单链表的最小和最大质数
原文：https://www.geeksforgeeks.org/minimum-and-maximum-prime-numbers-of-a-singly-linked-list/
给定一个包含N个节点的单链表，任务是查找最小和最大质数。
示例：
Input : List = 15 -> 16 -> 6 -> 7 -> 17
Output : Minimum : 7
Maximum : 17
Input : List = 15 -> 3 -> 4 -> 2 -> 9
Output : Minimum : 2
Maximum : 3
方法：
-
这个想法是将链表遍历到最后，并将
max和min变量分别初始化为INT_MIN和INT_MAX。 -
检查当前节点是否为质数。 如是：
-
如果当前节点的值大于
max，则将当前节点的值分配给max。 -
如果当前节点的值小于
min，则将当前节点的值分配给min。
-
-
重复上述步骤，直到到达列表末尾。
以下是上述想法的实现：
C++
// C++ implementation to find minimum
// and maximum prime number of
// 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 maximum and minimum
// prime nodes in a linked list
void minmaxPrimeNodes(Node** head_ref)
{
int minimum = INT_MAX;
int maximum = INT_MIN;
Node* ptr = *head_ref;
while (ptr != NULL) {
// If current node is prime
if (isPrime(ptr->data)) {
// Update minimum
minimum = min(minimum, ptr->data);
// Update maximum
maximum = max(maximum, ptr->data);
}
ptr = ptr->next;
}
cout << "Minimum : " << minimum << endl;
cout << "Maximum : " << maximum << endl;
}
// Driver program
int main()
{
// start with the empty list
Node* head = NULL;
// create the linked list
// 15 -> 16 -> 7 -> 6 -> 17
push(&head, 17);
push(&head, 7);
push(&head, 6);
push(&head, 16);
push(&head, 15);
minmaxPrimeNodes(&head);
return 0;
}
Java
// Java implementation to find minimum
// and maximum prime number of
// the singly linked list
class GFG
{
// 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 maximum and minimum
// prime nodes in a linked list
static void minmaxPrimeNodes(Node head_ref)
{
int minimum = Integer.MAX_VALUE;
int maximum = Integer.MIN_VALUE;
Node ptr = head_ref;
while (ptr != null)
{
// If current node is prime
if (isPrime(ptr.data))
{
// Update minimum
minimum = Math.min(minimum, ptr.data);
// Update maximum
maximum = Math.max(maximum, ptr.data);
}
ptr = ptr.next;
}
System.out.println("Minimum : " + minimum );
System.out.println("Maximum : " + maximum );
}
// Driver code
public static void main(String args[])
{
// start with the empty list
Node head = null;
// create the linked list
// 15 . 16 . 7 . 6 . 17
head = push(head, 17);
head = push(head, 7);
head = push(head, 6);
head = push(head, 16);
head = push(head, 15);
minmaxPrimeNodes(head);
}
}
// This code is contributed by Arnab Kundu
Python3
# Python3 implementation to find minimum
# and maximum prime number of
# the singly linked list
# Structure of a Node
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Function to insert a node at the beginning
# of the singly Linked List
def push(head_ref, new_data) :
new_node = Node(0)
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
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 = i + 6
return True
# Function to find maximum and minimum
# prime nodes in a linked list
def minmaxPrimeNodes(head_ref) :
minimum = 999999999
maximum = -999999999
ptr = head_ref
while (ptr != None):
# If current node is prime
if (isPrime(ptr.data)):
# Update minimum
minimum = min(minimum, ptr.data)
# Update maximum
maximum = max(maximum, ptr.data)
ptr = ptr.next
print ("Minimum : ", minimum)
print ("Maximum : ", maximum)
# Driver code
# start with the empty list
head = None
# create the linked list
# 15 . 16 . 7 . 6 . 17
head = push(head, 17)
head = push(head, 7)
head = push(head, 6)
head = push(head, 16)
head = push(head, 15)
minmaxPrimeNodes(head)
# This code is contributed by Arnab Kundu
C
// C# implementation to find minimum
// and maximum prime number of
// 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 maximum and minimum
// prime nodes in a linked list
static void minmaxPrimeNodes(Node head_ref)
{
int minimum = int.MaxValue;
int maximum = int.MinValue;
Node ptr = head_ref;
while (ptr != null)
{
// If current node is prime
if (isPrime(ptr.data))
{
// Update minimum
minimum = Math.Min(minimum, ptr.data);
// Update maximum
maximum = Math.Max(maximum, ptr.data);
}
ptr = ptr.next;
}
Console.WriteLine("Minimum : " + minimum);
Console.WriteLine("Maximum : " + maximum);
}
// Driver code
public static void Main()
{
// start with the empty list
Node head = null;
// create the linked list
// 15 . 16 . 7 . 6 . 17
head = push(head, 17);
head = push(head, 7);
head = push(head, 6);
head = push(head, 16);
head = push(head, 15);
minmaxPrimeNodes(head);
}
}
// This code is contributed by Princi Singh
输出：
Minimum : 7
Maximum : 17
时间复杂度：O(n)，其中N是链表中节点的数量。
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