给定N个元素的数组。任务是将给定元素一个接一个地插入链表的中间位置。每个插入操作应占用O(1)时间复杂度。
例子：
Input: arr[] = {1, 2, 3, 4, 5}
Output: 1 -> 3 -> 5 -> 4 -> 2 -> NULL
1 -> NULL
1 -> 2 -> NULL
1 -> 3 -> 2 -> NULL
1 -> 3 -> 4 -> 2 -> NULL
1 -> 3 -> 5 -> 4 -> 2 -> NULL
Input: arr[] = {5, 4, 1, 2}
Output: 5 -> 1 -> 2 -> 4 -> NULL
方法：有两种情况：
- 列表中存在的元素数少于2。
- 列表中存在的元素数大于2。
- 已经存在的元素数甚至是N，然后将新元素插入到(N / 2)+ 1的中间位置。
- 已经存在的元素数量为奇数，然后在当前中间元素(N / 2)+ 2旁边插入新元素。
我们再增加一个指针“中间”，它存储当前中间元素的地址，还有一个计数器，用于计算元素总数。
如果链表中已经存在的元素少于2个，那么Middle总是指向第一个位置，然后在当前Middle后面插入新节点。
如果链表中已经存在的元素大于2，则我们在当前中间位置旁边插入新节点并增加计数器。
如果插入后元素的数量为奇数，则中间点指向新插入的节点，否则中间指针没有变化。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Node structure
struct Node {
int value;
struct Node* next;
};
// Class to represent a node
// of the linked list
class LinkedList {
private:
struct Node *head, *mid;
int count;
public:
LinkedList();
void insertAtMiddle(int);
void show();
};
LinkedList::LinkedList()
{
head = NULL;
mid = NULL;
count = 0;
}
// Function to insert a node in
// the middle of the linked list
void LinkedList::insertAtMiddle(int n)
{
struct Node* temp = new struct Node();
struct Node* temp1;
temp->next = NULL;
temp->value = n;
// If the number of elements
// already present are less than 2
if (count < 2) {
if (head == NULL) {
head = temp;
}
else {
temp1 = head;
temp1->next = temp;
}
count++;
// mid points to first element
mid = head;
}
// If the number of elements already present
// are greater than 2
else {
temp->next = mid->next;
mid->next = temp;
count++;
// If number of elements after insertion
// are odd
if (count % 2 != 0) {
// mid points to the newly
// inserted node
mid = mid->next;
}
}
}
// Function to print the nodes
// of the linked list
void LinkedList::show()
{
struct Node* temp;
temp = head;
// Initializing temp to head
// Iterating and printing till
// The end of linked list
// That is, till temp is null
while (temp != NULL) {
cout << temp->value << " -> ";
temp = temp->next;
}
cout << "NULL";
cout << endl;
}
// Driver code
int main()
{
// Elements to be inserted one after another
int arr[] = { 1, 2, 3, 4, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
LinkedList L1;
// Insert the elements
for (int i = 0; i < n; i++)
L1.insertAtMiddle(arr[i]);
// Print the nodes of the linked list
L1.show();
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Node ure
static class Node
{
int value;
Node next;
};
// Class to represent a node
// of the linked list
static class LinkedList
{
Node head, mid;
int count;
LinkedList()
{
head = null;
mid = null;
count = 0;
}
// Function to insert a node in
// the middle of the linked list
void insertAtMiddle(int n)
{
Node temp = new Node();
Node temp1;
temp.next = null;
temp.value = n;
// If the number of elements
// already present are less than 2
if (count < 2)
{
if (head == null)
{
head = temp;
}
else
{
temp1 = head;
temp1.next = temp;
}
count++;
// mid points to first element
mid = head;
}
// If the number of elements already present
// are greater than 2
else
{
temp.next = mid.next;
mid.next = temp;
count++;
// If number of elements after insertion
// are odd
if (count % 2 != 0)
{
// mid points to the newly
// inserted node
mid = mid.next;
}
}
}
// Function to print the nodes
// of the linked list
void show()
{
Node temp;
temp = head;
// Initializing temp to head
// Iterating and printing till
// The end of linked list
// That is, till temp is null
while (temp != null)
{
System.out.print( temp.value + " -> ");
temp = temp.next;
}
System.out.print( "null");
System.out.println();
}
}
// Driver code
