给定一个正整数，编写一个函数来确定它是否是2的幂。 例如：
null
Input : n = 4Output : Yes22 = 4Input : n = 7Output : NoInput : n = 32Output : Yes25 = 32
1. 一个简单的方法是简单地取基数2的对数，如果你得到一个整数，那么这个数就是2的幂
C++
// C++ Program to find whether a // no is power of two #include<bits/stdc++.h> 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" <<endl: cout<< "No" <<endl; isPowerOfTwo(64)? cout<< "Yes" <<endl: cout<< "No" <<endl; return 0; } // This code is contributed by Surendra_Gangwar |
C
// C Program to find whether a // no is power of two #include<stdio.h> #include<stdbool.h> #include<math.h> /* 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" ): printf ( "No" ); isPowerOfTwo(64)? printf ( "Yes" ): printf ( "No" ); 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 |
蟒蛇3
# 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) |
PHP
<?php // PHP Program to find // whether a no is // power of two // Function to check // Log base 2 function Log2( $x ) { return (log10( $x ) / log10(2)); } // Function to check // if x is power of 2 function isPowerOfTwo( $n ) { return ( ceil (Log2( $n )) == floor (Log2( $n ))); } // Driver Code if (isPowerOfTwo(31)) echo "Yes" ; else echo "No" ; if (isPowerOfTwo(64)) echo "Yes" ; else echo "No" ; // This code is contributed // by Sam007 ?> |
Javascript
<script> // 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. </script> |
输出：
NoYes
时间复杂性： O（1） 辅助空间： O（1）
2. 另一种解决方案是将该数字不断除以2，也就是说，Don=n/2迭代。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变成1，那么它是2的幂。
C++
#include <bits/stdc++.h> 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" : cout<< "No" ; isPowerOfTwo(64)? cout<< "Yes" : cout<< "No" ; return 0; } // This code is contributed by rathbhupendra |
C
#include<stdio.h> #include<stdbool.h> /* 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" ): printf ( "No" ); isPowerOfTwo(64)? printf ( "Yes" ): printf ( "No" ); 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. |
蟒蛇3
# 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 |
PHP
<?php // 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; } // Driver Code if (isPowerOfTwo(31)) echo "Yes" ; else echo "No" ; if (isPowerOfTwo(64)) echo "Yes" ; else echo "No" ; // This code is contributed // by Sam007 ?> |
Javascript
<script> /* 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" ); </script> |
输出：
NoYes
时间复杂性： O（原木） 2. n）
辅助空间： O（1）
3. 另一种方法是使用这个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 <bits/stdc++.h> 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" + "" ); else System.out.print( "False" + "" ); if (powerOf2(m) == true ) System.out.print( "True" + "" ); else System.out.print( "False" + "" ); } } // This code is contributed by Princi Singh |
蟒蛇3
# 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" + "" ); } else { Console.Write( "False" + "" ); } if (powerOf2(m)) { Console.Write( "True" ); } else { Console.Write( "False" ); } } } // This code is contributed by rutvik_56 |
Javascript
<script> // 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" + "" ); else document.write( "False" + "" ); if (powerOf2(m) == true ) document.write( "True" + "" ); else document.write( "False" + "" ); // This code contributed by shikhasingrajput </script> |
输出
TrueFalse
时间复杂性： O（原木） 2. n）
辅助空间： O（原木） 2. n）
4. 两个数的所有幂只有一个位集。所以数一数设定位的数量，如果你得到1，那么这个数字是2的幂。请看 在整数中计算集合位 用于计算设定位。
5. 如果我们用2个数的幂减去1，那么唯一设定位之后的所有未设定位都变为设定位；设定的位元会变为未设定。 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
所以，如果一个数n是2的幂，那么n和n-1的按位&将为零。根据n&（n-1）的值，我们可以说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，我们的表达式将变成n&（！n&（n-1））（由于 https://www.geeksforgeeks.org/program-to-find-whether-a-no-is-power-of-two/ 感谢穆罕默德加入本案）。
下面是这个方法的实现。
时间复杂度：O（1）
空间复杂性：O（1）
C++
#include <bits/stdc++.h> 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" : cout<< "No" ; isPowerOfTwo(64)? cout<< "Yes" : cout<< "No" ; return 0; } // This code is contributed by rathbhupendra |
C
#include<stdio.h> #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" ): printf ( "No" ); isPowerOfTwo(64)? printf ( "Yes" ): printf ( "No" ); 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 |
蟒蛇3
# 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 |
PHP
<?php // PHP program to efficiently // check for power for 2 // Function 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 && (!( $x & ( $x - 1))); } // Driver Code if (isPowerOfTwo(31)) echo "Yes" ; else echo "No" ; if (isPowerOfTwo(64)) echo "Yes" ; else echo "No" ; // This code is contributed by Sam007 ?> |
Javascript
<script> // 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 </script> |
输出：
NoYes
时间复杂性： O（1）
辅助空间： O（1）
6. .另一种方法是使用逻辑查找给定数字的最右边的位集。
C++
#include <iostream> 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" : cout << "No" ; isPowerofTwo(128) ? cout << "Yes" : cout << "No" ; 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. |
蟒蛇3
# 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
<script> // 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 </script> |
输出
NoYes
时间复杂性： O（1）
空间复杂性： O（1）
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