查找N的位的可能排列

给定整数N ，任务是查找N的位是否可以交替排列，即0101…或10101… 。假设N表示为32位整数。
例子：

Input: N = 23 
Output: No 
“00000000000000000000000000010111” is the binary representation of 23 
and the required permutation of bits is not possible.
Input: N = 524280 
Output: Yes 
binary(524280) = “00000000000001111111111111111000” which can be 
rearranged to “01010101010101010101010101010101”. 
 

方法：由于给定整数必须用32位表示，并且1的数量必须等于其二进制表示形式中的0的数量，才能满足给定条件。因此，N中的置位位数必须为16，可以使用__builtin_popcount()轻松计算出
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
const int TOTAL_BITS = 32;
 
// Function that returns true if it is
// possible to arrange the bits of
// n in alternate fashion
bool isPossible(int n)
{
 
    // To store the count of 1s in the
    // binary representation of n
    int cnt = __builtin_popcount(n);
 
    // If the number set bits and the
    // number of unset bits is equal
    if (cnt == TOTAL_BITS / 2)
        return true;
    return false;
}
 
// Driver code
int main()
{
    int n = 524280;
 
    if (isPossible(n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
     
class GFG
{
 
static int TOTAL_BITS = 32;
 
// Function that returns true if it is
// possible to arrange the bits of
// n in alternate fashion
static boolean isPossible(int n)
{
 
    // To store the count of 1s in the
    // binary representation of n
    int cnt = Integer.bitCount(n);
 
    // If the number set bits and the
    // number of unset bits is equal
    if (cnt == TOTAL_BITS / 2)
        return true;
    return false;
}
 
// Driver code
static public void main (String []arr)
{
    int n = 524280;
 
    if (isPossible(n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
TOTAL_BITS = 32;
 
# Function that returns true if it is
# possible to arrange the bits of
# n in alternate fashion
def isPossible(n) :
 
    # To store the count of 1s in the
    # binary representation of n
    cnt = bin(n).count('1');
 
    # If the number set bits and the
    # number of unset bits is equal
    if (cnt == TOTAL_BITS // 2) :
        return True;
         
    return False;
 
# Driver code
if __name__ == "__main__" :
 
    n = 524280;
 
    if (isPossible(n)) :
        print("Yes");
    else :
        print("No");
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;
     
class GFG
{
static int TOTAL_BITS = 32;
 
static int CountBits(int value)
{
    int count = 0;
    while (value != 0)
    {
        count++;
        value &= value - 1;
    }
    return count;
}
 
// Function that returns true if it is
// possible to arrange the bits of
// n in alternate fashion
static bool isPossible(int n)
{
 
    // To store the count of 1s in the
    // binary representation of n
    int cnt = CountBits(n);
 
    // If the number set bits and the
    // number of unset bits is equal
    if (cnt == TOTAL_BITS / 2)
        return true;
    return false;
}
 
// Driver code
public static void Main (String []arr)
{
    int n = 524280;
 
    if (isPossible(n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by Mohit kumar

Javascript

// Javascript implementation of the approach
 
const TOTAL_BITS = 32;
 
function CountBits(value)
{
    let count = 0;
    while (value != 0)
    {
        count++;
        value &= value - 1;
    }
    return count;
}
 
// Function that returns true if it is
// possible to arrange the bits of
// n in alternate fashion
function isPossible(n)
{
 
    // To store the count of 1s in the
    // binary representation of n
    let cnt = CountBits(n);
 
    // If the number set bits and the
    // number of unset bits is equal
    if (cnt == parseInt(TOTAL_BITS / 2))
        return true;
    return false;
}
 
// Driver code
    let n = 524280;
 
    if (isPossible(n))
        document.write("Yes");
    else
        document.write("No");
 
// This code is contributed by subhammahato348.

输出：

Yes