Escribe una función que, para un no n dado, encuentre un número p que sea mayor o igual que n y sea la potencia más pequeña de 2.
Ejemplos:
Entrada: n = 5
Salida: 8Entrada: n = 17
Salida: 32Entrada: n = 32
Salida: 32
Hay muchas soluciones para esto. Tomemos el ejemplo de 17 para explicar algunos de ellos.
Método 1 (usando el registro del número)
1. Calculate Position of set bit in p(next power of 2):
pos = ceil(lgn) (ceiling of log n with base 2)
2. Now calculate p:
p = pow(2, pos)
Ejemplo :
Let us try for 17
pos = 5
p = 32
Método 2 (Al obtener la posición de solo el bit establecido en el resultado)
/* If n is a power of 2 then return n */
1 If (n & !(n&(n-1))) then return n
2 Else keep right shifting n until it becomes zero
and count no of shifts
a. Initialize: count = 0
b. While n ! = 0
n = n>>1
count = count + 1
/* Now count has the position of set bit in result */
3 Return (1 << count)
Ejemplo :
Let us try for 17
count = 5
p = 32
C++
// C++ program to find
// smallest power of 2
// greater than or equal to n
#include <bits/stdc++.h>
using namespace std;
unsigned int nextPowerOf2(unsigned int n)
{
unsigned count = 0;
// First n in the below condition
// is for the case where n is 0
if (n && !(n & (n - 1)))
return n;
while( n != 0)
{
n >>= 1;
count += 1;
}
return 1 << count;
}
// Driver Code
int main()
{
unsigned int n = 0;
cout << nextPowerOf2(n);
return 0;
}
// This code is contributed by rathbhupendra
C
#include<stdio.h>
unsigned int nextPowerOf2(unsigned int n)
{
unsigned count = 0;
// First n in the below condition
// is for the case where n is 0
if (n && !(n & (n - 1)))
return n;
while( n != 0)
{
n >>= 1;
count += 1;
}
return 1 << count;
}
// Driver Code
int main()
{
unsigned int n = 0;
printf("%d", nextPowerOf2(n));
return 0;
}
Java
// Java program to find
// smallest power of 2
// greater than or equal to n
import java.io.*;
class GFG
{
static int nextPowerOf2(int n)
{
int count = 0;
// First n in the below
// condition is for the
// case where n is 0
if (n > 0 && (n & (n - 1)) == 0)
return n;
while(n != 0)
{
n >>= 1;
count += 1;
}
return 1 << count;
}
// Driver Code
public static void main(String args[])
{
int n = 0;
System.out.println(nextPowerOf2(n));
}
}
// This article is contributed
// by Anshika Goyal.
Python3
def nextPowerOf2(n): count = 0 # First n in the below # condition is for the # case where n is 0 if (n and not(n & (n - 1))): return n while( n != 0): n >>= 1 count += 1 return 1 << count # Driver Code n = 0 print(nextPowerOf2(n)) # This code is contributed # by Smitha Dinesh Semwal
C#
// C# program to find smallest
// power of 2 greater than
// or equal to n
using System;
class GFG
{
static int nextPowerOf2(int n)
{
int count = 0;
// First n in the below
// condition is for the
// case where n is 0
if (n > 0 && (n & (n - 1)) == 0)
return n;
while(n != 0)
{
n >>= 1;
count += 1;
}
return 1 << count;
}
// Driver Code
public static void Main()
{
int n = 0;
Console.WriteLine(nextPowerOf2(n));
}
}
// This code is contributed by anuj_67.
PHP
<?php
// PHP program to find smallest
// power of 2 greater than or
// equal to n
function nextPowerOf2($n)
{
$count = 0;
// First n in the below condition
// is for the case where n is 0
if ($n && !($n&($n - 1)))
return $n;
while($n != 0)
{
$n >>= 1;
$count += 1;
}
return 1 << $count;
}
// Driver Code
$n = 0;
echo (nextPowerOf2($n));
// This code is contributed by vt_m
?>
Javascript
<script>
// JavaScript program to find
// smallest power of 2
// greater than or equal to n
function nextPowerOf2(n)
{
var count = 0;
// First n in the below condition
// is for the case where n is 0
if (n && !(n & (n - 1)))
return n;
while( n != 0)
{
n >>= 1;
count += 1;
}
return 1 << count;
}
// Driver Code
var n = 0;
document.write(nextPowerOf2(n));
</script>
Producción
1
Complejidad temporal: O(log n)
Espacio auxiliar: O(1)
Método 3 (Cambie el resultado uno por uno)
Gracias a coderyogi por sugerir este método. Este método es una variación del método 2 donde, en lugar de contar, cambiamos el resultado uno por uno en un ciclo.
