How do you find middle element of LinkedList in one pass is a programming
question often asked to Java and non Java programmers in telephonic Interview.
This question is similar to checking
palindrome or calculating
factorial, where Interviewer some time also ask to write code. In order to answer this question candidate must be familiar with LinkedList data structure
i.e. In case of singly LinkedList, each node of Linked List contains data and
pointer, which is address of next Linked List and last element of Singly Linked
List points towards null. Since in order to find middle element of Linked List
you need to find length of LinkedList, which is counting elements till end i.e.
until you find the last element on Linked List. What makes this data structure
Interview question interesting is that you need to

*find middle element of**LinkedList**in one pass*and you don’t know length of LinkedList. This is where candidates logical ability puts into test, whether he is familiar with space and time trade off or not etc. As if you think carefully you can solve this problem by using two pointers as mentioned in my last post on How to find length of Singly Linked List in Java. By using two pointers, incrementing one at each iteration and other at every second iteration. When first pointer will point at end of Linked List, second pointer will be pointing at middle node of Linked List. In fact this two pointer approach can solve multiple similar problems e.g. How to find 3^{rd}element from last in a Linked List in one Iteration or How to find nth element from last in a Linked List. In this Java programming tutorial we will see a Java program which finds middle element of Linked List in one Iteration.##
__Java program to find middle element of
LinkedList in one pass__

Here is complete Java program to find middle node of Linked List in Java.
Remember LinkedList class here is our custom class and don’t confuse this class
with java.util.LinkedList
which is a popular Collection class
in Java. In this Java program, our class LinkedList represent a linked list
data structure which contains collection of node and has head and tail. Each Node
contains data and address part. Main method of LinkedListTest class is used to
simulate the problem, where we created Linked List and added few elements on it
and then iterate over them to find middle element of Linked List in one pass in
Java.

**import**test.LinkedList.Node;

/**

*

**Java program to find middle element of linked list in one pass**.

* In order to find middle element of linked list we need to find length first

* but since we can only traverse linked list one time, we will use two pointers

* one which we will increment on each iteration while other which will be

* incremented every second iteration. so when first pointer will point to the

* end of linked list, second will be pointing to the middle element of linked list

* @author

*/

**public**

**class**LinkedListTest {

**public**

**static**

**void**main(

**String**args[]) {

*//creating LinkedList with 5 elements including head*

**LinkedList**linkedList =

**new**

**LinkedList**();

**LinkedList**.

**Node**head = linkedList.head();

linkedList.add(

**new**

**LinkedList**.

**Node**("1"));

linkedList.add(

**new**

**LinkedList**.

**Node**("2"));

linkedList.add(

**new**

**LinkedList**.

**Node**("3"));

linkedList.add(

**new**

**LinkedList**.

**Node**("4"));

*//finding middle element of LinkedList in single pass*

**LinkedList**.

**Node**current = head;

**int**length = 0;

**LinkedList**.

**Node**middle = head;

while(current.next() !=

**null**){

length++;

if(length%2 ==0){

middle = middle.next();

}

current = current.next();

}

if(length%2 == 1){

middle = middle.next();

}

**System**.out.println("length of LinkedList: " + length);

**System**.out.println("middle element of LinkedList : " + middle);

}

}

**class**

**LinkedList**{

**private**

**Node**head;

**private**

**Node**tail;

**public**

**LinkedList**(){

**this**.head =

**new**

**Node**("head");

tail = head;

}

**public**

**Node**head(){

**return**head;

}

**public**

**void**add(

**Node**node){

tail.next = node;

tail = node;

}

**public**

**static**

**class**

**Node**{

**private**

**Node**next;

**private**

**String**data;

**public**

**Node**(

**String**data){

**this**.data = data;

}

**public**

**String**data() {

**return**data;

}

**public**

**void**setData(

**String**data) {

**this**.data = data;

}

**public**

**Node**next() {

**return**next;

}

**public**

**void**setNext(

**Node**next) {

**this**.next = next;

}

**public**

**String**toString(){

**return**

**this**.data;

}

}

}

**Output:**

length of LinkedList: 4

middle element of LinkedList : 2

That’s all on

**How to find middle element of LinkedList in one pass.**As I said this is a good interview question to separate programmers from non programmers. Also technique mentioned here to find middle node of LinkedList can be used to find 3rd element from Last or nth element from last in a LinkedList as well.
Other

**Java programming tutorials**from Javarevisited blog
## 31 comments :

Maybe my mind is still not awaken after Christmas laziness, but... does this code really return middle element? I think it returns last element with even index. In this code it is solved like you were incrementing index every second time. If you update reference to current every second time, nothing good will come out of this. I'd set middle = middle.next() in loop and also ... initialize length as 0 (I don't consider head element as real element of the list (length should return number of elements carrying data). Greetings!

