给定一个链表和一个整数M ，任务是将链表的最后M个节点附加到最前面。
例子：
Input: List = 4 -> 5 -> 6 -> 1 -> 2 -> 3 -> NULL, M = 3
Output: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL
Input: List = 8 -> 7 -> 0 -> 4 -> 1 -> NULL, M = 2
Output: 4 -> 1 -> 8 -> 7 -> 0 -> NULL
方法：找到列表中最后M个节点中的第一个节点，该节点将是新的头节点，因此使上一个节点的下一个指针为NULL，并将原始列表的最后一个节点指向原始列表的头。最后，打印更新的列表。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Class for a node of
// the linked list
struct Node
{
// Data and the pointer
// to the next node
int data;
Node *next;
Node(int data)
{
this->data = data;
this->next = NULL;
}
};
// Function to print the linked list
void printList(Node* node)
{
while (node != NULL)
{
cout << (node->data) << " -> ";
node = node->next;
}
cout << "NULL";
}
// Recursive function to return the
// count of nodes in the linked list
int cntNodes(Node* node)
{
if (node == NULL)
return 0;
return (1 + cntNodes(node->next));
}
// Function to update and print
// the updated list nodes
void updateList(Node* head, int m)
{
// Total nodes in the list
int cnt = cntNodes(head);
if (cnt != m && m < cnt)
{
// Count of nodes to be skipped
// from the beginning
int skip = cnt - m;
Node *prev = NULL;
Node *curr = head;
// Skip the nodes
while (skip > 0)
{
prev = curr;
curr = curr->next;
skip--;
}
// Change the pointers
prev->next = NULL;
Node *tempHead = head;
head = curr;
// Find the last node
while (curr->next != NULL)
curr = curr->next;
// Connect it to the head
// of the sub list
curr->next = tempHead;
}
// Print the updated list
printList(head);
}
// Driver code
int main()
{
// Create the list
Node* head = new Node(4);
head->next = new Node(5);
head->next->next = new Node(6);
head->next->next->next = new Node(1);
head->next->next->next->next = new Node(2);
head->next->next->next->next->next = new Node(3);
int m = 3;
updateList(head, m);
return 0;
}
// This code is contributed by rutvik_56 Java
// Java implementation of the approach
class GFG {
// Class for a node of
// the linked list
static class Node {
// Data and the pointer
// to the next node
int data;
Node next;
Node(int data)
{
this.data = data;
this.next = null;
}
}
// Function to print the linked list
static void printList(Node node)
{
while (node != null) {
System.out.print(node.data + " -> ");
node = node.next;
}
System.out.print("NULL");
}
// Recursive function to return the
// count of nodes in the linked list
static int cntNodes(Node node)
{
if (node == null)
return 0;
return (1 + cntNodes(node.next));
}
// Function to update and print
// the updated list nodes
static void updateList(Node head, int m)
{
// Total nodes in the list
int cnt = cntNodes(head);
if (cnt != m && m < cnt) {
// Count of nodes to be skipped
// from the beginning
int skip = cnt - m;
Node prev = null;
Node curr = head;
// Skip the nodes
while (skip > 0) {
prev = curr;
curr = curr.next;
skip--;
}
// Change the pointers
prev.next = null;
Node tempHead = head;
head = curr;
// Find the last node
while (curr.next != null)
curr = curr.next;
// Connect it to the head
// of the sub list
curr.next = tempHead;
}
// Print the updated list
printList(head);
}
// Driver code
public static void main(String[] args)
{
// Create the list
Node head = new Node(4);
head.next = new Node(5);
head.next.next = new Node(6);
head.next.next.next = new Node(1);
head.next.next.next.next = new Node(2);
head.next.next.next.next.next = new Node(3);
int m = 3;
updateList(head, m);
}
}Python3
# Python3 implementation of the approach
# Class for a node of
# the linked list
class newNode:
# Constructor to initialize the node object
def __init__(self, data):
self.data = data
self.next = None
# Function to print the linked list
def printList(node):
while (node != None):
print(node.data,"->", end=" ")
node = node.next
print("NULL")
# Recursive function to return the
# count of nodes in the linked list
def cntNodes(node):
if (node == None):
return 0
return (1 + cntNodes(node.next))
# Function to update and print
# the updated list nodes
def updateList(head, m):
# Total nodes in the list
cnt = cntNodes(head)
if (cnt != m and m < cnt):
# Count of nodes to be skipped
# from the beginning
skip = cnt - m
prev = None
curr = head
# Skip the nodes
while (skip > 0):
prev = curr
curr = curr.next
skip-=1
# Change the pointers
prev.next = None
tempHead = head
head = curr
# Find the last node
while (curr.next != None):
curr = curr.next
# Connect it to the head
# of the sub list
curr.next = tempHead
# Print the updated list
printList(head)
# Driver code
# Create the list
head = newNode(4)
head.next = newNode(5)
head.next.next = newNode(6)
head.next.next.next = newNode(1)
head.next.next.next.next = newNode(2)
head.next.next.next.next.next = newNode(3)
m = 3
updateList(head, m)
# This code is contributed by shubhamsingh10C#
// C# implementation of the approach
using System;
class GFG
{
// Class for a node of
// the linked list
class Node
{
// Data and the pointer
// to the next node
public int data;
public Node next;
public Node(int data)
{
this.data = data;
this.next = null;
}
}
// Function to print the linked list
static void printList(Node node)
{
while (node != null)
{
Console.Write(node.data + " -> ");
node = node.next;
}
Console.Write("NULL");
}
// Recursive function to return the
// count of nodes in the linked list
static int cntNodes(Node node)
{
if (node == null)
return 0;
return (1 + cntNodes(node.next));
}
// Function to update and print
// the updated list nodes
static void updateList(Node head, int m)
{
// Total nodes in the list
int cnt = cntNodes(head);
if (cnt != m && m < cnt)
{
// Count of nodes to be skipped
// from the beginning
int skip = cnt - m;
Node prev = null;
Node curr = head;
// Skip the nodes
while (skip > 0)
{
prev = curr;
curr = curr.next;
skip--;
}
// Change the pointers
prev.next = null;
Node tempHead = head;
head = curr;
// Find the last node
while (curr.next != null)
curr = curr.next;
// Connect it to the head
// of the sub list
curr.next = tempHead;
}
// Print the updated list
printList(head);
}
// Driver code
public static void Main(String[] args)
{
// Create the list
Node head = new Node(4);
head.next = new Node(5);
head.next.next = new Node(6);
head.next.next.next = new Node(1);
head.next.next.next.next = new Node(2);
head.next.next.next.next.next = new Node(3);
int m = 3;
updateList(head, m);
}
}
// This code is contributed by PrinciRaj1992输出：
1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL