编写一个单行函数,以整数的二进制表示形式从右到左返回第一个1的位置。
I/P 18, Binary Representation 010010
O/P 2
I/P 19, Binary Representation 010011
O/P 1
算法:(示例12(1100))
令I / P为12(1100)
1.取给定数字的二进制补码,因为所有位都被还原
从右到左的第一个’1’除外(0100)
2进行一点与原始的否,这将返回与否
只需要一个(0100)
3取no的log2,您将得到(位置– 1)(2)
4加1(3)
解释 –
(n&〜(n-1))始终返回包含最右边的设置位为1的二进制数。
如果N = 12(1100),则将返回4(100)
在这里log2将返回您,我们可以用2的幂表示该数字的次数。
对于所有只包含最右边的置1的二进制数,如2,4,8,16,32…。
我们会发现最右边设置位的位置始终等于log2(Number)+1
程序:
C++
// C++ program for Position
// of rightmost set bit
#include
#include
using namespace std;
class gfg
{
public:
unsigned int getFirstSetBitPos(int n)
{
return log2(n & -n) + 1;
}
};
// Driver code
int main()
{
gfg g;
int n = 12;
cout << g.getFirstSetBitPos(n);
return 0;
}
// This code is contributed by SoM15242
C
// C program for Position
// of rightmost set bit
#include
#include
unsigned int getFirstSetBitPos(int n)
{
return log2(n & -n) + 1;
}
// Driver code
int main()
{
int n = 12;
printf("%u", getFirstSetBitPos(n));
getchar();
return 0;
}
Java
// Java Code for Position of rightmost set bit
class GFG {
public static int getFirstSetBitPos(int n)
{
return (int)((Math.log10(n & -n)) / Math.log10(2)) + 1;
}
// Drive code
public static void main(String[] args)
{
int n = 12;
System.out.println(getFirstSetBitPos(n));
}
}
// This code is contributed by Arnav Kr. Mandal
Python3
# Python Code for Position
# of rightmost set bit
import math
def getFirstSetBitPos(n):
return math.log2(n&-n)+1
# driver code
n = 12
print(int(getFirstSetBitPos(n)))
# This code is contributed
# by Anant Agarwal.
C#
// C# Code for Position of rightmost set bit
using System;
class GFG {
public static int getFirstSetBitPos(int n)
{
return (int)((Math.Log10(n & -n))
/ Math.Log10(2)) + 1;
}
// Driver method
public static void Main()
{
int n = 12;
Console.WriteLine(getFirstSetBitPos(n));
}
}
// This code is contributed by Sam007
PHP
Javascript
C++
// C++ program to find the
// position of first rightmost
// set bit in a given number.
#include
using namespace std;
// Function to find rightmost
// set bit in given number.
int getFirstSetBitPos(int n)
{
return ffs(n);
}
// Driver function
int main()
{
int n = 12;
cout << getFirstSetBitPos(n) << endl;
return 0;
}
C++
// C++ program to find the first
// rightmost set bit using XOR operator
#include
using namespace std;
// function to find the rightmost set bit
int PositionRightmostSetbit(int n)
{
// Position variable initialize with 1
// m variable is used to check the set bit
int position = 1;
int m = 1;
while (!(n & m)) {
// left shift
m = m << 1;
position++;
}
return position;
}
// Driver Code
int main()
{
int n = 16;
// function call
cout << PositionRightmostSetbit(n);
return 0;
}
Java
// Java program to find the
// first rightmost set bit
// using XOR operator
class GFG {
// function to find
// the rightmost set bit
static int PositionRightmostSetbit(int n)
{
// Position variable initialize
// with 1 m variable is used to
// check the set bit
int position = 1;
int m = 1;
while ((n & m) == 0) {
// left shift
m = m << 1;
position++;
}
return position;
}
// Driver Code
public static void main(String[] args)
{
int n = 16;
// function call
System.out.println(PositionRightmostSetbit(n));
}
}
// This code is contributed
// by Smitha
Python3
# Python3 program to find
# the first rightmost set
# bit using XOR operator
# function to find the
# rightmost set bit
def PositionRightmostSetbit(n):
