Find Middle of the Linked List
Last Updated :
04 Sep, 2024
Given a singly linked list, the task is to find the middle of the linked list. If the number of nodes are even, then there would be two middle nodes, so return the second middle node.
Example:
Input: linked list: 1->2->3->4->5
Output: 3
Explanation: There are 5 nodes in the linked list and there is one middle node whose value is 3.
Input: linked list = 10 -> 20 -> 30 -> 40 -> 50 -> 60
Output: 40
Explanation: There are 6 nodes in the linked list, so we have two middle nodes: 30 and 40, but we will return the second middle node which is 40.
[Naive Approach] By Counting Nodes - O(n) time and O(1) space:
The idea is to traversing the entire linked list once to count the total number of nodes. After determining the total count, traverse the list again and stop at the (count/2)th node to return its value. This method requires two passes through the linked list to find the middle element.
C++
// C++ program to illustrate how to find the middle element
// by counting the number of nodes
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node* next;
Node(int x) {
this->data = x;
this->next = nullptr;
}
};
// Helper function to find length of linked list
int getLength(Node* head) {
// Variable to store length
int length = 0;
// Traverse the entire linked list and increment length by
// 1 for each node
while (head) {
length++;
head = head->next;
}
// Return the number of nodes in the linked list
return length;
}
// Function to find the middle element of the linked list
int getMiddle(Node* head) {
// Finding length of the linked list
int length = getLength(head);
// traverse till we reached half of length
int mid_index = length / 2;
while (mid_index--) {
head = head->next;
}
// Head now will be pointing to the middle element
return head->data;
}
int main() {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40 -> 50 -> 60
Node* head = new Node(10);
head->next = new Node(20);
head->next->next = new Node(30);
head->next->next->next = new Node(40);
head->next->next->next->next = new Node(50);
head->next->next->next->next->next = new Node(60);
cout << getMiddle(head);
return 0;
}
// C program to illustrate how to find the middle element
// by counting the number of nodes
#include <stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node* next;
};
// Helper function to find length of linked list
int getLength(struct Node* head) {
// Variable to store length
int length = 0;
// Traverse the entire linked list and increment length
// by 1 for each node
while (head) {
length++;
head = head->next;
}
// Return the number of nodes in the linked list
return length;
}
// Function to find the middle element of the linked list
int getMiddle(struct Node* head) {
// Finding length of the linked list
int length = getLength(head);
// Traverse till we reach half of the length
int mid_index = length / 2;
while (mid_index--) {
head = head->next;
}
// Head now will be pointing to the middle element
return head->data;
}
struct Node* createNode(int x) {
struct Node* newNode
= (struct Node*)malloc(sizeof(struct Node));
newNode->data = x;
newNode->next = NULL;
return newNode;
}
int main() {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40 -> 50 -> 60
struct Node* head = createNode(10);
head->next = createNode(20);
head->next->next = createNode(30);
head->next->next->next = createNode(40);
head->next->next->next->next = createNode(50);
head->next->next->next->next->next = createNode(60);
printf("%d\n", getMiddle(head));
return 0;
}
// Java program to illustrate how to find the middle element
// by counting the number of nodes
// A singly linked list node
class Node {
int data;
Node next;
// Constructor to initialize a new node with data
Node(int x) {
this.data = x;
this.next = null;
}
}
class GfG {
// Helper function to find length of linked list
static int getLength(Node head) {
// Variable to store length
int length = 0;
// Traverse the entire linked list and increment
// length by 1 for each node
while (head != null) {
length++;
head = head.next;
}
// Return the number of nodes in the linked list
return length;
}
// Function to get the middle value of the linked list
static int getMiddle(Node head) {
// Finding the length of the list
int length = getLength(head);
// traverse till we reached half of length
int mid_index = length / 2;
while (mid_index > 0) {
head = head.next;
mid_index--;
}
// temp will be storing middle element
return head.data;
}
public static void main(String[] args) {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40 -> 50 -> 60
Node head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
head.next.next.next.next = new Node(50);
head.next.next.next.next.next = new Node(60);
System.out.println(getMiddle(head));
}
}
# Java program to illustrate how to find the middle element
# by counting the number of nodes
class Node:
def __init__(self, x):
self.data = x
self.next = None
# Helper function to find length of linked list
def getLength(head):
# Variable to store length
length = 0
# Traverse the entire linked list and increment length
# by 1 for each node
while head:
length += 1
head = head.next
