创建具有O和X的交替矩形的矩阵 _数据结构

本文概述

C ++
Java
Python3
C#
的PHP

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。
【创建具有O和X的交替矩形的矩阵】例子：

Input: m = 3, n = 3 Output: Following matrix X X X X 0 X X X XInput: 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 XInput: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 XInput: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 codeis 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# 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 XOutput for m = 4, n = 4 X X X X X 0 0 X X 0 0 X X X X XOutput 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。