简单递归问题
递归生成
*
**
***
****
****
***
**
*
我的代码如下，不知道有没有更简便的方法？
public static void dd(int begin,int level){
if (level>0){
b(begin);
System.out.print("\n");
dd(begin+1,level-1);
b(begin);
System.out.print("\n");
}
}

public static void b(int n){
if (n>0){
System.out.print("*");
b(n-1);
}
}

[解决办法]

Java code

public static void dd(int n)    {        for(int i=0;i<n;i++)        {            for(int j=0;j<=i+1;j++)            {                System.out.print("*");            }            System.out.print("\n");        }        for(int i=n;i>0;i--)        {            for(int j=0;j<i+1;j++)            {                System.out.print("*");            }            System.out.print("\n");        }    }
[解决办法]
这样行不？
public class test{
public test(int len){
System.out.println(f(len,len,len));
}
public String f(int x,int y,int len){
return y<0?"":(x<=len/2-Math.abs(y-len/2-1)?"*":" ")+(x==len?"\n":"")+f(x==0?len:x-1,x==0?y-1:y,len);
}
public static void main(String args[]){
new test(7);
}
}
[解决办法]
[code=java] 
 public class PrintSign{ 
  private static final int N = 5; 
 public static void main(String[] args) { 
     for(int i=1; i <=N; i++) { 
        System.out.println(PrintSign.Sign(i)); 
      } 
     for(int i=N; i>0; i--) { 
        System.out.println(PrintSign.Sign(i)); 
      } 
    
   } 
   public static String Sign(int n) { 
     if (n == 1) 
      return "*"; 
     else 
       return ("*" + Sign(n-1)); 
  
  } 
 } 
 [/code]
[解决办法]
Java code
public   class   PrintSign{       private   static   final   int   N   =   5; public   static   void   main(String[]   args)   {                 for(int   i=1;   i <=N;   i++)   {                           System.out.println(PrintSign.Sign(i));                   }                 for(int   i=N;   i> 0;   i--)   {                           System.out.println(PrintSign.Sign(i));                   }                     }         public   static   String   Sign(int   n)   {                 if   (n   ==   1)                       return   "*";                 else                         return   ("*"   +   Sign(n-1));           } }