给定正整数N ,任务是找到可以通过以N的二进制表示形式从LSB开始翻转连续的设置位获得的数字。
例子:
Input: N = 39
Output: 32
Explanation:
Binary representation of (39)10 = (100111)2
After flipping all consecutive set bits starting from the LSB, the number obtained is (100000)
Binary representation of (32)10 is (100000)2
Therefore, the number obtained is 32.
Input: N = 4
Output: 4
Explanation:
Binary representation of (4)10 = (100)2
Since the LSB is not set, the number remains unchanged.
天真的方法:最简单的方法是通过对N进行1的逻辑AND(&)运算直到N&1不为0并保持计数,从LSB开始查找连续的设置位数。 设置位的位数。然后,简单地将N移位设置位的数量即可。打印获得的号码作为所需答案。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the number after
// converting 1s from end to 0s
int findNumber(int N)
{
// Count of 1s
int count = 0;
// AND operation of N and 1
while ((N & 1) == 1) {
N = N >> 1;
count++;
}
// Left shift N by count
return N << count;
}
// Driver Code
int main()
{
int N = 39;
// Function Call
cout << findNumber(N);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to find the number after
// converting 1s from end to 0s
static int findNumber(int N)
{
// Count of 1s
int count = 0;
// AND operation of N and 1
while ((N & 1) == 1)
{
N = N >> 1;
count++;
}
// Left shift N by count
return N << count;
}
// Driver Code
public static void main (String[] args)
{
int N = 39;
// Function Call
System.out.println(findNumber(N));
}
}
// This code is contributed by AnkThon
Python3
# Python3 program for the above approach
# Function to find the number after
# converting 1s from end to 0s
def findNumber(N):
# Count of 1s
count = 0
# AND operation of N and 1
while ((N & 1) == 1):
N = N >> 1
count += 1
# Left shift N by count
return N << count
# Driver Code
if __name__ == "__main__":
N = 39
# Function Call
print(findNumber(N))
# This code is contributed by AnkThon
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to find the number after
// converting 1s from end to 0s
static int findNumber(int N)
{
// Count of 1s
int count = 0;
// AND operation of N and 1
while ((N & 1) == 1)
{
N = N >> 1;
count++;
}
// Left shift N by count
return N << count;
}
// Driver Code
public static void Main()
{
int N = 39;
// Function Call
Console.WriteLine(findNumber(N));
}
}
// This code is contributed by code_hunt
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the number after
// converting 1s from end to 0s
int findNumber(int N)
{
// Return the logical AND
// of N and (N + 1)
return N & (N + 1);
}
// Driver Code
int main()
{
int N = 39;
// Function Call
cout << findNumber(N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the number after
// converting 1s from end to 0s
static int findNumber(int N)
{
// Return the logical AND
// of N and (N + 1)
return N & (N + 1);
}
// Driver Code
public static void main(String[] args)
{
int N = 39;
// Function Call
System.out.print(findNumber(N));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program for the above approach
# Function to find the number after
# converting 1s from end to 0s
def findNumber(N):
# Return the logical AND
# of N and (N + 1)
return N & (N + 1)
# Driver Code
if __name__ == '__main__':
N = 39
# Function Call
print(findNumber(N))
# This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
class GFG
{
// Function to find the number after
// converting 1s from end to 0s
static int findNumber(int N)
{
// Return the logical AND
// of N and (N + 1)
return N & (N + 1);
}
// Driver Code
public static void Main(String[] args)
{
int N = 39;
// Function Call
Console.Write(findNumber(N));
}
}
// This code is contributed by 29AjayKumar
32
时间复杂度: O(log N)
辅助空间: O(1)
高效方法:要优化上述方法,请找到N和(N + 1)的逻辑与(& ) 。关键的观察结果是,对数字加1会使LSB中的每个连续设置位变为0 。因此, N&(N + 1)给出所需的数字。
插图:
N = 39, therefore (N+1)=40
⇒ N = 39 = (100111)
⇒ N+1 = 40 = (101000)
Performing Logical AND(&) operation:
1 0 0 1 1 1
& 1 0 1 0 0 0
----------------
1 0 0 0 0 0 ⇒ 32
----------------
Will this always work? Add 1 to N:
1 0 0 1 1 1
+ 1
-------------
1 0 1 0 0 0
--------------
It can be clearly seen that the continous set bits from the LSB becomnes unset.
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the number after
// converting 1s from end to 0s
int findNumber(int N)
{
// Return the logical AND
// of N and (N + 1)
return N & (N + 1);
}
// Driver Code
int main()
{
int N = 39;
// Function Call
cout << findNumber(N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the number after
// converting 1s from end to 0s
static int findNumber(int N)
{
// Return the logical AND
// of N and (N + 1)
return N & (N + 1);
}
// Driver Code
public static void main(String[] args)
{
int N = 39;
// Function Call
System.out.print(findNumber(N));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program for the above approach
# Function to find the number after
# converting 1s from end to 0s
def findNumber(N):
# Return the logical AND
# of N and (N + 1)
return N & (N + 1)
# Driver Code
if __name__ == '__main__':
N = 39
# Function Call
print(findNumber(N))
# This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
class GFG
{
// Function to find the number after
// converting 1s from end to 0s
static int findNumber(int N)
{
// Return the logical AND
// of N and (N + 1)
return N & (N + 1);
}
// Driver Code
public static void Main(String[] args)
{
int N = 39;
// Function Call
Console.Write(findNumber(N));
}
}
// This code is contributed by 29AjayKumar
32
时间复杂度: O(log N)
辅助空间: O(1)