C++
// C++ program to find smallest
// power of 2 greater than or
// equal to n
#include<bits/stdc++.h>
using namespace std;
unsigned int nextPowerOf2(unsigned int n)
{
unsigned int p = 1;
if (n && !(n & (n - 1)))
return n;
while (p < n)
p <<= 1;
return p;
}
// Driver Code
int main()
{
unsigned int n = 5;
cout << nextPowerOf2(n);
return 0;
}
// This code is contributed by rathbhupendra
C
#include<stdio.h>
unsigned int nextPowerOf2(unsigned int n)
{
unsigned int p = 1;
if (n && !(n & (n - 1)))
return n;
while (p < n)
p <<= 1;
return p;
}
// Driver Code
int main()
{
unsigned int n = 5;
printf("%d", nextPowerOf2(n));
return 0;
}
Java
// Java program to find smallest
// power of 2 greater than or
// equal to n
import java.io.*;
class GFG
{
static int nextPowerOf2(int n)
{
int p = 1;
if (n > 0 && (n & (n - 1)) == 0)
return n;
while (p < n)
p <<= 1;
return p;
}
// Driver Code
public static void main(String args[])
{
int n = 5;
System.out.println(nextPowerOf2(n));
}
}
// This article is contributed
// by Anshika Goyal.
Python3
def nextPowerOf2(n): p = 1 if (n and not(n & (n - 1))): return n while (p < n) : p <<= 1 return p; # Driver Code n = 5 print(nextPowerOf2(n)); # This code is contributed by # Smitha Dinesh Semwal
C#
// C# program to find smallest
// power of 2 greater than or
// equal to n
using System;
class GFG
{
static int nextPowerOf2(int n)
{
int p = 1;
if (n > 0 && (n & (n - 1)) == 0)
return n;
while (p < n)
p <<= 1;
return p;
}
// Driver Code
public static void Main()
{
int n = 5;
Console.Write(nextPowerOf2(n));
}
}
// This code is contributed by Smitha.
PHP
<?php
function nextPowerOf2($n)
{
$p = 1;
if ($n && !($n & ($n - 1)))
return $n;
while ($p < $n)
$p <<= 1;
return $p;
}
// Driver Code
$n = 5;
echo ( nextPowerOf2($n));
// This code is contributed by vt_m.
?>
Javascript
<script>
// Program to find smallest
// power of 2 greater than or
// equal to n
function nextPowerOf2( n)
{
p = 1;
if (n && !(n & (n - 1)))
return n;
while (p < n)
p <<= 1;
return p;
}
// Driver Code
n = 5;
document.write (nextPowerOf2(n));
//This code is contributed by simranarora5sos
</script>
Producción
8
Complejidad de tiempo: O(log(n))
Espacio auxiliar: O(1)
Método 4 (Personalizado y Rápido)
1. Subtract n by 1
n = n -1
2. Set all bits after the leftmost set bit.
/* Below solution works only if integer is 32 bits */
n = n | (n >> 1);
n = n | (n >> 2);
n = n | (n >> 4);
n = n | (n >> 8);
n = n | (n >> 16);
3. Return n + 1
Ejemplo :
Steps 1 & 3 of above algorithm are to handle cases
of power of 2 numbers e.g., 1, 2, 4, 8, 16,
Let us try for 17(10001)
step 1
n = n - 1 = 16 (10000)
step 2
n = n | n >> 1
n = 10000 | 01000
n = 11000
n = n | n >> 2
n = 11000 | 00110
n = 11110
n = n | n >> 4
n = 11110 | 00001
n = 11111
n = n | n >> 8
n = 11111 | 00000
n = 11111
n = n | n >> 16
n = 11110 | 00000
n = 11111
step 3: Return n+1
We get n + 1 as 100000 (32)
Programa:
C++
// C++ program to
// Finds next power of two
// for n. If n itself is a
// power of two then returns n
#include <bits/stdc++.h>
using namespace std;
unsigned int nextPowerOf2(unsigned int n)
{
n--;
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
n++;
return n;
}
// Driver Code
int main()
{
unsigned int n = 5;
cout << nextPowerOf2(n);
return 0;
}
// This code is contributed by SHUBHAMSINGH10
C
#include <stdio.h>
// Finds next power of two
// for n. If n itself is a
// power of two then returns n
unsigned int nextPowerOf2(unsigned int n)
{
n--;
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
n++;
return n;
}
// Driver Code
int main()
{
unsigned int n = 5;
printf("%d", nextPowerOf2(n));
return 0;
}
Java
// Java program to find smallest
// power of 2 greater than or
// equal to n
import java.io.*;
class GFG
{
// Finds next power of two
// for n. If n itself is a
// power of two then returns n
static int nextPowerOf2(int n)
{
n--;
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
n++;
return n;
}
// Driver Code
public static void main(String args[])
{
int n = 5;
System.out.println(nextPowerOf2(n));
}
}
// This article is contributed
// by Anshika Goyal.
Python3
# Finds next power of two # for n. If n itself is a # power of two then returns n def nextPowerOf2(n): n -= 1 n |= n >> 1 n |= n >> 2 n |= n >> 4 n |= n >> 8 n |= n >> 16 n += 1 return n # Driver program to test # above function n = 5 print(nextPowerOf2(n)) # This code is contributed # by Smitha
C#
// C# program to find smallest
// power of 2 greater than or
// equal to n
using System;
class GFG
{
// Finds next power of two
// for n. If n itself is a
// power of two then returns n
static int nextPowerOf2(int n)
{
n--;
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
n++;
return n;
}
// Driver Code
public static void Main()
{
int n = 5;
Console.WriteLine(nextPowerOf2(n));
}
}
// This code is contributed by anuj_67.
