所以我被问到这个问题,我只能解决代码的顶部,我被困在底部。
编写一个名为EmptyDiamond.java的Java程序,其中包含一个采用
的方法
整数N并在2N - 1行打印空菱形,如下所示
当n = 3时的样本输出。
1
2 2
3 3
2 2
1
到目前为止,这是我的代码:
public static void shape(int n){
//TOP PART
for(int i=1; i<=(n-1) ; i++){
System.out.print(" ");
}
System.out.println(1);
for(int i=2; i<=n; i++){
for(int j=1; j<=(n-i); j++){
System.out.print(" ");
}
System.out.print(i);
for(int j=1; j<=2*i-n+1; j++){
System.out.print(" ");
}
System.out.println(i);
}
//BOTTOM PART(The messed up part)
for(int i=n+1; i<=2*n-2; i++){
for(int j=1; j<=n-i; j++){
System.out.print(" ");
}
System.out.print(i);
for(int j=1; j<=n; j++){
System.out.print(" ");
}
System.out.print(i);
}
for(int i=1; i<=(n-1) ; i++){
System.out.print(" ");
}
System.out.println(1);
}
public static void main(String[]args){
shape(4);
}
答案 0 :(得分:0)
以下是打印空钻石的程序:
int n = 3; //change the value of n to increase the size of diamond
int upperCount = 1;
for(int i=n; i>=1; i--){
for(int j=i; j>=1; j--){
System.out.print(" ");
}
System.out.print(upperCount);
for(int j=0; j<=upperCount-2; j++){
System.out.print(" ");
}
for(int j=0; j<=upperCount-2; j++){
System.out.print(" ");
}
if(upperCount!=1){
System.out.print(upperCount);
}
upperCount++;
System.out.print("\n");
}
int lowerCount = n-1;
for(int i=1; i<=n-1; i++){
for(int j=0; j<=i; j++){
System.out.print(" ");
}
System.out.print(lowerCount);
for(int j=0; j<=lowerCount-2; j++){
System.out.print(" ");
}
for(int j=0; j<=lowerCount-2; j++){
System.out.print(" ");
}
if(lowerCount!=1){
System.out.print(lowerCount);
}
lowerCount--;
System.out.print("\n");
}
在代码的底部进行以下更改:
int lowerCount = n-1;
for(int i=n-1; i>=2; i--){
for(int j=1; j<=(n-i); j++){
System.out.print(" ");
}
System.out.print(i);
for(int j=1; j<=lowerCount; j++){
System.out.print(" ");
}
System.out.print(i);
lowerCount-=2;
}
答案 1 :(得分:0)
您可以在从 -n
到 n
的行和列上使用两个嵌套的 for 循环打印带有数字的空菱形形状。当 iAbs + jAbs == n
:
int n = 2;
for (int i = -n; i <= n; i++) {
// absolute value of 'i'
int iAbs = Math.abs(i);
for (int j = -n; j <= n; j++) {
// absolute value of 'j'
int jAbs = Math.abs(j);
// empty diamond shape
System.out.print(iAbs + jAbs == n ? jAbs + 1 : " ");
if (j < n) {
System.out.print(" ");
} else {
System.out.println();
}
}
}
输出:
1
2 2
3 3
2 2
1
你可以分别定义width和height:
int m = 4;
int n = 2;
int max = Math.max(m, n);
for (int i = -m; i <= m; i++) {
// absolute value of 'i'
int iAbs = Math.abs(i);
for (int j = -n; j <= n; j++) {
// absolute value of 'j'
int jAbs = Math.abs(j);
// empty diamond shape
System.out.print(iAbs + jAbs == max ? jAbs + 1 : " ");
if (j < n) {
System.out.print(" ");
} else {
System.out.println();
}
}
}
输出:
1
2 2
3 3
3 3
2 2
1
另见:
• Filling a 2d array with numbers in a rhombus form
• How to print a diamond of random numbers?
