给定一个整数数组，其中所有元素均小于10 ^ 6。
任务是找到数组中最大质数和最小质数之间的差。
例子：
Input : Array = 1, 2, 3, 5
Output : Difference is 3
Explanation :
The largest prime number in the array is 5 and the smallest is 2
So, the difference is 3
Input : Array = 3, 5, 11, 17
Output : Difference is 14一个简单的方法：
在基本方法中，我们将检查数组的每个元素是否为素数。
然后，选择最大和最小的质数并打印差异。
高效的方法：
有效的方法与基本方法非常相似。
我们将尝试通过创建Eratosthenes筛网来检查数字是否为素数，以检查O(1)时间是否为素数。
然后，我们将选择最大和最小的质数并打印差异。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define MAX 1000000
bool prime[MAX + 1];
void SieveOfEratosthenes()
{
// Create a boolean array "prime[0..n]" and initialize
// all the entries as true. A value in prime[i] will
// finally be false if 'i' is Not a prime, else true.
memset(prime, true, sizeof(prime));
// 1 is not prime
prime[1] = false;
for (int p = 2; p * p <= MAX; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= MAX; i += p)
prime[i] = false;
}
}
}
int findDiff(int arr[], int n)
{
// initial min max value
int min = MAX + 2, max = -1;
for (int i = 0; i < n; i++) {
// check if the number is prime or not
if (prime[arr[i]] == true) {
// set the max and min values
if (arr[i] > max)
max = arr[i];
if (arr[i] < min)
min = arr[i];
}
}
return (max == -1) ? -1 : (max - min);
}
// Driver code
int main()
{
// create the sieve
SieveOfEratosthenes();
int n = 4;
int arr[n] = { 1, 2, 3, 5 };
int res = findDiff(arr, n);
if (res == -1)
cout << "No prime numbers" << endl;
else
cout << "Difference is " << res << endl;
return 0;
} Java
// java implementation of the approach
import java.io.*;
class GFG {
static int MAX = 1000000;
static boolean prime[] = new boolean[MAX + 1];
static void SieveOfEratosthenes()
{
// Create a boolean array "prime[0..n]" and initialize
// all the entries as true. A value in prime[i] will
// finally be false if 'i' is Not a prime, else true.
//memset(prime, true, sizeof(prime));
for(int i=0;i max)
max = arr[i];
if (arr[i] < min)
min = arr[i];
}
}
return (max == -1)? -1 : (max - min);
}
// Driver code
public static void main (String[] args) {
// create the sieve
SieveOfEratosthenes();
int n = 4;
int arr[] = { 1, 2, 3, 5 };
int res = findDiff(arr, n);
if (res == -1)
System.out.print( "No prime numbers") ;
else
System.out.println( "Difference is " + res);
}
}
// This code is contributed by inder_verma.. Python 3
# Python 3 implementation of the approach
MAX = 1000000
# Create a boolean array "prime[0..n]" and initialize
# all the entries as true. A value in prime[i] will
# finally be false if 'i' is Not a prime, else true
prime = [True]*(MAX+1)
def SieveOfEratosthenes():
# 1 is not prime
prime[1] = False
p = 2
c=0
while (p * p <= MAX) :
c+= 1
# If prime[p] is not changed, then it is a prime
if (prime[p] == True) :
# Update all multiples of p
for i in range( p * 2, MAX+1 , p):
prime[i] = False
p += 1
def findDiff(arr, n):
# initial min max value
min = MAX + 2
max = -1
for i in range(n) :
# check if the number is prime or not
if (prime[arr[i]] == True) :
# set the max and min values
#print("arra ",arr[i])
#print("MAX ",max)
#print(" MIN ",min)
if (arr[i] > max):
max = arr[i]
if (arr[i] < min):
min = arr[i]
#print(" max ",max)
return -1 if (max == -1) else (max - min)
# Driver code
if __name__ == "__main__":
# create the sieve
SieveOfEratosthenes()
n = 4
arr = [ 1, 2, 3, 5 ]
res = findDiff(arr, n)
if (res == -1):
print("No prime numbers")
else:
print("Difference is " ,res )
# this code is contributed by
# ChitraNayalC#
// C# implementation of the approach
using System;
class GFG
{
static int MAX = 1000000;
static bool []prime = new bool[MAX + 1];
static void SieveOfEratosthenes()
{
// Create a boolean array "prime[0..n]"
// and initialize all the entries as
// true. A value in prime[i] will
// finally be false if 'i' is Not a
// prime, else true.
// memset(prime, true, sizeof(prime));
for(int i = 0; i < MAX + 1; i++)
prime[i] = true;
// 1 is not prime
prime[1] = false;
for (int p = 2; p * p <= MAX; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * 2; i <= MAX; i += p)
prime[i] = false;
}
}
}
static int findDiff(int []arr, int n)
{
// initial min max value
int min = MAX + 2, max = -1;
for (int i = 0; i < n; i++)
{
// check if the number is prime or not
if (prime[arr[i]] == true)
{
// set the max and min values
if (arr[i] > max)
max = arr[i];
if (arr[i] < min)
min = arr[i];
}
}
return (max == -1) ? -1 : (max - min);
}
// Driver code
public static void Main ()
{
// create the sieve
SieveOfEratosthenes();
int n = 4;
int []arr = { 1, 2, 3, 5 };
int res = findDiff(arr, n);
if (res == -1)
Console.WriteLine( "No prime numbers") ;
else
Console.WriteLine( "Difference is " + res);
}
}
// This code is contributed by inder_vermaJavascript
输出：
Difference is 3如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。