上一个较小的整数，其置位位数少

给定一个正整数’n’，在其二进制表示中具有’x’个置位位数。问题在于找到前一个较小的整数(小于n的最大整数)，在其二进制表示形式中具有(x-1)个设置位。
注意： 1 <= n
例子 ：

Input : 8
Output : 0
(8)10 = (1000)2
is having 1 set bit.

(0)10 = (0)2
is having 0 set bit and is the previous smaller.

Input : 25
Output : 24

步骤如下：

1. 在n的二进制表示中找到最右置位的位置(考虑位置1的最后一位，位置2的最后一位，依此类推)。让位置用pos表示。请参阅这篇文章。
2. 关闭或取消设置位置pos的位。请参阅这篇文章。

C++

// C++ implementation to find the previous
// smaller integer with one less number of
// set bits
#include
using namespace std;
 
// function to find the position of
// rightmost set bit.
int getFirstSetBitPos(int n)
{
    return log2(n & -n) + 1;
}
 
// function to find the previous smaller
// integer
int previousSmallerInteger(int n)
{
    // position of rightmost set bit of n
    int pos = getFirstSetBitPos(n);
 
    // turn off or unset the bit at
    // position 'pos'
    return (n & ~(1 << (pos - 1)));
}
 
// Driver program
int main()
{
    int n = 25;
    cout << previousSmallerInteger(n);
    return 0;
}

Java

// Java implementation to find the previous
// smaller integer with one less number of
// set bits
class GFG {
     
    // function to find the position of
    // rightmost set bit.
    static int getFirstSetBitPos(int n)
    {
        return (int)(Math.log(n & -n) / Math.log(2)) + 1;
    }
 
    // function to find the previous smaller
    // integer
    static int previousSmallerInteger(int n)
    {
         
        // position of rightmost set bit of n
        int pos = getFirstSetBitPos(n);
 
        // turn off or unset the bit at
        // position 'pos'
        return (n & ~(1 << (pos - 1)));
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int n = 25;
        System.out.print("Previous smaller Integer ="
                     + previousSmallerInteger(n));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 implementation to find
# the previous smaller integer with
# one less number of set bits
import math
 
# Function to find the position 
# of rightmost set bit.
def getFirstSetBitPos(n):
 
    return (int)(math.log(n & -n) /
                 math.log(2)) + 1
 
# Function to find the 
# previous smaller integer
def previousSmallerInteger(n):
 
    # position of rightmost set bit of n
    pos = getFirstSetBitPos(n)
 
    # turn off or unset the bit 
    # at position 'pos'
    return (n & ~(1 << (pos - 1)))
     
# Driver code
n = 25
print("Previous small Integer = ",
       previousSmallerInteger(n))
        
# This code is contributed by Anant Agarwal.

C#

// C# implementation to find the previous
// smaller integer with one less number of
// set bits
using System;
 
    class GFG {
         
    // function to find the position of
    // rightmost set bit.
    static int getFirstSetBitPos(int n)
    {
        return (int)(Math.Log(n & -n) /
                           Math.Log(2)) + 1;
    }
      
    // function to find the previous smaller
    // integer
    static int previousSmallerInteger(int n)
    {
         
        // position of rightmost set bit of n
        int pos = getFirstSetBitPos(n);
      
        // turn off or unset the bit at
        // position 'pos'
        return (n & ~(1 << (pos - 1)));
    }
     
    // Driver code
    public static void Main()
    {
        int n = 25;
         
        Console.WriteLine("Previous small Integer ="
                      + previousSmallerInteger(n));
    }
}
 
// This code is contributed by anant321.

Javascript

 
// Javascript implementation to find the previous
// smaller integer with one less number of
// set bits
 
// function to find the position of
// rightmost set bit.
function getFirstSetBitPos(n)
{
    return parseInt(Math.log(n & -n)/Math.log(2)) + 1;
}
 
// function to find the previous smaller
// integer
function previousSmallerInteger(n)
{
    // position of rightmost set bit of n
    var pos = getFirstSetBitPos(n);
 
    // turn off or unset the bit at
    // position 'pos'
    return (n & ~(1 << (pos - 1)));
}
 
// Driver program
var n = 25;
document.write("Previous smaller Integer = " + previousSmallerInteger(n));
     

输出 ：

Previous smaller integer = 24