Java递归输出数字模式

Question

要求的输出：

    5
   454
  34543
 2345432
123454321

我该如何使用递归呢？ 我有代码的想法是：

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);
}

我对if语句和System.out.println（）中的返回值感到困惑，因为数字5不会减少/增加，因为它将保持不变，例如它将垂直保持为5，我该如何处理问题？ 我的代码只是为了说明我正在做的例子。

Answer 1

也许这就是您想要的：

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);
    }
}

印刷品：

    5
   454
  34543
 2345432
123454321

Answer 2

我认为只是通过参数和前一个String（这是前一行）传递num：

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);

}