drawPyramid()函式的遞回版本

我是 C 的新手。我試圖實作一個串行drawPyramid功能。現在，我想drawPyramid_rec為我自己的實踐實作這個函式的遞回版本。但是，我被困了幾個小時。不知道如何處理每一行中的前導空格......我覺得我必須以某種方式存盤n第一個遞回呼叫中的值。或者也許不可能實作drawPyramid? 請幫忙！

#include <stdio.h>

void drawPyramid(int n);

int main(void)
{
    int n;
    do
    {
        printf("Height: ");
        scanf("%i", &n);
    }
    while (n > 8 || n < 1);

    drawPyramid(n);

    return 0;
}

void drawPyramid(int n)
{
    for (int height = 1; height <= n;   height)
    {
        for (int column = (n - height); column >= 1; --column)
        {
            printf(" "); // putchar(' ');
        }

        for (int column = 1; column <= height;   column)
        {
            putchar('#'); // printf("#");
        }

        printf("  ");

        for (int column = 1; column <= height;   column)
        {
            printf("#");
        }

        printf("\n");
    }
}

輸出：

Height: 5
    #  #
   ##  ##
  ###  ###
 ####  ####
#####  #####

uj5u.com熱心網友回復：

遞回函式可以通過以下方式查找，如下面的演示程式所示。

#include <stdio.h>
#include <limits.h>

FILE * drawPyramid( unsigned int n, unsigned int m, FILE *fp )
{
    const char c = '#';

    if ( INT_MAX < n ) n = INT_MAX;

    if ( m < n )
    {
        fprintf( fp, "%*c", n - m , c );
        for ( unsigned int i = 1; i < m   1; i   )
        {
            fputc( c, fp );
        }

        fprintf( fp, "%*c", 3, c );
        for ( unsigned int i = 1; i < m   1; i   )
        {
            fputc( c, fp );
        }
        
        fputc( '\n', fp );

        drawPyramid( n, m   1, fp );
    }

    return fp;
}

int main(void) 
{
    while ( 1 )
    {
        printf( "Enter the height of the pyramid (0 - exit): " );

        unsigned int n;

        if ( scanf( "%u", &n ) != 1 || n == 0 ) break;

        putchar( '\n' );

        drawPyramid( n, 0, stdout );

        putchar( '\n' );;
    }

    return 0;
}

程式輸出可能看起來像

Enter the height of the pyramid (0 - exit): 1

#  #

Enter the height of the pyramid (0 - exit): 2

 #  #
##  ##

Enter the height of the pyramid (0 - exit): 3

  #  #
 ##  ##
###  ###

Enter the height of the pyramid (0 - exit): 4

   #  #
  ##  ##
 ###  ###
####  ####

Enter the height of the pyramid (0 - exit): 5

    #  #
   ##  ##
  ###  ###
 ####  ####
#####  #####

Enter the height of the pyramid (0 - exit): 6

     #  #
    ##  ##
   ###  ###
  ####  ####
 #####  #####
######  ######

Enter the height of the pyramid (0 - exit): 7

      #  #
     ##  ##
    ###  ###
   ####  ####
  #####  #####
 ######  ######
#######  #######

Enter the height of the pyramid (0 - exit): 8

       #  #
      ##  ##
     ###  ###
    ####  ####
   #####  #####
  ######  ######
 #######  #######
########  ########

Enter the height of the pyramid (0 - exit): 9

        #  #
       ##  ##
      ###  ###
     ####  ####
    #####  #####
   ######  ######
  #######  #######
 ########  ########
#########  #########

Enter the height of the pyramid (0 - exit): 10

         #  #
        ##  ##
       ###  ###
      ####  ####
     #####  #####
    ######  ######
   #######  #######
  ########  ########
 #########  #########
##########  ##########

Enter the height of the pyramid (0 - exit): 0

uj5u.com熱心網友回復：

我覺得我必須以某種方式n在第一次遞回呼叫中存盤 的值。

是的，您必須保留 的值n，即金字塔的高度。為此，您可以向函式添加一個drawPyramid永遠不會更改它的額外引數。

void drawPyramid_recursive(int n, int height)
{
    if (height == 0) // base case
    {
        return;
    }

    drawPyramid_recursive(n, height - 1);

    for (int column = (n - height); column >= 1; --column)
    {
        printf(" "); // putchar(' ');
    }

    for (int column = 1; column <= height;   column)
    {
        putchar('#'); // printf("#");
    }

    printf("  ");

    for (int column = 1; column <= height;   column)
    {
        printf("#");
    }

    printf("\n");
}

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