PHP
<?php
// PHP program to find smallest
// power of 2 greater than or
// equal to n
// Finds next power of
// two for n. If n itself
// is a power of two then
// returns n
function nextPowerOf2($n)
{
$n--;
$n |= $n >> 1;
$n |= $n >> 2;
$n |= $n >> 4;
$n |= $n >> 8;
$n |= $n >> 16;
$n++;
return $n;
}
// Driver Code
$n = 5;
echo nextPowerOf2($n);
// This code contributed by Ajit
?>
Javascript
<script>
// Javascript program to find smallest
// power of 2 greater than or
// equal to n
// Finds next power of
// two for n. If n itself
// is a power of two then
// returns n
function nextPowerOf2(n)
{
n -= 1
n |= n >> 1
n |= n >> 2
n |= n >> 4
n |= n >> 8
n |= n >> 16
n += 1
return n
}
// Driver Code
n = 5;
document.write (nextPowerOf2(n));
// This code is contributed by bunnyram19
</script>
Producción
8
Complejidad de tiempo: O(log(n))
Espacio auxiliar: O(1)
Enfoque eficiente:
si el número dado es la potencia de dos, entonces es el número requerido; de lo contrario, configure solo el bit izquierdo del bit más significativo que nos da el número requerido.
C++
// C++ program to find
// smallest power of 2
// greater than or equal to n
#include <iostream>
using namespace std;
long long nextPowerOf2(long long N)
{
// if N is a power of two simply return it
if (!(N & (N - 1)))
return N;
// else set only the left bit of most significant bit
return 0x8000000000000000 >> (__builtin_clzll(N) - 1);
}
// Driver Code
int main()
{
long long n = 5;
cout << nextPowerOf2(n);
return 0;
}
// This code is contributed by Kasina Dheeraj.
C
// C program to find
// smallest power of 2
// greater than or equal to n
#include <stdio.h>
long long nextPowerOf2(long long N)
{
// if N is a power of two simply return it
if (!(N & (N - 1)))
return N;
// else set only the left bit of most significant bit
return 0x8000000000000000 >> (__builtin_clzll(N) - 1);
}
// Driver Code
int main()
{
long long n = 5;
printf("%lld", nextPowerOf2(n));
return 0;
}
// This code is contributed by Kasina Dheeraj.
Java
// Java program to find
// smallest power of 2
// greater than or equal to n
import java.io.*;
class GFG {
static long nextPowerOf2(long N)
{
// if N is a power of two simply return it
if ((N & (N - 1)) == 0)
return N;
// else set only the left bit of most significant
// bit as in Java right shift is filled with most
// significant bit we consider
return 0x4000000000000000L
>> (Long.numberOfLeadingZeros(N) - 2);
}
// Driver Code
public static void main(String args[])
{
long n = 5;
System.out.println(nextPowerOf2(n));
}
}
// This code is contributed
// by Kasina Dheeraj.
Python3
# Python program to find
# smallest power of 2
# greater than or equal to n
#include <iostream>
def nextPowerOf2(N):
# if N is a power of two simply return it
if not (N & (N - 1)):
return N
# else set only the left bit of most significant bit
return int("1" + (len(bin(N)) - 2) * "0", 2)
# Driver Code
n = 5
print(nextPowerOf2(n))
# this code is contributed by phasing17
C#
// C# program to find
// smallest power of 2
// greater than or equal to n
using System;
using System.Linq;
class GFG {
static int nextPowerOf2(int N)
{
// if N is a power of two simply return it
if ((N & (N - 1)) == 0)
return N;
// else set only the left bit of most significant
// bit
return Convert.ToInt32(
"1"
+ new string('0',
Convert.ToString(N, 2).Length),
2);
}
// Driver Code
public static void Main(string[] args)
{
int n = 5;
// Function call
Console.WriteLine(nextPowerOf2(n));
}
}
// This code is contributed
// by phasing17
Javascript
// JavaScript program to find
// smallest power of 2
// greater than or equal to n
function nextPowerOf2(N)
{
// if N is a power of two simply return it
if (!(N & (N - 1)))
return N;
// else set only the left bit of most significant bit
return parseInt("1" + "0".repeat(N.toString(2).length), 2);
}
// Driver Code
let n = 5;
console.log(nextPowerOf2(n));
// this code is contributed by phasing17
Producción
8
Complejidad de tiempo: O (1) ya que contar ceros a la izquierda puede causar una complejidad de tiempo de O (64) como máximo.
Espacio Auxiliar: O(1)
Publicación relacionada:
potencia más alta de 2 menor o igual que el número dado
Referencias:
http://en.wikipedia.org/wiki/Power_of_2
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA