创建具有O和X的交替矩形的矩阵

2021年5月14日15:53:17 发表评论 931 次浏览

本文概述

编写代码, 输入两个数字m和n并创建一个大小为m x n(m行n列)的矩阵, 其中每个元素均为X或0。必须交替填充Xs和0s, 矩阵应具有Xs的最外面的矩形, 然后是0s的矩形, 然后是Xs的矩形, 依此类推。

例子：

Input: m = 3, n = 3
Output: Following matrix 
X X X
X 0 X
X X X

Input: m = 4, n = 5
Output: Following matrix
X X X X X
X 0 0 0 X
X 0 0 0 X
X X X X X

Input:  m = 5, n = 5
Output: Following matrix
X X X X X
X 0 0 0 X
X 0 X 0 X
X 0 0 0 X
X X X X X

Input:  m = 6, n = 7
Output: Following matrix
X X X X X X X
X 0 0 0 0 0 X
X 0 X X X 0 X
X 0 X X X 0 X
X 0 0 0 0 0 X
X X X X X X X

强烈建议你最小化浏览器, 然后自己尝试。

在Shreepartners Gurgaon的校园招聘中提出了这个问题。我遵循以下方法。

1)使用代码打印矩阵在螺旋形式。

2)而不是打印数组, 而是在数组中插入元素" X"或" 0"。

以下是上述方法的实施。

C ++

#include <stdio.h>
  
//Function to print alternating rectangles of 0 and X
void fill0X( int m, int n)
{
     /*  k - starting row index
         m - ending row index
         l - starting column index
         n - ending column index
         i - iterator    */
     int i, k = 0, l = 0;
  
     //Store given number of rows and columns for later use
     int r = m, c = n;
  
     //A 2D array to store the output to be printed
     char a[m][n];
     char x = 'X' ; //Iniitialize the character to be stoed in a[][]
  
     //Fill characters in a[][] in spiral form. Every iteration fills
     //one rectangle of either Xs or Os
     while (k <m && l <n)
     {
         /* Fill the first row from the remaining rows */
         for (i = l; i <n; ++i)
             a[k][i] = x;
         k++;
  
         /* Fill the last column from the remaining columns */
         for (i = k; i <m; ++i)
             a[i][n-1] = x;
         n--;
  
         /* Fill the last row from the remaining rows */
         if (k <m)
         {
             for (i = n-1; i>= l; --i)
                 a[m-1][i] = x;
             m--;
         }
  
         /* Print the first column from the remaining columns */
         if (l <n)
         {
             for (i = m-1; i>= k; --i)
                 a[i][l] = x;
             l++;
         }
  
         //Flip character for next iteration
         x = (x == '0' )? 'X' : '0' ;
     }
  
     //Print the filled matrix
     for (i = 0; i <r; i++)
     {
         for ( int j = 0; j <c; j++)
             printf ( "%c " , a[i][j]);
         printf ( "\n" );
     }
}
  
/* Driver program to test above functions */
int main()
{
     puts ( "Output for m = 5, n = 6" );
     fill0X(5, 6);
  
     puts ( "\nOutput for m = 4, n = 4" );
     fill0X(4, 4);
  
     puts ( "\nOutput for m = 3, n = 4" );
     fill0X(3, 4);
  
     return 0;
}

Java

//Java code to demonstrate the working.
  
import java.io.*;
  
class GFG {
  
//Function to print alternating
//rectangles of 0 and X
  static void fill0X( int m, int n)
{
     /* k - starting row index
         m - ending row index
         l - starting column index
         n - ending column index
         i - iterator */
     int i, k = 0 , l = 0 ;
  
     //Store given number of rows
         //and columns for later use
     int r = m, c = n;
  
     //A 2D array to store
         //the output to be printed
     char a[][] = new char [m][n];
  
         //Iniitialize the character
         //to be stoed in a[][]
     char x = 'X' ; 
  
     //Fill characters in a[][] in spiral
         //form. Every iteration fills
     //one rectangle of either Xs or Os
     while (k <m && l <n)
     {
         /* Fill the first row from the remaining rows */
         for (i = l; i <n; ++i)
             a[k][i] = x;
         k++;
  