# Position variable initialize
# with 1 m variable is used
# to check the set bit
position = 1
m = 1
while (not(n & m)) :
# left shift
m = m << 1
position += 1
return position
# Driver Code
n = 16
# function call
print(PositionRightmostSetbit(n))
# This code is contributed
# by Smitha
C#
// C# program to find the
// first rightmost set bit
// using XOR operator
using System;
class GFG {
// function to find
// the rightmost set bit
static int PositionRightmostSetbit(int n)
{
// Position variable initialize
// with 1 m variable is used to
// check the set bit
int position = 1;
int m = 1;
while ((n & m) == 0) {
// left shift
m = m << 1;
position++;
}
return position;
}
// Driver Code
static public void Main()
{
int n = 16;
// function call
Console.WriteLine(
PositionRightmostSetbit(n));
}
}
// This code is contributed
// by @ajit
PHP
Javascript
C++
// C++ implementation of above approach
#include
using namespace std;
#define INT_SIZE 32
int Right_most_setbit(int num)
{
int pos = 1;
// counting the position of first set bit
for (int i = 0; i < INT_SIZE; i++) {
if (!(num & (1 << i)))
pos++;
else
break;
}
return pos;
}
int main()
{
int num = 18;
int pos = Right_most_setbit(num);
cout << pos << endl;
return 0;
}
// This approach has been contributed by @vivek kumar9
Java
//Java implementation of above approach
public class GFG {
static int INT_SIZE = 32;
static int Right_most_setbit(int num)
{
int pos = 1;
// counting the position of first set bit
for (int i = 0; i < INT_SIZE; i++) {
if ((num & (1 << i))== 0)
pos++;
else
break;
}
return pos;
}
//Driver code
public static void main(String[] args) {
int num = 18;
int pos = Right_most_setbit(num);
System.out.println(pos);
}
}
Python3
# Python 3 implementation of above approach
INT_SIZE = 32
def Right_most_setbit(num) :
pos = 1
# counting the position of first set bit
for i in range(INT_SIZE) :
if not(num & (1 << i)) :
pos += 1
else :
break
return pos
if __name__ == "__main__" :
num = 18
pos = Right_most_setbit(num)
print(pos)
# This code is contributed by ANKITRAI1
C#
// C# implementation of above approach
using System;
class GFG {
static int INT_SIZE = 32;
static int Right_most_setbit(int num)
{
int pos = 1;
// counting the position
// of first set bit
for (int i = 0; i < INT_SIZE; i++)
{
if ((num & (1 << i))== 0)
pos++;
else
break;
}
return pos;
}
// Driver code
static public void Main ()
{
int num = 18;
int pos = Right_most_setbit(num);
Console.WriteLine(pos);
}
}
PHP
Javascript
C++
// C++ program for above approach
#include
using namespace std;
// Program to find position of
// rightmost set bit
int PositionRightmostSetbit(int n)
{
int p=1;
// Iterate till number>0
while(n > 0)
{
// Checking if last bit is set
if(n&1){
return p;
}
// Increment position and right shift number
p++;
n=n>>1;
}
// set bit not found.
return -1;
}
// Driver Code
int main()
{
int n=18;
// Function call
int pos=Last_set_bit(n);
if(pos!=-1)
cout<
Java
// Java program for above approach
import java.io.*;
class GFG{
// Function to find position of
// rightmost set bit
public static int Last_set_bit(int n)
{
int p = 1;
// Iterate till number>0
while (n > 0)
{
// Checking if last bit is set
if ((n & 1) > 0)
{
return p;
}
// Increment position and
// right shift number
p++;
n = n >> 1;
}
// set bit not found.
