给定数字的二进制表示形式中的前导零数目

给定整数n，则输出no。以二进制形式表示的前导零。

前导零是数字二进制形式的第一个非零数字之前的任何0数字。
例子：

Input : 16
Output :27
As Binary(16) = (00000000000000000000000000010000)

Input :33
Output :26
As Binary(16)=(00000000000000000000000000100001)

解决方案1：天真的方法是转换no。转换为二进制形式，然后计算否。前导零。它使用昂贵的除法运算。

C++

// C++ program of number of leading zeros in 
// binary representation of a given number
#include 
using namespace std;
  
// Function to count the no. of leading zeros
int countZeros(unsigned int x)
{
    // Keep shifting x by one until leftmost bit
    // does not become 1.
    int total_bits = sizeof(x) * 8;
    int res = 0;
    while ( !(x & (1 << (total_bits - 1))) )
    {
        x = (x << 1);
        res++;
    }
  
    return res;
}
  
// Main function
int main()
{
    int x = 101;
    cout << countZeros(x);
    return 0;
}

Java

// Java program of number of leading zeros in 
// binary representation of a given number
class GFG 
{
static byte sizeofInt = 8;
  
// Function to count the no. of leading zeros
static int countZeros(int x)
{
    // Keep shifting x by one until leftmost bit
    // does not become 1.
    int total_bits = sizeofInt * 8;
    int res = 0;
    while ((x & (1 << (total_bits - 1))) == 0)
    {
        x = (x << 1);
        res++;
    }
  
    return res;
}
  
// Driver Code
public static void main(String[] args) 
{
    int x = 101;
    System.out.println(countZeros(x));
}
}
  
// This code is contributed by PrinciRaj1992

Python3

# Python3 program of number of 
# leading zeros in binary 
# representation of a given number
  
# Function to count the 
# no. of leading zeros
def countZeros(x):
      
    # Keep shifting x by one until 
    # leftmost bit does not become 1.
    total_bits = 32
    res = 0
    while ((x & (1 << (total_bits - 1))) == 0):
        x = (x << 1)
        res += 1
  
    return res
  
# Driver Code
x = 101
print(countZeros(x))
  
# This code is contributed 
# by Mohit Kumar

C#

// C# program of number of leading zeros in 
// binary representation of a given number
using System;
  
class GFG 
{
static byte sizeofInt = 8;
  
// Function to count the 
// no. of leading zeros
static int countZeros(int x)
{
    // Keep shifting x by one until 
    // leftmost bit does not become 1.
    int total_bits = sizeofInt * 8;
    int res = 0;
    while ((x & (1 << (total_bits - 1))) == 0)
    {
        x = (x << 1);
        res++;
    }
    return res;
}
  
// Driver Code
public static void Main(String[] args) 
{
    int x = 101;
    Console.WriteLine(countZeros(x));
}
}
  
// This code is contributed by Rajput-Ji

输出：

25

解决方案2：一种有效的方法是使用按位右移运算来实现相同的目的。算法中的步骤为：
设x为我们的编号。然后

unsigned y;
    int n = 32;
    y = x >>16; if (y != 0) {n = n -16; x = y;}
    y = x >> 8; if (y != 0) {n = n - 8; x = y;}
    y = x >> 4; if (y != 0) {n = n - 4; x = y;}
    y = x >> 2; if (y != 0) {n = n - 2; x = y;}
    y = x >> 1; if (y != 0) return n - 2;
    return n - x;

以上方法仅以12到20条指令执行。

C++

// C++ program of number of leading zeros in 
// binary representation of a given number
#include 
using namespace std;
  