public static void main(String args[])
{
// Elements to be inserted one after another
int arr[] = { 1, 2, 3, 4, 5 };
int n = arr.length;
LinkedList L1=new LinkedList();
// Insert the elements
for (int i = 0; i < n; i++)
L1.insertAtMiddle(arr[i]);
// Print the nodes of the linked list
L1.show();
}
}
// This code is contributed by Arnab KunduPython3
# Python3 implementation of the approach
# Node ure
class Node:
def __init__(self):
self.value = 0
self.next = None
# Class to represent a node
# of the linked list
class LinkedList:
def __init__(self) :
self.head = None
self.mid = None
self.count = 0
# Function to insert a node in
# the middle of the linked list
def insertAtMiddle(self , n):
temp = Node()
temp1 = None
temp.next = None
temp.value = n
# If the number of elements
# already present are less than 2
if (self.count < 2):
if (self.head == None) :
self.head = temp
else:
temp1 = self.head
temp1.next = temp
self.count = self.count + 1
# mid points to first element
self.mid = self.head
# If the number of elements already present
# are greater than 2
else:
temp.next = self.mid.next
self.mid.next = temp
self.count = self.count + 1
# If number of elements after insertion
# are odd
if (self.count % 2 != 0):
# mid points to the newly
# inserted node
self.mid = self.mid.next
# Function to print the nodes
# of the linked list
def show(self):
temp = None
temp = self.head
# Initializing temp to self.head
# Iterating and printing till
# The end of linked list
# That is, till temp is None
while (temp != None) :
print( temp.value, end = " -> ")
temp = temp.next
print( "None")
# Driver code
# Elements to be inserted one after another
arr = [ 1, 2, 3, 4, 5]
n = len(arr)
L1 = LinkedList()
# Insert the elements
for i in range(n):
L1.insertAtMiddle(arr[i])
# Print the nodes of the linked list
L1.show()
# This code is contributed by Arnab KunduC#
// C# implementation of the approach
using System;
class GFG
{
// Node ure
public class Node
{
public int value;
public Node next;
};
// Class to represent a node
// of the linked list
public class LinkedList
{
public Node head, mid;
public int count;
public LinkedList()
{
head = null;
mid = null;
count = 0;
}
// Function to insert a node in
// the middle of the linked list
public void insertAtMiddle(int n)
{
Node temp = new Node();
Node temp1;
temp.next = null;
temp.value = n;
// If the number of elements
// already present are less than 2
if (count < 2)
{
if (head == null)
{
head = temp;
}
else
{
temp1 = head;
temp1.next = temp;
}
count++;
// mid points to first element
mid = head;
}
// If the number of elements already present
// are greater than 2
else
{
temp.next = mid.next;
mid.next = temp;
count++;
// If number of elements after insertion
// are odd
if (count % 2 != 0)
{
// mid points to the newly
// inserted node
mid = mid.next;
}
}
}
// Function to print the nodes
// of the linked list
public void show()
{
Node temp;
temp = head;
// Initializing temp to head
// Iterating and printing till
// The end of linked list
// That is, till temp is null
while (temp != null)
{
Console.Write( temp.value + " -> ");
temp = temp.next;
}
Console.Write( "null");
Console.WriteLine();
}
}
// Driver code
public static void Main(String []args)
{
// Elements to be inserted one after another
int []arr = { 1, 2, 3, 4, 5 };
int n = arr.Length;
LinkedList L1=new LinkedList();
// Insert the elements
for (int i = 0; i < n; i++)
L1.insertAtMiddle(arr[i]);
// Print the nodes of the linked list
L1.show();
}
}
// This code contributed by Rajput-Ji输出：
1 -> 3 -> 5 -> 4 -> 2 -> NULL