比较两个数字的二进制表示形式中的前导零

给定两个整数x和y。使用按位运算比较并打印其中一个具有更多前导零。如果两者都没有。有相同的编号的前导零，打印“等于”。

注意：-前导零是该数字的二进制表示形式中第一个非零数字之前的任何0数字。

例子：

Input : 10, 16
Output :10
Explanation: If we represent the no.s using 8 bit only then 
Binary(10) = 00001010
Binary(16) = 00010000
Clearly, 10 has 4 leading zeros and 16 has 3 leading zeros

Input : 10, 12
Output : Equal
Binary(10) = 00001010
Binary(12) = 00001100
Both have equal no. of leading zeros.

解决方案1 ：天真的方法是首先找到数字的二进制表示形式，然后计算数字。前导零。

解决方案2：找到小于给定数字的最大2的幂，然后比较这些2的幂以确定答案。

解决方案3：一种有效的方法是按位XOR和AND运算符。

Case 1: If both have same no. of leading zeros then (x^y) <= (x & y) because same number of leading 0s would cause a 1 at higher position in x & y.

Case 2 : If we do negation of y and do bitwise AND with x, we get a one at higher position than in y when y has more number of leading 0s.

Case 3: Else x has more leading zeros

C++

// CPP program to find the number with more
// leading zeroes.
#include 
using namespace std;
  
// Function to compare the no. of leading zeros
void LeadingZeros(int x, int y)
{
    // if both have same no. of leading zeros
    if ((x ^ y) <= (x & y)) 
        cout << "\nEqual";
  
    // if y has more leading zeros
    else if ((x & (~y)) > y) 
        cout << y;
  
    else 
        cout << x;
}
  
// Main Function
int main()
{
    int x = 10, y = 16;
    LeadingZeros(x, y);
    return 0;
}

Java

// Java program to find the number 
// with more leading zeroes. 
class GFG
{
// Function to compare the no. 
// of leading zeros 
static void LeadingZeros(int x, int y) 
{ 
    // if both have same no. of 
    // leading zeros 
    if ((x ^ y) <= (x & y)) 
        System.out.print("\nEqual");
  
    // if y has more leading zeros 
    else if ((x & (~y)) > y) 
        System.out.print(y); 
  
    else
        System.out.print(x); 
} 
  
// Driver Code
public static void main (String[] args) 
{
    int x = 10, y = 16; 
    LeadingZeros(x, y); 
}
} 
  
// This code is contributed by Smitha

Python3

# Python 3 program to find the number 
# with more leading zeroes.
  
# Function to compare the no. of
# leading zeros
def LeadingZeros(x, y):
      
    # if both have same no. of
    # leading zeros
    if ((x ^ y) <= (x & y)): 
        print("Equal")
  
    # if y has more leading zeros
    elif ((x & (~y)) > y) :
        print(y)
  
    else:
        print(x)
  
# Driver Code
if __name__ == '__main__':
    x = 10
    y = 16
    LeadingZeros(x, y)
  
# This code is contributed 
# by Surendra_Gangwar

C#

// C# program to find the number 
// with more leading zeroes. 
using System;
  
class GFG
{
// Function to compare the no. 
// of leading zeros 
static void LeadingZeros(int x, int y) 
{ 
    // if both have same no. of 
    // leading zeros 
    if ((x ^ y) <= (x & y)) 
        Console.WriteLine("\nEqual"); 
  
    // if y has more leading zeros 
    else if ((x & (~y)) > y) 
        Console.WriteLine(y); 
  
    else
        Console.WriteLine(x); 
} 
  
// Driver Code 
static public void Main ()
{
    int x = 10, y = 16; 
    LeadingZeros(x, y); 
}
} 
  
// This code is contributed by ajit

PHP

 $y) 
        echo $y;
  
    else
        echo $x;
}
  
// Driver Code
$x = 10;
$y = 16;
LeadingZeros($x, $y);
      
// This code is contributed by ajit
?>

输出：

10

时间复杂度： O(1)
空间复杂度： O(1)