# Return the number of nodes in the linked list
return length
# Function to find the middle element of the linked list
def getMiddle(head):
# Finding length of the linked list
length = getLength(head)
# Traverse till we reach half of the length
mid_index = length // 2
while mid_index:
head = head.next
mid_index -= 1
# Head now will be pointing to the middle element
return head.data
def main():
# Create a hard-coded linked list:
# 10 -> 20 -> 30 -> 40 -> 50 -> 60
head = Node(10)
head.next = Node(20)
head.next.next = Node(30)
head.next.next.next = Node(40)
head.next.next.next.next = Node(50)
head.next.next.next.next.next = Node(60)
print(getMiddle(head))
if __name__ == "__main__":
main()
// C# program to illustrate how to find the middle element
// by counting the number of nodes
using System;
class Node {
public int data;
public Node next;
public Node(int x) {
this.data = x;
this.next = null;
}
}
class GfG {
// Helper function to find length of linked list
static int getLength(Node head) {
// Variable to store length
int length = 0;
// Traverse the entire linked list and increment
// length by 1 for each node
while (head != null) {
length++;
head = head.next;
}
// Return the number of nodes in the linked list
return length;
}
// Function to find the middle element of the linked
// list
static int getMiddle(Node head) {
// Finding length of the linked list
int length = getLength(head);
// Traverse till we reach half of the length
int mid_index = length / 2;
while (mid_index-- > 0) {
head = head.next;
}
// Head now will be pointing to the middle element
return head.data;
}
static void Main() {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40 -> 50 -> 60
Node head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
head.next.next.next.next = new Node(50);
head.next.next.next.next.next = new Node(60);
Console.WriteLine(getMiddle(head));
}
}
// Javascript program to illustrate how to find the middle
// element by counting the number of nodes
class Node {
constructor(x) {
this.data = x;
this.next = null;
}
}
// Helper function to find length of linked list
function getLength(head) {
// Variable to store length
let length = 0;
// Traverse the entire linked list and increment length
// by 1 for each node
while (head) {
length++;
head = head.next;
}
// Return the number of nodes in the linked list
return length;
}
// Function to find the middle element of the linked list
function getMiddle(head) {
// Finding length of the linked list
const length = getLength(head);
// Traverse till we reach half of the length
let mid_index = Math.floor(length / 2);
while (mid_index-- > 0) {
head = head.next;
}
// Head now will be pointing to the middle element
return head.data;
}
function main() {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40 -> 50 -> 60
const head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
head.next.next.next.next = new Node(50);
head.next.next.next.next.next = new Node(60);
console.log(getMiddle(head));
}
main();
Time Complexity: O(2 * n) = O(n) where n is the number of nodes in the linked list.
Auxiliary Space: O(1)
[Expected Approach] By Tortoise and Hare Algorithm - O(n) time and O(1) space:
We can use the Hare and Tortoise Algorithm to find the middle of the linked list. Traverse linked list using a slow pointer (slow_ptr) and a fast pointer (fast_ptr ). Move the slow pointer to the next node(one node forward) and the fast pointer to the next of the next node(two nodes forward). When the fast pointer reaches the last node or NULL, then the slow pointer will reach the middle of the linked list.
In case of odd number of nodes in the linked list, slow_ptr will reach the middle node when fast_ptr will reach the last node and in case of even number of nodes in the linked list, slow_ptr will reach the middle node when fast_ptr will become NULL.
Below is the working of above approach:
Code Implementation:
C++
// C++ program to illustrate how to find the middle element
// by using Floyd's Cycle Finding Algorithm
#include <iostream>
using namespace std;
class Node {
public:
int data;
Node* next;
Node(int x) {
this->data = x;
this->next = nullptr;
}
};
// Function to get the middle of the linked list
int getMiddle(Node* head) {
// Initialize the slow and fast pointer to the head of
// the linked list
Node* slow_ptr = head;
Node* fast_ptr = head;
while (fast_ptr != NULL && fast_ptr->next != NULL) {
// Move the fast pointer by two nodes
fast_ptr = fast_ptr->next->next;
// Move the slow pointer by one node
slow_ptr = slow_ptr->next;
}
// Return the slow pointer which is currenlty pointing
// to the middle node of the linked list
return slow_ptr->data;
}
int main() {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40 -> 50 -> 60
Node* head = new Node(10);
head->next = new Node(20);
head->next->next = new Node(30);
head->next->next->next = new Node(40);
head->next->next->next->next = new Node(50);
head->next->next->next->next->next = new Node(60);
cout << getMiddle(head) << endl;
return 0;
}
#include <stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node* next;
};
// Function to get the middle of the linked list
int getMiddle(struct Node* head) {
