链表中素数的阶乘之和
给定一个由 N 个整数组成的链表，任务是找出链表中每个素元素的阶乘之和。 例:
输入:L1 = 4->6->2->12->3 T3】输出:8 T6】说明: 质数为 2 和 3，遂为 2！+ 3!= 2 + 6 = 8. 输入:L1 = 7->4->5 输出: 5160 解释: 质数为 7 和 5，故名 7！+ 5!= 5160.
方法:按照以下步骤解决上述问题:
- 实现一个函数阶乘(n) ，使得找到 N 的阶乘。
- 初始化一个变量和= 0 。现在，遍历给定的列表，检查每个节点是否是素数。
- 如果节点是质数，则更新 sum = sum +阶乘(节点)，否则将节点移到下一个。
- 最后打印计算出的和。
以下是上述方法的实现:
C++
// C++ implementation to fine Sum of
// prime factorials in a 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)
{
// allocate node
Node* new_node
= (Node*)malloc(
sizeof(struct Node));
// put in the data
new_node->data = new_data;
// link the old list
// off the new node
new_node->next = (*head_ref);
// move the head to point
// to the new node
(*head_ref) = new_node;
}
// Function to return the factorial of N
int factorial(int n)
{
int f = 1;
for (int i = 1; i <= n; i++) {
f *= i;
}
return f;
}
// Function to check if number is prime
bool isPrime(int n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
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 return the sum of
// factorials of the LL elements
int sumFactorial(Node* head_1)
{
// To store the required sum
Node* ptr = head_1;
int s = 0;
while (ptr != NULL) {
// Add factorial of all the elements
if (isPrime(ptr->data)) {
s += factorial(ptr->data);
ptr = ptr->next;
}
else
ptr = ptr->next;
}
return s;
}
// Driver code
int main()
{
Node* head1 = NULL;
push(&head1, 4);
push(&head1, 6);
push(&head1, 2);
push(&head1, 12);
push(&head1, 3);
cout << sumFactorial(head1);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation to find Sum of
// prime factorials in a Linked list
import java.util.*;
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)
{
// allocate node
Node new_node = new Node();
// put in the data
new_node.data = new_data;
// link the old list
// off the new node
new_node.next = (head_ref);
// move the head to point
// to the new node
(head_ref) = new_node;
return head_ref;
}
// Function to return
// the factorial of n
static int factorial(int n)
{
int f = 1;
for (int i = 1; i <= n; i++) {
f *= i;
}
return f;
}
// Function to check if number is prime
static boolean isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
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 return the sum of
// factorials of the LL elements
static int sumFactorial(Node head_1)
{
Node ptr = head_1;
// To store the required sum
int s = 0;
while (ptr != null) {
// Add factorial of
// all the elements
if (isPrime(ptr.data)) {
s += factorial(ptr.data);
ptr = ptr.next;
}
else {
ptr = ptr.next;
}
}
return s;
}
// Driver Code
public static void main(String args[])
{
Node head1 = null;
head1 = push(head1, 4);
head1 = push(head1, 6);
head1 = push(head1, 2);
head1 = push(head1, 12);
head1 = push(head1, 3);
int ans = sumFactorial(head1);
System.out.println(ans);
}
}
Python 3
# Python implementation of the approach
class Node:
def __init__(self, data):
self.data = data
self.next = next
# Function to insert a
# node at the beginning
# of the singly Linked List
def push( head_ref, new_data) :
# allocate node
new_node = Node(0)
# put in the data
new_node.data = new_data
# link the old list
# off the new node
new_node.next = (head_ref)
# move the head to
# point to the new node
(head_ref) = new_node
return head_ref
def factorial(n):
f = 1;
for i in range(1, n + 1):
f *= i;
return f;
# prime for def
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
# Function to return the sum of
# factorials of the LL elements
def sumFactorial(head_ref1):
# To store the required sum
s = 0;
ptr1 = head_ref1
while (ptr1 != None) :
# Add factorial of all the elements
if(isPrime(ptr1.data)):
s += factorial(ptr1.data);
ptr1 = ptr1.next
else:
ptr1 = ptr1.next
return s;
# Driver code
# start with the empty list
head1 = None
# create the linked list
head1 = push(head1, 4)
head1 = push(head1, 6)
head1 = push(head1, 2)
head1 = push(head1, 12)
head1 = push(head1, 3)
ans = sumFactorial(head1)
print(ans)
C
// C# implementation to find Sum of
// prime factorials in a Linked list
using System;
class GFG{
// Node of the singly linked list
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)
{
// Allocate node
Node new_node = new Node();
// Put in the data
new_node.data = new_data;
// Link the old list
// off the new node
new_node.next = (head_ref);
// Move the head to point
// to the new node
(head_ref) = new_node;
return head_ref;
}
// Function to return
// the factorial of n
static int factorial(int n)
{
int f = 1;
for(int i = 1; i <= n; i++)
{
f *= i;
}
return f;
}
// Function to check if number
// is prime
static bool isPrime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
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 return the sum of
// factorials of the LL elements
static int sumFactorial(Node head_1)
{
Node ptr = head_1;
// To store the required sum
int s = 0;
while (ptr != null)
{
// Add factorial of
// all the elements
if (isPrime(ptr.data))
{
s += factorial(ptr.data);
ptr = ptr.next;
}
else
{
ptr = ptr.next;
}
}
return s;
}
// Driver Code
public static void Main(String []args)
{
Node head1 = null;
head1 = push(head1, 4);
head1 = push(head1, 6);
head1 = push(head1, 2);
head1 = push(head1, 12);
head1 = push(head1, 3);
int ans = sumFactorial(head1);
Console.WriteLine(ans);
}
}
// This code is contributed by Amit Katiyar
java 描述语言
<script>
// JavaScript implementation to find Sum of
// prime factorials in a Linked list
// Node of the singly linked list
class Node {
constructor() {
this.data = 0;
this.next = null;
}
}
// Function to insert a node
// at the beginning of the
// singly Linked List
function push(head_ref, new_data)
{
// Allocate node
var new_node = new Node();
// Put in the data
new_node.data = new_data;
// Link the old list
// off the new node
new_node.next = head_ref;
// Move the head to point
// to the new node
head_ref = new_node;
return head_ref;
}
// Function to return
// the factorial of n
function factorial(n) {
var f = 1;
for (var i = 1; i <= n; i++) {
f *= i;
}
return f;
}
// Function to check if number
// is prime
function isPrime(n) {
// Corner cases
if (n <= 1) return false;
if (n <= 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (var i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0) return false;
return true;
}
// Function to return the sum of
// factorials of the LL elements
function sumFactorial(head_1) {
var ptr = head_1;
// To store the required sum
var s = 0;
while (ptr != null)
{
// Add factorial of
// all the elements
if (isPrime(ptr.data))
{
s += factorial(ptr.data);
ptr = ptr.next;
} else {
ptr = ptr.next;
}
}
return s;
}
// Driver Code
var head1 = null;
head1 = push(head1, 4);
head1 = push(head1, 6);
head1 = push(head1, 2);
head1 = push(head1, 12);
head1 = push(head1, 3);
var ans = sumFactorial(head1);
document.write(ans);
// This code is contributed by rdtank.
</script>
Output:
8
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