Dados dos números enteros n y d. La tarea es encontrar el número entre 0 an que contiene el dígito específico d.
Ejemplos:
Input : n = 20
d = 5
Output : 5 15
Input : n = 50
d = 2
Output : 2 12 20 21 22 23 24 25 26 27 28 29 32 42
Enfoque 1:
tome un ciclo de 0 a n y verifique cada número uno por uno, si el número contiene el dígito d, imprímalo; de lo contrario, aumente el número. Continúe este proceso hasta que finalice el ciclo.
C++
// C++ program to print the number which
// contain the digit d from 0 to n
#include <bits/stdc++.h>
using namespace std;
// Returns true if d is present as digit
// in number x.
bool isDigitPresent(int x, int d)
{
// Break loop if d is present as digit
while (x > 0)
{
if (x % 10 == d)
break;
x = x / 10;
}
// If loop broke
return (x > 0);
}
// function to display the values
void printNumbers(int n, int d)
{
// Check all numbers one by one
for (int i = 0; i <= n; i++)
// checking for digit
if (i == d || isDigitPresent(i, d))
cout << i << " ";
}
// Driver code
int main()
{
int n = 47, d = 7;
printNumbers(n, d);
return 0;
}
Java
// Java program to print the number which
// contain the digit d from 0 to n
class GFG
{
// Returns true if d is present as digit
// in number x.
static boolean isDigitPresent(int x, int d)
{
// Break loop if d is present as digit
while (x > 0)
{
if (x % 10 == d)
break;
x = x / 10;
}
// If loop broke
return (x > 0);
}
// function to display the values
static void printNumbers(int n, int d)
{
// Check all numbers one by one
for (int i = 0; i <= n; i++)
// checking for digit
if (i == d || isDigitPresent(i, d))
System.out.print(i + " ");
}
// Driver code
public static void main(String[] args)
{
int n = 47, d = 7;
printNumbers(n, d);
}
}
Python3
# Python3 program to print the number which
# contain the digit d from 0 to n
# Returns true if d is present as digit
# in number x.
def isDigitPresent(x, d):
# Break loop if d is present as digit
while (x > 0):
if (x % 10 == d):
break
x = x / 10
# If loop broke
return (x > 0)
# function to display the values
def printNumbers(n, d):
# Check all numbers one by one
for i in range(0, n+1):
# checking for digit
if (i == d or isDigitPresent(i, d)):
print(i,end=" ")
# Driver code
n = 47
d = 7
print("The number of values are")
printNumbers(n, d)
#This code is contributed by
#Smitha Dinesh Semwal
C#
// C# program to print the number which
// contain the digit d from 0 to n
using System;
class GFG {
// Returns true if d is present as digit
// in number x.
static bool isDigitPresent(int x, int d)
{
// Break loop if d is present as digit
while (x > 0)
{
if (x % 10 == d)
break;
x = x / 10;
}
// If loop broke
return (x > 0);
}
// function to display the values
static void printNumbers(int n, int d)
{
// Check all numbers one by one
for (int i = 0; i <= n; i++)
// checking for digit
if (i == d || isDigitPresent(i, d))
Console.Write(i + " ");
}
// Driver code
public static void Main()
{
int n = 47, d = 7;
printNumbers(n, d);
}
}
// This code contribute by parashar.
PHP
<?php
// PHP program to print the number which
// contain the digit d from 0 to n
// Returns true if d is present as digit
// in number x.
function isDigitPresent($x, $d)
{
// Break loop if d is
// present as digit
while ($x > 0)
{
if ($x % 10 == $d)
break;
$x = $x / 10;
}
// If loop broke
return ($x > 0);
}
// function to display the values
function printNumbers($n, $d)
{
// Check all numbers one by one
for ($i = 0; $i <= $n; $i++)
// checking for digit
if ($i == $d || isDigitPresent($i, $d))
echo $i , " ";
}
// Driver Code
$n = 47;
$d = 7;
printNumbers($n, $d);
// This code contributed by ajit.