// Initialize the slow and fast pointer to the
// head of the linked list
struct Node* slow_ptr = head;
struct Node* fast_ptr = head;
// Traverse the linked list
while (fast_ptr != NULL && fast_ptr->next != NULL) {
// Move the fast pointer by two nodes
fast_ptr = fast_ptr->next->next;
// Move the slow pointer by one node
slow_ptr = slow_ptr->next;
}
// Return the slow pointer which is currently pointing
// to the middle node of the linked list
return slow_ptr->data;
}
struct Node* createNode(int x) {
struct Node* newNode =
(struct Node*)malloc(sizeof(struct Node));
newNode->data = x;
newNode->next = NULL;
return newNode;
}
int main() {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40 -> 50 -> 60
struct Node* head = createNode(10);
head->next = createNode(20);
head->next->next = createNode(30);
head->next->next->next = createNode(40);
head->next->next->next->next = createNode(50);
head->next->next->next->next->next = createNode(60);
printf("%d\n", getMiddle(head));
return 0;
}
// Java program to illustrate how to find the middle element
// by using Floyd's Cycle Finding Algorithm
class Node {
int data;
Node next;
Node(int x) {
this.data = x;
this.next = null;
}
}
class GfG {
// Function to get the middle of the linked list
static int getMiddle(Node head) {
// Initialize the slow and fast pointer to the head
// of the linked list
Node slow_ptr = head;
Node fast_ptr = head;
while (fast_ptr != null && fast_ptr.next != null) {
// Move the fast pointer by two nodes
fast_ptr = fast_ptr.next.next;
// Move the slow pointer by one node
slow_ptr = slow_ptr.next;
}
// Return the slow pointer which is currently
// pointing to the middle node of the linked list
return slow_ptr.data;
}
public static void main(String[] args) {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40 -> 50 -> 60
Node head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
head.next.next.next.next = new Node(50);
head.next.next.next.next.next = new Node(60);
System.out.println(getMiddle(head));
}
}
# Python program to illustrate how to find the middle element
# by using Floyd's Cycle Finding Algorithm
class Node:
def __init__(self, x):
self.data = x
self.next = None
# Function to get the middle of the linked list
def getMiddle(head):
# Initialize the slow and fast pointer to the
# head of the linked list
slow_ptr = head
fast_ptr = head
while fast_ptr is not None and fast_ptr.next is not None:
# Move the fast pointer by two nodes
fast_ptr = fast_ptr.next.next
# Move the slow pointer by one node
slow_ptr = slow_ptr.next
# Return the slow pointer which is currently pointing to the
# middle node of the linked list
return slow_ptr.data
def main():
# Create a hard-coded linked list:
# 10 -> 20 -> 30 -> 40 -> 50 -> 60
head = Node(10)
head.next = Node(20)
head.next.next = Node(30)
head.next.next.next = Node(40)
head.next.next.next.next = Node(50)
head.next.next.next.next.next = Node(60)
print(getMiddle(head))
if __name__ == "__main__":
main()
// C# program to illustrate how to find the middle element
// by using Floyd's Cycle Finding Algorithm
using System;
class Node {
public int data;
public Node next;
public Node(int x) {
this.data = x;
this.next = null;
}
}
class GfG {
// Function to get the middle of the linked list
static int getMiddle(Node head) {
// Initialize the slow and fast pointer to the head
// of the linked list
Node slow_ptr = head;
Node fast_ptr = head;
while (fast_ptr != null && fast_ptr.next != null) {
// Move the fast pointer by two nodes
fast_ptr = fast_ptr.next.next;
// Move the slow pointer by one node
slow_ptr = slow_ptr.next;
}
// Return the slow pointer which is currently
// pointing to the middle node of the linked list
return slow_ptr.data;
}
// Driver code
static void Main() {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40 -> 50 -> 60
Node head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
head.next.next.next.next = new Node(50);
head.next.next.next.next.next = new Node(60);
Console.WriteLine(getMiddle(head));
}
}
// Javascript program to illustrate how to find the middle
// element by using Floyd's Cycle Finding Algorithm
class Node {
constructor(new_data) {
this.data = new_data;
this.next = null;
}
}
// Function to get the middle of the linked list
function getMiddle(head) {
// Initialize the slow and fast pointer to the head of
// the linked list
let slow_ptr = head;
let fast_ptr = head;
while (fast_ptr !== null && fast_ptr.next !== null) {
// Move the fast pointer by two nodes
fast_ptr = fast_ptr.next.next;
// Move the slow pointer by one node
slow_ptr = slow_ptr.next;
}
// Return the slow pointer which is currently pointing
// to the middle node of the linked list
return slow_ptr.data;
}
function main() {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40 -> 50 -> 60
const head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
head.next.next.next.next = new Node(50);
head.next.next.next.next.next = new Node(60);
console.log(getMiddle(head));
}
main();
OutputMiddle Value Of Linked List is: 40
Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(1)
Finding middle element in a Linked list
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