答案 2 :(得分:0)
使用作为 Java-11 一部分引入的 String#repeat
,您可以使用单循环来实现。
public class Main {
public static void main(String[] args) {
int n = 3;
for (int i = 1 - n; i < n; i++) {
int x = Math.abs(i);
System.out.println(" ".repeat(x) + (n - x)
+ " ".repeat(Math.abs((n - x) * 2 - 3))
+ ((i == 1 - n || i == n - 1) ? "" : (n - x)));
}
}
}
输出:
1
2 2
3 3
2 2
1
您只需将空间量增加一个字符即可打印钻石的变体:
public class Main {
public static void main(String[] args) {
int n = 3;
for (int i = 1 - n; i < n; i++) {
int x = Math.abs(i);
System.out.println(" ".repeat(x) + (n - x)
+ " ".repeat(Math.abs((n - x) * 2 - 3))
+ ((i == 1 - n || i == n - 1) ? "" : (n - x)));
}
}
}
输出:
1
2 2
3 3
2 2
1
答案 3 :(得分:0)
替代解决方案:
public static void main(String[] args) {
int n = 7;
for (int i = -n; i <= n; i++) {
for (int j = -n; j <= n; j++) {
// edge of the diamond
int edge = Math.abs(i) + Math.abs(j);
// diamond shape with numbers
if (edge == n) System.out.print(n - Math.abs(i) + 1);
// beyond the edge && in chessboard order || vertical borders
else if (edge > n && (i + j) % 2 != 0 || Math.abs(j) == n)
System.out.print("*");
// empty part
else System.out.print(" ");
}
System.out.println();
}
}
输出:
** * * 1 * * **
* * * 2 2 * * *
** * 3 3 * **
* * 4 4 * *
** 5 5 **
* 6 6 *
*7 7*
8 8
*7 7*
* 6 6 *
** 5 5 **
* * 4 4 * *
** * 3 3 * **
* * * 2 2 * * *
** * * 1 * * **
答案 4 :(得分:0)
可能有点晚了,但是因为您的消息的底部只是镜像的第一部分,所以您可以使用 Stack 以相反的顺序打印消息:
public static void main(String[] args) {
int maxNumber = 3;
Stack<String> message = new Stack<>();
// upper part
for(int row = 0; row < maxNumber; row++) {
int prefix = maxNumber - (row + 1);
int spaces = row >= 2 ? row * 2 - 1 : row;
String line = getLine(row, prefix, spaces);
System.out.println(line);
if(row != maxNumber - 1)
message.add(line);
}
// bottom part
while(!message.isEmpty())
System.out.println(message.pop());
}
public static String getLine(int row, int prefix, int spaces) {
StringBuilder line = new StringBuilder("_".repeat(prefix));
line.append(row+1);
if(row != 0) {
line.append("_".repeat(spaces));
line.append(row+1);
}
return line.toString();
}
输出:
__1
_2_2
3___3
_2_2
__1
您当然可以使用任何您想要填充堆栈的方法(即生成消息的上部),就像这个问题建议的方法一样。我描述的上半部分包含第一行(包含)到中间行(不包含)。
答案 5 :(得分:0)
解决方案:名为 EmptyDiamond.java
的 Java 程序包含一个方法,该方法接受一个整数 n
并在 2n − 1
行上打印一个空菱形。
public class EmptyDiamond {
public static void main(String[] args) {
shape(3); // Change n to increase size of diamond
}
public static void shape(int n) {
int max = 2 * n - 1; // length of the diamond - top to bottom
int loop = 0; // with of each loop. initialized with 0
for (int i = 1; i <= max; i++) {
int val = 0;
if (i <= n) {
loop = n + i - 1;// e.g. when i = 2 and n = 3 loop 4 times
val = i; // value to be printed in each loop ascending
} else {
loop = n + (max - i); //e.g. when i = 4 and n = 3 loop 4 times
val = max - i + 1; // value to be printed in each loop descending
}
for (int j = 1; j <= loop; j++) {
// (value end of loop)
// || (value in the beginning when i <= n)
// || (value in the beginning when i > n)
if (j == loop
|| j == (n - i + 1)
|| j == (n - val + 1)) {
System.out.print(val); // Print values
} else {
System.out.print(" "); // Print space
}
}
System.out.println(); // Print next line
}
}
}
当n = 3
时输出:
1
2 2
3 3
2 2
1
答案 6 :(得分:0)
仅流功能:
public static void printEmptyDiamond(int n) {
IntStream.range(1, 2*n)
.map(i-> i > n ? 2*n-i : i)
.mapToObj(i -> " ".repeat(n-i) + i + (i>1 ? " ".repeat(2*(i-1)-1)+i : ""))
.forEach(System.out::println);
}
示例输出 (printEmptyDiamond(7)
):
1
2 2
3 3
4 4
5 5
6 6
7 7
6 6
5 5
4 4
3 3
2 2
1
附说明:
public static void printEmptyDiamond(int n) {
IntStream.range(1, 2*n)
.map(i-> i > n? 2*n-i : i) // numbers from 1 to n ascending, then descending to 1 again
.mapToObj(i -> " ".repeat(n-i) // leading spaces
+ i // leading number
+ (i>1 ? // only when number is > 1
" ".repeat(2*(i-1)-1) // middle spaces
+ i // trailing number
: ""))
.forEach (System.out::println);
}
答案 7 :(得分:0)
我这样做是为了好玩,这是代码:
import java.util.Scanner;
public class Diamond {
public static void main(String[] args) {
Scanner read = new Scanner(System.in);
int num = read.nextInt();
read.nextLine();
//TOP
for(int i = 1;i<=num;i++) {
//LEFT
for(int k = i; k<num;k++) {
if ( k % 2 == 0 ) {
System.out.print(" ");
}
else {
System.out.print(" ");
}
}
if(i>1) {
for(int j =1;j<=i;j++) {
if (j==1 || j== i) {
for(int u=0;u<j;u++) {
System.out.print(" ");
}
System.out.print(i);
}
else {
System.out.print(" ");
}
}
System.out.println("");
}
else {
System.out.println(" "+i);
}
}
//BOTTOM
for(int i = num-1;i>0;i--) {
for(int k = i; k<num;k++) {
if ( k % 2 == 0 ) {
System.out.print(" ");
}
else {
System.out.print(" ");
}
}
if(i>1) {
for(int j =1;j<=i;j++) {
if (j==1 || j== i) {
for(int u=0;u<j;u++) {
System.out.print(" ");
}
System.out.print(i);
}
else {
System.out.print(" ");
}
}
System.out.println("");
}
else {
System.out.println(" "+i);
}
}
}
}
和输出:
7
1
2 2
3 3
4 4
5 5
6 6
7 7
6 6
5 5
4 4
3 3
2 2
1
看过其他答案后,我可以跳过一大堆循环。只是决定在最后振作起来并尽快完成。