检查一个数字是否有两个相邻的设定位
给定一个数，你必须检查是否有一对相邻的设置位。 例:
Input : N = 67
Output : Yes
There is a pair of adjacent set bit
The binary representation is 100011
Input : N = 5
Output : No
一个简单的解决方案是遍历所有位。对于每个设置的位，检查下一个位是否也被设置。 一个有效的解决方案是将数字移位 1，然后按位“与”。如果按位“与”非零，则有两个相邻的设置位。否则不行。
C++
// CPP program to check
// if there are two
// adjacent set bits.
#include <iostream>
using namespace std;
bool adjacentSet(int n)
{
return (n & (n >> 1));
}
// Driver Code
int main()
{
int n = 3;
adjacentSet(n) ?
cout << "Yes" :
cout << "No";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to check
// if there are two
// adjacent set bits.
class GFG
{
static boolean adjacentSet(int n)
{
int x = (n & (n >> 1));
if(x > 0)
return true;
else
return false;
}
// Driver code
public static void main(String args[])
{
int n = 3;
if(adjacentSet(n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Sam007.
Python 3
# Python 3 program to check if there
# are two adjacent set bits.
def adjacentSet(n):
return (n & (n >> 1))
# Driver Code
if __name__ == '__main__':
n = 3
if (adjacentSet(n)):
print("Yes")
else:
print("No")
# This code is contributed by
# Shashank_Sharma
C
// C# program to check
// if there are two
// adjacent set bits.
using System;
class GFG
{
static bool adjacentSet(int n)
{
int x = (n & (n >> 1));
if(x > 0)
return true;
else
return false;
}
// Driver code
public static void Main ()
{
int n = 3;
if(adjacentSet(n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by Sam007.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to check
// if there are two
// adjacent set bits.
function adjacentSet($n)
{
return ($n & ($n >> 1));
}
// Driver Code
$n = 3;
adjacentSet($n) ?
print("Yes") :
print("No");
// This code is contributed by Sam007.
?>
java 描述语言
<script>
// Javascript program to check
// if there are two
// adjacent set bits.
function adjacentSet(n)
{
let x = (n & (n >> 1));
if(x > 0)
return true;
else
return false;
}
// driver program
let n = 3;
if(adjacentSet(n))
document.write("Yes");
else
document.write("No");
</script>
输出:
Yes
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