// Function to count the no. of leading zeros
int countZeros(int x)
{
    unsigned y;
    int n = 32;
    y = x >> 16;
    if (y != 0) {
        n = n - 16;
        x = y;
    }
    y = x >> 8;
    if (y != 0) {
        n = n - 8;
        x = y;
    }
    y = x >> 4;
    if (y != 0) {
        n = n - 4;
        x = y;
    }
    y = x >> 2;
    if (y != 0) {
        n = n - 2;
        x = y;
    }
    y = x >> 1;
    if (y != 0)
        return n - 2;
    return n - x;
}
  
// Main function
int main()
{
    int x = 101;
    cout << countZeros(x);
    return 0;
}

Java

// Java program of number of leading zeros in 
// binary representation of a given number
import java.io.*;
  
class GFG {
    // Function to count the no. of leading zeros 
static int countZeros(int x) 
{ 
    int y; 
    int n = 32; 
    y = x >> 16; 
    if (y != 0) { 
        n = n - 16; 
        x = y; 
    } 
    y = x >> 8; 
    if (y != 0) { 
        n = n - 8; 
        x = y; 
    } 
    y = x >> 4; 
    if (y != 0) { 
        n = n - 4; 
        x = y; 
    } 
    y = x >> 2; 
    if (y != 0) { 
        n = n - 2; 
        x = y; 
    } 
    y = x >> 1; 
    if (y != 0) 
        return n - 2; 
    return n - x; 
} 
  
// Main function 
    public static void main (String[] args) {
    int x = 101; 
    System.out.println (countZeros(x)); 
    }
//This code is contributed by @Tushil.    
}

Python3

# Python3 program of number of leading zeros in
# binary representation of a given number
  
  
# Function to count the no. of leading zeros
def countZeros(x):
    n = 32;
    y = x >> 16;
    if (y != 0):
        n = n - 16;
        x = y;
  
    y = x >> 8;
    if (y != 0):
        n = n - 8;
        x = y;
  
    y = x >> 4;
    if (y != 0):
        n = n - 4;
        x = y;
  
    y = x >> 2;
    if (y != 0):
        n = n - 2;
        x = y;
  
    y = x >> 1;
    if (y != 0):
        return n - 2;
    return n - x;
  
  
# Main function
def main():
    x = 101;
    print(countZeros(x))
  
  
if __name__ == '__main__':
    main()

C＃

// C# program of number of leading zeros in 
// binary representation of a given number
using System;
  
class GFG
{
// Function to count the no. of 
// leading zeros 
static int countZeros(int x) 
{ 
    int y; 
    int n = 32; 
    y = x >> 16; 
      
    if (y != 0) 
    { 
        n = n - 16; 
        x = y; 
    } 
    y = x >> 8; 
      
    if (y != 0)
    { 
        n = n - 8; 
        x = y; 
    } 
    y = x >> 4; 
      
    if (y != 0) 
    { 
        n = n - 4; 
        x = y; 
    } 
    y = x >> 2; 
      
    if (y != 0)
    { 
        n = n - 2; 
        x = y; 
    } 
    y = x >> 1; 
      
    if (y != 0) 
        return n - 2; 
    return n - x; 
} 
  
// Driver Code
static public void Main ()
{
    int x = 101; 
    Console.WriteLine(countZeros(x)); 
}
}
  
// This code is contributed by ajit

的PHP

> 16;
    if ($y != 0)
    {
        $n = $n - 16;
        $x = $y;
    }
    $y = $x >> 8;
    if ($y != 0)
    {
        $n = $n - 8;
        $x = $y;
    }
    $y = $x >> 4;
    if ($y != 0) 
    {
        $n = $n - 4;
        $x = $y;
    }
    $y = $x >> 2;
    if ($y != 0) {
        $n = $n - 2;
        $x = $y;
    }
    $y = $x >> 1;
    if ($y != 0)
        return $n - 2;
    return $n - $x;
}
  
// Driver Code
$x = 101;
echo countZeros($x);
  
// This code is contributed 
// by Akanksha Rai

输出：

25

时间复杂度：此方法的时间复杂度为O(1)
空间复杂度：此方法的空间复杂度为O(1)