整个事情转换成一个自动循环。
这应该有效。您可能只需要大多数方法和printDiamond(_)；
import java.util.Scanner;
public class StarsTry {
public static void main(String[] args) {
int reader;
Scanner kBoard = new Scanner(System.in);
do {
System.out.println("Insert a number of rows: ");
reader = kBoard.nextInt();
printDiamond(reader);
} while (reader != 0);
}
public static void printStars(int n) {
if (n >= 1) {
System.out.print("*");
printStars(n - 1);
}
}
public static void printTopTriangle(int rows) {
int x = 1;
for (int j = (rows - 1); j >= 0; j--, x += 2) {
printSpaces(j);
printStars(x);
System.out.print("\n");
}
}
public static void printSpaces(int n) {
if (n >= 1) {
System.out.print(" ");
printSpaces(n - 1);
}
}
public static void printBottomTriangle(int rows, int startSpaces) {
int x = 1 + (2 * (rows - 1));
for (int j = startSpaces; j <= (rows) && x > 0; j++, x -= 2) {
printSpaces(j);
printStars(x);
System.out.print("\n");
}
}
public static void printBottomTriangle(int rows) {
int x = 1 + (2 * (rows - 1));
for (int j = 0; j <= (rows - 1) && x > 0; j++, x -= 2) {
printSpaces(j);
printStars(x);
System.out.print("\n");
}
}
public static void printDiamond(int rows) {
printTopTriangle((int) rows / 2 + 1);
printBottomTriangle((int) rows / 2, 1);
}
}
import java.util.Scanner;
public class MakeDiamond {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (true) {
System.out.println("Let's Creat Diamonds");
System.out.println("If number increases Diamonds gets bigger. Please input number lager than 1 : ");
int user_input = sc.nextInt(); //gets user's input
System.out.println("");
int x = user_input;
int front_space = -5;
for (int i = 0; i < 2 * user_input + 1; i++) {
for (int a = front_space; a < Math.abs(i - user_input); a++) {
System.out.print(" "); //Change a bit if diamonds are not in good shape
}
if (i < user_input + 1) {
for (int b = 0; b < 2 * i + 1; b++) {
System.out.print("* "); //Change a bit if diamonds are not in good shape
}
} else if (i > user_input) {
for (int c = 0; c < 2 * x - 1; c++) {
System.out.print("* "); //Change a bit if diamonds are not in good shape
}
x--;
}
System.out.print('\n');
}
System.out.println("\nRun Again? 1 = Run, 2 = Exit : ");
int restart = sc.nextInt();
System.out.println("");
if (restart == 2) {
System.out.println("Exit the Program.");
System.exit(0);
sc.close();
}
}
}
}
使用 while 或 for 循环制作菱形时。我认为使用“Math.abs”将是最简单的方法。
您可以通过扫描仪输入数字，当输入数字增加时，钻石会变大。
我使用 Eclipse 制作了这个程序。因此，空间会因您的运行环境而异。像另一个 IDE、CMD 或终端。如果钻石形状不好。换个空格就好了。
您可以按如下方式打印星号菱形（数学运算符） ：
int m = 4;
int n = 4;
for (int i = -m; i <= m; i++) {
for (int j = -n; j <= n; j++) {
int val = Math.abs(i) + Math.abs(j);
System.out.print(val > Math.max(m, n) ? " " : "∗");
if (j < n) {
System.out.print(" ");
} else {
System.out.println();
}
}
}
输出：
∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗
∗
public class Main {
public static void main(String[] args) {
int number = 23,l =1,diff = 1,rem = number/2,rep = 0;
for(int i=1;i<=number;i++){
if(i < number/2 +1){
for(int k=rem;k>=1;k--)
System.out.print(" ");
for(int j=1;j<=l;j++)
System.out.print("*");
diff = 2;
rem -= 1;
}
if(i >= number/2 +1){
for(int k=0;k<rep;k++)
System.out.print(" ");
for(int j=1;j<=l;j++)
System.out.print("*");
diff =3;
rep += 1;
}
System.out.println();
l = diff == 2 ? (l + 2) : (l - 2);
}
}
}
//仅适用于奇数...