@Pio Jac, you are right if we don't consider head as real element. In that case initialing length with 0 make sense. Also In order to handle odd length e.g. 3, 5 or 7 where middle element is not updated, we need to handle once loop finished. I have updated code. Thanks for pointing out, it seems, its me, who has Christmas laziness :)

Can you please write code for How to find 3rd element from last in Singular Linked List, I didn't get your point on saying that same technique can be used to find Nth element from last in singly linked list. Sorry.

I was asked to find middle item in a linked list in single pass and when I asked what is length of linked list they say, you need to find that as well. I didn't thought that I can use two pointers or two variables to keep track of middle and last item, something I missed. By the way What is big O for best case and worst case for this solution ?

In what aspect does this separate the programmers from nonprogrammers, even when you yourself didn't elaborate a complete algorithm?

What is a programmer and what is a nonprogrammer?

@Anonymous, aspect here is coding, converting logic to code. algorithm here is simple where one pointer is incremented on each iteration, while other is incremented in every second iteration. let me know which part you didn't understand.

Actually, I would dispute the assertion that this is being done "in one pass". You're making two separate passes through the list - it's just that you're doing them concurrently and not consecutively.

instead of checking length why not you just increment one pointer twice and other pointer only one. this would make your linked list middle element checking code simpler.

I have to agree with Ken, you just just do two traversals within one loop.

@Guido, Can you please elaborate? I think node.next() doesn't traverse the list it just like incrementing pointer, don't agree?

@Javin: I understand it just increments the pointer.

But as far as I can see, the code keeps track of two elements (i.e. pointers) in the loop: middle and current.

So even though it is executed within one for-loop, I do see two traversals:

* current looping all the way till the end

* middle looping till the middle

That is why I agreed with the statement of Ken that I would not call this one pass. Could just be a word thing and that with one-pass here it is meant you are only allowed to use one for-loop.

Ken and Guido.. Good Catch.. I had the same question when I read the article. As you said, if it is a wording issue(like using one for-loop), it clears my doubt.

- Durai

Hi,

The simple way to find the middle element is written below

import java.util.ArrayList;

import java.util.Collections;

import java.util.Comparator;

import java.util.HashMap;

import java.util.LinkedList;

import java.util.Set;

import javax.swing.text.html.HTMLDocument.Iterator;

import javax.xml.soap.Node;

public class test2 {

public static void main(String args[])

{

LinkedList l=new LinkedList();

l.add(2);

l.add(3);

l.add(9);

l.add(4);

java.util.Iterator i1=l.iterator();

java.util.Iterator i2=l.iterator();

int i=0;

int middle = 0;

while(i1.hasNext())

{

if(i==0)

{

i1.next();

i=1;

}

else if(i==1)

{

if(i1.hasNext())

{

i1.next();

middle=(Integer) i2.next();

i=0;

}

}

}

System.out.println("middle"+middle);

}

}

This extra code:

if(length%2 == 1){

middle = middle.next();

}

is not required..because your code to find middle element will give the same result without this piece of code anyway

Ken is correct. Unless I am missing something, this isn't one pass, it is 1.5. If you demand an answer for a single pass, a student should respond that it isn't possible and that only 1.5 is possible. The idea that answering this 'incorrectly' could have cost someone a job is slightly disconcerting.

Guys please comment on my approach, I am also doing in 1.5 pass. Please suggest if it is efficient or can be improved ?

package my.data.structures;

class LlNode{

int data;

LlNode next;

LlNode prev;

}

class Linked{

LlNode start;

public Linked()

{

start = null;

}

public void insertLl(int d){

LlNode node = new LlNode();

node.data = d;

if (start == null)

{

start = node;

start.next = null;

start.prev = null;

}

else{

LlNode current;

LlNode previous;

current = start;

while(true){

previous = current;

if(current.next==null)

{

current.next = node;

node.prev = previous;

node.next = null;

return;

}

else{

current = current.next;

}

}

}

}

public void displayLl(){

LlNode current;

current = start;

while(current != null){

System.out.print(current.data+"---->");

current = current.next;

}

}

public void findmid() {

// TODO Auto-generated method stub

LlNode current;

