以降序对数组的质数进行排序

给定一个整数数组“ arr”，任务是按照其相对位置降序排列该数组中的所有质数，即，其他元素的其他位置不得受到影响。
例子：

Input: arr[] = {2, 5, 8, 4, 3}
Output: 5 3 8 4 2

Input: arr[] = {10, 12, 2, 6, 5}
Output: 10 12 5 6 2

方法：

• 创建一个筛子以检查元素在O(1)中是否为质数。
• 遍历数组并检查数字是否为素数。如果是素数，请将其存储在向量中。
• 然后，按降序对向量进行排序。
• 再次遍历数组，并用矢量元素一一替换素数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
bool prime[100005];
 
void SieveOfEratosthenes(int n)
{
 
    memset(prime, true, sizeof(prime));
 
    // false here indicates
    // that it is not prime
    prime[1] = false;
 
    for (int p = 2; p * p <= n; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) {
 
            // Update all multiples of p,
            // set them to non-prime
            for (int i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }
}
 
// Function that sorts
// all the prime numbers
// from the array in descending
void sortPrimes(int arr[], int n)
{
    SieveOfEratosthenes(100005);
 
    // this vector will contain
    // prime numbers to sort
    vector v;
 
    for (int i = 0; i < n; i++) {
 
        // if the element is prime
        if (prime[arr[i]])
            v.push_back(arr[i]);
    }
 
    sort(v.begin(), v.end(), greater());
 
    int j = 0;
 
    // update the array elements
    for (int i = 0; i < n; i++) {
        if (prime[arr[i]])
            arr[i] = v[j++];
    }
}
 
// Driver code
int main()
{
 
    int arr[] = { 4, 3, 2, 6, 100, 17 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    sortPrimes(arr, n);
 
    // print the results.
    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
    static boolean prime[] = new boolean[100005];
 
    static void SieveOfEratosthenes(int n)
    {
 
        Arrays.fill(prime, true);
 
        // false here indicates
        // that it is not prime
        prime[1] = false;
 
        for (int p = 2; p * p <= n; p++)
        {
 
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p]) {
 
                // Update all multiples of p,
                // set them to non-prime
                for (int i = p * 2; i < n; i += p)
                {
                    prime[i] = false;
                }
            }
        }
    }
 
    // Function that sorts
    // all the prime numbers
    // from the array in descending
    static void sortPrimes(int arr[], int n)
    {
        SieveOfEratosthenes(100005);
 
        // this vector will contain
        // prime numbers to sort
        Vector v = new Vector();
 
        for (int i = 0; i < n; i++)
        {
 
            // if the element is prime
            if (prime[arr[i]])
            {
                v.add(arr[i]);
            }
        }
        Comparator comparator = Collections.reverseOrder();
        Collections.sort(v, comparator);
 
        int j = 0;
 
        // update the array elements
        for (int i = 0; i < n; i++)
        {
            if (prime[arr[i]])
            {
                arr[i] = v.get(j++);
            }
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {4, 3, 2, 6, 100, 17};
        int n = arr.length;
 
        sortPrimes(arr, n);
 
        // print the results.
        for (int i = 0; i < n; i++)
        {
            System.out.print(arr[i] + " ");
        }
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
def SieveOfEratosthenes(n):
 
    # false here indicates
    # that it is not prime
    prime[1] = False
    p = 2
    while p * p <= n:
 
        # If prime[p] is not changed,
        # then it is a prime
        if prime[p]:
 
            # Update all multiples of p,
            # set them to non-prime
            for i in range(p * 2, n + 1, p):
                prime[i] = False
         
        p += 1
 
# Function that sorts all the prime
# numbers from the array in descending
def sortPrimes(arr, n):
 
    SieveOfEratosthenes(100005)
 
    # This vector will contain
    # prime numbers to sort
    v = []
    for i in range(0, n):
 
        # If the element is prime
        if prime[arr[i]]:
            v.append(arr[i])
 
    v.sort(reverse = True)
    j = 0
 
    # update the array elements
    for i in range(0, n):
        if prime[arr[i]]:
            arr[i] = v[j]
            j += 1
             
    return arr
     
# Driver code
if __name__ == "__main__":
 
    arr = [4, 3, 2, 6, 100, 17]
    n = len(arr)
     
    prime = [True] * 100006
    arr = sortPrimes(arr, n)
 
    # print the results.
    for i in range(0, n):
        print(arr[i], end = " ")
     
# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    static bool []prime = new bool[100005];
 
    static void SieveOfEratosthenes(int n)
    {
 
        for(int i = 0; i < 100005; i++)
            prime[i] = true;
 
        // false here indicates
        // that it is not prime
        prime[1] = false;
 
        for (int p = 2; p * p <= n; p++)
        {
 
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p])
            {
 
                // Update all multiples of p,
                // set them to non-prime
                for (int i = p * 2; i < n; i += p)
                {
                    prime[i] = false;
                }
            }
        }
    }
 
    // Function that sorts
    // all the prime numbers
    // from the array in descending
    static void sortPrimes(int []arr, int n)
    {
        SieveOfEratosthenes(100005);
 
        // this vector will contain
        // prime numbers to sort
        List v = new List();
 
        for (int i = 0; i < n; i++)
        {
 
            // if the element is prime
            if (prime[arr[i]])
            {
                v.Add(arr[i]);
            }
        }
        v.Sort();
        v.Reverse();
 
        int j = 0;
 
        // update the array elements
        for (int i = 0; i < n; i++)
        {
            if (prime[arr[i]])
            {
                arr[i] = v[j++];
            }
        }
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {4, 3, 2, 6, 100, 17};
        int n = arr.Length;
 
        sortPrimes(arr, n);
 
        // print the results.
        for (int i = 0; i < n; i++)
        {
            Console.Write(arr[i] + " ");
        }
    }
}
 
// This code contributed by Rajput-Ji

Javascript

 
// Javascript implementation of the approach
 
var prime = Array(100005).fill(true);
 
function SieveOfEratosthenes( n)
{
 
    // false here indicates
    // that it is not prime
    prime[1] = false;
 
    for (var p = 2; p * p <= n; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) {
 
            // Update all multiples of p,
            // set them to non-prime
            for (var i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }
}
 
// Function that sorts
// all the prime numbers
// from the array in descending
function sortPrimes(arr, n)
{
    SieveOfEratosthenes(100005);
 
    // this vector will contain
    // prime numbers to sort
    var v = [];
 
    for (var i = 0; i < n; i++) {
 
        // if the element is prime
        if (prime[arr[i]])
            v.push(arr[i]);
    }
 
    v.sort((a,b)=>b-a)
 
    var j = 0;
 
    // update the array elements
    for (var i = 0; i < n; i++) {
        if (prime[arr[i]])
            arr[i] = v[j++];
    }
}
 
// Driver code
var arr = [4, 3, 2, 6, 100, 17 ];
var n = arr.length;
sortPrimes(arr, n);
// print the results.
for (var i = 0; i < n; i++) {
    document.write( arr[i] + " ");
}
 
 

输出：

4 17 3 6 100 2