Java 以倒置的直角形式打印奇数

我试着取一个数字并打印它是奇数，如下所示：

    if i take 5 as a number it should give this: 

    1 3 5
    3 5
    5

   and if i take 9 it should do the same thing:

  1 3 5 7 9
  3 5 7 9
  5 7 9
  7 9
  9

这就是我到目前为止所拥有的，我被困住了。我无法在3之后打印5，并在三角形中以5结尾：

public class first{
    static void afficher(int a){
    for(int i=1;i<=a;i++){
        if(i%2!=0){
            System.out.printf("%d",i);
        }
    }
    System.out.println();

    for(int j=3;j<=a-2;j++){
        if(j%2!=0){
            System.out.printf("%d",j);
        }
    }
}




     public static void main(String[]args){
        afficher(5);


    }

}

---

如果这样打印二维曲面，则该算法的时间复杂度为^2。因此，有两个嵌套的FOR：

通过不检查数字是否为奇数，而是采取步骤2，可以稍微优化算法

---

请参阅。

打印原因如下：

1 3 5 -> your i loop runs here (from 1 to 5)
3     -> your j loop runs here (from 3 to (less than OR equal to 5))

因此，我建议如下：

对通用值使用2个嵌套循环：

 i running from 1 to the input number increasing by 2
 j running from i to the input number increasing by 2 also ending with line change'/n'  

检查输入的数字是否为奇数

---

必须使用嵌套for循环来解决此问题。检查以下代码

public class OddNumberLoop {

 public static void main(String[] args) {

    Scanner inpupt = new Scanner(System.in);

    System.out.print("Input the starting number : ");
    int start = inpupt.nextInt();
    for(int i = 1 ; i <= start; i += 2){
        for(int x = i; x <= start; x += 2) System.out.print(x+ " ");
        System.out.println();
    }

 }
}

---

因为您正在打印曲面，所以在for循环中需要两个嵌套的for.a for循环？

 i running from 1 to the input number increasing by 2
 j running from i to the input number increasing by 2 also ending with line change'/n'  

public class OddNumberLoop {

 public static void main(String[] args) {

    Scanner inpupt = new Scanner(System.in);

    System.out.print("Input the starting number : ");
    int start = inpupt.nextInt();
    for(int i = 1 ; i <= start; i += 2){
        for(int x = i; x <= start; x += 2) System.out.print(x+ " ");
        System.out.println();
    }

 }
}