📜  检查数字是否有其他模式的位|套装1

📅  最后修改于: 2021-05-25 03:57:01             🧑  作者: Mango

给定一个整数n> 0,任务是确定该整数的位表示形式是否具有备用模式。例如-5具有替代图案,即101。
如果打印纸有其他样式,则打印“是”,否则打印“否”。在这里,替代模式可以像0101或1010。

例子:

Input :  15
Output :  No
Explanation: Binary representation of 15 is 1111.

Input :  10
Output :  Yes
Explanation: Binary representation of 10 is 1010.

一种简单的方法是找到其二进制等效项,然后检查其位。

C++
// C++ program to find if a number has alternate
// bit pattern
#include
using namespace std;
  
// Returns true if n has alternate bit pattern
// else false
bool findPattern(int n)
{
    // Store last bit
    int prev = n % 2;
    n = n/2;
  
    // Traverse through remaining bits
    while (n > 0)
    {
        int curr = n % 2;
  
        // If current bit is same as previous
        if (curr == prev)
           return false;
  
        prev = curr;
        n = n / 2;
    }
  
    return true;
}
  
// Driver code
int main()
{
    int n = 10;
    if (findPattern(n))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}


Java
// Java program to find if a number has alternate
// bit pattern
  
class Test
{
    // Returns true if n has alternate bit pattern
    // else false
    static boolean findPattern(int n)
    {
        // Store last bit
        int prev = n % 2;
        n = n/2;
       
        // Traverse through remaining bits
        while (n > 0)
        {
            int curr = n % 2;
       
            // If current bit is same as previous
            if (curr == prev)
               return false;
       
            prev = curr;
            n = n / 2;
        }
       
        return true;
    }
      
    // Driver method
    public static void main(String args[])
    {
        int n = 10;
        System.out.println(findPattern(n) ? "Yes" : "No");
       
    }
}


Python3
# Python3 program to find if a number 
# has alternate bit pattern
  
# Returns true if n has alternate 
# bit pattern else false
def findPattern(n):
  
    # Store last bit
    prev = n % 2
    n = n // 2
  
    # Traverse through remaining bits
    while (n > 0):
      
        curr = n % 2
  
        # If current bit is same as previous
        if (curr == prev):
            return False
  
        prev = curr
        n = n // 2
      
    return True
  
# Driver code
n = 10
print("Yes") if(findPattern(n)) else print("No")
  
# This code is contributed by Anant Agarwal.


C#
// Program to find if a number
// has alternate bit pattern
using System;
  
class Test {
  
    // Returns true if n has alternate
    // bit pattern else returns false
    static bool findPattern(int n)
    {
        // Store last bit
        int prev = n % 2;
        n = n / 2;
  
        // Traverse through remaining bits
        while (n > 0) {
            int curr = n % 2;
  
            // If current bit is same as previous
            if (curr == prev)
                return false;
  
            prev = curr;
            n = n / 2;
        }
  
        return true;
    }
  
    // Driver method
    public static void Main()
    {
        int n = 10;
        Console.WriteLine(findPattern(n) ? "Yes" : "No");
    }
}
  
// This code is contributed by Anant Agarwal.


PHP
 0)
    {
        $curr = $n % 2;
  
        // If current bit is 
        // same as previous
        if ($curr == $prev)
        return false;
  
        $prev = $curr;
        $n = floor($n / 2);
    }
  
    return true;
}
  
    // Driver code
    $n = 10;
    if (findPattern($n))
        echo "Yes";
    else
        echo "No";
          
    return 0;
  
// This code is contributed by nitin mittal. 
?>


输出:

Yes

参考:
http://stackoverflow.com/questions/38690278/program-to-check-whether-the-given-integer-has-an-alternate-pattern