return -1;
}
// Driver Code
public static void main(String[] args)
{
int n = 18;
// Function call
int pos = Last_set_bit(n);
if (pos != -1)
System.out.println(pos);
else
System.out.println("0");
}
}
// This code is contributed by RohitOberoi
C#
// C# program for above approach
using System;
class GFG{
// Function to find position of
// rightmost set bit
public static int Last_set_bit(int n)
{
int p = 1;
// Iterate till number>0
while (n > 0)
{
// Checking if last bit is set
if ((n & 1) > 0)
{
return p;
}
// Increment position and
// right shift number
p++;
n = n >> 1;
}
// Set bit not found.
return -1;
}
// Driver Code
public static void Main(string[] args)
{
int n = 18;
// Function call
int pos = Last_set_bit(n);
if (pos != -1)
Console.WriteLine(pos);
else
Console.WriteLine("0");
}
}
// This code is contributed by AnkThon
Javascript
3
使用ffs()函数: ffs()函数返回第一个最低有效位的索引。索引从ffs()函数从1开始。
例如:
n = 12 = 1100
在上面的示例中,ffs(n)返回最右边的设置位索引3。
C++
// C++ program to find the
// position of first rightmost
// set bit in a given number.
#include
using namespace std;
// Function to find rightmost
// set bit in given number.
int getFirstSetBitPos(int n)
{
return ffs(n);
}
// Driver function
int main()
{
int n = 12;
cout << getFirstSetBitPos(n) << endl;
return 0;
}
3
使用XOR和&运算符:
将m初始化为1,以从最右边的位开始的位检查其XOR。将m左移一个,直到找到第一个设置位,因为当我们对m执行&运算时,第一个设置位给出一个数字。
C++
// C++ program to find the first
// rightmost set bit using XOR operator
#include
using namespace std;
// function to find the rightmost set bit
int PositionRightmostSetbit(int n)
{
// Position variable initialize with 1
// m variable is used to check the set bit
int position = 1;
int m = 1;
while (!(n & m)) {
// left shift
m = m << 1;
position++;
}
return position;
}
// Driver Code
int main()
{
int n = 16;
// function call
cout << PositionRightmostSetbit(n);
return 0;
}
Java
// Java program to find the
// first rightmost set bit
// using XOR operator
class GFG {
// function to find
// the rightmost set bit
static int PositionRightmostSetbit(int n)
{
// Position variable initialize
// with 1 m variable is used to
// check the set bit
int position = 1;
int m = 1;
while ((n & m) == 0) {
// left shift
m = m << 1;
position++;
}
return position;
}
// Driver Code
public static void main(String[] args)
{
int n = 16;
// function call
System.out.println(PositionRightmostSetbit(n));
}
}
// This code is contributed
// by Smitha
Python3
# Python3 program to find
# the first rightmost set
# bit using XOR operator
# function to find the
# rightmost set bit
def PositionRightmostSetbit(n):
# Position variable initialize
# with 1 m variable is used
# to check the set bit
position = 1
m = 1
while (not(n & m)) :
# left shift
m = m << 1
position += 1
return position
# Driver Code
n = 16
# function call
print(PositionRightmostSetbit(n))
# This code is contributed
# by Smitha
C#
// C# program to find the
// first rightmost set bit
// using XOR operator
using System;
class GFG {
// function to find
// the rightmost set bit
static int PositionRightmostSetbit(int n)
{
// Position variable initialize
// with 1 m variable is used to
// check the set bit
int position = 1;
int m = 1;
while ((n & m) == 0) {
// left shift
m = m << 1;
position++;
}
return position;
}
// Driver Code
static public void Main()
{
int n = 16;
// function call
Console.WriteLine(
PositionRightmostSetbit(n));
}
}
// This code is contributed
// by @ajit
的PHP
Java脚本
5
这种方法是由mubashshir ahmad贡献的
