由 N 中最右边的设置位形成的数字
原文:https://www . geesforgeks . org/number-由 n 中最右边的 set 位构成/
给定一个整数 N ，任务是找到一个取 N 中最右边的置位形成的整数 M ，即 M 中唯一的置位将是 N 中最右边的置位，其余位将被取消置位。 举例:
输入: N = 7 输出: 1 7 = 111，最后一个设置位形成的数字为 001 即 1。 输入: N = 10 输出:2 10 = 1010->0010 = 2 输入: N = 16 输出: 16
进场:
- 存储x = n&(n–1)，这将在 n 中从右侧取消第一个设置位。
- 现在，更新 n = n ^ x 以设置已更改的位，并取消设置所需整数的所有其他位。
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the integer formed
// by taking the rightmost set bit in n
int firstSetBit(int n)
{
// n & (n - 1) unsets the first set
// bit from the right in n
int x = n & (n - 1);
// Take xor with the original number
// The position of the 'changed bit'
// will be set and rest will be unset
return (n ^ x);
}
// Driver code
int main()
{
int n = 12;
cout << firstSetBit(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class geeks
{
// Function to return the integer formed
// by taking the rightmost set bit in n
public static int firstSetBit(int n)
{
// n & (n - 1) unsets the first set
// bit from the right in n
int x = n & (n-1);
// Take xor with the original number
// The position of the 'changed bit'
// will be set and rest will be unset
return (n ^ x);
}
// Driver Code
public static void main (String[] args)
{
int n = 12;
System.out.println(firstSetBit(n));
}
}
// This code is contributed by
// sanjeev2552
Python 3
# Python3 implementation of the approach
# Function to return the integer formed
# by taking the rightmost set bit in n
def firstSetBit(n):
# n & (n - 1) unsets the first set
# bit from the right in n
x = n & (n - 1)
# Take xor with the original number
# The position of the 'changed bit'
# will be set and rest will be unset
return (n ^ x)
# Driver code
n = 12
print(firstSetBit(n))
# This code is contributed by mohit kumar 29
C
// C# implementation of the approach
using System;
class geeks
{
// Function to return the integer formed
// by taking the rightmost set bit in n
public static int firstSetBit(int n)
{
// n & (n - 1) unsets the first set
// bit from the right in n
int x = n & (n-1);
// Take xor with the original number
// The position of the 'changed bit'
// will be set and rest will be unset
return (n ^ x);
}
// Driver Code
public static void Main ()
{
int n = 12;
Console.WriteLine(firstSetBit(n));
}
}
// This code is contributed by
// anuj_67..
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to return the integer formed
// by taking the rightmost set bit in n
function firstSetBit(n)
{
// n & (n - 1) unsets the first set
// bit from the right in n
let x = n & (n - 1);
// Take xor with the original number
// The position of the 'changed bit'
// will be set and rest will be unset
return (n ^ x);
}
// Driver code
let n = 12;
document.write(firstSetBit(n));
// This code is contributed by Manoj.
</script>
Output:
4
时间复杂度: O(1)
辅助空间: O(1)
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