为每个数组元素找到最接近的 2 的幂

给定一个大小为N的数组arr[] ，任务是为每个数组元素打印最接近的 2 的幂。
笔记： 如果碰巧有两个最近的 2 次幂，请考虑较大的那个。

例子：

Input: arr[] = {5, 2, 7, 12} 
Output: 4 2 8 16 
Explanation: 
The nearest power of arr[0] ( = 5) is 4. 
The nearest power of arr[1] ( = 2) is 2. 
The nearest power of arr[2] ( = 7) is 8. 
The nearest power of arr[3] ( = 12) are 8 and 16. Print 16, as it is the largest.

Input: arr[] = {31, 13, 64} 
Output: 32 16 64

处理方法：按照以下步骤解决问题：

• 从左到右遍历数组。
• 对于每个数组元素，找到最接近的大于和小于它的 2 的幂，即计算pow(2, log ₂ (arr[i]))和pow(2, log ₂ (arr[i]) + 1) 。
• 计算这两个值与当前数组元素的差值，并按照问题陈述中的指定打印最接近的值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to find nearest power of two
// for every element in the given array
void nearestPowerOfTwo(int arr[], int N)
{
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Calculate log of the
        // current array element
        int lg = log2(arr[i]);
        int a = pow(2, lg);
        int b = pow(2, lg + 1);
 
        // Find the nearest
        if ((arr[i] - a) < (b - arr[i]))
            cout << a << " ";
        else
            cout << b << " ";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 5, 2, 7, 12 };
    int N = sizeof(arr) / sizeof(arr[0]);
    nearestPowerOfTwo(arr, N);
    return 0;
}

Java

// Java program to implement the above approach
import java.io.*;
 
class GFG {
 
    // Function to find the nearest power of two
    // for every element of the given array
    static void nearestPowerOfTwo(int[] arr, int N)
    {
 
        // Traverse the array
        for (int i = 0; i < N; i++) {
 
            // Calculate log of the
            // current array element
            int lg = (int)(Math.log(arr[i])
                           / Math.log(2));
 
            int a = (int)(Math.pow(2, lg));
            int b = (int)(Math.pow(2, lg + 1));
 
            // Find the nearest
            if ((arr[i] - a) < (b - arr[i]))
                System.out.print(a + " ");
            else
                System.out.print(b + " ");
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 5, 2, 7, 12 };
        int N = arr.length;
        nearestPowerOfTwo(arr, N);
    }
}

Python3

# Python program to implement the above approach
import math
 
# Function to find the nearest power
# of two for every array element
def nearestPowerOfTwo(arr, N):
 
    # Traverse the array
    for i in range(N):
 
        # Calculate log of current array element
        lg = (int)(math.log2(arr[i]))
 
        a = (int)(math.pow(2, lg))
        b = (int)(math.pow(2, lg + 1))
 
        # Find the nearest
        if ((arr[i] - a) < (b - arr[i])):
            print(a, end = " ")
        else:
            print(b, end = " ")
 
 
# Driver Code
arr = [5, 2, 7, 12]
N = len(arr)
nearestPowerOfTwo(arr, N)

C#

// C# program to implement the above approach
using System;
 
class GFG {
 
    // Function to find nearest power of two
    // for every array element
    static void nearestPowerOfTwo(int[] arr, int N)
    {
 
        // Traverse the array
        for (int i = 0; i < N; i++) {
 
            // Calculate log of the
            // current array element
            int lg = (int)(Math.Log(arr[i])
                           / Math.Log(2));
 
            int a = (int)(Math.Pow(2, lg));
            int b = (int)(Math.Pow(2, lg + 1));
 
            // Find the nearest
            if ((arr[i] - a) < (b - arr[i]))
                Console.Write(a + " ");
            else
                Console.Write(b + " ");
        }
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int[] arr = { 5, 2, 7, 12 };
        int N = arr.Length;
        nearestPowerOfTwo(arr, N);
    }
}

Javascript

 
// JavaScript program to implement
// the above approach
 
    // Function to find the nearest power of two
    // for every element of the given array
    function nearestPowerOfTwo(arr , N) {
 
        // Traverse the array
        for (i = 0; i < N; i++) {
 
            // Calculate log of the
            // current array element
            var lg = parseInt( (Math.log(arr[i]) / Math.log(2)));
            var a = parseInt( (Math.pow(2, lg)));
            var b = parseInt( (Math.pow(2, lg + 1)));
 
            // Find the nearest
            if ((arr[i] - a) < (b - arr[i]))
                document.write(a + " ");
            else
                document.write(b + " ");
        }
    }
 
    // Driver Code
     
        var arr = [ 5, 2, 7, 12 ];
        var N = arr.length;
        nearestPowerOfTwo(arr, N);
 
// This code is contributed by todaysgaurav
 

输出：

4 2 8 16

时间复杂度： O(N)
辅助空间： O(1)

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