在 Java 中使用循环的数字模式
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Number Patterns using loops in Java
提问by dev
I have been trying different variations of for loops and have no clue how to make these patterns:
我一直在尝试 for 循环的不同变体，但不知道如何制作这些模式：
Pattern 1
模式一
54321
5432
543
54
5
Pattern 2
模式2
1
12
123
1234
12345
Pattern 3
模式3
12345
2345
345
45
5
Pattern 4
图案 4
1
123
12345
123
1
My code that almost matched pattern 1 is the following but doesn't work like the example above.
我几乎匹配模式 1 的代码如下，但不像上面的例子那样工作。
for (int i = 1 ; i <= rows ; i++) {
for (int j = (rows + 1 - i) ; j > 0 ; j-- ) {
System.out.print(j);
}
System.out.print("\n");
}
采纳答案by Sentimental
public class PrintPattern {
public static void main(String[] args){
printPattern1();
printPattern2();
printPattern3();
printPattern4();
}
public static void printPattern1(){
for (int i = 0; i<5; i++){
for(int j = 5; j>i; j--)
System.out.print(j);
System.out.println();
}
}
public static void printPattern2(){
for (int i = 0; i<5; i++){
for(int k = 0; k<4-i; k++)
System.out.print(" ");
for(int j = 1; j<=i+1; j++)
System.out.print(j);
System.out.println();
}
}
public static void printPattern3(){
for (int i = 0; i<5; i++){
for(int k = 0; k<i; k++)
System.out.print(" ");
for(int j = i+1; j<=5; j++)
System.out.print(j);
System.out.println();
}
}
public static void printPattern4(){
for (int i = 0; i<5; i++){
for(int k = 0; k<Math.abs(2-i); k++)
System.out.print(" ");
for(int j = 1; j<=5-2*Math.abs(2-i); j++)
System.out.print(j);
for (int p = 0; p<Math.abs(2-i); p++)
System.out.print(" ");
System.out.println();
}
}
}
}
回答by mikebolt
Your inner for loop
你的内部 for 循环
for (int j = (rows + 1 - i) ; j > 0 ; j-- ){
System.out.print(j);
}
will always count down to 1, because it keeps going until j is zero. Also, the number that the current row starts on will depend on the current row, because you used i in your assignment to j. To get pattern 1 both of those things will have to change.
将始终倒数到 1，因为它会一直持续到 j 为零。此外，当前行开始的编号将取决于当前行，因为您在分配给 j 时使用了 i。要获得模式 1，这两件事都必须改变。
回答by Elliott Frisch
Since you posted an attempt for pattern one, I'll tell you a solution for pattern one -
既然你发布了模式一的尝试，我会告诉你模式一的解决方案 -
int rows = 5; // <- Start at 5.
for (int i = rows; i > 0; i--) { // <- Use decrementing loop(s).
for (int j = rows; j > rows - i; j--) { // <- Start at 5 (again)
System.out.print(j);
}
System.out.println();
}
Output is pattern 1 in your question,
输出是您问题中的模式 1，
54321
5432
543
54
5
回答by Gee858eeG
public static void main(String[] args) {
int rows = 5;
System.out.println("------ PATTERN 1 ------");
for (int i = 1 ; i <= rows ; i++){
for (int j = rows; j >= i ; j--){
System.out.print(j);
}
System.out.println();
}
System.out.println("\n------ PATTERN 2 ------");
for (int i = 1 ; i <= rows ; i++){
int k;
for (k = rows ; k > i; k--){
System.out.print(" ");
}
for (int j = 1; j <= k ; j++){
System.out.print(j);
}
System.out.println();
}
System.out.println("\n------ PATTERN 3 ------");
for (int i = rows ; i >= 1 ; i--){
int k;
for (k = rows ; k > i; k--){
System.out.print(" ");
}
for (int j = 1; j <= k ; j++){
System.out.print(j);
}
System.out.println();
}
System.out.println("\n------ PATTERN 4 ------");
int whitespaces = rows/2;
for (int i = 1 ; i <= rows; i++){
// absolute value of whitespaces
int abs_whitespaces =
(whitespaces < 0 ? -whitespaces : whitespaces);
for (int j = 0 ; j < abs_whitespaces ; j++){
System.out.print(" ");
}
for (int j = 1 ; j <= rows - 2 * abs_whitespaces ; j++){
System.out.print(j);
}
whitespaces-=1;
System.out.println();
}
}
回答by anshul
The first program is:
第一个程序是：
class timepass {
public static void main() {
for (int a = -1;a<=5;a++) {
for(int b = 5; b >= a;b--) {
System.out.print("*");
}
System.out.println();
// enter code here
}
}
}
The second program is:
第二个程序是：
class timepass {
public static void main() {
for(int i = 1;i<= 6;i++) {
for(int j = 1;j<= i ;j++) {
System.out.print("*");
}
System.out.println();
}
}
}
回答by Sohrab Alam
Try this:
尝试这个：
public class NumberPattern {
public static void main(String[] args) {
for (int i = 1; i <= 3; i++) {
for (int k = 2; k >= i; k--) {
System.out.print(" ");
}
for (int j = 1; j <= (2 * i - 1); j++) {
System.out.print(j);
}
System.out.println();
}
for (int i = 1; i <= 2; i++) {
for (int k = i; k > 0; k--) {
System.out.print(" ");
}
if (i % 2 != 0) {
for (int j = 1; j <= (2 * i + 1); j++) {
System.out.print(j);
}
} else {
for (int j = 1; j <= (i / 2); j++) {
System.out.print(j);
}
}
System.out.println();
}
}
}
}
回答by Harish Yellapragada
import java.util.Scanner;
public class Triangle {
public static void main(String[ ] args){
Scanner scan = new Scanner(System.in);
System.out.println("enter no.of line you need");
int n = scan.nextInt();
for(int i=1;i<=n;i++){
for(int j=5;j>=i;j--){
System.out.print(j);
}
System.out.println(" ");
}}}
