给定一个数字N，任务是打印以下图案：-
null
例如：
Input : N = 4Output : 4 4 4 4 4 4 4 4 3 3 3 3 3 4 4 3 2 2 2 3 4 4 3 2 1 2 3 4 4 3 2 2 2 3 4 4 3 3 3 3 3 4 4 4 4 4 4 4 4Input : N = 2Output : 2 2 2 2 1 2 2 2 2
方法1： 常见的观察结果是，这样形成的正方形大小为（2*N-1）x（2*N-1）。用N填充第一行和第列，最后一行和第列，然后逐渐减少N，并以同样的方式填充剩余的行和第列。填充两行两列后，每次减少N。
以下是上述方法的实施情况：
C++
// C++ program to print the // spiral pattern #include <bits/stdc++.h> using namespace std; // Function to print the pattern void pattern( int value) { // Declare a square matrix int row = 2 * value - 1; int column = 2 * value - 1; int arr[row][column]; int i, j, k; for (k = 0; k < value; k++) { // store the first row // from 1st column to last column j = k; while (j < column - k) { arr[k][j] = value - k; j++; } // store the last column // from top to bottom i = k + 1; while (i < row - k) { arr[i][row - 1 - k] = value - k; i++; } // store the last row // from last column to 1st column j = column - k - 2; while (j >= k) { arr[column - k - 1][j] = value - k; j--; } // store the first column // from bottom to top i = row - k - 2; while (i > k) { arr[i][k] = value - k; i--; } } // print the pattern for (i = 0; i < row; i++) { for (j = 0; j < column; j++) { cout << arr[i][j] << " " ; } cout << endl; } } // Driver code int main() { int n = 5; pattern(n); return 0; } |
JAVA
// Java program to print // the spiral pattern class GFG { // Function to print the pattern static void pattern( int value) { // Declare a square matrix int row = 2 * value - 1 ; int column = 2 * value - 1 ; int [][] arr = new int [row][column]; int i, j, k; for (k = 0 ; k < value; k++) { // store the first row // from 1st column to last column j = k; while (j < column - k) { arr[k][j] = value - k; j++; } // store the last column // from top to bottom i = k + 1 ; while (i < row - k) { arr[i][row - 1 - k] = value - k; i++; } // store the last row // from last column // to 1st column j = column - k - 2 ; while (j >= k) { arr[column - k - 1 ][j] = value - k; j--; } // store the first column // from bottom to top i = row - k - 2 ; while (i > k) { arr[i][k] = value - k; i--; } } // print the pattern for (i = 0 ; i < row; i++) { for (j = 0 ; j < column; j++) { System.out.print(arr[i][j] + " " ); } System.out.println(); } } // Driver code public static void main(String[] args) { int n = 5 ; pattern(n); } } // This code is contributed // by ChitraNayal |
Python3
# Python3 program to print # the spiral pattern # Function to print the pattern def pattern(value): # Declare a square matrix row = 2 * value - 1 column = 2 * value - 1 arr = [[ 0 for i in range (row)] for j in range (column)] for k in range ( value): # store the first row # from 1st column to # last column j = k while (j < column - k): arr[k][j] = value - k j + = 1 # store the last column # from top to bottom i = k + 1 while (i < row - k): arr[i][row - 1 - k] = value - k i + = 1 # store the last row # from last column # to 1st column j = column - k - 2 while j > = k : arr[column - k - 1 ][j] = value - k j - = 1 # store the first column # from bottom to top i = row - k - 2 while i > k : arr[i][k] = value - k i - = 1 # print the pattern for i in range (row): for j in range (column): print (arr[i][j], end = " " ) print () # Driver code if __name__ = = "__main__" : n = 5 pattern(n) # This code is contributed # by ChitraNayal |
C#
// C# program to print // the spiral pattern using System; class GFG { // Function to print the pattern static void pattern( int value) { // Declare a square matrix int row = 2 * value - 1; int column = 2 * value - 1; int [, ] arr = new int [row, column]; int i, j, k; for (k = 0; k < value; k++) { // store the first row // from 1st column to // last column j = k; while (j < column - k) { arr[k, j] = value - k; j++; } // store the last column // from top to bottom i = k + 1; while (i < row - k) { arr[i, row - 1 - k] = value - k; i++; } // store the last row // from last column // to 1st column j = column - k - 2; while (j >= k) { arr[column - k - 1, j] = value - k; j--; } // store the first column // from bottom to top i = row - k - 2; while (i > k) { arr[i, k] = value - k; i--; } } // print the pattern for (i = 0; i < row; i++) { for (j = 0; j < column; j++) { Console.Write(arr[i, j] + " " ); } Console.Write( "" ); } } // Driver code public static void Main() { int n = 5; pattern(n); } } // This code is contributed // by ChitraNayal |
PHP
<?php // PHP program to print // the spiral pattern // Function to print the pattern function pattern( $value ) { // Declare a square matrix $row = 2 * $value - 1; $column = 2 * $value - 1; $arr = array ( array ()); for ( $k = 0; $k < $value ; $k ++) { // store the first row // from 1st column to // last column $j = $k ; while ( $j < $column - $k ) { $arr [ $k ][ $j ] = $value - $k ; $j ++; } // store the last column // from top to bottom $i = $k + 1; while ( $i < $row - $k ) { $arr [ $i ][ $row - 1 - $k ] = $value - $k ; $i ++; } // store the last row // from last column // to 1st column $j = $column - $k - 2; while ( $j >= $k ) { $arr [ $column - $k - 1][ $j ] = $value - $k ; $j --; } // store the first column // from bottom to top $i = $row - $k - 2; while ( $i > $k ) { $arr [ $i ][ $k ] = $value - $k ; $i --; } } // print the pattern for ( $i = 0; $i < $row ; $i ++) { for ( $j = 0; $j < $column ; $j ++) { echo $arr [ $i ][ $j ] . " " ; } echo "" ; } } // Driver code $n = 5; pattern( $n ); // This code is contributed // by ChitraNayal ?> |
