N与2的任意幂之间的最小绝对差

给定正整数N ，任务是找到N与2的幂之间的最小绝对差。

例子：

Input: N = 3 
Output: 1 
Smaller power of 2 nearest to 3 is 2, abs(3 – 2) = 1 
Higher power of 2 nearest to 3 is 4, abs(4 – 3) = 1

Input: N = 6 
Output: 2 

方法：

1. 找到小于或等于N的2的最高幂，并将其存储在变量low中。
2. 找到大于或等于N的2的最小幂并将其存储在变量high中。
3. 现在，答案将是max(N-low，high-N) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the highest power
// of 2 less than or equal to n
int prevPowerof2(int n)
{
    int p = (int)log2(n);
    return (int)pow(2, p);
}
 
// Function to return the smallest power
// of 2 greater than or equal to n
int nextPowerOf2(int n)
{
    int p = 1;
    if (n && !(n & (n - 1)))
        return n;
 
    while (p < n)
        p <<= 1;
 
    return p;
}
 
// Function that returns the minimum
// absolute difference between n
// and any power of 2
int minDiff(int n)
{
    int low = prevPowerof2(n);
    int high = nextPowerOf2(n);
 
    return min(n - low, high - n);
}
 
// Driver code
int main()
{
    int n = 6;
 
    cout << minDiff(n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
     
    // Function to return the highest power
    // of 2 less than or equal to n
    static int prevPowerof2(int n)
    {
        int p = (int)(Math.log(n) / Math.log(2));
         
        return (int)Math.pow(2, p);
    }
     
    // Function to return the smallest power
    // of 2 greater than or equal to n
    static int nextPowerOf2(int n)
    {
        int p = 1;
        if ((n == 0) && !((n & (n - 1)) == 0))
            return n;
     
        while (p < n)
            p <<= 1;
     
        return p;
    }
     
    // Function that returns the minimum
    // absolute difference between n
    // and any power of 2
    static int minDiff(int n)
    {
        int low = prevPowerof2(n);
        int high = nextPowerOf2(n);
     
        return Math.min(n - low, high - n);
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 6;
     
        System.out.println(minDiff(n));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
from math import log
 
# Function to return the highest power
# of 2 less than or equal to n
def prevPowerof2(n):
    p = int(log(n))
    return pow(2, p)
 
# Function to return the smallest power
# of 2 greater than or equal to n
def nextPowerOf2(n):
    p = 1
    if (n and (n & (n - 1)) == 0):
        return n
 
    while (p < n):
        p <<= 1
 
    return p
 
# Function that returns the minimum
# absolute difference between n
# and any power of 2
def minDiff(n):
    low = prevPowerof2(n)
    high = nextPowerOf2(n)
 
    return min(n - low, high - n)
 
# Driver code
n = 6
 
print(minDiff(n))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function to return the highest power
    // of 2 less than or equal to n
    static int prevPowerof2(int n)
    {
        int p = (int)(Math.Log(n) / Math.Log(2));
         
        return (int)Math.Pow(2, p);
    }
     
    // Function to return the smallest power
    // of 2 greater than or equal to n
    static int nextPowerOf2(int n)
    {
        int p = 1;
        if ((n == 0) && !((n & (n - 1)) == 0))
            return n;
     
        while (p < n)
            p <<= 1;
     
        return p;
    }
     
    // Function that returns the minimum
    // absolute difference between n
    // and any power of 2
    static int minDiff(int n)
    {
        int low = prevPowerof2(n);
        int high = nextPowerOf2(n);
     
        return Math.Min(n - low, high - n);
    }
     
    // Driver code
    public static void Main (String []args)
    {
        int n = 6;
     
        Console.WriteLine(minDiff(n));
    }
}
 
// This code is contributed by Arnab Kundu

Javascript

 
 
// Javascript implementation of the approach
 
// Function to return the highest power
// of 2 less than or equal to n
function prevPowerof2(n)
{
    var p = parseInt(Math.log(n)/Math.log(2));
    return parseInt(Math.pow(2, p));
}
 
// Function to return the smallest power
// of 2 greater than or equal to n
function nextPowerOf2(n)
{
    var p = 1;
    if (n && !(n & (n - 1)))
        return n;
 
    while (p < n)
        p <<= 1;
 
    return p;
}
 
// Function that returns the minimum
// absolute difference between n
// and any power of 2
function minDiff(n)
{
    var low = prevPowerof2(n);
    var high = nextPowerOf2(n);
 
    return Math.min(n - low, high - n);
}
 
// Driver code
var n = 6;
document.write(minDiff(n));
 
// This code is contributed by noob2000.

输出

2

时间复杂度： O(log N)
辅助空间： O(1)