编写一个代码，输入两个数字m和n，并创建一个大小为m x n（m行和n列）的矩阵，其中每个元素都是x或0。X和0必须交替填充，矩阵应该有最外面的X矩形，然后是0矩形，然后是X矩形，依此类推。
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例如：
Input: m = 3, n = 3Output: Following matrix X X XX 0 XX X XInput: m = 4, n = 5Output: Following matrixX X X X XX 0 0 0 XX 0 0 0 XX X X X XInput: m = 5, n = 5Output: Following matrixX X X X XX 0 0 0 XX 0 X 0 XX 0 0 0 XX X X X XInput: m = 6, n = 7Output: Following matrixX X X X X X XX 0 0 0 0 0 XX 0 X X X 0 XX 0 X X X 0 XX 0 0 0 0 0 XX X X X X X X
我们强烈建议尽量减少浏览器，并先自己尝试。 这个问题是在什里帕特纳·古尔冈的校园招聘中提出的。我遵循以下方法。 1) 使用 螺旋形打印矩阵的代码 . 2) 在数组中交替插入元素“X”或“0”，而不是打印数组。
以下是上述方法的实现。
C++
#include <bits/stdc++.h> using namespace std; // 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' ; // Initialize the character to be stored 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++) cout << " " << a[i][j]; cout << "" ; } } /* Driver program to test above functions */ int main() { puts ( "Output for m = 5, n = 6" ); fill0X(5, 6); puts ( "Output for m = 4, n = 4" ); fill0X(4, 4); puts ( "Output for m = 3, n = 4" ); fill0X(3, 4); return 0; } // This code is contributed by shivanisinghss2110 |
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' ; // Initialize the character to be stored 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 ( "" ); } } /* Driver program to test above functions */ int main() { puts ( "Output for m = 5, n = 6" ); fill0X(5, 6); puts ( "Output for m = 4, n = 4" ); fill0X(4, 4); puts ( "Output 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]; // Initialize the character // to be stored 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 print alternating 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' # Initialize the character to # be stored 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]; // Initialize the character // to be stored 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. |
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 // Initialize the character // to be stored 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 "" ; } } // Driver Code echo "Output for m = 5, n = 6" ; fill0X(5, 6); echo "Output for m = 4, n = 4" ; fill0X(4, 4); echo "Output for m = 3, n = 4" ; fill0X(3, 4); // This code is contributed by ChitraNayal. ?> |
Javascript
<script> // JavaScript code to demonstrate the working. // 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 */ let i, k = 0, l = 0; // Store given number of rows // and columns for later use let r = m, c = n; // A 2D array to store // the output to be printed let a = new Array(m); for (let i = 0; i < m; i++) { a[i] = new Array(n); } // Initialize the character // to be stored in a[][] let 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 (let j = 0; j < c; j++) { document.write(a[i][j] + " " ); } document.write( "<br>" ); } } // Driver Code document.write( "Output for m = 5, n = 6 <br>" ); fill0X(5, 6); document.write( "Output for m = 4, n = 4<br>" ); fill0X(4, 4); document.write( "Output for m = 3, n = 4<br>" ); fill0X(3, 4); // This code is contributed by rag2127 </script> |
输出：
Output for m = 5, n = 6X X X X X XX 0 0 0 0 XX 0 X X 0 XX 0 0 0 0 XX X X X X XOutput for m = 4, n = 4X X X XX 0 0 XX 0 0 XX X X XOutput for m = 3, n = 4X X X XX 0 0 XX X X X
时间复杂度：O（mn） 辅助空间：O（mn）
请建议是否有更好的解决方案，在空间和时间方面更有效。 本文由 迪帕克·比什特 。如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请发表评论
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