Tenemos que imprimir un patrón en el que la columna del medio contenga solo 1, las columnas del lado derecho contengan un dígito constante mayor que 1 y las columnas del lado izquierdo contengan un dígito constante mayor que 1. Cada fila debe verse como un palíndromo .
Ejemplos:
Input : 3
Output :
1
2 1 2
1
Input : 5
Output :
1
2 1 2
3 2 1 2 3
2 1 2
1
A continuación se muestra la implementación para imprimir el siguiente patrón:
C++
// CPP program to print pattern
#include <bits/stdc++.h>
using namespace std;
void display()
{
int n = 5;
int space = n / 2, num = 1;
// Outer for loop for
// number of rows
for (int i = 1; i <= n; i++)
{
// Inner for loop for
// printing space
for (int j = 1; j <= space; j++)
cout<<" ";
// Logic for printing
// the pattern for everyline
int count = num / 2 + 1;
for (int k = 1; k <= num; k++)
{
cout<<count;
// Value of count decrements
// in every cycle
if (k <= num /2 )
count--;
// Value of count will
// increment in every cycle
else
count++;
}
cout<<"\n";
// Before reaching half Space
// is decreased by 1 and num
// is increased by 2
if (i <= n / 2)
{
space = space - 1;
num = num + 2;
}
// After reaching to half
// space is increased by 1
// and num is decreased by 2
else
{
space = space + 1;
num = num - 2;
}
}
}
// Driver Code
int main()
{
display();
return 0;
}
// This code is contributed by Nikita tiwari.
Java
// Java program to print above pattern
import java.util.Scanner;
class Pattern
{
void display()
{
int n = 5;
int space = n / 2, num = 1;
// Outer for loop for
// number of rows
for (int i = 1; i <= n; i++)
{
// Inner for loop
// for printing space
for (int j = 1; j <= space; j++)
System.out.print(" ");
// Logic for printing
// the pattern for everyline
int count = num / 2 + 1;
for (int k = 1; k <= num; k++)
{
System.out.print(count);
// Value of count decrements
// in every cycle
if (k <= num /2 )
count--;
// Value of count will increment
// in every cycle
else
count++;
}
System.out.println();
// Before reaching half Space is decreased
// by 1 and num is increased by 2
if (i <= n / 2)
{
space = space - 1;
num = num + 2;
}
// After reaching to half space is increased
// by 1 and num is decreased by 2
else
{
space = space + 1;
num = num - 2;
}
}
}
// Driver Code
public static void main(String[] args)
{
Pattern p = new Pattern();
p.display();
}
}
Python3
# Python 3 program to
# print above pattern
def display() :
n = 5
space = n // 2
num = 1
# Outer for loop for
# number of rows
for i in range(1, n+1) :
# Inner for loop for
# printing space
for j in range(1, space+1) :
print(" ", end = "")
# Logic for printing the
# pattern for everyline
count = num // 2 + 1
for k in range(1, num+1) :
print(count, end = "")
# Value of count decrements
# in every cycle
if (k <= num // 2 ) :
count = count -1
# Value of count will
# increment in every cycle
else :
count = count + 1
print()
# Before reaching half Space
# is decreased by 1 and num
# is increased by 2
if (i <= n // 2) :
space = space - 1
num = num + 2
# After reaching to half
# space is increased by 1
# and num is decreased by 2
else :
space = space + 1
num = num - 2
# Driver Code
display()
#This code is contributed by Nikita Tiwari.
C#
// C# program to print above pattern
using System;
class Pattern
{
void display()
{
int n = 5;
int space = n / 2, num = 1;
// Outer for loop for
// number of rows
for (int i = 1; i <= n; i++)
{
// Inner for loop
// for printing space
for (int j = 1; j <= space; j++)
Console.Write(" ");
// Logic for printing
// the pattern for everyline
int count = num / 2 + 1;
for (int k = 1; k <= num; k++)
{
Console.Write(count);
// Value of count decrements
// in every cycle
if (k <= num /2 )
count--;
// Value of count will increment
// in every cycle
else
count++;
}
Console.WriteLine();
// Before reaching half Space is decreased
// by 1 and num is increased by 2
if (i <= n / 2)
{
space = space - 1;
num = num + 2;
}
// After reaching to half space is increased
// by 1 and num is decreased by 2
else
{
space = space + 1;
num = num - 2;
}
}
}
// Driver Code
public static void Main()
{
Pattern p = new Pattern();
p.display();
}
}
// This code is contributed by vt_m.
PHP
<?php
// php program to print
// above pattern
function display($n)
{
$space = $n / 2;
$num = 1;
// Outer for loop for
// number of rows
for ($i = 1; $i <= $n; $i++)
{
// Inner for loop
// for printing space
for ($j = 1; $j <= $space; $j++)
echo " ";
// Logic for printing
// the pattern for everyline
$count = $num / 2 + 1;
for ($k = 1; $k <= $num; $k++)
{
echo floor($count);
// Value of count decrements
// in every cycle
if ($k <= $num /2 )
$count--;
// Value of count will increment
// in every cycle
else
$count++;
}
echo "\n";
// Before reaching half Space
// is decreased by 1 and
// num is increased by 2
if ($i <= $n / 2)
{
$space = $space - 1;
$num = $num + 2;
}
// After reaching to half
// space is increased by 1
// and num is decreased by 2
else
{
$space = $space + 1;
$num = $num - 2;
}
}
}
// Driver Code
$n = 5;
display($n);
// This code is contributed by mits
?>
Javascript
<script>
// JavaScript program to print pattern
function display() {
var n = 5;
var space = parseInt(n / 2),
num = 1;
// Outer for loop for
// number of rows
for (var i = 1; i <= n; i++) {
// Inner for loop for
// printing space
for (var j = 1; j <= space; j++)
document.write(" ");
// Logic for printing
// the pattern for everyline
var count = parseInt(num / 2 + 1);
for (var k = 1; k <= num; k++) {
document.write(count);
// Value of count decrements
// in every cycle
if (k <= num / 2) count--;
// Value of count will
// increment in every cycle
else count++;
}
document.write("<br>");
// Before reaching half Space
// is decreased by 1 and num
// is increased by 2
if (i <= n / 2) {
space = space - 1;
num = num + 2;
}
// After reaching to half
// space is increased by 1
// and num is decreased by 2
else {
space = space + 1;
num = num - 2;
}
}
}
// Driver Code
display();
</script>
Producción :
1 212 32123 212 1
Complejidad de tiempo: O(n 2 ), donde n representa la entrada dada.
Espacio auxiliar: O(1), no se requiere espacio adicional, por lo que es una constante.
Publicación traducida automáticamente
Artículo escrito por Bishal Kumar Dubey y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA