程序查找给定数字是否是 2 的幂

给定一个正整数，编写一个函数来判断它是否是 2 的幂。
例子 ：

Input : n = 4
Output : Yes
22 = 4

Input : n = 7
Output : No

Input : n = 32
Output : Yes
25 = 32

1.一个简单的方法是简单地取以 2 为底的数字的对数，如果你得到一个整数，那么这个数字就是 2 的幂

C++

// C++ Program to find whether a
// no is power of two
#include
using namespace std;
 
// Function to check if x is power of 2
bool isPowerOfTwo(int n)
{
   if(n==0)
   return false;
 
   return (ceil(log2(n)) == floor(log2(n)));
}
 
// Driver program
int main()
{
    isPowerOfTwo(31)? cout<<"Yes"<

---

C

// C Program to find whether a
// no is power of two
#include
#include
#include
 
/* Function to check if x is power of 2*/
bool isPowerOfTwo(int n)
{
   if(n==0)
   return false;
 
   return (ceil(log2(n)) == floor(log2(n)));
}
 
// Driver program
int main()
{
    isPowerOfTwo(31)? printf("Yes\n"): printf("No\n");
    isPowerOfTwo(64)? printf("Yes\n"): printf("No\n");
    return 0;
}
 
// This code is contributed by bibhudhendra

Java

// Java Program to find whether a
// no is power of two
class GFG
{
/* Function to check if x is power of 2*/
static boolean isPowerOfTwo(int n)
{
    if(n==0)
    return false;
 
return (int)(Math.ceil((Math.log(n) / Math.log(2)))) ==
       (int)(Math.floor(((Math.log(n) / Math.log(2)))));
}
 
// Driver Code
public static void main(String[] args)
{
    if(isPowerOfTwo(31))
    System.out.println("Yes");
    else
    System.out.println("No");
     
    if(isPowerOfTwo(64))
    System.out.println("Yes");
    else
    System.out.println("No");
}
}
 
// This code is contributed by mits

Python3

# Python3 Program to find
# whether a no is
# power of two
import math
 
# Function to check
# Log base 2
def Log2(x):
    if x == 0:
        return false;
 
    return (math.log10(x) /
            math.log10(2));
 
# Function to check
# if x is power of 2
def isPowerOfTwo(n):
    return (math.ceil(Log2(n)) ==
            math.floor(Log2(n)));
 
# Driver Code
if(isPowerOfTwo(31)):
    print("Yes");
else:
    print("No");
 
if(isPowerOfTwo(64)):
    print("Yes");
else:
    print("No");
     
# This code is contributed
# by mits

C#

// C# Program to find whether
// a no is power of two
using System;
 
class GFG
{
     
/* Function to check if
   x is power of 2*/
static bool isPowerOfTwo(int n)
{
 
    if(n==0)
     return false;
 
    return (int)(Math.Ceiling((Math.Log(n) /
                               Math.Log(2)))) ==
           (int)(Math.Floor(((Math.Log(n) /
                              Math.Log(2)))));
}
 
// Driver Code
public static void Main()
{
    if(isPowerOfTwo(31))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
     
    if(isPowerOfTwo(64))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

Javascript

// javascript Program to find whether a
// no is power of two
 
    /* Function to check if x is power of 2 */
    function isPowerOfTwo(n)
    {
        if (n == 0)
            return false;
 
        return parseInt( (Math.ceil((Math.log(n) / Math.log(2))))) == parseInt( (Math.floor(((Math.log(n) / Math.log(2))))));
    }
 
    // Driver Code
     
    if (isPowerOfTwo(31))
        document.write("Yes<br/>");
    else
        document.write("No<br/>");
 
    if (isPowerOfTwo(64))
        document.write("Yes<br/>");
    else
        document.write("No<br/>");
 
// This code is contributed by shikhasingrajput.

输出：

No
Yes

时间复杂度： O(1)
辅助空间： O(1)

2.另一种解决方案是不断将数字除以 2，即迭代地执行 n = n/2。在任何迭代中，如果 n%2 变为非零且 n 不是 1，则 n 不是 2 的幂。如果 n 变为 1，则它是 2 的幂。

C++

#include 
using namespace std;
 
/* Function to check if x is power of 2*/
bool isPowerOfTwo(int n)
{
    if (n == 0)
        return 0;
    while (n != 1)
    {
        if (n%2 != 0)
            return 0;
        n = n/2;
    }
    return 1;
}
 
/*Driver code*/
int main()
{
    isPowerOfTwo(31)? cout<<"Yes\n": cout<<"No\n";
    isPowerOfTwo(64)? cout<<"Yes\n": cout<<"No\n";
    return 0;
}
 
