检查给定数字是否稀疏

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📜 检查给定数字是否稀疏

📅 最后修改于: 2021-04-23 19:00:28 🧑 作者: Mango

如果在数字的二进制表示中未设置两个或更多连续位，则将数字称为稀疏数。编写函数以检查给定数字是否为稀疏。
例子：

Input:  x  = 72
Output: true
Explanation:  Binary representation of 72 is 01001000. 
There are no two consecutive 1's in binary representation

Input:  x  = 12
Output: false
Explanation:  Binary representation of 12 is 1100. 
Third and fourth bits (from end) are set.

我们强烈建议您单击此处并进行实践，然后再继续解决方案。

如果我们仔细观察，那么我们会注意到，如果我们可以使用“给定数”的二进制表示形式的二进制“与”运算(即给定数的一半)(即给定数的一半)来判断该数是否稀疏。如果数字稀疏，则AND运算符的结果将为0；如果数字稀疏，则AND运算符的结果将为非零。

照片5354-1

以下是上述想法的实现。

C++

// C++ program to check if n is sparse or not
#include
using namespace std;
 
// Return true if n is sparse, else false
bool checkSparse(int n)
{
    // n is not sparse if there is set
    // in AND of n and n/2
    if (n & (n>>1))
        return false;
 
    return true;
}
 
// Driver program
int main()
{
    cout << checkSparse(72) << endl;
    cout << checkSparse(12) << endl;
    cout << checkSparse(2) << endl;
    cout << checkSparse(3) << endl;
    return 0;
}

Java

// JAVA Code to Check if a
// given number is sparse or not
import java.util.*;
 
class GFG {
     
    // Return true if n is
    // sparse,else false
    static int checkSparse(int n)
    {
 
        // n is not sparse if there
        // is set in AND of n and n/2
        if ((n & (n>>1)) >=1)
            return 0;
      
        return 1;
    }
     
    // Driver code
    public static void main(String[] args)
    {
        System.out.println(checkSparse(72)) ;
        System.out.println(checkSparse(12)) ;
        System.out.println(checkSparse(2)) ;
        System.out.println(checkSparse(3)) ;
        }
    }
 
//This code is contributed by Arnav Kr. Mandal.

Python3

# Python program to check
# if n is sparse or not
 
# Return true if n is
# sparse, else false
def checkSparse(n):
 
    # n is not sparse if there is set
    # in AND of n and n/2
    if (n & (n>>1)):
        return 0
  
    return 1
     
# Driver code
print(checkSparse(72))
print(checkSparse(12))
print(checkSparse(2))
print(checkSparse(30))
 
# This code is contributed
# by Anant Agarwal.

C#

// C# Code to Check if a guven
// number is sparse or not
using System;
 
class GFG {
     
    // Return true if n is
    // sparse,else false
    static int checkSparse(int n)
    {
 
        // n is not sparse if there
        // is set in AND of n and n/2
        if ((n & (n >> 1)) >= 1)
            return 0;
     
        return 1;
    }
     
    // Driver code
    public static void Main()
    {
        Console.WriteLine(checkSparse(72));
        Console.WriteLine(checkSparse(12));
        Console.WriteLine(checkSparse(2));
        Console.WriteLine(checkSparse(3));
    }
}
 
// This code is contributed by Sam007.

PHP

> 1))
        return 0;
 
    return 1;
}
 
// Driver Code
echo checkSparse(72), "\n";
echo checkSparse(12), "\n";
echo checkSparse(2), "\n";
echo checkSparse(3), "\n";
 
// This code is contributed by Ajit.
?>

Javascript

输出：