Javascript
<script> // Javascript program to print the // spiral pattern // Function to print the pattern function pattern(value) { // Declare a square matrix let row = 2 * value - 1; let column = 2 * value - 1; let arr = new Array(row); for (let i = 0; i < row; i++) { arr[i] = new Array(column); } let i, j, k; for (k = 0; k < value; k++) { // Store the first row // from 1st column to last column j = k; while (j < column - k) { arr[k][j] = value - k; j++; } // Store the last column // from top to bottom i = k + 1; while (i < row - k) { arr[i][row - 1 - k] = value - k; i++; } // Store the last row // from last column to 1st column j = column - k - 2; while (j >= k) { arr[column - k - 1][j] = value - k; j--; } // Store the first column // from bottom to top i = row - k - 2; while (i > k) { arr[i][k] = value - k; i--; } } // Print the pattern for (i = 0; i < row; i++) { for (j = 0; j < column; j++) { document.write(arr[i][j] + " " ); } document.write( "<br>" ); } } // Driver code let n = 5; pattern(n); // This code is contributed by subham348 </script> |
输出：
5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 5 5 4 3 3 3 3 3 4 5 5 4 3 2 2 2 3 4 5 5 4 3 2 1 2 3 4 5 5 4 3 2 2 2 3 4 5 5 4 3 3 3 3 3 4 5 5 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5
方法2： 从 i=1 和 j=1 ，可以观察到所需矩阵的每个值 最大值（绝对值（i-n）、绝对值（j-n））+1 .
以下是上述方法的实施情况：
C++14
// C++ implementation of the approach #include <algorithm> #include <iostream> using namespace std; // Function to print the required pattern void pattern( int n) { // Calculating boundary size int p = 2 * n - 1; for ( int i = 1; i <= p; i++) { for ( int j = 1; j <= p; j++) { // Printing the values cout << max( abs (i - n), abs (j - n)) + 1 << " " ; } cout << endl; } } // Driver code int main() { int n = 5; pattern(n); return 0; } // This code is contributed by : Vivek kothari |
C
// C implementation of the approach #include <stdio.h> #include <stdlib.h> // Function Declaration int max( int , int ); // Function to print the required pattern void pattern( int n) { // Calculating boundary size int size = 2 * n - 1; for ( int i = 1; i <= size; i++) { for ( int j = 1; j <= size; j++) { // Printing the values printf ( "%d " , max( abs (i - n), abs (j - n)) + 1); } printf ( "" ); } } // Function to return maximum value int max( int val1, int val2) { if (val1 > val2) return val1; return val2; } // Driver code int main() { int n = 5; pattern(n); return 0; } // This code is contributed by Yuvaraj R |
JAVA
// Java implementation of the approach import java.io.*; import java.util.*; class GFG{ // Function to print the required pattern public static void pattern( int n) { // Calculating boundary size int size = 2 * n - 1 ; for ( int i = 1 ; i <= size; i++) { for ( int j = 1 ; j <= size; j++) { // Printing the values System.out.print(Math.max( Math.abs(i - n), Math.abs(j - n)) + 1 + " " ); } System.out.println(); } } // Driver code public static void main(String[] args) { int n = 5 ; pattern(n); } } // This code is contributed by Yuvaraj R |
Python3
# Python3 implementation of the approach # Function to print the required pattern def pattern(n): # Calculating boundary size p = 2 * n - 1 for i in range ( 1 , p + 1 ): for j in range ( 1 , p + 1 ): # Printing the values print ( max ( abs (i - n), abs (j - n)) + 1 , " " , end = "") print () # Driver code n = 5 pattern(n) # This code is contributed by subhammahato348 |
C#
// C# implementation of the approach using System; class GFG{ // Function to print the required pattern static void pattern( int n) { // Calculating boundary size int size = 2 * n - 1; for ( int i = 1; i <= size; i++) { for ( int j = 1; j <= size; j++) { // Printing the values Console.Write(Math.Max(Math.Abs(i - n), Math.Abs(j - n)) + 1 + " " ); } Console.WriteLine(); } } // Driver code public static void Main() { int n = 5; pattern(n); } } // This code is contributed by subhammahato348 |
Javascript
<script> // Javascript implementation of the approach // Function to print the required pattern function pattern(n) { // Calculating boundary size let p = 2 * n - 1; for (let i = 1; i <= p; i++) { for (let j = 1; j <= p; j++) { // Printing the values document.write((Math.max(Math.abs(i - n), Math.abs(j - n)) + 1) + " " ); } document.write( "<br>" ); } } // Driver code let n = 5; pattern(n); </script> |
输出
5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 5 5 4 3 3 3 3 3 4 5 5 4 3 2 2 2 3 4 5 5 4 3 2 1 2 3 4 5 5 4 3 2 2 2 3 4 5 5 4 3 3 3 3 3 4 5 5 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5
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