非重复素数
给定一个包含重复质数和非质数的数组 arr[] ，任务是找出只出现一次的质数。
示例:
输入: arr[] = {2，3，4，6，7，9，7，23，21，2，3} 输出: 23 解释: 在给定数组中，23 是唯一出现一次的素数。
输入: arr[] = {17，19，7，5，29，5，2，2，7，17，19} 输出: 29 说明: 在给定数组中，29 是唯一出现一次的素数。
天真方法:要解决上面提到的问题，解决方法是检查每个元素是否是质数。如果它是质数，那么检查它是否只出现一次。一旦找到一个出现一次的主元素，就打印它。
时间复杂度: O(N 2 )
- 在散列表中使用筛预计算并存储质数。
- 还要创建一个 HashMap 来存储数字及其频率。
- 逐一遍历数组中的所有元素，然后:
- 使用 O(1)中的筛散列表检查当前数字是否为质数。
- 如果数字是质数，那么增加它在 HashMap 中的频率。
- 遍历 HashMap，打印所有频率为 1 的数字。
下面是上述方法的实现:
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find
// Non-repeating Primes
import java.util.*;
class GFG {
// Function to find count of prime
static Vector<Boolean> findPrimes(
int arr[], int n)
{
// Find maximum value in the array
int max_val = Arrays
.stream(arr)
.max()
.getAsInt();
// Find and store all prime numbers
// up to max_val using Sieve
// Create a boolean array "prime[0..n]".
// A value in prime[i]
// will finally be false
// if i is Not a prime, else true.
Vector<Boolean> prime
= new Vector<>(max_val + 1);
for (int i = 0; i < max_val + 1; i++)
prime.add(i, Boolean.TRUE);
// Remaining part of SIEVE
prime.add(0, Boolean.FALSE);
prime.add(1, Boolean.FALSE);
for (int p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed,
// then it is a prime
if (prime.get(p) == true) {
// Update all multiples of p
for (int i = p * 2;
i <= max_val;
i += p)
prime.add(i, Boolean.FALSE);
}
}
return prime;
}
// Function to print
// Non-repeating primes
static void nonRepeatingPrimes(
int arr[], int n)
{
// Precompute primes using Sieve
Vector<Boolean> prime
= findPrimes(arr, n);
// Create HashMap to store
// frequency of prime numbers
HashMap<Integer, Integer> mp
= new HashMap<>();
// Traverse through array elements and
// Count frequencies of all primes
for (int i = 0; i < n; i++) {
if (prime.get(arr[i]))
if (mp.containsKey(arr[i]))
mp.put(arr[i],
mp.get(arr[i]) + 1);
else
mp.put(arr[i], 1);
}
// Traverse through map and
// print non repeating primes
for (Map.Entry<Integer, Integer>
entry : mp.entrySet()) {
if (entry.getValue() == 1)
System.out.println(
entry.getKey());
}
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 3, 4, 6, 7, 9,
7, 23, 21, 3 };
int n = arr.length;
nonRepeatingPrimes(arr, n);
}
}
Python 3
# Python3 program to find
# Non-repeating Primes
# Function to find count of prime
def findPrimes( arr, n):
# Find maximum value in the array
max_val = max(arr)
# Find and store all prime numbers
# up to max_val using Sieve
# Create a boolean array "prime[0..n]".
# A value in prime[i]
# will finally be false
# if i is Not a prime, else true.
prime = [True for i in range(max_val + 1)]
# Remaining part of SIEVE
prime[0] = False
prime[1] = False
p = 2
while(p * p <= max_val):
# 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_val + 1, p):
prime[i] = False
p += 1
return prime;
# Function to print
# Non-repeating primes
def nonRepeatingPrimes(arr, n):
# Precompute primes using Sieve
prime = findPrimes(arr, n);
# Create HashMap to store
# frequency of prime numbers
mp = dict()
# Traverse through array elements and
# Count frequencies of all primes
for i in range(n):
if (prime[arr[i]]):
if (arr[i] in mp):
mp[arr[i]] += 1
else:
mp[arr[i]] = 1
# Traverse through map and
# print non repeating primes
for entry in mp.keys():
if (mp[entry] == 1):
print(entry);
# Driver code
if __name__ == '__main__':
arr = [ 2, 3, 4, 6, 7, 9, 7, 23, 21, 3]
n = len(arr)
nonRepeatingPrimes(arr, n);
# This code is contributed by pratham76.
C
// C# program to find
// Non-repeating Primes
using System;
using System.Collections;
using System.Linq;
using System.Collections.Generic;
class GFG{
// Function to find count of prime
static List<bool> findPrimes(int []arr, int n)
{
// Find maximum value in the array
int max_val = arr.Max();
// Find and store all prime numbers
// up to max_val using Sieve
// Create a boolean array "prime[0..n]".
// A value in prime[i]
// will finally be false
// if i is Not a prime, else true.
List<bool> prime = new List<bool>(max_val + 1);
for(int i = 0; i < max_val + 1; i++)
prime.Add(true);
// Remaining part of SIEVE
prime[0] = false;
prime[1] = false;
for(int p = 2; p * p <= max_val; 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_val;
i += p)
prime[i] = false;
}
}
return prime;
}
// Function to print
// Non-repeating primes
static void nonRepeatingPrimes(int []arr, int n)
{
// Precompute primes using Sieve
List<bool> prime = findPrimes(arr, n);
// Create HashMap to store
// frequency of prime numbers
Dictionary<int,
int> mp = new Dictionary<int,
int>();
// Traverse through array elements and
// Count frequencies of all primes
for(int i = 0; i < n; i++)
{
if (prime[arr[i]])
if (mp.ContainsKey(arr[i]))
mp[arr[i]]++;
else
mp.Add(arr[i], 1);
}
// Traverse through map and
// print non repeating primes
foreach(KeyValuePair<int, int> entry in mp)
{
if (entry.Value == 1)
Console.WriteLine(entry.Key);
}
}
// Driver code
public static void Main(string[] args)
{
int []arr = { 2, 3, 4, 6, 7, 9,
7, 23, 21, 3 };
int n = arr.Length;
nonRepeatingPrimes(arr, n);
}
}
// This code is contributed by rutvik_56
Output:
2
23
时间复杂度:O(O(n * log(log(n)))) 辅助空间: O(K)，其中 K 是数组中最大的值。
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