         /* Fill the last column from the remaining columns */
         for (i = k; i <m; ++i)
             a[i][n- 1 ] = x;
         n--;
  
         /* Fill the last row from the remaining rows */
         if (k <m)
         {
             for (i = n- 1 ; i>= l; --i)
                 a[m- 1 ][i] = x;
             m--;
         }
  
         /* Print the first column
                 //from the remaining columns */
         if (l <n)
         {
             for (i = m- 1 ; i>= k; --i)
                 a[i][l] = x;
             l++;
         }
  
         //Flip character for next iteration
         x = (x == '0' )? 'X' : '0' ;
     }
  
     //Print the filled matrix
     for (i = 0 ; i <r; i++)
     {
         for ( int j = 0 ; j <c; j++)
             System.out.print(a[i][j] + " " );
         System.out.println();
     }
}
  
/* Driver program to test above functions */
public static void main (String[] args) {
  
     System.out.println( "Output for m = 5, n = 6" );
     fill0X( 5 , 6 );
  
     System.out.println( "Output for m = 4, n = 4" );
     fill0X( 4 , 4 );
  
     System.out.println( "Output for m = 3, n = 4" );
     fill0X( 3 , 4 );
          
     }
}
  
//This code  is contributed by vt_m

Python3

# Python3 program to Create a matrix with
# alternating rectangles of O and X
  
# Function to pralternating rectangles 
# of 0 and X 
def fill0X(m, n):
      
     # k - starting row index 
     # m - ending row index 
     # l - starting column index 
     # n - ending column index 
     # i - iterator 
     i, k, l = 0 , 0 , 0
  
     # Store given number of rows and 
     # columns for later use 
     r = m
     c = n 
  
     # A 2D array to store the output 
     # to be printed 
     a = [[ None ] * n for i in range (m)] 
     x = 'X' # Iniitialize the character to
             # be stoed in a[][] 
  
     # Fill characters in a[][] in spiral form. 
     # Every iteration fills one rectangle of 
     # either Xs or Os 
     while k <m and l <n:
          
         # Fill the first row from the 
         # remaining rows
         for i in range (l, n):
             a[k][i] = x 
         k + = 1
  
         # Fill the last column from 
         # the remaining columns
         for i in range (k, m):
             a[i][n - 1 ] = x 
         n - = 1
  
         # Fill the last row from the 
         # remaining rows 
         if k <m:
             for i in range (n - 1 , l - 1 , - 1 ):
                 a[m - 1 ][i] = x 
             m - = 1
  
         # Print the first column from 
         # the remaining columns 
         if l <n:
             for i in range (m - 1 , k - 1 , - 1 ):
                 a[i][l] = x 
             l + = 1
  
         # Flip character for next iteration
         x = 'X' if x = = '0' else '0'
  
     # Print the filled matrix 
     for i in range (r):
         for j in range (c):
             print (a[i][j], end = " " )
         print ()
  
# Driver Code
if __name__ = = '__main__' :
      
     print ( "Output for m = 5, n = 6" ) 
     fill0X( 5 , 6 ) 
  
     print ( "Output for m = 4, n = 4" ) 
     fill0X( 4 , 4 ) 
  
     print ( "Output for m = 3, n = 4" ) 
     fill0X( 3 , 4 )
      
# This code is contributed by pranchalK

C#

//C# code to demonstrate the working.
using System;
  
class GFG {
  
     //Function to print alternating
     //rectangles of 0 and X
     static void fill0X( int m, int n)
     {
          
         /* k - starting row index
         m - ending row index
         l - starting column index
         n - ending column index
         i - iterator */
         int i, k = 0, l = 0;
      
         //Store given number of rows
         //and columns for later use
         int r = m, c = n;
      
         //A 2D array to store
         //the output to be printed
         char [, ]a = new char [m, n];
      
         //Iniitialize the character
         //to be stoed in a[][]
         char x = 'X' ; 
      
         //Fill characters in a[][] in spiral
         //form. Every iteration fills
         //one rectangle of either Xs or Os
         while (k <m && l <n)
         {
              
             /* Fill the first row from the
             remaining rows */
             for (i = l; i <n; ++i)
                 a[k, i] = x;
             k++;
      