使用String#repeat作为 Java-11 的一部分引入，您可以在单个循环中使用单个语句来完成。
public class Main {
public static void main(String[] args) {
final int MID_ROW_NUM = 5;
for (int i = 1 - MID_ROW_NUM; i < MID_ROW_NUM; i++) {
System.out.println(" ".repeat(Math.abs(i)) + "*".repeat((MID_ROW_NUM - Math.abs(i)) * 2 - 1));
}
}
}
输出：
*
***
*****
*******
*********
*******
*****
***
*
通过更改空间，您还可以打印菱形的变体：
public class Main {
public static void main(String[] args) {
final int MID_ROW_NUM = 5;
for (int i = 1 - MID_ROW_NUM; i < MID_ROW_NUM; i++) {
System.out.println(" ".repeat(Math.abs(i)) + "* ".repeat((MID_ROW_NUM - Math.abs(i)) * 2 - 1));
}
}
}
输出：
*
* * *
* * * * *
* * * * * * *
* * * * * * * * *
* * * * * * *
* * * * *
* * *
*
尝试这个
public class Main {
public static void main(String[] args) {
int n = 50;
int space = n - 1;
for (int i = 0; i < n; i++) {
for (int j = 0; j < space; j++) {
System.out.print(" ");
}
for (int j = 0; j <= i; j++) {
System.out.print("* ");
}
System.out.println("");
space--;
}
space = 0;
for (int i = n; i > 0; i--) {
for (int j = 0; j < space; j++) {
System.out.print(" ");
}
for (int j = 0; j < i; j++) {
System.out.print("* ");
}
System.out.println("");
space++;
}
}
}
public class Main {
private static int l =1;
public static void main(String[] args) {
int number =9;
for(int i=1;i<=2*number -1;i++){
if(i<=number){
for(int j=1;j<=(number-i);j++)
System.out.print(" ");
for(int j=1;j<=i;j++)
System.out.print("* ");
}
if(i>number){
for(int j=1;j<=i-number;j++)
System.out.print(" ");
for(int j=1;j<=number-l;j++)
System.out.print("* ");
l += 1;
}
System.out.println();
}
}
}
package com.DiamondPrintingProgram;
import java.util.Scanner;
public class DiamondPrintingProgram {
public static void main(String[] args) {
int num = getInput();
int middle = (int) num / 2 + 1;
printOutput(num,middle);
}
public static int getInput() {
Scanner sc = new Scanner(System.in);
int num;
System.out.print("Enter a odd number: ");
while (true) {
num = sc.nextInt();
if (num % 2 != 0) {
break;
}
System.out.print("Please Enter a ODD NUMBER: ");
}
return num;
}
private static void printOutput(int num, int middle){
char asterisk = '*';
for (int j = 0; j < num; j++) {
for (int i = 1; i <= num; i++) {
if (j < middle) {
if ((i < (middle - j) || i > (middle + j))) {
System.out.print(' ');
} else {
System.out.print(asterisk);
}
} else {
if ((i < (j - middle + 2)) || (i > (2 * num - j - middle))) {
System.out.print(' ');
} else {
System.out.print(asterisk);
}
}
}
System.out.println();
}
}
}
我在我的大学里有确切的课业，这也需要我在 3 个 for 循环中完成。
我就是这样做的。
为了简单的解释，我将钻石分为两部分。
| 不。行数 | 不。空间 | 不。星星 | 总数 插槽数 |
|---|---|---|---|
| 1 | 4 | 1 | 5 |
| 2 | 3 | 3 | 6 |
| 3 | 2 | 5 | 7 |
| 4 | 1 | 7 | 8 |
| 5 | 0 | 9 | 9 |
| 6 | 1 | 7 | 8 |
| 7 | 2 | 5 | 7 |
| 8 | 3 | 3 | 6 |
| 9 | 4 | 1 | 5 |
我想找到那个号码。插槽和编号。每行的空格数，然后分配编号。星星真的很容易。
并考虑到没有。插槽中，第 1 - 5 行和第 6 - 9 行将成为两个独立的组（即 middleLine）。
号的等式。前半部分的插槽数将是 numberOfLines(ie i) + (middleLine - 1)，其中当 maxNumberOfLines 为 9 时，middleLine - 1 将为 4。
号的等式。后半部分的插槽数将是 middleLine（即替换 I）+（middleLine - 1）（即与上述相同）-（i - middleLine），其中当 i = 6 时，i - middleLine 将为 -1。
对于空间，前半部分将是 middleLine - i，后半部分将是 i - middleLine，它们恰好处于负关系（或关于它们的斜率对称）。
public class printOutDiamondWith3Loops {
public static void main(String[] args) {
int userInput = 9;
double maxNumberOfLines = userInput;
// double type is used instead of integer type in order to prevent removal of remainder when a division performed.
double middleLine = Math.ceil(maxNumberOfLines/2);
// Print out the diamond.
for (int i = 1; i <= maxNumberOfLines; i++) {
// Determine the number of lines, which is also the maximum number of slots (the line in the middle).
if (i <= middleLine) {
// Separate the whole diamond into two parts(as mentioned above).
for (int j = 1; j <= i + ((middleLine - 1)); j++) {
// Determine the no. of slots in each line from line 1 to 5.
if (j <= middleLine - i) {
// Determine the no. of spaces and stars.
System.out.print(" ");
} else {
System.out.print("*");
}
}
} else { // i > middleLine
for (int k = 1; k <= (middleLine + (middleLine - 1)) - (i - middleLine); k++) {
// Determine the no. of slots in each line from line 6 to 9.
// For better understanding, I did not simplify the above condition.
// Noticeably, the first middleLine in above for loop is a replacement of i.
if (k <= i - middleLine) {
// Determine the no. of spaces and stars.
System.out.print(" ");
} else {
System.out.print("*");
}
}
}
System.out.println();
}
}
有了这样的框架，做进一步的改变就容易多了，比如让用户输入号码。他们想要的线条。
希望这个答案可以帮助你。
我可以借给你一个更详细的版本，虽然不一定需要（上面的解释已经解释了这些概念）：print-Out-Diamond-With-3-Loops-Advanced-Version.java
package practice;
import java.util.Scanner;
public class Practice {
public static void main(String[] args) {
for (int i = 0; i <= 10; i++) {
if (i <= 5) {
for (int k = 1; k <= 5 - i; k++) {
System.out.print(" ");
}
for (int j = 0; j <= i; j++) {
System.out.print(" *");
}
}
if (i > 5) {
for (int k = 0; k <= i - 6; k++) {
System.out.print(" ");
}
for (int j = 0; j <= 10 - i; j++) {
System.out.print(" *");
}
}
System.out.println();
}
}
}
class Inc_Dec {
public static void main(String[] args) {
int le = 11;
int c = 0;
int j1 = (le / 2) + 1;
int j2 = le - j1;
for (int i = 1; i <= le; i++) {
if (c < j1) {
for (int k = (j1 - i); k > 0; k--) {
System.out.print(" ");
}
for (int j = 1; j <= i; j++) {
System.out.print("*" + " ");
}
c++;
System.out.println();
} else {
for (int k = (i - j1); k > 0; k--) {
System.out.print(" ");
}
for (int j = (le - i + 1); j > 0; j--) {
System.out.print("*" + " ");
}
System.out.println();
}
}
}
}