比较两个数字的二进制表示中的前导零
给定两个整数 x 和 y。使用按位运算比较并打印哪一个有更多的前导零。如果两个数字的前导零数量相同，则打印“相等”。 注意:-前导零是数字二进制记数法中第一个非零数字之前的任何 0 数字。 例:
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 :天真的方法是先找到数字的二进制表示，然后计算前导零的个数。 解 2: 找到比给定数字小的最大二次幂，比较这些二次幂决定答案。 解决方案 3: 一种有效的方法是按位异或和与运算符。
情况 1:如果两者都有相同数量的前导零，那么(x^y) <= (x & y) because same number of leading 0s would cause a 1 at higher position in x & y. 情况 2:如果我们对 y 进行求反并与 x 按位“与”,当 y 有更多数量的前导 0 时，我们会在比 y 更高的位置得到 1。 情况 3:否则 x 有更多的前导零
C++
// CPP program to find the number with more
// leading zeroes.
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find the number
// with more leading zeroes.
// Function to compare the no.
// of leading zeros
function LeadingZeros($x, $y)
{
// if both have same no. of
// leading zeros
if (($x ^ $y) <= ($x & $y))
echo "\nEqual";
// if y has more leading zeros
else if (($x & (~$y)) > $y)
echo $y;
else
echo $x;
}
// Driver Code
$x = 10;
$y = 16;
LeadingZeros($x, $y);
// This code is contributed by ajit
?>
java 描述语言
<script>
// Javascript program to find the number with more
// leading zeroes.
// Function to compare the no. of leading zeros
function LeadingZeros(x, y)
{
// if both have same no. of leading zeros
if ((x ^ y) <= (x & y))
document.write("<br>Equal");
// if y has more leading zeros
else if ((x & (~y)) > y)
document.write(y);
else
document.write(x);
}
// Main Function
let x = 10, y = 16;
LeadingZeros(x, y);
</script>
Output:
10
时间复杂度:O(1) T3】空间复杂度: O(1)
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