给定一个包含N个节点的单链接列表，任务是找到列表中所有可能的节点的总和，该节点包含具有恰好三个不同因素的值。
例子：
Input: 1 -> 2 -> 4 -> 5
Output: 4
Explanation:
Factors of 2 are {1, 2}
Factors of 3 are {1, 3}
Factors of 4 are {1, 2, 4}
Factors of 5 are {1, 5}
Since only 4 contains three distinct factors. Therefore, the required output is 4.
Input: 1 -> 6 -> 8 -> 10
Output: 0
方法：这个想法基于这样一个事实，即如果一个数字是质数，那么该数字的平方恰好包含三个不同的因子。请按照以下步骤解决问题：
- 初始化一个变量，例如nodeSum ，以存储包含具有恰好三个不同因素的值的节点之和。
- 遍历链表，对于每个节点，检查当前节点的平方根是否为质数。如果发现为true，则将nodeSum增加当前节点的值。
- 最后，打印nodeSum的值。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Structure of a
// singly Linked List
struct Node {
// Stores data value
// of a Node
int data;
// Stores pointer
// to next Node
struct Node* next;
};
// Function to insert a node at the
// beginning of the singly Linked List
void push(Node** head_ref,
int new_data)
{
// Create a new Node
Node* new_node = new Node;
// Insert the data into
// the Node
new_node->data = new_data;
// Insert pointer to
// the next Node
new_node->next = (*head_ref);
// Update head_ref
(*head_ref) = new_node;
}
// Function to check if a number
// is a prime number or not
bool CheckPrimeNumber(int n)
{
// Base Case
if (n <= 1)
return false;
if (n <= 3)
return true;
// If n is multiple of 2 or 3
if (n % 2 == 0 || n % 3 == 0)
return false;
// Iterate over the range
// [5, sqrt(N)]
for (int i = 5; i * i <= n;
i = i + 6) {
// If n is multiple of
// i or (i + 2)
if (n % i == 0 || n % (i + 2) == 0)
return false;
}
return true;
}
// Function to check if number is
// a perfect square number or not
bool checkperfect_square(int n)
{
// Stores square root of n
long double sr = sqrt(n);
// If n is a
// perfect square number
if (sr - floor(sr) == 0) {
return true;
}
return false;
}
// Function to check if n is square
// of a prime number or not
bool is_square(int n)
{
// If n is a perfect square
if (checkperfect_square(n)) {
// Stores sqrt of n
int root = sqrt(n);
// If root is a prime number
if (CheckPrimeNumber(root)) {
return true;
}
}
return false;
}
// Function to find sum of node values
// which contains three distinct factors
int calculate_sum(struct Node* head)
{
// Stores head of
// the linked list
Node* curr = head;
// Stores sum of nodes whose data value
// contains three distinct factors
int nodeSum = 0;
// Traverse the linked list
while (curr) {
// If data value of curr is
// square of a prime number
if (is_square(curr->data)) {
// Update nodeSum
nodeSum += curr->data;
}
// Update curr
curr = curr->next;
}
// Return sum
return nodeSum;
}
// Driver Code
int main()
{
// Stores head node of
// the linked list
Node* head = NULL;
// Insert all data values
// in the linked list
push(&head, 5);
push(&head, 4);
push(&head, 2);
push(&head, 1);
cout << calculate_sum(head);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Structure of a
// singly Linked List
static class Node
{
// Stores data value
// of a Node
int data;
// Stores pointer
// to next Node
Node next;
};
static Node head = null;
// Function to insert a node at the
// beginning of the singly Linked List
static Node push(int new_data)
{
// Create a new Node
Node new_node = new Node();
// Insert the data into
// the Node
new_node.data = new_data;
// Insert pointer to
// the next Node
new_node.next = head;
// Update head_ref
return head = new_node;
}
// Function to check if a number
// is a prime number or not
static boolean CheckPrimeNumber(int n)
{
// Base Case
if (n <= 1)
return false;
if (n <= 3)
return true;
// If n is multiple of 2 or 3
if (n % 2 == 0 || n % 3 == 0)
return false;
// Iterate over the range
// [5, Math.sqrt(N)]
for(int i = 5; i * i <= n; i = i + 6)
{
// If n is multiple of
// i or (i + 2)
if (n % i == 0 || n % (i + 2) == 0)
return false;
}
return true;
}
// Function to check if number is
// a perfect square number or not
static boolean checkperfect_square(int n)
{
// Stores square root of n
long sr = (long)Math.sqrt(n);
// If n is a
// perfect square number
if (sr - Math.floor(sr) == 0)
{
return true;
}
return false;
}
// Function to check if n is square
// of a prime number or not
static boolean is_square(int n)
{
// If n is a perfect square
if (checkperfect_square(n))
{
// Stores sqrt of n
int root = (int)Math.sqrt(n);
// If root is a prime number
if (CheckPrimeNumber(root))
{
return true;
}
}
return false;
}
// Function to find sum of node values
// which contains three distinct factors
static int calculate_sum(Node head)
{
// Stores head of
// the linked list
Node curr = head;
// Stores sum of nodes whose data value
// contains three distinct factors
int nodeSum = 0;
// Traverse the linked list
while (curr.next != null)
{
// If data value of curr is
// square of a prime number
if (is_square(curr.data))
{
// Update nodeSum
nodeSum += curr.data;
}
// Update curr
curr = curr.next;
}
