Dado un número entero N , la tarea es imprimir todos los números hasta N en orden lexicográfico .
Ejemplos:
Entrada: N = 15
Salida:
1 10 11 12 13 14 15 2 3 4 5 6 7 8 9Entrada: N = 19
Salida:
1 10 11 12 13 14 15 16 17 18 19 2 3 4 5 6 7 8 9
Enfoque:
Para resolver el problema, siga los pasos a continuación:
- Iterar de 1 a N y almacenar todos los números en forma de strings.
- Ordene el vector que contiene las strings.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ Program to implement the
// above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print all the
// numbers up to n in
// lexicographical order
void lexNumbers(int n)
{
vector<string> s;
for (int i = 1; i <= n; i++) {
s.push_back(to_string(i));
}
sort(s.begin(), s.end());
vector<int> ans;
for (int i = 0; i < n; i++)
ans.push_back(stoi(s[i]));
for (int i = 0; i < n; i++)
cout << ans[i] << " ";
}
// Driver Program
int main()
{
int n = 15;
lexNumbers(n);
return 0;
}
Java
// Java Program to implement the
// above approach
import java.util.*;
class GFG{
// Function to print all the
// numbers up to n in
// lexicographical order
static void lexNumbers(int n)
{
Vector<String> s = new Vector<String>();
for (int i = 1; i <= n; i++)
{
s.add(String.valueOf(i));
}
Collections.sort(s);
Vector<Integer> ans = new Vector<Integer>();
for (int i = 0; i < n; i++)
ans.add(Integer.valueOf(s.get(i)));
for (int i = 0; i < n; i++)
System.out.print(ans.get(i) + " ");
}
// Driver Program
public static void main(String[] args)
{
int n = 15;
lexNumbers(n);
}
}
// This code is contributed by sapnasingh4991
Python3
# Python3 program to implement # above approach # Function to print all the # numbers up to n in # lexicographical order def lexNumbers(n): s = [] for i in range(1, n + 1): s.append(str(i)) s.sort() ans = [] for i in range(n): ans.append(int(s[i])) for i in range(n): print(ans[i], end = ' ') # Driver Code if __name__ == "__main__": n = 15 lexNumbers(n) # This code is contributed by Ediga_Manisha
C#
// C# program to implement the
// above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to print all the
// numbers up to n in
// lexicographical order
static void lexNumbers(int n)
{
List<String> s = new List<String>();
for(int i = 1; i <= n; i++)
{
s.Add(String.Join("", i));
}
s.Sort();
List<int> ans = new List<int>();
for(int i = 0; i < n; i++)
ans.Add(Int32.Parse(s[i]));
for(int i = 0; i < n; i++)
Console.Write(ans[i] + " ");
}
// Driver code
public static void Main(String[] args)
{
int n = 15;
lexNumbers(n);
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// Javascript Program to implement the
// above approach
// Function to print all the
// numbers up to n in
// lexicographical order
function lexNumbers(n)
{
let s = [];
for (let i = 1; i <= n; i++)
{
s.push(i.toString());
}
s.sort();
let ans = [];
for (let i = 0; i < n; i++)
ans.push(parseInt(s[i]));
for (let i = 0; i < n; i++)
document.write(ans[i] + " ");
}
// Driver Program
let n = 15;
lexNumbers(n);
// This code is contributed by avanitrachhadiya2155
</script>
Producción
1 10 11 12 13 14 15 2 3 4 5 6 7 8 9
Complejidad de tiempo: O(N log N) // la complejidad de tiempo de la función de clasificación es NlogN
Complejidad espacial: O(N) // porque se usa un vector extra de string s
Otro enfoque:
- Usando DFS: siempre multiplique la temperatura por 10 hasta que la temperatura * 10 sea mayor que n
- Incremente la temperatura en 1 cuando el último dígito de la temperatura no sea igual a 9
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
void dfs(int temp, int n, vector<int> &sol);
void lexNumbers(int n)
{
vector<int> sol;
dfs(1, n, sol);
cout << "[" << sol[0];
for (int i = 1; i < sol.size(); i++)
cout << ", "<< sol[i];
cout << "]";
}
void dfs(int temp, int n, vector<int> &sol)
{
if (temp > n)
return;
sol.push_back(temp);
dfs(temp * 10, n, sol);
if (temp % 10 != 9)
dfs(temp + 1, n, sol);
}
int main()
{
int n = 15;
lexNumbers(n);
return 0;
}
// This Code is contributed by ShubhamSingh10
Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
public static void lexNumbers(int n)
{
List<Integer> sol = new ArrayList<>();
dfs(1, n, sol);
System.out.println(sol);
}
public static void dfs(int temp, int n,
List<Integer> sol)
{
if (temp > n)
return;
sol.add(temp);
dfs(temp * 10, n, sol);
if (temp % 10 != 9)
dfs(temp + 1, n, sol);
}
public static void main(String[] args)
{
int n = 15;
lexNumbers(n);
}
}
Python3
# Python program for the above approach
def lexNumbers(n):
sol = []
dfs(1, n, sol)
print("[", sol[0], end= "", sep ="")
for i in range(1,n):
print(", ", sol[i], end= "", sep ="")
print("]")
def dfs(temp, n, sol):
if (temp > n):
return
sol.append(temp)