             /* Fill the last column from the
             remaining columns */
             for (i = k; i <m; ++i)
                 a[i, n-1] = x;
             n--;
      
             /* Fill the last row from the
             remaining rows */
             if (k <m)
             {
                 for (i = n-1; i>= l; --i)
                     a[m-1, i] = x;
                 m--;
             }
      
             /* Print the first column
             from the remaining columns */
             if (l <n)
             {
                 for (i = m-1; i>= k; --i)
                     a[i, l] = x;
                 l++;
             }
      
             //Flip character for next
             //iteration
             x = (x == '0' )? 'X' : '0' ;
         }
      
         //Print the filled matrix
         for (i = 0; i <r; i++)
         {
             for ( int j = 0; j <c; j++)
                 Console.Write(a[i, j] + " " );
             Console.WriteLine();
         }
     }
      
     /* Driver program to test 
     above functions */
     public static void Main ()
     {
         Console.WriteLine( "Output for"
                     + " m = 5, n = 6" );
         fill0X(5, 6);
      
         Console.WriteLine( "Output for"
                     + " m = 4, n = 4" );
         fill0X(4, 4);
      
         Console.WriteLine( "Output for"
                     + " m = 3, n = 4" );
         fill0X(3, 4);
     }
}
  
//This code is contributed by Sam007.

的PHP

<?php
//PHP program to Create a matrix with
//alternating rectangles of O and X
  
//Function to print alternating
//rectangles of 0 and X
function fill0X( $m , $n )
{
      
     /* k - starting row index
         m - ending row index
         l - starting column index
         n - ending column index
         i - iterator */
     $k = 0;
     $l = 0;
  
     //Store given number of rows 
     //and columns for later use
     $r = $m ; 
     $c = $n ;
  
     //A 2D array to store the 
     //output to be printed
     //Iniitialize the character
     //to be stoed in a[][]
     $x = 'X' ; 
  
     //Fill characters in a[][] in 
     //spiral form. Every iteration fills
     //one rectangle of either Xs or Os
     while ( $k <$m && $l <$n )
     {
          
         /* Fill the first row from 
            the remaining rows */
         for ( $i = $l ; $i <$n ; ++ $i )
             $a [ $k ][ $i ] = $x ;
         $k ++;
  
         /* Fill the last column from
            the remaining columns */
         for ( $i = $k ; $i <$m ; ++ $i )
             $a [ $i ][ $n - 1] = $x ;
         $n --;
  
         /* Fill the last row from 
            the remaining rows */
         if ( $k <$m )
         {
             for ( $i = $n - 1; $i>= $l ; -- $i )
                 $a [ $m - 1][ $i ] = $x ;
             $m --;
         }
  
         /* Print the first column from
            the remaining columns */
         if ( $l <$n )
         {
             for ( $i = $m - 1; $i>= $k ; -- $i )
                 $a [ $i ][ $l ] = $x ;
             $l ++;
         }
  
         //Flip character for
         //next iteration
         $x = ( $x == '0' )? 'X' : '0' ;
     }
  
     //Print the filled matrix
     for ( $i = 0; $i <$r ; $i ++)
     {
         for ( $j = 0; $j <$c ; $j ++)
             echo ( $a [ $i ][ $j ]. " " );
         echo "\n" ;
     }
}
  
//Driver Code
echo "Output for m = 5, n = 6\n" ;
fill0X(5, 6);
  
echo "\nOutput for m = 4, n = 4\n" ;
fill0X(4, 4);
  
echo "\nOutput for m = 3, n = 4\n" ;
fill0X(3, 4);
  
//This code is contributed by ChitraNayal.
?>

输出如下：

Output for m = 5, n = 6
X X X X X X
X 0 0 0 0 X
X 0 X X 0 X
X 0 0 0 0 X
X X X X X X

Output for m = 4, n = 4
X X X X
X 0 0 X
X 0 0 X
X X X X

Output for m = 3, n = 4
X X X X
X 0 0 X
X X X X

时间复杂度：O(mn)

辅助空间：O(mn)

请建议是否有人在空间和时间方面有更好的解决方案, 并且效率更高。

本文作者：迪帕克·比什特(Deepak Bisht)。如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请发表评论。

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