LlNode mid;

current = start;

mid = start;

int midnode, i,j;

midnode = 1;

int cur;

cur = 0;

i=0;j=0;

while(current != null){

if(cur==1)

{

System.out.println("\nMid Node is at : "+midnode);

return;

}

else{

mid = mid.next;

midnode++;

for (i=0;i<2;i++){

if(current.next == null){

System.out.println("No mid, length is even");

return;

}

current = current.next;

}

if(current.next == null){

cur=1;

}

}

}

}

}

public class LinkedList{

public static void main(String[] args) {

Linked l = new Linked();

// TODO Auto-generated method stub

l.insertLl(1);

l.insertLl(2);

l.insertLl(3);

l.insertLl(4);

l.insertLl(5);

l.insertLl(6);

l.insertLl(7);

l.displayLl();

l.findmid();

}

}

import java.util.LinkedList;

public class LinkedListTest {

/**

* @param args

*/

public static void main(String[] args) {

// TODO Auto-generated method stub

LinkedList node = new LinkedList();

node.add(4);

node.add(5);

node.add(7);

node.add(10);

node.add(2);

node.add(1);

node.add(22);

LinkedListTest obj =new LinkedListTest();

float len = (float) node.size();

float midInd = 0 ;

if(obj.evenOdd(len))

{

midInd= len/2;

}else

{

midInd = (float) ((len/2)- 0.5);

}

int j = (int)midInd;

j = node.get(j);

System.out.print(midInd + " "+j);

}

public boolean evenOdd(float i)

{

if ( i % 2 == 0 )

{

return true;

}else

{

return false;

}

}

}

The solution presented here obviously does more than one pass and hence isn't valid. It doesn't matter if all you do is chasing the pointers. You are still traversing 1.5 times and could just as well do a count run and then stop in the middle for the next one with the same effort.

A java LinkedList is a doubly linked list (http://docs.oracle.com/javase/7/docs/api/java/util/LinkedList.html) so the whole premise that you would be working on a singly linked list is wrong too. The correct solution would be to advance from both sides (front and back) of the doubly linked list simultaneously until you meet in the middle element (happens for uneven list length) or the next element matches the current element of the other (happens for even list lengths).

public void middle(){

node slow=start.next;

node fast=start.next;

while(fast.next!=null)

{

slow=slow.next;

fast=fast.next.next;

}

System.out.println(slow.data);

}

10->9->8->7->6->5->4->3->2->1->

5

if(length%2 == 1){

middle = middle.next();

}

Code is not required.

It works without that irrespective of even or odd number of nodes.

THIS IS IN C

================================================================================

//C PROGRAM FOR PRINTING THE MIDDLE ELEMENT OF THE LIST

#include

#include

struct node

{

int data;

struct node *link;

};

typedef struct node *NODE;

NODE getnode()

{

NODE X;

X=(NODE)malloc(sizeof(struct node));

if(X==NULL)

{

printf("NO NODES TO BE CREATED \n");

exit(0);

}

return X;

}

NODE insert(NODE root,int item)

{

NODE temp=getnode();

temp->data = item;

temp->link=root;

return temp;

}

NODE display(NODE root)

{

NODE temp=root;

while(temp!=NULL)

{

printf("%d-->",temp->data);

temp = temp->link;

}

}

void middle(NODE root)

{

NODE cur,prev;

if(root==NULL)

{

printf("NO NODES TO BE DISPALYED \n");

}

else

{

prev=root;

cur=root;

while(cur->link!=NULL)

{

prev=prev->link;

cur=cur->link->link;

}

if(cur->link==NULL)

printf("THE MIDDLE ELEMENT IS %d \n",prev->data);

}

}

int main()

{

NODE root;

int item,ch;

while(1)

{

printf("1:INSERT 2:DISPLAY 3:MIDDLE 4:EXIT \n");

printf("ENTER UR CHOICE \n");

scanf("%d",&ch);

switch(ch)

{

case 1:printf("ENTER THE ITEM U WANT TO INSERT \n");

scanf("%d",&item);

root=insert(root,item);

break;

case 2:display(root);

break;

case 3:middle(root);

break;

case 4:exit(0);

break;

}

}

}

==============================================================================

simple way to display middle element

List llTest=new LinkedList();

llTest.add(23);

llTest.add(25);

llTest.add(34);

llTest.add(21);

llTest.add(29);

llTest.add(31);

llTest.add(39);

System.out.println(llTest);

System.out.println(" element of LinkedList is : " + ((LinkedList) llTest).get(llTest.size()/2));

System.out.println("size:"+(llTest.size()));