// This code is contributed by rathbhupendra

C

#include
#include
 
/* Function to check if x is power of 2*/
bool isPowerOfTwo(int n)
{
  if (n == 0)
    return 0;
  while (n != 1)
  {
      if (n%2 != 0)
         return 0;
      n = n/2;
  }
  return 1;
}
 
/*Driver program to test above function*/
int main()
{
  isPowerOfTwo(31)? printf("Yes\n"): printf("No\n");
  isPowerOfTwo(64)? printf("Yes\n"): printf("No\n");
  return 0;
}

Java

// Java program to find whether
// a no is power of two
import java.io.*;
 
class GFG {
 
    // Function to check if
    // x is power of 2
    static boolean isPowerOfTwo(int n)
    {
        if (n == 0)
            return false;
         
        while (n != 1)
        {
            if (n % 2 != 0)
                return false;
            n = n / 2;
        }
        return true;
    }
 
    // Driver program
    public static void main(String args[])
    {
        if (isPowerOfTwo(31))
            System.out.println("Yes");
        else
            System.out.println("No");
 
        if (isPowerOfTwo(64))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by Nikita tiwari.

Python3

# Python program to check if given
# number is power of 2 or not
 
# Function to check if x is power of 2
def isPowerOfTwo(n):
    if (n == 0):
        return False
    while (n != 1):
            if (n % 2 != 0):
                return False
            n = n // 2
             
    return True
 
# Driver code
if(isPowerOfTwo(31)):
    print('Yes')
else:
    print('No')
if(isPowerOfTwo(64)):
    print('Yes')
else:
    print('No')
 
# This code is contributed by Danish Raza

C＃

// C# program to find whether
// a no is power of two
using System;
 
class GFG
{
     
    // Function to check if
    // x is power of 2
    static bool isPowerOfTwo(int n)
    {
        if (n == 0)
            return false;
         
        while (n != 1) {
            if (n % 2 != 0)
                return false;
                 
            n = n / 2;
        }
        return true;
    }
 
    // Driver program
    public static void Main()
    {
        Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No");
        Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No");
 
    }
}
 
// This code is contributed by Sam007

Javascript

 
    /* Function to check if x is power of 2*/
    function isPowerOfTwo(n)
    {
        if (n == 0)
            return 0;
        while (n != 1)
        {
            if (n%2 != 0)
                return 0;
            n = n/2;
        }
        return 1;
    } 
     
    isPowerOfTwo(31)? document.write("Yes" + "</br>"): document.write("No" + "</br>");
    isPowerOfTwo(64)? document.write("Yes"): document.write("No");
 

输出 ：

No
Yes

时间复杂度： O(log 2 n)

辅助空间： O(1)

3.另一种方法是使用这种简单的递归解决方案。它使用与上述迭代解决方案相同的逻辑，但使用递归而不是迭代。

C++

// C++ program for above approach
#include 
using namespace std;
 
// Function which checks whether a
// number is a power of 2
bool powerOf2(int n)
{
    // base cases
    // '1' is the only odd number
    // which is a power of 2(2^0)
    if (n == 1)
      return true;
     
    // all other odd numbers are not powers of 2
    else if (n % 2 != 0 || n ==0)
      return false;
     
    // recursive function call
    return powerOf2(n / 2);
}
 
// Driver Code
int main()
{
    int n = 64;//True
    int m = 12;//False
 
    if (powerOf2(n) == 1)
      cout << "True" << endl;
 
    else cout << "False" << endl;
 
    if (powerOf2(m) == 1)
      cout << "True" << endl;
 
    else
      cout << "False" << endl;
}
 
//code contributed by Moukthik a.k.a rowdyninja

Java

// Java program for
// the above approach
import java.util.*;
class GFG{
 
// Function which checks
// whether a number is a
// power of 2
static boolean powerOf2(int n)
{
  // base cases
  // '1' is the only odd number
  // which is a power of 2(2^0)
  if (n == 1)
    return true;
 
  // all other odd numbers are
  // not powers of 2
  else if (n % 2 != 0 ||
           n ==0)
    return false;
 
  // recursive function call
  return powerOf2(n / 2);
}
 
// Driver Code
public static void main(String[] args)
{
  //True
  int n = 64;
   
  //False
  int m = 12;
 
  if (powerOf2(n) == true)
    System.out.print("True" + "\n");
  else System.out.print("False" + "\n");
 
  if (powerOf2(m) == true)
    System.out.print("True" + "\n");
  else
    System.out.print("False" + "\n");
}
}
 
