通过将所有位翻转到最右边的设置位的左侧而形成的数字

给定整数N ，任务是将所有位翻转到最右边的设置位的左侧，并打印生成的数字。

例子：

Input: N = 10 
Output: 6 
Explanation: 
10 (1010 in binary) 
flipping all bits left to rightmost set bit (index 2) 
-> 6 (0110 in binary)

Input: N = 120 
Output: 8 
Explanation: 
120 (1111000 in binary) 
flipping all bits left to rightmost set bit (index 3) 
-> 8 (0001000 in binary)

天真的方法：
为了解决上述问题，我们将按照以下步骤操作：

• 将给定的整数转换为二进制形式，并将每个位存储到数组中。
• 遍历数组并在第一次出现1之后中断。
• 将所有位翻转到该索引的左侧。将该二进制序列转换为十进制数，然后将其返回。

时间复杂度： O(N)
辅助空间： O(N)

高效方法：
为了优化上述方法，我们将尝试使用按位运算符。

1. 我们从最右边找到设置的位位置，即LSB和总位数。
2. 将给定数字与所有置位位数等于总位数的数字进行异或。
3. 我们不想从设置的位翻转最右边的位。因此，我们再次对所有设置位的总设置位等于firstbit的数字进行XOR运算。

下面是上述方法的实现：

C++

// C++ program to find the
// integer formed after flipping
// all bits to the left of the
// rightmost set bit
#include 
using namespace std;
 
int totCount;
int firstCount;
 
// Function to get the total count
void getTotCount(int num)
{
    totCount = 1;
    firstCount = 1;
    int temp = 1;
     
    // Moving until we get
    // the rightmost set bit
    while ((num & temp) == 0)
    {
        temp = temp << 1;
        totCount += 1;
    }
     
    firstCount = totCount;
    temp = num >> totCount;
     
    // To get total number
    // of bits in a number
    while (temp != 0)
    {
        totCount += 1;
        temp = temp >> 1;
    }
}
     
// Function to find the integer formed
// after flipping all bits to the left
// of the rightmost set bit
int flipBitsFromRightMostSetBit(int num)
{
     
    // Find the total count of bits and
    // the rightmost set bit
    getTotCount(num);
     
    // XOR given number with the
    // number which has is made up
    // of only totbits set
    int num1 = num ^ ((1 << totCount) - 1);
     
    // To avoid flipping the bits
    // to the right of the set bit,
    // take XOR with the number
    // made up of only set firstbits
    num1 = num1 ^ ((1 << firstCount) - 1);
     
    return num1;
}
 
// Driver Code
int main()
{
    int n = 120;
     
    cout << flipBitsFromRightMostSetBit(n)
         << endl;
 
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

Java

// Java program to find the
// integer formed after flipping
// all bits to the left of the
// rightmost set bit
import java.util.*;
 
class GFG{
     
static int totCount;
static int firstCount;
 
// Function to get the total count
static void getTotCount(int num)
{
    totCount = 1;
    firstCount = 1;
    int temp = 1;
     
    // Moving until we get
    // the rightmost set bit
    while ((num & temp) == 0)
    {
        temp = temp << 1;
        totCount += 1;
    }
     
    firstCount = totCount;
    temp = num >> totCount;
     
    // To get total number
    // of bits in a number
    while (temp != 0)
    {
        totCount += 1;
        temp = temp >> 1;
    }
}
     
// Function to find the integer formed
// after flipping all bits to the left
// of the rightmost set bit
static int flipBitsFromRightMostSetBit(int num)
{
     
    // Find the total count of bits and
    // the rightmost set bit
    getTotCount(num);
     
    // XOR given number with the
    // number which has is made up
    // of only totbits set
    int num1 = num ^ ((1 << totCount) - 1);
     
    // To avoid flipping the bits
    // to the right of the set bit,
    // take XOR with the number
    // made up of only set firstbits
    num1 = num1 ^ ((1 << firstCount) - 1);
     
    return num1;
}
         
// Driver Code
public static void main (String[] args)
{
    int n = 120;
     
