只使用递归从星星中创建一个三角形

注意在您的迭代方法中,您有两个计数器:第一个是您所在的行 line ,第二个是你在这条线上的位置 x y 和 十

你也可以注意到三角形中的每一条连续的线都是前一条线加上一颗星。如果递归函数为前一行返回一个星号字符串,那么下一行总是该字符串加上一个星号。所以,你的代码应该是这样的:

public String printTriangle (int count) {
    if( count <= 0 ) return "";

    String p = printTriangle(count - 1);
    p = p + "*";
    System.out.println(p);

    return p;
 }

#!/usr/bin/env python

def printTriangle(n):
    if n > 1:
        printTriangle(n - 1)
    # now that we reached 1, we can start printing out the stars 
    # as we climb out the stack ...
    print '*' * n

if __name__ == '__main__':
    printTriangle(5)

输出如下所示:

$ python 2717111.py
*
**
***
****
*****

void printStars(int count) {
    if (count == 0) return;

    System.out.print("*");
    printStars(count - 1);
}
printStars(5);    //Prints 5 stars

您还可以使用单个(不太优雅的)递归来完成,如下所示:

public static void printTriangle (int leftInLine, int currLineSize, int leftLinesCount) {
    if (leftLinesCount == 0)
        return;
    if (leftInLine == 0){ //Completed current line?
        System.out.println();
        printTriangle(currLineSize+1, currLineSize+1, leftLinesCount-1);
    }else{
        System.out.print("*");
        printTriangle(leftInLine-1,currLineSize,leftLinesCount);
    }
}

public static void printTriangle(int size){
    printTriangle(1, 1, size);
}

其思想是方法参数表示完整的绘图状态。

该方法以恒星数作为参数。我们叫它n。

1. 使用n-1递归调用自身。

2. 打印一条有n个星星的线。

如果n==0,请确保不执行任何操作。

package playground.tests;

import junit.framework.TestCase;

public class PrintTriangleTest extends TestCase {
    public void testPrintTriangle() throws Exception {
        assertEquals("*\n**\n***\n****\n*****\n", printTriangleRecursive(5, 0, 0));
    }

    private String printTriangleRecursive(int count, int line, int character) {
        if (line == count)
            return "";
        if (character > line)
            return "\n" + printTriangleRecursive(count, line + 1, 0);
        return "*" + printTriangleRecursive(count, line, character + 1);
    }

}

    void trianglePrint(int rows){
            int static currentRow = 1;
            int static currentStar = 1;

            // enter new line in this condition
            // (star > currentrow)  

            if (currentStar > currentRow ){
                currentStar = 1;
                currentRow++;
                cout << endl;
            }

            if (currentRow > rows){
                return; // finish
            }

            cout << "*";
            currentStar++;

            trianglePrint(rows);
        }

#include<iostream>
using namespace std;
 void ll__(int x){
  char static c = '0'; // character c will determine when to put newline
  if(!x){
    if(c=='1'){
      cout<<'\n';
    }
    return;
  }
  if(c=='0'){
    ll__(x-1); //  rows to be called in the stack and differentiated by character '0'
  }
  if(x==1 && c=='0'){
   cout<<'*';
  }else{  // columns to be printed in every row as per the row number in the stack
   c = '1';
   ll__(x-1);
   cout<< '*';
  }
}

int main(){
//writes code here
 ll__(5);
 exit(0);
}

public void printTriangle(int count)
{    
    if (count == 0) return;
    printTriangle(count - 1);
    for (int x = 1; x <= count; x++) { 
        System.out.print("*"); 
    }
    System.out.print("\n"); 
}

所以,你需要创建一个小的块。那个街区需要什么信息?只是最大值。但是递归需要知道它在哪条线上。。。你会得到一个构造函数,比如:

public void printTriangle (int current, int max)

public void printTriangle (int current, int max)
{ 
    if (current <= max) 
    { 
         // Draw the line of stars...
         for (int x=0; x<current; x++)
         {
             System.out.print("*")
         }
         // add a newline
         System.out.print("\n"); 

         // Do it again for the next line, but make it 1-bigger
         printTriangle(current + 1, max);
    } 
} 

现在,你要做的就是启动它:

printTriangle(1, 5);

我想应该这样做

public void printTriangle (int count) {
   if(count >= 0) {
      printTriangle(count-1);
          for(int i = 0; i < count; i++) {
              System.out.print("*" + " ");
          }
      System.out.println(); 
   }
}