检查两个数字是否仅在一位位置不同

给定两个非负整数a和b 。问题是检查两个数字是否仅在一位位置不同。
例子：

Input : a = 13, b = 9
Output : Yes
(13)10 = (1101)2
(9)10 = (1001)2
Both the numbers differ at one bit position only, i.e,
differ at the 3rd bit from the right.

Input : a = 15, b = 8
Output : No

方法：以下是步骤：

1. 计算num = a ^ b。
2. 检查num是否为2的幂。请参阅这篇文章。

C++

// C++ implementation to check whether the two
// numbers differ at one bit position only
#include 
using namespace std;
 
// function to check if x is power of 2
bool isPowerOfTwo(unsigned int x)
{
    // First x in the below expression is
    // for the case when x is 0
    return x && (!(x & (x - 1)));
}
 
// function to check whether the two numbers
// differ at one bit position only
bool differAtOneBitPos(unsigned int a,
                       unsigned int b)
{
    return isPowerOfTwo(a ^ b);
}
 
// Driver program to test above
int main()
{
    unsigned int a = 13, b = 9;
    if (differAtOneBitPos(a, b))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java implementation to check whether the two 
// numbers differ at one bit position only
import java.io.*;
import java.util.*;
 
class GFG {
      // function 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);
      }
 
      // function to check whether the two numbers
      // differ at one bit position only
      static boolean differAtOneBitPos(int a, int b)
      {
             return isPowerOfTwo(a ^ b);
      }
      // Driver code
      public static void main(String args[])
      {
             int a = 13, b = 9;
             if (differAtOneBitPos(a, b) == true)
                 System.out.println("Yes");
             else
                 System.out.println("No");
      }
}
 
// This code is contributed by rachana soma

Python3

# Python3 implementation to check whether the two
# numbers differ at one bit position only
 
# 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)))
 
# function to check whether the two numbers
# differ at one bit position only
def differAtOneBitPos( a , b ):
    return isPowerOfTwo(a ^ b)
     
# Driver code to test above
a = 13
b = 9
if (differAtOneBitPos(a, b)):
    print("Yes")
else:
    print( "No")
 
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# implementation to check whether the two
// numbers differ at one bit position only
using System;
class GFG
{
     
    // function 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);
    }
 
    // function to check whether the two numbers
    // differ at one bit position only
    static bool differAtOneBitPos(int a, int b)
    {
        return isPowerOfTwo(a ^ b);
    }
     
    // Driver code
    public static void Main()
    {
        int a = 13, b = 9;
        if (differAtOneBitPos(a, b) == true)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by ihritik

Javascript

 
// JavaScript program to check whether the two 
// numbers differ at one bit position only
 
      // 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!= 0 && ((x & (x - 1)) == 0);
      }
   
      // function to check whether the two numbers
      // differ at one bit position only
      function differAtOneBitPos(a, b)
      {
             return isPowerOfTwo(a ^ b);
      }
  
// Driver code         
        let a = 13, b = 9;
             if (differAtOneBitPos(a, b) == true)
                 document.write("Yes");
             else
                 document.write("No");
        
       // This code is contributed by code_hunt.

输出：

Yes

时间复杂度： O(1)。