在末尾的第n个节点后插入一个节点
在给定的单链表中在距离末尾的第n个节点之后插入节点x。 确保列表包含距离末尾的第n个节点。 另外1 <= n。
例子：
Input : list: 1->3->4->5
n = 4, x = 2
Output : 1->2->3->4->5
4th node from the end is 1 and
insertion has been done after this node.
Input : list: 10->8->3->12->5->18
n = 2, x = 11
Output : 10->8->3->12->5->11->18
方法 1（使用列表的长度）：
查找链表的长度，即列表中的节点数。 设为len。 现在从头开始遍历列表，从第一个节点到第len - n + 1个节点，然后在该节点之后插入新节点。 此方法需要两次遍历列表。
C++
// C++ implementation to insert a node after
// the n-th node from the end
#include <bits/stdc++.h>
using namespace std;
// structure of a node
struct Node {
int data;
Node* next;
};
// function to get a new node
Node* getNode(int data)
{
// allocate memory for the node
Node* newNode = (Node*)malloc(sizeof(Node));
// put in the data
newNode->data = data;
newNode->next = NULL;
return newNode;
}
// function to insert a node after the
// nth node from the end
void insertAfterNthNode(Node* head, int n, int x)
{
// if list is empty
if (head == NULL)
return;
// get a new node for the value 'x'
Node* newNode = getNode(x);
Node* ptr = head;
int len = 0, i;
// find length of the list, i.e, the
// number of nodes in the list
while (ptr != NULL) {
len++;
ptr = ptr->next;
}
// traverse up to the nth node from the end
ptr = head;
for (i = 1; i <= (len - n); i++)
ptr = ptr->next;
// insert the 'newNode' by making the
// necessary adjustment in the links
newNode->next = ptr->next;
ptr->next = newNode;
}
// function to print the list
void printList(Node* head)
{
while (head != NULL) {
cout << head->data << " ";
head = head->next;
}
}
// Driver program to test above
int main()
{
// Creating list 1->3->4->5
Node* head = getNode(1);
head->next = getNode(3);
head->next->next = getNode(4);
head->next->next->next = getNode(5);
int n = 4, x = 2;
cout << "Original Linked List: ";
printList(head);
insertAfterNthNode(head, n, x);
cout << "\nLinked List After Insertion: ";
printList(head);
return 0;
}
Java
// Java implementation to insert a node after
// the n-th node from the end
class GfG
{
// structure of a node
static class Node
{
int data;
Node next;
}
// function to get a new node
static Node getNode(int data)
{
// allocate memory for the node
Node newNode = new Node();
// put in the data
newNode.data = data;
newNode.next = null;
return newNode;
}
// function to insert a node after the
// nth node from the end
static void insertAfterNthNode(Node head, int n, int x)
{
// if list is empty
if (head == null)
return;
// get a new node for the value 'x'
Node newNode = getNode(x);
Node ptr = head;
int len = 0, i;
// find length of the list, i.e, the
// number of nodes in the list
while (ptr != null)
{
len++;
ptr = ptr.next;
}
// traverse up to the nth node from the end
ptr = head;
for (i = 1; i <= (len - n); i++)
ptr = ptr.next;
// insert the 'newNode' by making the
// necessary adjustment in the links
newNode.next = ptr.next;
ptr.next = newNode;
}
// function to print the list
static void printList(Node head)
{
while (head != null)
{
System.out.print(head.data + " ");
head = head.next;
}
}
// Driver code
public static void main(String[] args)
{
// Creating list 1->3->4->5
Node head = getNode(1);
head.next = getNode(3);
head.next.next = getNode(4);
head.next.next.next = getNode(5);
int n = 4, x = 2;
System.out.print("Original Linked List: ");
printList(head);
insertAfterNthNode(head, n, x);
System.out.println();
System.out.print("Linked List After Insertion: ");
printList(head);
}
}
// This code is contributed by prerna saini
Python
# Python implementation to insert a node after
# the n-th node from the end
# Linked List node
class Node:
def __init__(self, data):
self.data = data
self.next = None
# function to get a new node
def getNode(data) :
# allocate memory for the node
newNode = Node(0)
# put in the data
newNode.data = data
newNode.next = None
return newNode
# function to insert a node after the
# nth node from the end
def insertAfterNthNode(head, n, x) :
# if list is empty
if (head == None) :
return
# get a new node for the value 'x'
newNode = getNode(x)
