给定一个非负数n 。在n的二进制表示中找到最右边的未设置位的位置,请考虑位置1的最后一位,位置2的最后一位,依此类推。如果n的二进制表示中不存在0 ,则为0 。然后打印“ -1”。
例子:
Input : n = 9
Output : 2
(9)10 = (1001)2
The position of rightmost unset bit in the binary
representation of 9 is 2.
Input : n = 32
Output : 1
方法:以下是步骤:
- 如果n = 0,则返回1。
- 如果n的所有位都置1,则返回-1。请参阅这篇文章。
- 否则,不按给定数字执行按位运算(等于1的补码的运算)。使其为num =〜n。
- 获取num的最右边设置位的位置。这将是n的最右边未设置位的位置。
C++
// C++ implementation to get the position of rightmost unset bit
#include
using namespace std;
// function to find the position
// of rightmost set bit
int getPosOfRightmostSetBit(int n)
{
return log2(n&-n)+1;
}
// function to get the position of rightmost unset bit
int getPosOfRightMostUnsetBit(int n)
{
// if n = 0, return 1
if (n == 0)
return 1;
// if all bits of 'n' are set
if ((n & (n + 1)) == 0)
return -1;
// position of rightmost unset bit in 'n'
// passing ~n as argument
return getPosOfRightmostSetBit(~n);
}
// Driver program to test above
int main()
{
int n = 9;
cout << getPosOfRightMostUnsetBit(n);
return 0;
}
Java
// Java implementation to get the
// position of rightmost unset bit
class GFG {
// function to find the position
// of rightmost set bit
static int getPosOfRightmostSetBit(int n)
{
return (int)((Math.log10(n & -n)) / Math.log10(2)) + 1;
}
// function to get the position
// of rightmost unset bit
static int getPosOfRightMostUnsetBit(int n) {
// if n = 0, return 1
if (n == 0)
return 1;
// if all bits of 'n' are set
if ((n & (n + 1)) == 0)
return -1;
// position of rightmost unset bit in 'n'
// passing ~n as argument
return getPosOfRightmostSetBit(~n);
}
// Driver code
public static void main(String arg[])
{
int n = 9;
System.out.print(getPosOfRightMostUnsetBit(n));
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python3 implementation to get the position
# of rightmost unset bit
# import library
import math as m
# function to find the position
# of rightmost set bit
def getPosOfRightmostSetBit(n):
return (m.log(((n & - n) + 1),2))
# function to get the position ot rightmost unset bit
def getPosOfRightMostUnsetBit(n):
# if n = 0, return 1
if (n == 0):
return 1
# if all bits of 'n' are set
if ((n & (n + 1)) == 0):
return -1
# position of rightmost unset bit in 'n'
# passing ~n as argument
return getPosOfRightmostSetBit(~n)
# Driver program to test above
n = 13;
ans = getPosOfRightMostUnsetBit(n)
#rounding the final answer
print (round(ans))
# This code is contributed by Saloni Gupta.
C#
// C# implementation to get the
// position of rightmost unset bit
using System;
class GFG
{
// function to find the position
// of rightmost set bit
static int getPosOfRightmostSetBit(int n)
{
return (int)((Math.Log10(n & -n)) / Math.Log10(2)) + 1;
}
// function to get the position
// of rightmost unset bit
static int getPosOfRightMostUnsetBit(int n) {
// if n = 0, return 1
if (n == 0)
return 1;
// if all bits of 'n' are set
if ((n & (n + 1)) == 0)
return -1;
// position of rightmost unset bit in 'n'
// passing ~n as argument
return getPosOfRightmostSetBit(~n);
}
// Driver code
public static void Main()
{
int n = 9;
Console.Write(getPosOfRightMostUnsetBit(n));
}
}
// This code is contributed by Sam007
PHP
Javascript
输出:
2