// This code is contributed by Princi Singh

Python3

# Python program for above approach
 
# function which checks whether a
# number is a power of 2
def powerof2(n):
   
    # base cases
    # '1' is the only odd number
    # which is a power of 2(2^0)
    if n == 1:
        return True
     
    # all other odd numbers are not powers of 2
    elif n%2 != 0 or n == 0:
        return False
     
    #recursive function call
    return powerof2(n/2)
   
# Driver Code
if __name__ == "__main__":
   
  print(powerof2(64)) #True
  print(powerof2(12)) #False
   
#code contributed by Moukthik a.k.a rowdyninja

C＃

// C# program for above approach
using System;
 
class GFG{
     
// Function which checks whether a
// number is a power of 2
static bool powerOf2(int n)
{
     
    // Base cases
    // '1' is the only odd number
    // which is a power of 2(2^0)
    if (n == 1)
      return true;
      
    // All other odd numbers
    // are not powers of 2
    else if (n % 2 != 0 || n == 0)
      return false;
      
    // Recursive function call
    return powerOf2(n / 2);
}
 
// Driver code
static void Main()
{
   
    int n = 64;//True
    int m = 12;//False
     
    if (powerOf2(n))
    {
        Console.Write("True" + "\n");
    }
    else
    {
        Console.Write("False" + "\n");
    }
     
    if (powerOf2(m))
    {
        Console.Write("True");
    }
    else
    {
        Console.Write("False");
    }
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// javascript program for
// the above approach
 
// Function which checks
// whether a number is a
// power of 2
 
function powerOf2(n)
{
  // base cases
  // '1' is the only odd number
  // which is a power of 2(2^0)
  if (n == 1)
    return true;
 
  // all other odd numbers are
  // not powers of 2
  else if (n % 2 != 0 ||
           n ==0)
    return false;
 
  // recursive function call
  return powerOf2(n / 2);
}
 
// Driver Code
//True
var n = 64;
 
//False
var m = 12;
 
if (powerOf2(n) == true)
  document.write("True" + "\n");
else document.write("False" + "\n");
 
if (powerOf2(m) == true)
  document.write("True" + "\n");
else
  document.write("False" + "\n");
   
// This code contributed by shikhasingrajput
 

输出

True
False

时间复杂度： O(log 2 n)

辅助空间： O(log 2 n)

4.两个数的所有幂只有一位集合。所以数数。设置位的数量，如果您得到 1，则该数字是 2 的幂。请参阅以整数计数设置位以计算设置位。

5.如果我们将 2 的幂减去 1，则唯一设置位之后的所有未设置位都将被设置；设置位变为未设置。
例如对于 4 ( 100) 和 16(10000)，我们在减去 1 后得到以下结果
3 –> 011
15 –> 01111

因此，如果数字 n 是 2 的幂，则 n 和 n-1 的按位 & 将为零。我们可以根据 n&(n-1) 的值说 n 是 2 的幂或不是 2 的幂。当 n 为 0 时，表达式 n&(n-1) 将不起作用。为了处理这种情况，我们的表达式将变为 n& (!n&(n-1)) （感谢 https://www.geeksforgeeks.org/program - 查找是否不属于二的幂/穆罕默德添加这个案例）。

下面是这个方法的实现。

时间复杂度：O(1)

空间复杂度：O(1)

C++

#include 
using namespace std;
#define bool int
 
/* Function to check if x is power of 2*/
bool isPowerOfTwo (int x)
{
    /* First x in the below expression is for the case when x is 0 */
    return x && (!(x&(x-1)));
}
 
/*Driver code*/
int main()
{
    isPowerOfTwo(31)? cout<<"Yes\n": cout<<"No\n";
    isPowerOfTwo(64)? cout<<"Yes\n": cout<<"No\n";
    return 0;
}
 
// This code is contributed by rathbhupendra

C

#include
#define bool int
 
/* Function to check if x is power of 2*/
bool isPowerOfTwo (int x)
{
  /* First x in the below expression is for the case when x is 0 */
  return x && (!(x&(x-1)));
}
 
/*Driver program to test above function*/
int main()
{
  isPowerOfTwo(31)? printf("Yes\n"): printf("No\n");
  isPowerOfTwo(64)? printf("Yes\n"): printf("No\n");
  return 0;
}

Java

// Java program to efficiently
// check for power for 2
 
class Test
{
    /* Method to check if x is power of 2*/
    static boolean isPowerOfTwo (int x)
    {
      /* First x in the below expression is
        for the case when x is 0 */
        return x!=0 && ((x&(x-1)) == 0);
         
