大于或等于n的2的最小幂

编写一个函数，对于给定的否n，它找到一个大于或等于n且为2的最小幂的数字p。

例子 ：

Input : n = 5
    Output: 8     

    Input  : n = 17
    Output : 32     

    Input  : n = 32
    Output : 32     

有很多解决方案。让我们以17为例来解释其中的一些。

方法1(使用数字的对数)

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) 

例子 ：

Let us try for 17
            pos = 5
            p   = 32    

方法2(通过获取结果中仅设置位的位置)

/* 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)  

例子 ：

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 
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
 
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

>= 1;
    $count += 1;
}
 
return 1 << $count;
}
 
// Driver Code
$n = 0;
echo (nextPowerOf2($n));
 
// This code is contributed by vt_m
?>

Javascript

 
// 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));
     

输出 ：

1

方法3(将结果一一移位)
感谢coderyogi建议这种方法。此方法是方法2的一种变体，其中我们不循环获取结果，而不是获取计数。

C++

// C++ program to find smallest
// power of 2 greater than or
// equal to n
#include
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
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.

Java脚本

// 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

输出 ：

8

时间复杂度： O(lgn)

方法4(自定义和快速)

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

例子 ：

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)

程序：

C++

// C++ program to 
// Finds next power of two
// for n. If n itself is a
// power of two then returns n
#include 
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 
// 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.

的PHP

> 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
?>

Java脚本

 
// 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
 

输出 ：

8

时间复杂度： O(lgn)
相关文章：
小于或等于给定数的2的最高幂
参考 ：
http://en.wikipedia.org/wiki/Power_of_2