java - 如何以更干净的方式使用递归在沙漏下面进行编码？

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最近，我遇到了一个关于在 java 中使用递归编码沙漏的小测验。我设法按如下方式对其进行编码，但希望接受是否有更好的编码方法。

下面是我完成的编码部分。

 public class Demo {
        int initialInput = 0;
        int noOfAsteriskForNextLine= 0 ;
        int noOfSpace =1;
        Boolean isReversed = false;`enter code here`
        public static void main (String[] args){
            Demo d = new Demo();
            d.run();
        }
        public void printLineOfAsterisk(int noOfAsterisk){
            if(initialInput== 0){
                initialInput=noOfAsterisk;
            }

            if(noOfAsteriskForNextLine==0){
                noOfAsteriskForNextLine=noOfAsterisk;
            }

                for(int i=0;i<noOfSpace;i++){
                System.out.print(" ");
            }
            for(int i=0;i<noOfAsterisk;i++){
                System.out.print(" *");
            }
            System.out.print("\n");
                     if(isReversed&&initialInput==noOfAsterisk){
                return;
            }
            if(isReversed||noOfAsteriskForNextLine==1){
                isReversed =true;
                noOfAsteriskForNextLine++;
                noOfSpace--;
             printLineOfAsterisk(noOfAsteriskForNextLine);

            }else{
            noOfSpace++;
            noOfAsteriskForNextLine--;
            printLineOfAsterisk(noOfAsteriskForNextLine);
        }
    }
        public void run(){
            printLineOfAsterisk(4);
        }

    }

预期输出如下，取决于传入的参数

  * * * *
   * * *
    * *
     *
    * *
   * * *
  * * * *

Answer 1

试试这个。这里 offset 就是 no。在星星之前给予的空间。最初，该函数将使用 offset = 0 调用，这意味着在第一行中不会打印空格。

starCount 是编号。打印空格后在行中打印的星星数。

increment 只是增加或减少编号。空格和星号最初是 1 意味着没有。下一次迭代中的空格数将增加 1 和 no。星星数量将减少 1。

请参阅代码中的注释以获取更多说明

import java.util.*;

public class Main
{
        public static void hourGlass(int offset, int starCount, int increment) {
              // This is the terminating condition. 
              // If offset is negative and increment is also negative, it means that no. of spaces were positive before this function call.
              // This implies that the lower half of the hourglass is completed,
              // So just return from here without executing anything
              if(offset < 0 && increment < 0)
                     return;

              // Print spaces
              for(int i=0; i<offset; i++) 
                     System.out.print(" ");
              // Print stars with a space 
              for(int i=0; i<starCount; i++) 
                     System.out.print("* ");
              // Print a new line 
              System.out.println();

              // If no. of stars is 1, then it means the upper half is completed.
              // So, here the increment is set to -1 
              // So that from next recursive call onwards stars will increase and spaces will decrease by 1
              if(starCount == 1)
                     increment = -1;

              // Call this function recursively
              hourGlass(offset + increment, starCount - increment, increment);
        }

        public static void main(String[] args)
        {
               // first argument is no. of spaces which will be zero initially
               // second arguments is number of starts, change it according to your need
               // third argument is just to increment/decrement stars & spaces
               Main.hourGlass(0, 5, 1);
        }
}

输出:

* * * * *
 * * * *
  * * *
   * *
    *
   * *
  * * *
 * * * *
* * * * *