    System.out.println(
        flipBitsFromRightMostSetBit(n));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to find the
# integer formed after flipping
# all bits to the left of the
# rightmost set bit
 
# Function to get the total count
def getTotCount(num):
    totCount = 1
    firstCount = 1
     
    temp = 1
     
    # Moving until we get
    # the rightmost set bit
    while (not(num & temp)):
        temp = temp << 1
        totCount += 1
    firstCount = totCount
     
    temp = num >> totCount
     
    # To get total number
    # of bits in a number
    while (temp):
        totCount += 1
        temp = temp >> 1
     
    return totCount, firstCount
     
     
# Function to find the integer formed
# after flipping all bits to the left
# of the rightmost set bit
def flipBitsFromRightMostSetBit(num):
     
    # Find the total count of bits and
    # the rightmost set bit
    totbit, firstbit = getTotCount(num)
     
     
    # XOR given number with the
    # number which has is made up
    # of only totbits set
     
    num1 = num ^ ((1 << totbit) - 1)
     
    # To avoid flipping the bits
    # to the right of the set bit,
    # take XOR with the number
    # made up of only set firstbits
     
    num1 = num1 ^ ((1 << firstbit) - 1)
     
    return num1
     
 
if __name__=='__main__':
    n = 120
    print(flipBitsFromRightMostSetBit(n))

C#

// C# program to find the
// integer formed after flipping
// all bits to the left of the
// rightmost set bit
using System;
 
class GFG{
     
static int totCount;
static int firstCount;
 
// Function to get the total count
static void getTotCount(int num)
{
    totCount = 1;
    firstCount = 1;
    int temp = 1;
     
    // Moving until we get
    // the rightmost set bit
    while ((num & temp) == 0)
    {
        temp = temp << 1;
        totCount += 1;
    }
     
    firstCount = totCount;
    temp = num >> totCount;
     
    // To get total number
    // of bits in a number
    while (temp != 0)
    {
        totCount += 1;
        temp = temp >> 1;
    }
}
     
// Function to find the integer formed
// after flipping all bits to the left
// of the rightmost set bit
static int flipBitsFromRightMostSetBit(int num)
{
     
    // Find the total count of bits and
    // the rightmost set bit
    getTotCount(num);
     
    // XOR given number with the
    // number which has is made up
    // of only totbits set
    int num1 = num ^ ((1 << totCount) - 1);
     
    // To avoid flipping the bits
    // to the right of the set bit,
    // take XOR with the number
    // made up of only set firstbits
    num1 = num1 ^ ((1 << firstCount) - 1);
     
    return num1;
}
         
// Driver Code
public static void Main (string[] args)
{
    int n = 120;
     
    Console.Write(
        flipBitsFromRightMostSetBit(n));
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// JavaScript program to find the
// integer formed after flipping
// all bits to the left of the
// rightmost set bit
let totCount;
let firstCount;
 
// Function to get the total count
function getTotCount(num)
{
    totCount = 1;
    firstCount = 1;
    let temp = 1;
     
    // Moving until we get
    // the rightmost set bit
    while ((num & temp) == 0)
    {
        temp = temp << 1;
        totCount += 1;
    }
     
    firstCount = totCount;
    temp = num >> totCount;
     
    // To get total number
    // of bits in a number
    while (temp != 0)
    {
        totCount += 1;
        temp = temp >> 1;
    }
}
     
// Function to find the leteger formed
// after flipping all bits to the left
// of the rightmost set bit
function flipBitsFromRightMostSetBit(num)
{
     
    // Find the total count of bits and
    // the rightmost set bit
    getTotCount(num);
     
    // XOR given number with the
    // number which has is made up
    // of only totbits set
    let num1 = num ^ ((1 << totCount) - 1);
     
    // To avoid flipping the bits
    // to the right of the set bit,
    // take XOR with the number
    // made up of only set firstbits
    num1 = num1 ^ ((1 << firstCount) - 1);
     
    return num1;
}
     
// Driver Code
let n = 120;
 
document.write(flipBitsFromRightMostSetBit(n));
 
// This code is contributed by sanjoy_62
 

输出：

8

时间复杂度： O(N)
辅助空间： O(1)