获取最右边未置位位的位置

原文:https://www . geesforgeks . org/get-position-最右侧-unset-bit/

给定一个非负数 n 。在 n 的二进制表示中找到最右边未设置位的位置，考虑位置 1 的最后一位，位置 2 的第二个最后一位，以此类推。如果 n 的二进制表示中没有0。然后打印“-1”。 示例:

Input : n = 9
Output : 2
(9)<sub>10</sub> = (1001)2
The position of rightmost unset bit in the binary
representation of 9 is 2.

Input : n = 32
Output : 1

进场:以下为步骤:

  1. 如果 n = 0，返回 1。
  2. 如果 n 的所有位都被置位，返回-1。参考这篇帖子。
  3. 否则对给定的数字执行按位非运算(相当于 1 的补码的运算)。让它成为 num = ~n。
  4. 获取号最右边设定位的位置。这将是 n 最右边未置位的位置。

C++

// C++ implementation to get the position of rightmost unset bit
#include <bits/stdc++.h>

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 语言(一种计算机语言，尤用于创建网站)

// 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.

Python 3

# 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

服务器端编程语言（Professional Hypertext Preprocessor 的缩写）

<?php
// PHP implementation to get the
// position of rightmost unset bit

// function to find the position
// of rightmost set bit
function getPosOfRightmostSetBit( $n)
{
    return ceil(log($n &- $n) + 1);
}

// function to get the position
// of rightmost unset bit
function 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 Code
    $n = 9;
    echo getPosOfRightMostUnsetBit($n);

// This code is contributed by anuj_67.
?>

java 描述语言

<script>
// JavaScript implementation to get the position of rightmost unset bit

// function to find the position
// of rightmost set bit
function getPosOfRightmostSetBit(n)
{
    return Math.log2(n&-n)+1;
}

// function to get the position of rightmost unset bit
function 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
    let n = 9;
    document.write(getPosOfRightMostUnsetBit(n));

// This code is contributed by Manoj.
</script>

输出:

2

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