使用左移(<<):用1初始化pos,最多迭代INT_SIZE(Here 32)并检查是否设置了位,如果设置了位则中断循环,否则增加pos。
C++
// C++ implementation of above approach
#include
using namespace std;
#define INT_SIZE 32
int Right_most_setbit(int num)
{
int pos = 1;
// counting the position of first set bit
for (int i = 0; i < INT_SIZE; i++) {
if (!(num & (1 << i)))
pos++;
else
break;
}
return pos;
}
int main()
{
int num = 18;
int pos = Right_most_setbit(num);
cout << pos << endl;
return 0;
}
// This approach has been contributed by @vivek kumar9
Java
//Java implementation of above approach
public class GFG {
static int INT_SIZE = 32;
static int Right_most_setbit(int num)
{
int pos = 1;
// counting the position of first set bit
for (int i = 0; i < INT_SIZE; i++) {
if ((num & (1 << i))== 0)
pos++;
else
break;
}
return pos;
}
//Driver code
public static void main(String[] args) {
int num = 18;
int pos = Right_most_setbit(num);
System.out.println(pos);
}
}
Python3
# Python 3 implementation of above approach
INT_SIZE = 32
def Right_most_setbit(num) :
pos = 1
# counting the position of first set bit
for i in range(INT_SIZE) :
if not(num & (1 << i)) :
pos += 1
else :
break
return pos
if __name__ == "__main__" :
num = 18
pos = Right_most_setbit(num)
print(pos)
# This code is contributed by ANKITRAI1
C#
// C# implementation of above approach
using System;
class GFG {
static int INT_SIZE = 32;
static int Right_most_setbit(int num)
{
int pos = 1;
// counting the position
// of first set bit
for (int i = 0; i < INT_SIZE; i++)
{
if ((num & (1 << i))== 0)
pos++;
else
break;
}
return pos;
}
// Driver code
static public void Main ()
{
int num = 18;
int pos = Right_most_setbit(num);
Console.WriteLine(pos);
}
}
的PHP
Java脚本
输出 :
2
另一种使用右Shift(>>)的方法:
初始化pos = 1。重复执行直到数字> 0,然后在每个步骤中检查是否设置了最后一位。如果设置了最后一位,则返回当前位置,否则将pos加1,向右移n 1。
C++
// C++ program for above approach
#include
using namespace std;
// Program to find position of
// rightmost set bit
int PositionRightmostSetbit(int n)
{
int p=1;
// Iterate till number>0
while(n > 0)
{
// Checking if last bit is set
if(n&1){
return p;
}
// Increment position and right shift number
p++;
n=n>>1;
}
// set bit not found.
return -1;
}
// Driver Code
int main()
{
int n=18;
// Function call
int pos=Last_set_bit(n);
if(pos!=-1)
cout<
Java
// Java program for above approach
import java.io.*;
class GFG{
// Function to find position of
// rightmost set bit
public static int Last_set_bit(int n)
{
int p = 1;
// Iterate till number>0
while (n > 0)
{
// Checking if last bit is set
if ((n & 1) > 0)
{
return p;
}
// Increment position and
// right shift number
p++;
n = n >> 1;
}
// set bit not found.
return -1;
}
// Driver Code
public static void main(String[] args)
{
int n = 18;
// Function call
int pos = Last_set_bit(n);
if (pos != -1)
System.out.println(pos);
else
System.out.println("0");
}
}
// This code is contributed by RohitOberoi
C#
// C# program for above approach
using System;
class GFG{
// Function to find position of
// rightmost set bit
public static int Last_set_bit(int n)
{
int p = 1;
// Iterate till number>0
while (n > 0)
{
// Checking if last bit is set
if ((n & 1) > 0)
{
return p;
}
// Increment position and
// right shift number
p++;
n = n >> 1;
}
// Set bit not found.
return -1;
}
// Driver Code
public static void Main(string[] args)
{
int n = 18;
// Function call
int pos = Last_set_bit(n);
if (pos != -1)
Console.WriteLine(pos);
else
Console.WriteLine("0");
}
}
// This code is contributed by AnkThon
Java脚本
2