单链表的质数节点数
原文：https://www.geeksforgeeks.org/count-of-prime-nodes-of-a-singly-linked-list/
给定一个包含N个节点的单链表，任务是查找质数的总数。
示例：
Input: List = 15 -> 5 -> 6 -> 10 -> 17
Output: 2
5 and 17 are the prime nodes
Input: List = 29 -> 3 -> 4 -> 2 -> 9
Output: 3
2, 3 and 29 are the prime nodes
方法：想法是将链表遍历到最后并检查当前节点是否为质数。 如果是，则将计数增加 1 并继续执行相同操作，直到遍历所有节点。
下面是上述方法的实现：
C++
// C++ implementation to find count of prime numbers
// in the singly linked list
#include <bits/stdc++.h>
using namespace std;
// Node of the singly linked list
struct Node {
int data;
Node* next;
};
// Function to insert a node at the beginning
// of the singly Linked List
void push(Node** head_ref, int new_data)
{
Node* new_node = new Node;
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
// Function to check if a number is prime
bool isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Function to find count of prime
// nodes in a linked list
int countPrime(Node** head_ref)
{
int count = 0;
Node* ptr = *head_ref;
while (ptr != NULL) {
// If current node is prime
if (isPrime(ptr->data)) {
// Update count
count++;
}
ptr = ptr->next;
}
return count;
}
// Driver program
int main()
{
// start with the empty list
Node* head = NULL;
// create the linked list
// 15 -> 5 -> 6 -> 10 -> 17
push(&head, 17);
push(&head, 10);
push(&head, 6);
push(&head, 5);
push(&head, 15);
// Function call to print require answer
cout << "Count of prime nodes = "
<< countPrime(&head);
return 0;
}
Java
// Java implementation to find count of prime numbers
// in the singly linked list
class solution
{
// Node of the singly linked list
static class Node {
int data;
Node next;
}
// Function to insert a node at the beginning
// of the singly Linked List
static Node push(Node head_ref, int new_data)
{
Node new_node = new Node();
new_node.data = new_data;
new_node.next = ( head_ref);
( head_ref) = new_node;
return head_ref;
}
// Function to check if a number is prime
static boolean isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Function to find count of prime
// nodes in a linked list
static int countPrime(Node head_ref)
{
int count = 0;
Node ptr = head_ref;
while (ptr != null) {
// If current node is prime
if (isPrime(ptr.data)) {
// Update count
count++;
}
ptr = ptr.next;
}
return count;
}
// Driver program
public static void main(String args[])
{
// start with the empty list
Node head = null;
// create the linked list
// 15 . 5 . 6 . 10 . 17
head=push(head, 17);
head=push(head, 10);
head=push(head, 6);
head=push(head, 5);
head=push(head, 15);
// Function call to print require answer
System.out.print( "Count of prime nodes = "+ countPrime(head));
}
}
// This code is contributed by Arnab Kundu
Python3
# Python3 implementation to find count of
# prime numbers in the singly linked list
# Function to check if a number is prime
def isPrime(n):
# Corner cases
if n <= 1:
return False
if n <= 3:
return True
# This is checked so that we can skip
# middle five numbers in below loop
if n % 2 == 0 or n % 3 == 0:
return False
i = 5
while i * i <= n:
if n % i == 0 or n % (i + 2) == 0:
return False
i += 6
return True
# Link list node
class Node:
def __init__(self, data, next):
self.data = data
self.next = next
class LinkedList:
def __init__(self):
self.head = None
# Push a new node on the front of the list.
def push(self, new_data):
new_node = Node(new_data, self.head)
self.head = new_node
# Function to find count of prime
# nodes in a linked list
def countPrime(self):
count = 0
ptr = self.head
while ptr != None:
# If current node is prime
if isPrime(ptr.data):
# Update count
count += 1
ptr = ptr.next
return count
# Driver Code
if __name__ == "__main__":
# Start with the empty list
linkedlist = LinkedList()
# create the linked list
# 15 -> 5 -> 6 -> 10 -> 17
linkedlist.push(17)
linkedlist.push(10)
linkedlist.push(6)
linkedlist.push(5)
linkedlist.push(15)
# Function call to print require answer
print("Count of prime nodes =",
linkedlist.countPrime())
# This code is contributed by Rituraj Jain
C
// C# implementation to find count of prime numbers
// in the singly linked list
using System;
class GFG
{
// Node of the singly linked list
public class Node
{
public int data;
public Node next;
}
// Function to insert a node at the beginning
// of the singly Linked List
static Node push(Node head_ref, int new_data)
{
Node new_node = new Node();
new_node.data = new_data;
new_node.next = ( head_ref);
( head_ref) = new_node;
return head_ref;
}
// Function to check if a number is prime
static bool isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Function to find count of prime
// nodes in a linked list
static int countPrime(Node head_ref)
{
int count = 0;
Node ptr = head_ref;
while (ptr != null)
{
// If current node is prime
if (isPrime(ptr.data))
{
// Update count
count++;
}
ptr = ptr.next;
}
return count;
}
// Driver code
public static void Main(String []args)
{
// start with the empty list
Node head = null;
// create the linked list
// 15 . 5 . 6 . 10 . 17
head=push(head, 17);
head=push(head, 10);
head=push(head, 6);
head=push(head, 5);
head=push(head, 15);
// Function call to print require answer
Console.Write( "Count of prime nodes = "+ countPrime(head));
}
}
// This code has been contributed by 29AjayKumar
输出：
Count of prime nodes = 2
时间复杂度：O(N * sqrt(P))，其中N是链表的长度，P是列表中的最大元素
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