需要帮助集中文本输出吗 import java.util.Scanner; 公共级钻石{ 公共静态void main(字符串[]args){ 扫描仪扫描=新扫描仪(System.in); System.out.println(“请输入一个整数”); int lines=scan.nextInt(); 对于(int counter=1;counter
,您需要在每行之前打印一定数量的空格。然后,您需要另一个for循环来执行相反的操作。 请尝试以下代码:需要帮助集中文本输出吗 import java.util.Scanner; 公共级钻石{ 公共静态void main(字符串[]args){ 扫描仪扫描=新扫描仪(System.in); System.out.println(“请输入一个整数”); int lines=scan.nextInt(); 对于(int counter=1;counter,java,text,center,output,Java,Text,Center,Output,,您需要在每行之前打印一定数量的空格。然后,您需要另一个for循环来执行相反的操作。 请尝试以下代码: import java.util.Scanner; public class diamond { public static void main (String[] args){ Scanner scan = new Scanner(System.in); System.out.println("Please enter an integer");
import java.util.Scanner;
public class diamond {
public static void main (String[] args){
Scanner scan = new Scanner(System.in);
System.out.println("Please enter an integer");
int lines = scan.nextInt();
for(int counter = 1; counter <= lines; counter++)
{
if (counter%2 != 0)
{
for(int count2 = 1; count2 <= counter; count2++){
System.out.print("*");
}
System.out.println();
}
}
}
}
publicstaticvoidmain(字符串[]args){
扫描仪扫描=新扫描仪(System.in);
System.out.println(“请输入一个整数”);
int lines=scan.nextInt();
对于(int counter=1;counter我认为您应该准备一个字符串在每行上打印出来,这样您就可以确切地知道它有多少个字符,当行数增加时,删除字符串中心的两个“*”并在前面添加一个“”,然后再次打印出来。您需要计算在哪里打印空格(”
)星号在哪里。
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Please enter an integer ");
int lines = scan.nextInt();
for (int counter = 1; counter <= lines; counter++) {
if (counter % 2 != 0) {
for (int i = 0; i < lines - (counter / 2) - 3; i++) {
System.out.print(" ");
}
for (int count2 = 1; count2 <= counter; count2++) {
System.out.print("*");
}
System.out.println();
}
}
for (int counter = lines - 1; counter >= 1; counter--) {
if (counter % 2 != 0) {
for (int i = 0; i < lines - (counter / 2) - 3; i++) {
System.out.print(" ");
}
for (int count2 = 1; count2 <= counter; count2++) {
System.out.print("*");
}
System.out.println();
}
}
}