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字，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。
- 我们从最右边找到设置的位位置，即LSB和总位数。
- 将给定数字与所有置位位数等于总位数的数字进行异或。
- 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 offbeatPython3
# 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输出：
8
时间复杂度： O(N)
辅助空间： O(1)