ptr = head
len = 0
i = 0
# find length of the list, i.e, the
# number of nodes in the list
while (ptr != None) :
len = len + 1
ptr = ptr.next
# traverse up to the nth node from the end
ptr = head
i = 1
while ( i <= (len - n) ) :
ptr = ptr.next
i = i + 1
# insert the 'newNode' by making the
# necessary adjustment in the links
newNode.next = ptr.next
ptr.next = newNode
# function to print the list
def printList( head) :
while (head != None):
print(head.data ,end = " ")
head = head.next
# Driver code
# Creating list 1->3->4->5
head = getNode(1)
head.next = getNode(3)
head.next.next = getNode(4)
head.next.next.next = getNode(5)
n = 4
x = 2
print("Original Linked List: ")
printList(head)
insertAfterNthNode(head, n, x)
print()
print("Linked List After Insertion: ")
printList(head)
# This code is contributed by Arnab Kundu
C
// C# implementation to insert a node after
// the n-th node from the end
using System;
class GfG
{
// structure of a node
public class Node
{
public int data;
public Node next;
}
// function to get a new node
static Node getNode(int data)
{
// allocate memory for the node
Node newNode = new Node();
// put in the data
newNode.data = data;
newNode.next = null;
return newNode;
}
// function to insert a node after the
// nth node from the end
static void insertAfterNthNode(Node head, int n, int x)
{
// if list is empty
if (head == null)
return;
// get a new node for the value 'x'
Node newNode = getNode(x);
Node ptr = head;
int len = 0, i;
// find length of the list, i.e, the
// number of nodes in the list
while (ptr != null)
{
len++;
ptr = ptr.next;
}
// traverse up to the nth node from the end
ptr = head;
for (i = 1; i <= (len - n); i++)
ptr = ptr.next;
// insert the 'newNode' by making the
// necessary adjustment in the links
newNode.next = ptr.next;
ptr.next = newNode;
}
// function to print the list
static void printList(Node head)
{
while (head != null)
{
Console.Write(head.data + " ");
head = head.next;
}
}
// Driver code
public static void Main(String[] args)
{
// Creating list 1->3->4->5
Node head = getNode(1);
head.next = getNode(3);
head.next.next = getNode(4);
head.next.next.next = getNode(5);
int n = 4, x = 2;
Console.Write("Original Linked List: ");
printList(head);
insertAfterNthNode(head, n, x);
Console.WriteLine();
Console.Write("Linked List After Insertion: ");
printList(head);
}
}
// This code has been contributed by 29AjayKumar
输出：
Original Linked List: 1 3 4 5
Linked List After Insertion: 1 2 3 4 5
时间复杂度：O(n)，其中n是列表中节点的数量。
方法 2（单遍历）：
该方法使用两个指针，一个是slow_ptr，另一个是fast_ptr。 首先将fast_ptr移到从头开始的第n个节点。 使slow_ptr，指向列表的第一个节点。 现在，同时移动两个指针，直到fast_ptr指向最后一个节点。 此时，slow_ptr，将指向距离末尾的第n个节点。 在该节点之后插入新节点。 此方法需要单遍遍列表。
C++
// C++ implementation to insert a node after the
// nth node from the end
#include <bits/stdc++.h>
using namespace std;
// structure of a node
struct Node {
int data;
Node* next;
};
// function to get a new node
Node* getNode(int data)
{
// allocate memory for the node
Node* newNode = (Node*)malloc(sizeof(Node));
// put in the data
newNode->data = data;
newNode->next = NULL;
return newNode;
}
// function to insert a node after the
// nth node from the end
void insertAfterNthNode(Node* head, int n, int x)
{
// if list is empty
if (head == NULL)
return;
// get a new node for the value 'x'
Node* newNode = getNode(x);
// Initializing the slow and fast pointers
Node* slow_ptr = head;
Node* fast_ptr = head;
// move 'fast_ptr' to point to the nth node
// from the beginning
for (int i = 1; i <= n - 1; i++)
fast_ptr = fast_ptr->next;
// iterate until 'fast_ptr' points to the
// last node
while (fast_ptr->next != NULL) {
// move both the pointers to the
// respective next nodes
slow_ptr = slow_ptr->next;
fast_ptr = fast_ptr->next;
}
// insert the 'newNode' by making the
// necessary adjustment in the links
newNode->next = slow_ptr->next;
slow_ptr->next = newNode;
}
// function to print the list
void printList(Node* head)
{
while (head != NULL) {
cout << head->data << " ";
head = head->next;
