我试图打印出这种模式，但是我在排列数字并将它们分开以达到预期的输出时遇到了麻烦。
Expected Output:
1
2 1
3 2 1
4 3 2 1
5 4 3 2 1
6 5 4 3 2 1
1 2 3 4 5 6
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
这是我的代码。我不知道如何分配数字并按照输出指定的顺序输入正确的数字。
import java.util.Scanner;
public class patterns{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
for (int i = 6; i >= 1; i--)
{
for (int j = 1; j < i; j++)
{
System.out.print(" ");
}
for (int j = i; j <= 6; j++)
{
System.out.print(j);
}
System.out.println();
}
System.out.println();
for (int i = 1; i <= 6; i++)
{
for (int j = 1; j < i; j++)
{
System.out.print(" ");
}
for (int j = i; j <= 6; j++)
{
System.out.print(j);
}
System.out.println();
}
}
}
这是输出I＆＃39;获取
run:
6
56
456
3456
23456
123456
123456
23456
3456
456
56
6
答案 0 :(得分：0)
这里是您的模式的修改代码。
public class Patterns {
public static void main(String[] args) {
for (int i = 0; i < 6; i++) {
for (int j = 0; j < (5 - i) * 2; j++) {
System.out.print(" ");
}
for (int j = i + 1; j >= 1; j--) {
System.out.print(j + " ");
}
System.out.println();
}
System.out.println();
for (int i = 0; i < 6; i++) {
for (int j = 0; j < i * 2; j++) {
System.out.print(" ");
}
for (int j = 1; j <= 6 - i; j++) {
System.out.print(j + " ");
}
System.out.println();
}
}
}
输出
1
2 1
3 2 1
4 3 2 1
5 4 3 2 1
6 5 4 3 2 1
1 2 3 4 5 6
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
更新：修复空格错误
工作=＆GT;考虑模式的前半部分
1 line 0: 10 spaces = (5 - 0) * 2 spaces
2 1 line 1: 8 spaces = (5 - 1) * 2 spaces
3 2 1 line 2: 6 spaces = (5 - 2) * 2 spaces
4 3 2 1 line 3: 4 spaces = (5 - 3) * 2 spaces
5 4 3 2 1 line 4: 2 spaces = (5 - 4) * 2 spaces
6 5 4 3 2 1 line 5: 0 spaces = (5 - 5) * 2 spaces
考虑模式的后半部分
1 2 3 4 5 6 line 0: 0 spaces = 0 * 2 spaces
1 2 3 4 5 line 1: 2 spaces = 1 * 2 spaces
1 2 3 4 line 2: 4 spaces = 2 * 2 spaces
1 2 3 line 3: 6 spaces = 3 * 2 spaces
1 2 line 4: 8 spaces = 4 * 2 spaces
1 line 5: 10 spaces = 5 * 2 spaces
答案 1 :(得分：0)
你应该认为每个三角形块实际上有6x6个方格单元。所以这意味着你需要2个嵌套循环，每个循环迭代6次。
然后在内部，您将决定是否打印数字或空格，具体取决于某些条件。
for (int i = 6; i >= 1; i--)
{
for (int j = 1; j <= 6; j++)
{
if (j < i)
{
System.out.print(" ");
} else
{
System.out.print(7 - j + " ");
}
}
System.out.println("");
}
System.out.println("");
for (int i = 1; i <= 6; i++)
{
for (int j = 1; j <= 6; j++)
{
if (j < i)
{
System.out.print(" ");
} else
{
System.out.print(j-i+1 + " ");
}
}
System.out.println("");
}
1
2 1
3 2 1
4 3 2 1
5 4 3 2 1
6 5 4 3 2 1
1 2 3 4 5 6
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
答案 2 :(得分：0)
这样做的一种方法是：
import java.util.Scanner;
public class pattern{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
for (int i = 1; i <= 6; i++)
{
String val = "";
for (int j = i; j >= 1; j--)
{
val += j+ " ";
}
System.out.printf("%12s", val);
System.out.println();
}
System.out.println();
for (int i = 6; i >= 1; i--)
{
String val = "";
int ii=1;
for (int j = i; j >= 1; j--)
{
val += ii + " ";
ii++;
}
System.out.printf("%12s", val);
System.out.println();
}
}
输出：
1
2 1
3 2 1
4 3 2 1
5 4 3 2 1
6 5 4 3 2 1
1 2 3 4 5 6
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
要记住的一件事是，当您尝试格式化文本时，应使用printf而不是添加打印行。
答案 3 :(得分：0)
int a = 654321;
String str = new Integer(a).toString();
for (int i = str.length() - 1; i >= 0 ; i--) {
String g = String.format("%6s", str.substring(i));
System.out.println(g.replaceAll(".(?=.)", "$0 "));
}
System.out.println();
str = new StringBuffer(str).reverse().toString();
for (int i = str.length() - 1; i >= 0 ; i--) {
String g = String.format("%6s", str.substring(0, i+1));
System.out.println(g.replaceAll(".(?=.)", "$0 "));
}
答案 4 :(得分：0)
最好的方法是使用数组 - ＆gt;你循环次数减少了。 http://ideone.com/diYtvM干杯！
import java.util.Scanner;
public class patterns{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int[][] arr = new int[6][6];
for(int i = 0; i<6; i++)
arr[i][5] = 1;
for(int i = 1; i<6; i++)
for (int j = 4; j>=5-i; j--)
{
arr[i][j] = arr[i-1][j+1] + 1;
}
for(int i = 0; i<6; i++)
{
for (int j = 0; j<6; j++)
{
if(arr[i][j] > 0)
System.out.print(arr[i][j] + " ");
else
System.out.print(" ");
}
System.out.println("");
}
System.out.println("");
int z = 0;
for(int i = 5; i>=0; i--)
{
for(int k = 0; k<z; k++)
System.out.print(" ");
for (int j = 5; j>=0; j--)
{
if(arr[i][j] > 0)
System.out.print(arr[i][j] + " ");
}
System.out.println("");
z++;
}
}
}
OUT：
1
2 1
3 2 1
4 3 2 1
5 4 3 2 1
6 5 4 3 2 1
1 2 3 4 5 6
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1