    }
     
    // Driver method
    public static void main(String[] args)
    {
         System.out.println(isPowerOfTwo(31) ? "Yes" : "No");
         System.out.println(isPowerOfTwo(64) ? "Yes" : "No");
         
    }
}
// This program is contributed by Gaurav Miglani   

Python3

# Python program to check if given
# number is power of 2 or not
 
# Function to check if x is power of 2
def isPowerOfTwo (x):
 
    # First x in the below expression
    # is for the case when x is 0
    return (x and (not(x & (x - 1))) )
 
# Driver code
if(isPowerOfTwo(31)):
    print('Yes')
else:
    print('No')
     
if(isPowerOfTwo(64)):
    print('Yes')
else:
    print('No')
     
# This code is contributed by Danish Raza   

C＃

// C# program to efficiently
// check for power for 2
using System;
 
class GFG
{
    // Method to check if x is power of 2
    static bool isPowerOfTwo (int x)
    {
        // First x in the below expression 
        // is for the case when x is 0
        return x != 0 && ((x & (x - 1)) == 0);
         
    }
     
    // Driver method
    public static void Main()
    {
        Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No");
        Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No");
         
    }
}
 
// This code is contributed by Sam007

Javascript

 
// JavaScript program to efficiently
// check for power for 2  
 
/* Method to check if x is power of 2*/
    function isPowerOfTwo (x)
    {
      /* First x in the below expression is
        for the case when x is 0 */
        return x!=0 && ((x&(x-1)) == 0);
         
    }
     
// Driver method
document.write(isPowerOfTwo(31) ? "Yes" : "No");
document.write("<br>"+(isPowerOfTwo(64) ? "Yes" : "No"));
 
// This code is contributed by 29AjayKumar
 

输出 ：

No
Yes

时间复杂度： O(1)

辅助空间： O(1)

6 .另一种方法是使用逻辑找到给定数字的最右边的位集。

C++

#include 
using namespace std;
 
/* Function to check if x is power of 2*/
bool isPowerofTwo(long long n)
{
    if (n == 0)
        return 0;
    if ((n & (~(n - 1))) == n)
        return 1;
    return 0;
}
/*Driver code*/
int main()
{
 
    isPowerofTwo(30) ? cout << "Yes\n" : cout << "No\n";
    isPowerofTwo(128) ? cout << "Yes\n" : cout << "No\n";
    return 0;
}
// This code is contributed by Sachin

Java

// Java program of the above approach
import java.io.*;
 
class GFG {
   
    // Function to check if x is power of 2
    static boolean isPowerofTwo(int n)
    {
        if (n == 0)
            return false;
        if ((n & (~(n - 1))) == n)
            return true;
        return false;
    }
    public static void main(String[] args)
    {
        if (isPowerofTwo(30) == true)
            System.out.println("Yes");
        else
            System.out.println("No");
       
          if (isPowerofTwo(128) == true)
            System.out.println("Yes");
        else
            System.out.println("No");
                 
    }
}
 
// This code is contributed by rajsanghavi9.

Python3

# Python program of the above approach
 
# Function to check if x is power of 2*/
def isPowerofTwo(n):
 
    if (n == 0):
        return 0
    if ((n & (~(n - 1))) == n):
        return 1
    return 0
 
# Driver code
if(isPowerofTwo(30)):
    print('Yes')
else:
    print('No')
     
if(isPowerofTwo(128)):
    print('Yes')
else:
    print('No')
     
# This code is contributed by shivanisinghss2110

C＃

// C# program of the above approach
 
 
using System;
public class GFG {
   
    // Function to check if x is power of 2
    static bool isPowerofTwo(int n)
    {
        if (n == 0)
            return false;
        if ((n & (~(n - 1))) == n)
            return true;
        return false;
    }
    public static void Main(String[] args)
    {
        if (isPowerofTwo(30) == true)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
       
          if (isPowerofTwo(128) == true)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
                 
    }
}
 
// This code contributed by gauravrajput1

Javascript

// javascript program of the above approach
  
    // Function to check if x is power of 2
    function isPowerofTwo(n)
    {
        if (n == 0)
            return false;
        if ((n & (~(n - 1))) == n)
            return true;
        return false;
    }
 
        if (isPowerofTwo(30) == true)
            document.write("Yes<br/>");
        else
            document.write("No<br/>");
       
          if (isPowerofTwo(128) == true)
            document.write("Yes<br/>");
        else
            document.write("No<br/>");
                 
// This code is contributed by umadevi9616

输出

No
Yes

时间复杂度： O(1)

空间复杂度： O(1)