}
}
// Driver program to test above
int main()
{
// Creating list 1->3->4->5
Node* head = getNode(1);
head->next = getNode(3);
head->next->next = getNode(4);
head->next->next->next = getNode(5);
int n = 4, x = 2;
cout << "Original Linked List: ";
printList(head);
insertAfterNthNode(head, n, x);
cout << "\nLinked List After Insertion: ";
printList(head);
return 0;
}
Java
// Java implementation to
// insert a node after the
// nth node from the end
class GfG
{
// structure of a node
static class Node
{
int data;
Node next;
}
// function to get a new node
static Node getNode(int data)
{
// allocate memory for the node
Node newNode = new Node();
// put in the data
newNode.data = data;
newNode.next = null;
return newNode;
}
// function to insert a node after
// the nth node from the end
static void insertAfterNthNode(Node head,
int n, int x)
{
// if list is empty
if (head == null)
return;
// get a new node for the value 'x'
Node newNode = getNode(x);
// Initializing the slow
// and fast pointers
Node slow_ptr = head;
Node fast_ptr = head;
// move 'fast_ptr' to point to the
// nth node from the beginning
for (int i = 1; i <= n - 1; i++)
fast_ptr = fast_ptr.next;
// iterate until 'fast_ptr' points
// to the last node
while (fast_ptr.next != null)
{
// move both the pointers to the
// respective next nodes
slow_ptr = slow_ptr.next;
fast_ptr = fast_ptr.next;
}
// insert the 'newNode' by making the
// necessary adjustment in the links
newNode.next = slow_ptr.next;
slow_ptr.next = newNode;
}
// function to print the list
static void printList(Node head)
{
while (head != null)
{
System.out.print(head.data + " ");
head = head.next;
}
}
// Driver code
public static void main(String[] args)
{
// Creating list 1->3->4->5
Node head = getNode(1);
head.next = getNode(3);
head.next.next = getNode(4);
head.next.next.next = getNode(5);
int n = 4, x = 2;
System.out.println("Original Linked List: ");
printList(head);
insertAfterNthNode(head, n, x);
System.out.println();
System.out.println("Linked List After Insertion: ");
printList(head);
}
}
// This code is contributed by
// Prerna Saini.
C
// C# implementation to
// insert a node after the
// nth node from the end
using System;
class GfG
{
// structure of a node
public class Node
{
public int data;
public Node next;
}
// function to get a new node
static Node getNode(int data)
{
// allocate memory for the node
Node newNode = new Node();
// put in the data
newNode.data = data;
newNode.next = null;
return newNode;
}
// function to insert a node after
// the nth node from the end
static void insertAfterNthNode(Node head,
int n, int x)
{
// if list is empty
if (head == null)
return;
// get a new node for the value 'x'
Node newNode = getNode(x);
// Initializing the slow
// and fast pointers
Node slow_ptr = head;
Node fast_ptr = head;
// move 'fast_ptr' to point to the
// nth node from the beginning
for (int i = 1; i <= n - 1; i++)
fast_ptr = fast_ptr.next;
// iterate until 'fast_ptr' points
// to the last node
while (fast_ptr.next != null)
{
// move both the pointers to the
// respective next nodes
slow_ptr = slow_ptr.next;
fast_ptr = fast_ptr.next;
}
// insert the 'newNode' by making the
// necessary adjustment in the links
newNode.next = slow_ptr.next;
slow_ptr.next = newNode;
}
// function to print the list
static void printList(Node head)
{
while (head != null)
{
Console.Write(head.data + " ");
head = head.next;
}
}
// Driver code
public static void Main()
{
// Creating list 1->3->4->5
Node head = getNode(1);
head.next = getNode(3);
head.next.next = getNode(4);
head.next.next.next = getNode(5);
int n = 4, x = 2;
Console.WriteLine("Original Linked List: ");
printList(head);
insertAfterNthNode(head, n, x);
Console.WriteLine();
Console.WriteLine("Linked List After Insertion: ");
printList(head);
}
}
/* This code contributed by PrinciRaj1992 */
输出：
Original Linked List: 1 3 4 5
Linked List After Insertion: 1 2 3 4 5
时间复杂度：O(n)，其中n是列表中节点的数量。
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