// Return sum
return nodeSum;
}
// Driver Code
public static void main(String[] args)
{
// Stores head node of
// the linked list
// Insert all data values
// in the linked list
head = push(5);
head = push(4);
head = push(2);
head = push(1);
System.out.print(calculate_sum(head));
}
}
// This code is contributed by Rajput-JiPython3
# Python3 program to implement
# the above approach
import math
# Structure of a
# singly linked list
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
# Function to insert a nodeat the
# beginning of the singly linked list
def Push(self, new_data):
# Create new node
# and insert the data
# into the node
new_node = Node(new_data)
# Insert pointer to the
# next node
new_node.next = self.head
# Update head
self.head = new_node
# Function to find sum of node values
# which contains three distinct factors
def calculate_sum(self):
# Stores the head of linked list
curr = self.head
# Stores sum of nodes whose data
# value contains three distinct factors
nodeSum = 0
# Traverse the linked list
while(curr):
# If data value of curr is
# square of a prime number
if (is_square(curr.data)):
# Update nodeSum
nodeSum += curr.data
# Update curr
curr = curr.next
# Return Sum
return nodeSum
# Function to check if a number
# is a prime number or not
def CheckPrimeNumber(n):
# Base Class
if n <= 1:
return False
if n <= 3:
return True
# If n is multiple of 2 or 3
if (n % 2 == 0 or n % 3 == 0):
return False
# Iterate over the range
# [5,sqrt(N)]
i = 5
while ((i * i) <= n):
# If n is multiple of
# i or (i+2)
if (n % i == 0 or n % (i + 2) == 0):
return False
i += 6
return True
# Function to check if number is
# a perfect square number or not
def checkperfect_square(n):
# Stores square root of n
sr = math.sqrt(n)
# If n is a perfect
# square number
if (sr - math.floor(sr)) == 0:
return True
return False
# Function to check if n is square
# of a prime number or not
def is_square(n):
# If n is a perfect square
if (checkperfect_square(n)):
# Stores sqrt of n
root = math.sqrt(n)
# If root is a prime number
if (CheckPrimeNumber(root)):
return True
return False
# Driver Code
if __name__ == "__main__" :
LList = LinkedList()
# Insert all data values
# in the linked list
LList.Push(5)
LList.Push(4)
LList.Push(2)
LList.Push(1)
print(LList.calculate_sum())
# This code is contributed by VirusbuddahC#
// C# program to implement
// the above approach
using System;
class GFG{
// Structure of a
// singly Linked List
public class Node
{
// Stores data value
// of a Node
public int data;
// Stores pointer
// to next Node
public Node next;
};
static Node head = null;
// Function to insert a node at the
// beginning of the singly Linked List
static Node push(int new_data)
{
// Create a new Node
Node new_node = new Node();
// Insert the data into
// the Node
new_node.data = new_data;
// Insert pointer to
// the next Node
new_node.next = head;
// Update head_ref
return head = new_node;
}
// Function to check if a number
// is a prime number or not
static bool CheckPrimeNumber(int n)
{
// Base Case
if (n <= 1)
return false;
if (n <= 3)
return true;
// If n is multiple of 2 or 3
if (n % 2 == 0 || n % 3 == 0)
return false;
// Iterate over the range
// [5, Math.Sqrt(N)]
for(int i = 5; i * i <= n; i = i + 6)
{
// If n is multiple of
// i or (i + 2)
if (n % i == 0 || n % (i + 2) == 0)
return false;
}
return true;
}
// Function to check if number is
// a perfect square number or not
static bool checkperfect_square(int n)
{
// Stores square root of n
long sr = (long)Math.Sqrt(n);
// If n is a perfect square number
if (sr - Math.Floor((double)sr) == 0)
{
return true;
}
return false;
}
// Function to check if n is square
// of a prime number or not
static bool is_square(int n)
{
// If n is a perfect square
if (checkperfect_square(n))
{
// Stores sqrt of n
int root = (int)Math.Sqrt(n);
// If root is a prime number
if (CheckPrimeNumber(root))
{
return true;
}
}
return false;
}
// Function to find sum of node values
// which contains three distinct factors
static int calculate_sum(Node head)
{
// Stores head of
// the linked list
Node curr = head;
// Stores sum of nodes whose data value
// contains three distinct factors
int nodeSum = 0;
// Traverse the linked list
while (curr.next != null)
{
// If data value of curr is
// square of a prime number
if (is_square(curr.data))
{
// Update nodeSum
nodeSum += curr.data;
}
// Update curr
curr = curr.next;
}
// Return sum
return nodeSum;
}
// Driver Code
public static void Main(String[] args)
{
// Stores head node of
// the linked list
// Insert all data values
// in the linked list
head = push(5);
head = push(4);
head = push(2);
head = push(1);
Console.Write(calculate_sum(head));
}
}
// This code is contributed by aashish1995输出：
4时间复杂度： O(N *√X)，其中X是链表中的最大元素
辅助空间： O(1)