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📜  切换数字的第一位和最后一位

📅  最后修改于: 2021-04-26 18:37:49             🧑  作者: Mango

给定数字n,任务是仅切换数字的第一位和最后一位
例子 :

Input : 10
Output : 3
Binary representation of 10 is
1010. After toggling first and
last bits, we get 0011.

Input : 15
Output : 6

先决条件:找到给定号码的MSB。
1)生成一个数字,该数字包含已设置的第一位和最后一位。我们需要将所有中间位更改为0,并将转角位保持为1。
2)答案是生成的数字和原始数字的XOR。

C++
// CPP program to toggle first and last
// bits of a number
#include 
using namespace std;
 
// Returns a number which has same bit
// count as n and has only first and last
// bits as set.
int takeLandFsetbits(int n)
{
    // set all the bit of the number
    n |= n >> 1;
    n |= n >> 2;
    n |= n >> 4;
    n |= n >> 8;
    n |= n >> 16;
 
    // Adding one to n now unsets
    // all bits and moves MSB to
    // one place. Now we shift
    // the number by 1 and add 1.
    return ((n + 1) >> 1) + 1;
}
 
int toggleFandLbits(int n)
{
    // if number is 1
    if (n == 1)
        return 0;
 
    // take XOR with first and
    // last set bit number
    return n ^ takeLandFsetbits(n);
}
 
// Driver code
int main()
{
    int n = 10;
    cout << toggleFandLbits(n);
    return 0;
}


Java
// Java program to toggle first and last
// bits of a number
import java.io.*;
 
class GFG {
     
    // Returns a number which has same bit
    // count as n and has only first and last
    // bits as set.
    static int takeLandFsetbits(int n)
    {
        // set all the bit of the number
        n |= n >> 1;
        n |= n >> 2;
        n |= n >> 4;
        n |= n >> 8;
        n |= n >> 16;
     
        // Adding one to n now unsets
        // all bits and moves MSB to
        // one place. Now we shift
        // the number by 1 and add 1.
        return ((n + 1) >> 1) + 1;
    }
     
    static int toggleFandLbits(int n)
    {
        // if number is 1
        if (n == 1)
            return 0;
     
        // take XOR with first and
        // last set bit number
        return n ^ takeLandFsetbits(n);
    }
     
    // Driver code
    public static void main(String args[])
    {
        int n = 10;
        System.out.println(toggleFandLbits(n));
    }
}
 
/*This code is contributed by Nikita Tiwari.*/


Python3
# Python 3 program to toggle first
# and last bits of a number.
 
 
# Returns a number which has same bit
# count as n and has only first and last
# bits as set.
def takeLandFsetbits(n) :
     
# set all the bit of the number
    n = n | n >> 1
    n = n | n >> 2
    n = n | n >> 4
    n = n | n >> 8
    n = n | n >> 16
 
    # Adding one to n now unsets
    # all bits and moves MSB to
    # one place. Now we shift
    # the number by 1 and add 1.
    return ((n + 1) >> 1) + 1
     
def toggleFandLbits(n) :
    # if number is 1
    if (n == 1) :
        return 0
 
    # take XOR with first and
    # last set bit number
    return n ^ takeLandFsetbits(n)
 
# Driver code
n = 10
print(toggleFandLbits(n))
 
# This code is contributed by Nikita Tiwari.


C#
// C# program to toggle first and last
// bits of a number
using System;
 
class GFG {
      
    // Returns a number which has same bit
    // count as n and has only first and last
    // bits as set.
    static int takeLandFsetbits(int n)
    {
        // set all the bit of the number
        n |= n >> 1;
        n |= n >> 2;
        n |= n >> 4;
        n |= n >> 8;
        n |= n >> 16;
       
        // Adding one to n now unsets
        // all bits and moves MSB to
        // one place. Now we shift
        // the number by 1 and add 1.
        return ((n + 1) >> 1) + 1;
    }
       
    static int toggleFandLbits(int n)
    {
         
        // if number is 1
        if (n == 1)
            return 0;
       
        // take XOR with first and
        // last set bit number
        return n ^ takeLandFsetbits(n);
    }
       
    // Driver code
    public static void Main()
    {
         
        int n = 10;
         
        Console.WriteLine(toggleFandLbits(n));
    }
}
  
// This code is contributed by Anant Agarwal.


PHP
> 1;
    $n |= $n >> 2;
    $n |= $n >> 4;
    $n |= $n >> 8;
    $n |= $n >> 16;
 
    // Adding one to n now unsets
    // all bits and moves MSB to
    // one place. Now we shift
    // the number by 1 and add 1.
    return (($n + 1) >> 1) + 1;
}
 
function toggleFandLbits(int $n)
{
    // if number is 1
    if ($n == 1)
        return 0;
 
    // take XOR with first and
    // last set bit number
    return $n ^ takeLandFsetbits($n);
}
 
// Driver code
$n = 10;
echo toggleFandLbits($n);
 
// This code is contributed by mits
?>


Javascript


输出 :

3

时间复杂度为O(1)。