@pavan, linked list doesn't allow random access. You can cannot get element by calling get(index). This is true for our own linked list implementation and also for java.util.LinkedList, which is a doubly linked list.

public static void main(String[] args) {

// TODO Auto-generated method stub

LinkedList linked = new LinkedList();

Scanner scn = new Scanner(System.in);

System.out.println("How many numbers you want to add");

int n = scn.nextInt();

System.out.println("Start the element to enter in list");

for(int c= 0 ; c <n ; c++ ){

int s = scn.nextInt();

linked.add(c,s);

}

System.out.println(linked);

System.out.println(linked.size());

System.out.println(n/2);

int first = linked.get(n/2);

System.out.println(first);

}

// I assume we can remove first and remove last element in a loop

public static String findMidNodeInLinkedList(final LinkedList list) {

String mid = null;

while (list.peekFirst() != null) {

mid = list.removeFirst();

if (list.peekLast() == null) {

return mid;

}

list.removeLast();

}

return mid;

}

// Below there is only one pass, using one ListIterator from head, one desendingIterator from tail

public static String findMid2(LinkedList list) {

ListIterator iter1 = list.listIterator(0);

Iterator iter2 = list.descendingIterator();

String fromHead = null;

String fromTail = null;

while (iter1.hasNext()) {

fromHead = iter1.next();

if (fromHead == fromTail) {

return iter2.next();

}

fromTail = iter2.next();

if (fromHead == fromTail) {

return fromHead;

}

}

return fromHead;

}

Hi, my first Java. How to find middle element of LinkedList in one pass.

Iterator iterates through the LinkedList always starts at Z.size()/2. The "count" speaks it all.

class Numbers {

int N;

Numbers (int R) { N = R;}

public String toString() { return "" + N; }

}

public class FindMiddle {

public static void main(String[] args) {

LinkedList Z= new LinkedList<>();

Z.add(new Numbers(1)); Z.add(new Numbers(2)); Z.add(new Numbers(3)); Z.add(new Numbers(4)); Z.add(new Numbers(5));

Z.add(new Numbers(6)); Z.add(new Numbers(7)); Z.add(new Numbers(8)); Z.add(new Numbers(999));

print(Z.toString()); print(Z.size());

ListIterator it = Z.listIterator(Z.size()/2);

int count = 0;

while( it.hasNext()) {

it.next();

++count;

}

print(count);

}

}

/*OUTPUT

[1, 2, 9, 4, 5, 6, 7, 999, 999]

9

count is:5

*/

@Huy, good solution, try writing some JUnit tests as well to see if your program works on boundary conditions e.g. linked list with even number of elements, odd number of elements etc.

/**

*Find below code for finding the middle element by two pointer method

*/

Node nod1 = list;

Node nod2 = list;

while (nod2.next != null) {

if (nod2.next != null)

nod2 = nod2.next;

else

nod1 = nod1.next;

if (nod2.next != null)

nod2 = nod2.next;

else

nod1 = nod1.next;

}

System.out.println("middle by two pointer increment : " + nod1.value);

Hi jarvin, here's your request. find odd and even numbers in the list

class Numbers {

int N;

Numbers (int R) { N = R;}

public String toString() { return "" + N; }

int getValue() { return N; }

}

public class FindMiddle {

public static void main(String[] args) {

LinkedList Z= new LinkedList<>();Z.add(new Numbers(100)); Z.add(new Numbers(200)); Z.add(new Numbers(300)); Z.add(new Numbers(400)); Z.add(new Numbers(500));Z.add(new Numbers(600)); Z.add(new Numbers(700)); Z.add(new Numbers(800)); Z.add(new Numbers(999)); Z.add(new Numbers(9999));Z.add(new Numbers(99999999)); Z.add(new Numbers(99999999));

print(Z.toString()); print(Z.size());

ListIterator it = Z.listIterator();

int b;

if ( it.hasNext()) {

it.next();

for ( int i = 0; i < Z.size(); i++) {

b = Z.get(i).getValue();

if ( b % 2 == 0) {

print(b);

}

else

print(b);

}

}

}

}

/*OUTPUT

[100, 200, 300, 400, 500, 600, 700, 800, 999, 9999, 99999999, 99999999]

12

100

200

300

400

500

600

700

800

999

9999

99999999

99999999

*/

@Huy, Thanks for your code, but I think I suggested to write JUnit tests for your code to verify if the solution works for a linked list with even number of elements and a list with odd number of elements e.g. linked list with 5 elements.

nevermind, good practice anyway.

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