?>
Javascript
<script>
// JavaScript program to print the number which
// contain the digit d from 0 to n
// Returns true if d is present as digit
// in number x.
function isDigitPresent(x, d)
{
// Break loop if d is present as digit
while (x > 0)
{
if (x % 10 == d)
break;
x = x / 10;
}
// If loop broke
return (x > 0);
}
// Function to display the values
function printNumbers(n, d)
{
// Check all numbers one by one
for(let i = 0; i <= n; i++)
// Checking for digit
if (i == d || isDigitPresent(i, d))
document.write(i + " ");
}
// Driver code
let n = 47, d = 7;
printNumbers(n, d);
// This code is contributed by splevel62
</script>
Producción:
The number of values are 7 17 27 37 47
Complejidad de tiempo: O (nlogn)
Espacio Auxiliar: O(n)
Enfoque 2:
este enfoque utiliza cada número como una string y el dígito de verificación está presente o no. Este enfoque utiliza la función String.indexOf() para verificar si el carácter está presente en la string o no.
String.indexOf() >= 0 significa que el carácter está presente
y String.indexOf() = -1 significa que el carácter no está presente
C++
// CPP program to print the number which
// contain the digit d from 0 to n
#include<bits/stdc++.h>
using namespace std;
// function to display the values
void printNumbers(int n, int d)
{
// Converting d to character
string st = "";
st += to_string(d);
char ch = st[0];
string p = "";
p += ch;
// Loop to check each digit one by one.
for (int i = 0; i <= n; i++)
{
// initialize the string
st = "";
st = st + to_string(i);
int idx = st.find(p);
// checking for digit
if (i == d || idx!=-1)
cout << (i) << " ";
}
}
// Driver code
int main()
{
int n = 100, d = 5;
printNumbers(n, d);
}
// This code is contributed by
// Surendra_Gangwar
Java
// Java program to print the number which
// contain the digit d from 0 to n
public class GFG {
// function to display the values
static void printNumbers(int n, int d)
{
// Converting d to character
String st = "" + d;
char ch = st.charAt(0);
// Loop to check each digit one by one.
for (int i = 0; i <= n; i++) {
// initialize the string
st = "";
st = st + i;
// checking for digit
if (i == d || st.indexOf(ch) >= 0)
System.out.print(i + " ");
}
}
// Driver code
public static void main(String[] args)
{
int n = 100, d = 5;
printNumbers(n, d);
}
}
Python3
# Python 3 program to print the number # which contain the digit d from 0 to n def index(st, ch): for i in range(len(st)): if(st[i] == ch): return i; return -1 # function to display the values def printNumbers(n, d): # Converting d to character st = "" + str(d) ch = st[0] # Loop to check each digit one by one. for i in range(0, n + 1, 1): # initialize the string st = "" st = st + str(i) # checking for digit if (i == d or index(st, ch) >= 0): print(i, end = " ") # Driver code if __name__ == '__main__': n = 100 d = 5 printNumbers(n, d) # This code is contributed by # Shashank_Sharma
C#
// C# program to print the number which
// contain the digit d from 0 to n
using System;
class GFG
{
// function to display the values
static void printNumbers(int n, int d)
{
// Converting d to character
String st = "" + d;
char ch = st[0];
// Loop to check each digit one by one.
for (int i = 0; i < n; i++)
{
// initialize the string
st = "";
st = st + i;
// checking for digit
if (i == d || st.IndexOf(ch) >= 0)
Console.Write(i + " ");
}
}
// Driver code
public static void Main()
{
int n = 100, d = 5;
printNumbers(n, d);
}
}
/* This code contributed by PrinciRaj1992 */
Javascript
<script>
// Javascript program to print the number which
// contain the digit d from 0 to n
// Function to display the values
function printNumbers(n, d)
{
// Converting d to character
let st = "" + d;
let ch = st[0];
// Loop to check each digit one by one.
for(let i = 0; i < n; i++)
{
// Initialize the string
st = "";
st = st + i;
// Checking for digit
if (i == d || st.indexOf(ch) >= 0)
document.write(i + " ");
}
}
// Driver code
let n = 100, d = 5;
printNumbers(n, d);
// This code is contributed by decode2207
</script>
Producción:
5 15 25 35 45 50 51 52 53 54 55 56 57 58 59 65 75 85 95
Complejidad de tiempo : O(n)
Complejidad espacial : O(1)
Publicación traducida automáticamente
Artículo escrito por Manish_100 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA