Java递归到输出数字模式

所需输出：

    5
   454
  34543
 2345432
123454321

如何使用递归实现这一点？我的代码思想是：

public static void main(String[] args)
{
      System.out.println(func(5)); 
}
public static String func(int num)
{
     return num + "" +meth(num-1, num, num-1);
}

public static String meth(int start, int num, int end)
{

    if(start==1)
    {
        return "1";
    }
    System.out.println(start+num+end);

    return meth(start-1, num, end-1);
}

我不知道在if语句和System.out.println（）中返回什么， 因为数字5不会减少/增加，因为它将保持不变。例如，它将保持垂直方向上的5，我如何处理这个问题？

---

我的代码更像是一个说明，只是为了证明我正在这样做。

也许这就是您要寻找的：

public class Main {
    public static void main(String[] args) {
        startRecursion(5);
    }

    private static void startRecursion(int number) {
        String aligner = "";
        for (int i = 0; i < number - 1; i++) {
            aligner += " ";
        }
        recursion(String.valueOf(number), number, number, aligner);
    }

    private static void recursion(String value, int startNumber, int lastNumber, String aligner) {
        if (lastNumber < 1) {
            return;
        }

        if (lastNumber != startNumber) {
            value = lastNumber + value + lastNumber;
        }

        System.out.println(aligner + value);

        if (!aligner.isEmpty()) {
            aligner = aligner.substring(0, aligner.length() - 1);
        }

        recursion(value, startNumber, lastNumber - 1, aligner);
    }
}

---

我认为只需通过参数传递num和前一个字符串（即前一行）：

private静态字符串meth（int-num，String-previous）{
字符串空格=”；
对于（int i=0；i您正在调用尚未定义的方法
numberTree
）-我假设这是一个输入错误（应该是
meth
，反之亦然）？是的，没错，我的错误。现在错误已修复。
    5
   454
  34543
 2345432
123454321

private static String meth(int num,String previous) {

     String space="";
    for(int i=0; i<num; i++) space+=" ";
    //If number is negative, return empty String
    if(num<=0) return "";

    //if number is 1, we need to check if previous string is empty or not, because if is empty we need then umber only once, otherwise we need to add to the string
    else if(num==1){
        if(!previous.isEmpty()) return space+num+previous+num;
        else return space+num+"";
    }

    //Here is checked if previous is empty and we do the same as before with number one
    String currentRow=previous.isEmpty()? String.valueOf(num) : num+previous+num;

    //We return the current row (with the current number), and we add the next row (or tree level) passing the number-1 and the row we have
    return space+currentRow+"\n"+meth(num-1,currentRow);

}