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 JainC#
// 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 输出：
4 17 3 6 100 2