dfs(temp * 10, n, sol)
if (temp % 10 != 9):
dfs(temp + 1, n, sol)
n = 15
lexNumbers(n)
# This Code is contributed by ShubhamSingh10
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
public static void lexNumbers(int n)
{
List<int> sol = new List<int>();
dfs(1, n, sol);
Console.WriteLine("[" + string.Join(", ", sol) + "]");
}
public static void dfs(int temp, int n,
List<int> sol)
{
if (temp > n)
return;
sol.Add(temp);
dfs(temp * 10, n, sol);
if (temp % 10 != 9)
dfs(temp + 1, n, sol);
}
// Driver code
public static void Main()
{
int n = 15;
lexNumbers(n);
}
}
// This code is contributed by shubhamsingh10
Javascript
<script>
// JavaScript program for the above approach
function lexNumbers(n)
{
var sol = [];
dfs(1, n, sol);
document.write("["+ sol[0]);
for (var i = 1; i < sol.length; i++)
document.write(", "+ sol[i]);
document.write("]");
}
function dfs(temp, n, sol)
{
if (temp > n)
return;
sol.push(temp);
dfs(temp * 10, n, sol);
if (temp % 10 != 9)
dfs(temp + 1, n, sol);
}
var n = 15;
lexNumbers(n);
// This Code is contributed by ShubhamSingh10
</script>
Producción
[1, 10, 11, 12, 13, 14, 15, 2, 3, 4, 5, 6, 7, 8, 9]
Complejidad Temporal: O(N)
Espacio Auxiliar: O (1)
Cuando hay un rango: –
Dados dos números enteros L y R, la tarea es imprimir todos los números en el rango de L a R (inclusive) en orden lexicográfico.
Ejemplos:
Input: L = 9 , R = 21 Output: 10 11 12 13 14 15 16 17 18 19 20 21 9 Input: L = 1 , R= 13 Output: 1 10 11 12 13 2 3 4 5 6 7 8 9
Acercarse:
Para resolver el problema, siga los pasos a continuación:
- Iterar de L a R (inclusive) y almacenar todos los números en forma de strings.
- Ordene el vector que contiene las strings.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to implement the
// above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print all the
// numbers form l to r in
// lexicographical order
void lexNumbers(int l, int r)
{
vector<string> s;
for(int i = l; i <= r; i++)
{
s.push_back(to_string(i));
}
sort(s.begin(),s.end());
vector<int> ans;
for(int i = 0; i < s.size(); i++)
ans.push_back(stoi(s[i]));
for(int i = 0; i < s.size(); i++)
cout << ans[i] << " ";
}
// Driver code
int main()
{
int l = 9;
int r = 21;
lexNumbers(l, r);
}
// This code is contributed by ajaykr00kj
Java
// Java Program to implement the
// above approach
import java.util.*;
class GFG {
// Function to print all the
// numbers form l to r in
// lexicographical order
static void lexNumbers(int l, int r)
{
Vector<String> s = new Vector<String>();
for (int i = l; i <= r; i++) {
s.add(String.valueOf(i));
}
Collections.sort(s);
Vector<Integer> ans = new Vector<Integer>();
for (int i = 0; i < s.size(); i++)
ans.add(Integer.valueOf(s.get(i)));
for (int i = 0; i < s.size(); i++)
System.out.print(ans.get(i) + " ");
}
// Driver Program
public static void main(String[] args)
{
int l = 9;
int r = 21;
lexNumbers(l, r);
}
}
Python3
# Python 3 program to implement # the above approach # Function to print all the # numbers form l to r in # lexicographical order def lexNumbers(l, r): s = [] for i in range(l, r + 1): s.append(str(i)) s.sort() ans = [] for i in range(len(s)): ans.append(int(s[i])) for i in range(len(s)): print(ans[i], end = " ") # Driver code if __name__ == "__main__": l = 9 r = 21 lexNumbers(l, r) # This code is contributed by Chitranayal
C#
// C# program to implement the
// above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to print all the
// numbers form l to r in
// lexicographical order
static void lexNumbers(int l, int r)
{
List<String> s = new List<String>();
for (int i = l; i <= r; i++)
{
s.Add(String.Join("", i));
}
s.Sort();
List<int> ans = new List<int>();
for (int i = 0; i < s.Count; i++)
ans.Add(Int32.Parse(s[i]));
for (int i = 0; i < s.Count; i++)
Console.Write(ans[i] + " ");
}
// Driver Program
static public void Main()
{
int l = 9;
int r = 21;
lexNumbers(l, r);
}
}
// This code is contributed by Dharanendra L V
Javascript
<script>
// Javascript Program to implement the
// above approach
// Function to print all the
// numbers form l to r in
// lexicographical order
function lexNumbers(l,r)
{
let s = [];
for (let i = l; i <= r; i++) {
s.push((i).toString());
}
s.sort();
let ans = [];
for (let i = 0; i < s.length; i++)
ans.push(parseInt(s[i]));
for (let i = 0; i < s.length; i++)
document.write(ans[i] + " ");
}
// Driver Program
let l = 9;
let r = 21;
lexNumbers(l, r);
// This code is contributed by rag2127
</script>
Producción
10 11 12 13 14 15 16 17 18 19 20 21 9
Complejidad de Tiempo: O(N*logN)
Espacio Auxiliar: O (1)
Publicación traducida automáticamente
Artículo escrito por suyash ghuge y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA